Second Year Quantum Mechanics - Problems 1 Wave ......Second Year Quantum Mechanics - Problems 3...

Second Year Quantum Mechanics - Problems 1

Wave properties and the de Broglie relations

Paul Dauncey, 7 Oct 2011

1. Calculate the wavelengths associated with the following:

(a) Electrons of kinetic energy 100 eV.

(b) X-ray photons of energy 10 keV.

(c) Neutrons in thermal equilibrium with a reactor at a temperature 500 K.

(d) A 100 g golf ball, struck at 60 m s−1.

2. Monoenergetic neutrons from the source in Question 1(c) above are detected at a heightz above the source. The detected neutrons travel horizontally from the source, and thenpass through the interferometer shown for which the two possible paths, 1 and 2, have thesame length. Note, this interferometer is unusual in that particles travel equal distancesbut have different wavelengths on the two paths and hence a phase difference arises.

(a) Assuming the change of momentum is small, then show that the difference in mo-mentum of neutrons at the source (p) and at the detector (p′) is given by

δp = (p− p′) ≈ m2ngzλ

2πh,

where mn is the neutron mass and λ its de Broglie wavelength.

(b) Given that for a wavenumber k, the phase change when travelling a distance l isφ = kl (or equivalently 2πl/λ), then show that the phase difference for neutronsfollowing the two paths 1 and 2 is

δφ =m2

ngAλ

2πh2 radians,

where A is the area enclosed by the paths 1 and 2. Note the contributions from thetwo vertical sections of the paths are identical and hence cancel.

(c) The whole system is rotated slowly about the incident beam direction. What isobserved at the detector? If A = 10−3 m2, estimate the number of interference“fringes” that are swept out during one complete turn of the system.

1 eV= 1.6× 10−19 J kB = 1.38× 10−23 J K−1 h = 1.05× 10−34 J s c = 3× 108 ms−1

me = 9.1× 10−31 kg mn = 1840me g = 9.81 m s−2

1

Second Year Quantum Mechanics - Problems 2The Schrodinger equation and probability


1. In developing the time-dependent Schrodinger equation (TDSE) we took a “back to front”approach. We postulated a solution and looked for an appropriate wave equation thathas the important feature of agreeing with the de Broglie relations. It is therefore notsurprising that the postulated solution ψ = Ae−i(ωt−kx) satisfies the equation when V = 0,i.e. for free particles.

(a) Check the above wavefunction does indeed satisfy the free-particle Schrodinger equa-tion and is also consistent with the de Broglie relations.

(b) The solution describes waves that go from left to right. Show that ψ = Ae−i(ωt+kx) isalso a solution. Also, show that it describes waves that propagate from right to left;to do this, sketch the real part of the wavefunction as a function of kx for arbitraryA (e.g. let A = 1), for ωt = 0, ωt = π/4 and ωt = π/2.

(c) Show that, for a particle whose wavefunction is either of the two solutions ψ givenabove, the probability density |ψ|2 for finding the particle at any particular positionis the same everywhere in space, i.e. it is independent of x.

(d) Calculate the probability current density J , where

J =ih

2m

[(∂ψ∗

∂x

)ψ − ψ∗

(∂ψ

∂x

)]=

h

mIm

(ψ∗∂ψ

∂x

)for both of the solutions above. If k changes, how do |ψ|2 and J change?

2. The general time-dependent Schrodinger equation, i.e. with V (x) 6= 0, is linear so we canadd solutions.

(a) Check this; i.e. show that if ψ1(x, t) and ψ2(x, t) are solutions, then the superpositionstate ψs = αψ1+βψ2, where α and β are arbitrary complex numbers, is also a solution.

(b) As shown in the lectures, for the infinite square well potential (V = 0 for |x| ≤ a andV = ∞ elsewhere), the ground and first excited energy eigenstates within the wellare given by

u1(x) =1√a

cos(πx/2a), u2(x) =1√a

sin(πx/a)

Calculate the energies that these eigenstates correspond to, and hence find ∆E =E2 − E1.

(c) The full solutions of the time-dependent Schrodinger equation for these two energyeigenstates are ψ1(x, t) = u1(x)e−iE1t/h and ψ2(x, t) = u2(x)e−iE2t/h. Sketch theprobability density |ψn|2 as a function of x/a for each of the eigenstates. Also, findthe probability current density J in each case.

(d) A wavefunction is formed in a superposition of these two energy states with α = β =1/√

2. Calculate the probability density including the time dependence. Sketch thedensity as a function of x/a at ∆Et/h = 0, ∆Et/h = π/4 and ∆Et/h = π/2.

To follow up on this, visit the website http://www.falstad.com/qm1d/. This is a Javaapplet that allows you to simulate the time behaviour of superposition wavefunctions. Setit up for an infinite well and click on the first two little clock faces at the bottom of thescreen. This makes a superposition of the lowest two energy eigenstates, as you calculatedabove. Explore other features of the problem e.g. build other superpositions, etc.

1


Square wells and potential steps


1. The energies of the bound states in a finite square well potential can be found as outlined inthe lectures and in Handout 2, which you will need for the following. The plot below showsthe relevant functions for an electron confined in a 1-D finite square well potential with adepth of 3.4 eV and a total width of 1.4 nm. The solid line corresponds to γ2+k2 = 2mV/h2

and the dotted and dashed lines to γ = k tan(ka) and γ = −k cot(ka), which are for evenand odd parity solutions, respectively.

(a) How many bound states exist for an electron in the potential well described above?What parity do the highest and lowest energy states in the well have?

(b) Calculate the value of the solid line intercept on the y axis using the above formulae.Also find the value of the intercept on the x axis of the dashed line representing thelowest energy odd parity state. Are these consistent with approximate values you canread from the graph?

(c) Estimate a rough approximation to the values of k for the highest and lowest statesin the well from the above graph. Using these as starting values, find1 y = ka, andhence k and γ, to three significant figures for these two states, using the function foreven parity solutions (as described in Handout 2)

cos(y)− hy

a√

2mV= 0

1This can easily be done in Excel by evaluating the function for a range of values around your initial guessand finding which is closest to zero.

1

(d) We define Pin/Pout as the relative probability of finding the electron inside the wellcompared with finding it in the “classically forbidden region”, i.e. outside the well.Show that for the even parity states

Pin

Pout=

γ

k

[ka + 1

2 sin 2ka

cos2 ka

]Calculate values of Pin/Pout for the highest and the lowest states.

(e) Imagine distorting the well described above by keeping the width the same but makingit progressively shallower. Estimate how shallow the well would have to be to justsupport two bound states.

Visit http://www.falstad.com/qm1d/ and set the applet up for a finite square well tostudy this problem further.

2. An electron of kinetic energy E, moving in the direction of increasing x, approaches apotential step described by:

V (x) = 0, x ≤ 0V (x) = V0, x > 0

for a positive constant V0.

(a) For E > V0, calculate the coefficients of reflection R and transmission T at x = 0and check that R + T = 1. Plot out R and T as a function of y = (E − V0)/V0 for0 ≤ (E − V0)/V0 < 1.

(b) For a classical particle, what would you expect to happen in this situation?(c) Consider the case where the potential is negative for x > 0, i.e. V (x) = −V0. Again,

calculate R and T and express in terms of z = E/V0. Comment on your result.

3. Exam question: 2002 Question 1.

A particle of mass m is bound in a potential well defined by

V (x)→∞ x < 0V (x) = 0 0 ≤ x ≤ a

V (x) = W x > a.

Show that the particle energy E is a solution of the equation

γ = −k cot ka

where k2 = 2mE/h2 and γ2 = 2m(W − E)/h2. [12 marks]

Show, graphically or otherwise, that the particle can be bound only if

W >π2h2

8ma2.

[4 marks]

Show also, that if W is much larger than the spacing of the energy levels, then the groundstate energy is given approximately by

E =h2π2

2ma2.

[4 marks]

2


Simple harmonic oscillators


1. Most of the interesting operators in quantum mechanics are Hermitian. A pair of operatorsthat we will use later to solve the simple harmonic oscillator, (αx+ip) and (αx−ip), whereα is a real constant, turn out not to be Hermitian. Show that these operators are indeednon-Hermitian. (You will need to use the fact that x and p are Hermitian.)

2. The ground state of the simple harmonic oscillator (SHO), which has V (x) = mω2x2/2,has an energy eigenstate which can be written as

u0(x) = Ae−mωx2/2h

where A is a normalisation constant. (Evaluating A is not required for this question.)

(a) By direct substitution, show that u0 is indeed an eigenstate of the simple harmonicoscillator TISE and determine its energy eigenvalue.

(b) Operating on any function with an operator gives another (in general different) func-tion. Calculate the function resulting from acting on u0 with the (non-Hermitian)operator αx− ip, discussed in Qu. 1 above, with α = mω

u1 = (mωx− ip)u0

By direct substitution into the TISE, show u1 is also an energy eigenstate and deter-mine its eigenvalue.

(c) Calculate the result of operating on u1 with the other (non-Hermitian) operatordiscussed in Qu. 1, namely mωx+ ip. Show this is proportional to u0.

(d) The operators mωx± ip seem to move down and up between the energy eigenstates.To determine what happens if we try to lower the ground state, consider operatingon u0 with mωx+ ip. Comment on your result.

3. An ion in a harmonic ion trap sees a potential which is effectively that of a simple harmonicoscillator. It has a natural oscillation frequency given by ν = 1MHz. Ignoring any internalexcitations, it is known to be in a superposition of the n = 0, 1 and 2 SHO energy states.A measurement is then made and it is found to be in the n = 2 level.

(a) What is the energy of the ion after the measurement has been made (give your answerin eV)?

(b) Convert this to an effective temperature using E ∼ kBT/2. You should find that theion has to be very cold indeed in order for the quantised nature of its motion to berelevant.

(c) If another measurement is made immediately after the first what will the energy ofthe ion be measured to be? If the ion is perfectly isolated, would your answer changeif the second measurement was instead made a significant time after the first?

(d) In reality, the walls of the ion trap will emit (black-body like) radiation due to havingsome temperature. Qualitatively how would the outcome of the second measurementa significant time after the first depend on this temperature?

kB = 1.38× 10−23 JK−1

1


A particle of mass m in a harmonic potential given by V1(x) = 12mω

21x

2 is not in an energyeigenstate but rather is described by the normalised wavefunction

ψ =(

2mω1

πh

)1/4

e−mω1x2/h.

(i) By writing down the appropriate overlap integral show that the probability that ameasurement of the particle’s energy will give the result E = 3hω1/2 is zero.

[4 marks]Note: The lowest normalised energy eigenstates for the general harmonic oscillatorpotential V (x) = 1

2mω2x2 are

u0 = (mω/πh)1/4 exp(−mωx2/2h)u1 = (4/π)1/4(mω/h)3/4x exp(−mωx2/2h)u2 = (mω/4πh)1/4[2(mω/h)x2 − 1] exp(−mωx2/2h)

(ii) Evaluate the probability P1 that a measurement of the particle’s energy will give theresult E = 5hω1/2.

[6 marks]

(iii) The potential is suddenly made steeper so that it is described by the potential V2(x) =12mω

22x

2 with ω2 > ω1. We define P2 to be the probability that a measurement of theparticle’s energy made after this sudden change will yield the ground state energy ofthe new potential. Show that

P2 =√

2(ω1ω2)1/2

ω1 + ω2/2.

[5 marks]

(iv) Under what circumstances is P2 = 1? In this case what kind of function is ψ withrespect to V2(x)?

[5 marks]

You may require the standard integrals∫ ∞−∞

exp(−ax2) dx =√π

a,∫ ∞

−∞x2 exp(−ax2) dx =

12a

√π

a.

2


Expectation values and time dependencePaul Dauncey, 4 Nov 2011

1. The normalised ground state energy eigenfunction for a particle of mass m confined in asimple harmonic oscillator (SHO) potential is

u0(x) =(α√π

)1/2

exp(−α2x2/2) .

where α2 = mω0/h. A particle in this potential is described at some time by the normalisedwave function

ψ(x) =(β√π

)1/2

exp(−β2x2/2)

for some real constant β.

(a) What is the probability that a measurement of the particle’s energy will give theground state energy E0?

(b) Calculate the expectation value of the kinetic energy of the particle.

(c) Give your answers for Qu. 1(a) and 1(b) for the particular case of β = α. Evaluate theratio of the expectation value of the kinetic energy to the total energy in this case.How does this ratio compare with the ratio of the classical average of the kineticenergy to the total energy for the motion of a SHO?

Standard integrals:∫ ∞

−∞exp(−ax2)dx =

√π

a,

∫ ∞

−∞x2 exp(−ax2)dx =

12a

√π

a.

2. The ground state of the SHO is given in Qu. 1. The normalised first excited state energyeigenfunction is

u1(x) =(

2α√π

)1/2

αx exp(−α2x2/2)

(a) If the particle wave function at time t = 0 is

ψ(x) =(α√π

)1/2 (1√2

+ αx

)exp(−α2x2/2)

calculate 〈E〉, the expectation value of the particle’s energy at t = 0.

(b) Write down the particle’s wave function ψ(x, t) at a later time t.

(c) Show that 〈x〉, the expectation value of the particle’s position, executes simple har-monic motion with angular frequency ω0.

(d) Without further calculation, state the frequency with which 〈x〉 would vary if insteadthis had been a superposition of the ground state and the second excited state.

1


Show that the eigenvalues of an operator A are real when the operator is Hermitian: i.e.∫ ∞

−∞ψ∗1Aψ2 dx =

∫ ∞

−∞[Aψ1]∗ψ2 dx

for any wavefunctions ψ1(x), ψ2(x).

[4 marks]

Use the time-dependent Schrodinger equation, together with the Hermitian property of Hto derive the equation of motion:

d

dt〈A〉 =

i

h〈[H, A]〉

giving the time variation of the expectation value of a quantum mechanical operator Awhich does not itself depend explicitly on time, and where H is the Hamiltonian for thesystem.

[7 marks]

How is this equation relation to conservation laws? Give two examples.

[3 marks]

Show how the equation leads to a form of Newton’s law of motion for the rate of changeof the average momentum.

[6 marks]

2


Momentum space and the Uncertainty Principle

Paul Dauncey, 11 Nov 2011

1. A particle has a wave function given by

ψ(x) = 1√2a, |x| ≤ a

ψ(x) = 0 , |x| > a

(a) Calculate the RMS uncertainty in x.

(b) The overlap integral of momentum eigenstates with this wavefunction will give mea-surement amplitudes. In this case, since p is a continuous variable, the square of themodulus of the amplitude gives the probability density (rather than just the probabil-ity) for measuring the particle with momentum p. Taking the momentum eigenstatesas Aeipx/h, obtain an expression for this probability density.

(c) Show that the relative probability of measuring the particle with momentum p =πh/2a and with p = 0 is 4/π2.

(d) The above can be related to particles diffracted by a slit. Consider a beam of particlestravelling a long distance parallel to the z direction, with an effectively unique, largemomentum pz, incident on a slit with aperture between x = ±a so they are diffracted,i.e. get a spread of values in px. Show that the angle between the central maximumand the first diffraction minimum is given by θ ≈ λ/2a, where λ = 2πh/p is the deBroglie wavelength of the particle.

2. This question shows that the condition for equality in the Schwarz inequality amounts toa stipulation that only a Gaussian function (i.e. a function like |u|2 ∼ Ae−ax2

where Aand a are constants) can be a minimum uncertainty bound energy eigenstate.

Consider the operators x and p acting on a bound energy eigenstate u(x). Use the conditionQ′ψ = λR′ψ from Handout 5, with p ≡ Q and x ≡ R, to generate a simple differentialequation for u and then solve this. To get the answer you will have to insist that thefunction u you come up with is normalisable.

To be a minimum uncertainty state, the anticommutator part of the general uncertaintyrelation also has to be zero. This term is(⟨

12{x, p}

⟩− 〈x〉〈p〉

)2

Show for the Gaussian function you derived above that this is indeed zero.

You may need to use the standard integrals given at the end of the next question.


(i) Define the expectation value of an observable in terms of its corresponding operatorand the wave function ψ(x).

[2 marks]

(ii) A free particle moves in one dimension and is described by the wave function

ψ(x) = A exp(ikx) exp

(−ax2

2

).

1

(a) Normalise this wave function.[3 marks]

(b) What is the expectation value of the particle’s position?[1 mark]

(c) What is the root mean squared (rms) uncertainty in the particle’s position?[2 marks]

(d) What is the expectation value of the particle’s momentum?[4 marks]

(e) Verify that the uncertainty in the particle’s momentum is h√a/2.

[5 marks](f) Hence calculate the rms uncertainty product for this wave function.

[1 mark]

(iii) Comment briefly on the physical significance of the result obtained in (f) above.[2 marks]

Note the following integrals:∫ ∞

−∞exp(−y2) dy =

√π

∫ ∞

−∞y2 exp(−y2) dy =

√π

2

2


SHO ladder operators


1. Lowering and raising operators for the quantum harmonic oscillator can be defined as

A =1√2

(√mω

hx + i

p√mhω

)A† =

1√2

(√mω

hx− i

p√mhω

)respectively. The Hamiltonian operator can be written in terms of them in several ways

H = hω

(AA† − 1

2

)= hω

(A†A +

12

)=

hω

2

(AA† + A†A

)(a) Express x and p in terms of A and A†.

(b) Hence show that x2 and p2 can be written:

x2 =12

h

mω(A2 + AA† + A†A + A†2)

p2 = −12hmω(A2 − AA† − A†A + A†2)

(c) Use the raising and lowering properties of the ladder operators and the orthogonalityof the energy eigenstates to show that the expectation values of x, p, x2 and p2 for aparticle in an energy eigenstate are

〈x〉 = 0, 〈x2〉 =12

h

mω

⟨AA† + A†A

⟩〈p〉 = 0, 〈p2〉 =

12hmω

⟨AA† + A†A

⟩(d) Given that the energies of the eigenstates are En = (n+1/2)hω, use the above results

to show that the uncertainty product for the nth eigenstate is ∆x∆p = (n + 1/2)h.

2. Use the forms of the ladder operators from the above question in the following.

(a) Explain why the ground state wavefunction of the simple harmonic oscillator has tosatisfy

Aumin = 0

(b) Using one of the forms of H expressed in terms of the A and A† operators given inQu. 1, solve for the energy of this state.

(c) Use the standard explicit form of the x and p operators to write Aumin = 0 as adifferential equation and solve it. (Do not bother to normalise the solution.)

(d) If there was an upper bound to the energy of the eigenstates, then there would be anequivalent equation

A†umax = 0

Repeat the above procedure to find the energy and an explicit wavefunction for thisstate. Explain why, although it is mathematically acceptable, it is not physicallyacceptable.

1


The Hamiltonian for a one-dimensional simple harmonic oscillator is given by

H =hω

2(P 2 + X2).

where X ∝ x and P ∝ p. A pair of operators R+ and R− are defined as

R+ = P + iX and R− = P − iX.

where R+ is the adjoint of R− i.e.∫ ∞

−∞u∗nR−umdx =

∫ ∞

−∞(R+un)∗umdx .

(i) Using the relation[P , X] = −i

show thatR−R+ = (2H/hω) + 1 .

(It can similarly be shown that

R+R− = (2H/hω)− 1 .

DO NOT PROVE THIS)[4 marks]

(ii) Show that[R+, H/hω] = −R+ .

[4 marks]

(iii) If un is a normalised energy eigenfunction with eigenvalue En, i.e. Hun = Enun,show that R+un is also an eigenfunction of H with eigenvalue En + hω.

[5 marks]

(iv) By writing R+un = cn+1un+1 and using the fact that R+ is the adjoint of R−, showthat ∫ ∞

−∞u∗nR−R+undx = |cn+1|2 .

[3 marks]

(v) Using the the expression for R−R+ given in part (i) and the result of part (iv) showthat R+un =

√2(n + 1)un+1 (provided cn+1 is real).

[4 marks]

2


Perturbation solutions


1. An infinite square well can be defined through

V = ∞ for |x| > a, V = 0 for |x| < a

The energies of the eigenstates for a particle of mass m are

En =π2h2n2

8ma2

and the normalised eigenfunctions are

un(x) =1√a

cosnπx

2a, n = 1, 3, 5...

un(x) =1√a

sinnπx

2a, n = 2, 4, 6...

(a) Sketch the perturbed well given by

V = ∞ for |x| > a, V = δ for − a < x < 0, V = 0 for 0 < x < a,

where δ > 0 is a constant, and hence sketch the perturbation part of the potential,i.e. the part in addition to the original infinite square well potential.

(b) If δ � E1, the step at the bottom of the well can be treated using perturbationtheory. Assume that this is the case then use the first order correction to the energyas given in Handout 6

E(1)n =

∫ ∞

−∞u∗nH ′un dx

to calculate the approximate shift in the energy of the ground state.(c) Handout 6 also gives the first order correction to the energy eigenstates as

u(1)n (x) =

∑m6=n

anmum(x) =∑m6=n

(∫∞−∞ u∗mH ′un dx

En − Em

)um(x)

Find the contribution of u2(x) to the perturbed ground state wavefunction. Qualita-tively, what is the overall effect of the admixture of u2(x) in terms of the probabilityof finding the particle at x > 0 compared with x < 0?

(d) The perturbation expansion is in powers of δ so the first order corrections are onlyuseful at small δ. Consider the limit of large δ � E1 such that the potential can beconsidered to be approximately an infinite square well, but now only over the range0 < x < a. From the above information, deduce the exact energy of the ground statein this limit and compare with your perturbation calculation of the energy, if it isassumed to be valid for large δ.

You may need the following standard integrals in solving the above problems∫cos2 θdθ =

θ

2+

14

sin 2θ,

∫sin 2θ cos θdθ = −1

2cos θ − 1

6cos 3θ

∫cos 3θ cos θdθ =

14

sin 2θ +18

sin 4θ

1


A particle is in a one-dimensional potential given by V = 12mω2x2 + εx where the term εx

can be considered as a small perturbation to the quadratic term and ε is a real constant.

(i) In perturbation theory, first-order corrected energy eigenfunctions Un are given by

Un = un + u(1)n with

u(1)n =

∑m6=n

∫+∞−∞ u∗mH ′undx

En − Emum

Use the fact that ε x u0 = C u1, where C is a constant, and the orthogonality of theharmonic oscillator energy eigenfunctions to show that

u(1)0 =

C

E0 − E1u1.

[4 marks]

(ii) The normalised ground and first excited state energy eigenfunctions for the unper-turbed harmonic oscillator are given by

u0 = (α/π)1/4 exp(−αx2/2)u1 = (4/π)1/4α3/4x exp(−αx2/2)

where α = mω/h. Use these explicit forms of u0 and u1 to find an expression for Cin terms of ε and α. Hence show that

U0 = u0 − ε(2mω3h

)−1/2u1

[3 marks]

(iii) What is the expectation value of x for the state u0?[2 marks]

(iv) Show that the expectation value of x for state U0, is given by

〈x〉 = −ε/mω2

Considering the shape of the perturbed potential, comment on the direction of thisshift in 〈x〉.

[8 marks]

(v) Show that there is no first order shift to the energy of the ground state.[3 marks]

Depending on your method you may find the following standard integral useful∫ ∞

−∞y2e−y2

dy =√

π

2

2


Angular momentum

Paul Dauncey, 2 Dec 2011

1. This question constructs explicit, general eigenstates of angular momentum in Cartesians(x, y, z) rather than spherical polars. It should be answered using Cartesian coordinatesfor expressing the angular momentum operators. When appropriate exploit the xyz-cyclicproperties.

(a) Consider any function f(r)i. Verify that ∂f/∂x = f ′x/r, where f ′ = df/dr, and similarly for ∂f/∂y and

∂f/∂z.ii. Using this result, show that any function f(r) satisfies the eigenvalue equation

for each of the angular momentum operators Lx, Ly, Lz and L2. What are thecorresponding eigenvalues?

(b) Functions gx, gy and gz are defined as gi = rif(r), i.e.

gx = xf(r), gy = yf(r), gz = zf(r).

i. Show thatX0 = gx and X± =

(gy ± igz)√2

are eigenstates of Lx and determine the corresponding eigenvalues. Show thatthey are also eigenstates of L2 with eigenvalues 2h2, i.e. l = 1.

ii. From cyclic symmetry, the equivalent eigenstates of Lz with l = 1 are

Z0 = gz and Z± =(gx ± igy)√

2

Express the X− eigenstate of Lx as a superposition of these eigenstates of Lz.iii. A particle is in the Lx eigenstate X− when measurements are made of L2 and

of Lz. What are the possible results of these measurements? Assuming thefunctions are normalised, what are the probabilities of obtaining these results?What is the expectation value of Lz?

(c) The above implies that l = 2 states can be formed by combining functions gij =rirjf(r). There are six such terms, namely

gxx = x2f(r), gyy = y2f(r), gzz = z2f(r),gxy = xyf(r), gxz = xzf(r), gyz = yzf(r),

and hence six independent combinations. However, for l = 2, there are only 2l+1 = 5states, so one combination must give a different l value. Find this combination andstate which l value it corresponds to.


In cartesian co-ordinates the operator L2 = L · L, where L is the quantum mechanicaloperator for angular momentum, can be written as

L2 = L2x + L2

y + L2z,

where the components of L obey the commutators

[Lx, Ly] = ihLz, [Ly, Lz] = ihLx, [Lz, Lx] = ihLy.

1

(i) Show that

[L2x, Lz] = −ih(LxLy + LyLx).

You may assume without proof that

[L2y, Lz] = ih(LxLy + LyLx).

Finally by considering [L2z, Lz] show that [L2, Lz] = 0. By similar reasoning it is possi-

ble to show that [L2, Lx] = [L2, Ly] = 0 (do not prove this). What general conclusionsabout angular momentum can you draw from these commutation relations?

[7 marks]

(ii) Raising and lowering operators for angular momentum can be defined as

L± = Lx ± iLy.

Show that

[Lz, L−] = −hL−.

[3 marks]

(iii) If φm is an eigenfunction of Lz with eigenvalue mh show that (L−φm) is also aneigenfunction of Lz but with eigenvalue (m− 1)h.

[3 marks]

(iv) In cartesian coordinates

Lx = ypz − zpy, Ly = zpx − xpz, Lz = xpy − ypx.

Show that

φ−1 =(x− iy)√

2and φ0 = z,

are eigenfunctions of Lz with eigenvalues −h and 0 respectively.[3 marks]

(v) Verify that the effect of L− on φ0 is as predicted in part (iii) above.[4 marks]

2


Angular momentum and central potentials


1. (a) Show that the potential for an isotropic 3D harmonic oscillator (i.e. one that issymmetric in the three coordinates) takes on a particularly simple form.

(b) The energy eigenstates can be characterised by three quantum numbers, unxnynz ,where each one is associated with one coordinate. The 1D harmonic oscillator groundstate is

u0 = Ae−ax2/2

where a = mω0/h and A is a normalisation constant. Show the 3D ground state hasthe form

u000 = Be−ar2/2

and find its energy.

(c) Using spherical polar coordinates, normalise this wavefunction to find B.

(d) This state is an eigenstate of L2; what is its eigenvalue?

(e) What is the energy and degeneracy of the first excited state? Given that all thesedegenerate states have the same l, then deduce what that value of l must be.

Standard integral ∫ ∞

0x2e−ax2

dx =14

√π

a3

2. An electron is bound by the Coulomb field of a proton, forming a hydrogen atom. Itsnormalised wavefunction is

ψ = A(3u100 − 2u211 + 3u210 +√

3u21−1)

where unlmlare normalised hydrogen atom energy eigenstates with eigenvalues

En = −13.6 eVn2

and l and ml are the angular momentum quantum numbers.

(a) Find the normalisation constant A for the wavefunction ψ.

(b) Since the stateU2 = B(−2u211 + 3u210 +

√3u21−1)

is a sum of degenerate energy eigenstates, it is also an energy eigenstate. Find thenormalisation constant B and express ψ in terms of u100 and U2.

(c) Give these two possible outcomes for a measurement of the energy and calculate theirprobabilities.

(d) Check you would have got the same result for the energy probabilites as (c) if youhad taken the probability of E2 as the probability of getting the energy of any of thethree degenerate states.

1

(e) Calculate the possible outcomes and their probabilities for a measurement on ψ ofL2.

(f) Similarly, calculate the possible outcomes and their probabilities for a measurementon ψ of Lz.

(g) If the Lz measurement is performed after measuring L2 to be zero, how would youranswer to part (f) change?


(i) A pair of observables represented by operators Q and R have simultaneous eigen-states un with eigenvalues qn and rn respectively. By assuming ψ is some arbitrarysuperposition of the eigenstates un show that [Q, R] = 0.

[3 marks]

(ii) The operators L2 and Lz have the forms

L2 = −h2

[1

sin2 θ

∂2

∂φ2+

1sin θ

∂

∂θ

(sin θ

∂

∂θ

)]

Lz = −ih ∂

∂φ

in spherical polar co-ordinates. Verify that Y11 is a simultaneous eigenstate of L2 andLz with eigenvalues given by the integers l = 1 and ml = 1.

[8 marks]

(iii) A particle has the wavefunction

ψ(r) =R(r)√

2π

[sin θe−iφ

2+ cos θ

]

where R(r) is a normalised radial wavefunction. Express this wavefunction as asuperposition of eigenstates of L2 and Lz.

[3 marks]

(iv) Answer the following questions for a particle with the wavefunction ψ(r):

(a) If a measurement is made of L2 what will be the result?(b) What is the probability that the result of a measurement of Lz will be −h?(c) What is the expectation value of Lz?

[6 marks]

Spherical Harmonics

Y10 =√

34π

cos θ,

Y1±1 = ∓√

38π

sin θe±iφ

2


Spin


1. The Pauli spin matrices are

σx =(

0 11 0

), σy =

(0 −ii 0

), σz =

(1 00 −1

)and the spin component operators are defined in terms of these through Si = (h/2)σi.

(a) Verify that for any of the Pauli matrices σ2i = I. Hence, show the possible eigenvalues

of any of the matrices are ±1 and hence the eigenvalues of any of the Si are ±h/2.(b) The above means the eigenvalues of the y component spin operator

Sy =h

2

(0 −ii 0

)are ±h/2, so find the corresponding normalised spin eigenstates ηy+ and ηy−. Verifythe orthogonality relation η†y+ηy− = 0.

(c) If the spin state is χ =15

(34

), find the expectation values of Sz and Sy.

(d) Express this spin state as a linear superposition of the two eigenstates of Sy.Hence find the probability that a measurement of Sy yields the result −h/2.

(e) If Sy is measured and the result is−h/2, what is the spin state after this measurement?What is the expectation value of Sz in a subsequent measurement? Why is it differentfrom the expectation value before the measurement of Sy?

2. When an electron is in a magnetic field B pointing in the z direction, there is a spin-dependent term in the Hamiltonian corresponding to the interaction energy of the electronspin magnetic moment with the magnetic field

Hspin = −µ.B = −Bµz =e

meBSz =

2µB

hBSz

where µB = eh/2me is the Bohr magneton. The spin state at time t = 0 is given by

χ =15

(34

)(a) Find the spin state at a subsequent time t.(b) Find the expectation value of Sz and show it is constant as a function of time.(c) Show that the expectation values of Sx and Sy execute harmonic motion with an

angular frequency 2µBB/h.(d) Find the value of 〈Sx〉2 + 〈Sy〉2 + 〈Sz〉2.


(i) Verify the commutation relation for the spin operators:

[Sx, Sy] = ihSz,

where

Sx =h

2

(0 11 0

), Sy =

h

2

(0 −ii 0

), Sz =

h

2

(1 00 −1

).

[5 marks]

1

(ii) The magnetic interaction energy of an electron, with charge −e and mass m, in amagnetic field B is

E =e

mB.S.

For the case of B = B√2(0, 1, 1), show that the corresponding Hamiltonian is

H =eh

2m

B√2

(1 −ii −1

).

[3 marks]

(iii) Show that

u1 = N1

(1

i(√

2− 1)

)and u2 = N2

(1

−i(√

2 + 1)

)are eigenvectors of this Hamiltonian and hence obtain expressions for the correspond-ing eigenvalues E1 and E2.

[4 marks](The normalisation constants N1,2 = (4 ∓ 2

√2)−1/2, but you are NOT required to

verify these values.)

(iv) Verify that u1 and u2 are orthogonal.[1 mark]

(v) An electron is described by the spin state χ = 1√2

(1i

).

Obtain an expression for the expectation value of its magnetic energy in the B-fielddefined in (ii) above.

[3 marks]

(vi) Hence, or otherwise, show that the probabilities of measuring E1 and E2 for thiselectron are in the ratio:

Prob(E1)Prob(E2)

=√

2 + 1√2− 1

.

[4 marks]

2

Second Year Quantum Mechanics - Solutions 1

Wave properties and the de Broglie relations

Paul Dauncey, Released 14 Oct 2011

1. In each case we use the de Broglie relation: λ = h/p = 2πh/p

(a) Use p =√

2meE so

λ =2× 3.14× 1.05× 10−34

(2× 9.1× 10−31 × 100× 1.6× 10−19)1/2≈ 1.2× 10−10 m

(b) For photons p = E/c so

λ =2× 3.14× 1.05× 10−34 × 3× 108

104 × 1.6× 10−19≈ 1.2× 10−10 m

(c) For E = 3kBT/2, then p =√

2mnE =√

3mnkBT so

λ =2× 3.14× 1.05× 10−34

(3× 1840× 9.1× 10−31 × 1.38× 10−23 × 500)1/2≈ 1.1× 10−10 m

(d) Use p = mv so

λ =2× 3.14× 1.05× 10−34

0.1× 60≈ 10−34 m

2. (a) The neutron energy is conserved so the gain in potential energy mngz results in anequivalent loss of kinetic energy and hence a change in momentum. The value ofδp = p− p′ can be found using

mngz =p2

2mn− p′2

2mn=

p2 − p′2

2mn=

(p− p′)(p + p′)2mn

so

(p− p′)(p + p′)

2= m2

ngz

Since δp� p, then (p + p′)/2 ≈ p so

δp ≈ m2ngz

p=

m2ngzλ

2πh

Or alternatively

p′2

2mn=

p2

2mn−mngz =

p2

2mn

(1− 2m2

ngz

p2

)

so using a binomial expansion

p′ = p

(1− 2m2

ngz

p2

)1/2

≈ p

(1− 1

22m2

ngz

p2

)= p− m2

ngz

p

so that δp = m2ngz/p = m2

ngzλ/2πh as before.

(b) The neutron momentum is related to the wave number k by p = hk. Due to themomentum difference δp there is a corresponding difference in wave number (or wave-length) for the horizontal paths of length l. The phase difference δφ is given by thedifference in wave number through

δφ = φ− φ′ = lk − lk′ = lδk =lδp

h=

m2nglzλ

2πh2

1

Or alternatively, using φ = 2πl/λ

φ =2πl

λ=

lp

hand φ′ =

2πl

λ′=

lp′

h

so

δφ = φ− φ′ =l

h(p− p′) =

lδp

h=

m2nglzλ

2πh2

which gives the same result. Since lz = A, the area enclosed by the two paths, then

δφ =m2

ngAλ

2πh2

(c) For λ = 1.1× 10−10 m and A = 10−3 m2, this gives δφ ≈ 43 radians.As the interferometer is rotated about the beam axis, maxima and minima in theneutron counting rate will be observed. Rotation by π/2 (which results in no heightdifference between the arms, i.e. z = 0) will change the phase by 43 radians, hence≈ 43/2π ≈ 7 “fringes” will pass by during the rotation. Continuing to rotate theapparatus past z = 0 will reverse the direction of movement of the fringes but eachπ/2 will give another ≈ 7 fringes. Hence, a complete rotation will sweep through43× 4/2π ≈ 27 fringes.

Experimental test of Gravitationally induced Quantum Interference, performed at the FordMotor Company reactor, and reported in PHYSICS TODAY, December 1980.

2


The Schrodinger equation and probability


1. (a) The time-dependent Schrodinger equation for a free particle is

ih∂ψ

∂t= − h2

2m∂2ψ

∂x2

For the given solution ψ = Ae−i(ωt−kx)

∂ψ

∂t= −iωψ, ∂ψ

∂x= ikψ so

∂2ψ

∂x2= −k2ψ

Hence, this satisfies the free particle Schrodinger equation if

hω =h2k2

2m

The de Broglie relations are that E = hω and p = hk so this would mean

E =p2

2m

which is the correct relationship between E and p for a free particle (i.e. when V = 0)and so this solution is consistent with the de Broglie relations.

(b) For the solution ψ = Ae−i(ωt+kx), then

∂ψ

∂t= −iωψ and

∂ψ

∂x= −ikψ so

∂2ψ

∂x2= −k2ψ

which is the same as before and hence is also a solution. For A = 1, then

Re(ψ) = cos(kx+ ωt)

At ωt = 0 we have Re(ψ) = cos(kx), at ωt = π/4 we have Re(ψ) = cos(kx + π/4),and at ωt = π/2 we have Re(ψ) = cos(kx+ π/2). These are shown below.

1

(c) For either wavefunction

|ψ|2 = A∗ei(ωt∓kx) ×Ae−i(ωt∓kx) = |A|2

which has no spatial variation.(d) Using the derivative calculated above, then

J =h

mIm [ψ∗(±ikψ)] = ± hk

m|ψ|2 = ± p

m|A|2 = ±v|A|2

using the de Broglie relation for p. Hence, the current is proportional to the classicalvelocity of the particle and is in the same direction. It again has no spatial variation.Changing k will not affect |ψ|2 but J will change proportionally to k. This illustratesthe importance of considering both |ψ|2 and J .

2. (a) Each of the solutions ψn satisfy the Schrodinger equation and this can be written as(ih∂

∂t+h2

2m∂2

∂x2− V

)ψn = 0

Trying the superposition state ψs, then(ih∂

∂t+h2

2m∂2

∂x2− V

)ψs =

(ih∂

∂t+h2

2m∂2

∂x2− V

)(αψ1 + βψ2)

= α

(ih∂

∂t+h2

2m∂2

∂x2− V

)ψ1 + β

(ih∂

∂t+h2

2m∂2

∂x2− V

)ψ2

= 0

and hence ψs is indeed also a solution.(b) For u1, then

∂u1

∂x= − π

2a√a

sin(πx/2a) so∂2u1

∂x2= − π2

4a2√a

cos(πx/2a) = − π2

4a2u1

Hence, its energy eigenvalue is given by

− h2

2m∂2u1

∂x2=

π2h2

8a2mu1 = E1u1

so E1 = π2h2/8a2m. Similarly for u2

∂u2

∂x=

π

a√a

cos(πx/a) so∂2u2

∂x2= − π2

a2√a

sin(πx/2a) = −π2

a2u2

and so

− h2

2m∂2u2

∂x2=

π2h2

2a2mu2 = E2u2

so E2 = π2h2/2a2m = 4E1 and so ∆E = 3π2h2/8a2m.(c) The probability densities are

|ψ1|2 = u∗1(x)eiE1t/hu1(x)e−iE1t/h = |u1|2 =

1a

cos2(πx/2a)

|ψ2|2 = u∗2(x)eiE2t/hu2(x)e−iE2t/h = |u2|2 =

1a

sin2(πx/a)

These are shown below. Since the un are both real, and hence they (and theirderivatives) will have no imaginary component, then it is clear that J = 0 for both.This is true for any bound energy eigenstate as there is no overall motion.

2

(d) The superposition probability density is

|ψs|2 =12(ψ∗1 + ψ∗2)(ψ1 + ψ2) =

12

(u1e

iE1t/h + u2eiE2t/h

) (u1e

−iE1t/h + u2e−iE2t/h

)=

12

[|u1|2 + |u2|2 + u1u2

(ei∆Et/h + e−i∆Et/h

)]=

12

[|u1|2 + |u2|2 + 2u1u2 cos (∆Et/h)

]This is

|ψs|2 =12a

[cos2(πx/2a) + sin2(πx/a) + 2 cos(πx/2a) sin(πx/a) cos (∆Et/h)

]which is shown below for ∆Et/h = 0, ∆Et/h = π/4 and ∆Et/h = π/2.

Note, athough the probability density for the superposition changes with time, it doesnot look like the travelling wave in that there is not one peak which moves along thex axis. The two peaks increase and decrease in size while centred on effectively thesame position.

3


Square wells and potential steps


1. (a) There are five places where the tangent and cotangent curves cross the circle. Eachof these points corresponds to a bound state. There are therefore five bound states.The lowest state is even parity as it has the least number of turning points and theythen alternate, so both the first and the fifth are even parity states.

(b) The solid line y axis intercept, γ0, is given by the value of γ when k = 0 which isγ2

0 = 2mV/h2. This is

γ0 =√

2× 9.109× 10−31 × 3.4× 1.602× 10−19

1.055× 10−34= 9.44× 109 m−1 = 9.44 nm−1

which is consistent with the figure.The lowest odd parity state is the second line and the x axis intercept corresponds tocot(ka) = 0, for which k = π/2a. With a = 0.7 nm, then this is k = 2.24 nm−1, againconsistent with the figure. (Note the total width is 2a = 1.4 nm so a = 0.7 nm.)

(c) Approximate values for the lowest and highest states are k1 ∼ 2 nm−1 and k5 ∼9 nm−1. The equation to solve is

f(y) = cos(y)− y

γ0a= cos(y)− 0.1513y = 0

Numerically, this gives y1 = k1a = 1.363, so k1 = 1.95 nm−1. Also y5 = k5a = 6.481,so k5 = 9.26 nm−1. Using γ = k tan(ka), then these give γ1 = 9.25 nm−1 and γ5 =1.86 nm−1.

(d) From the handout, for the even parity states we have

uL = Beγx

uC =Be−γa

cos kacos kx

uR = Be−γx

The probability of being within the well is

Pin =∫ a

−a|uC |2 dx =

∫ a

−a

(Be−γa

cos ka

)2

cos2 kx dx

=B2e−2γa

cos2 ka

∫ a

−a

12

(1 + cos 2kx) dx =B2e−2γa

2 cos2 ka

[x +

12k

sin 2kx

]a−a

=B2e−2γa

k cos2 ka

(ka +

12

sin 2ka

)while

Pout =∫ −a

−∞(Beγx)2 dx +

∫ ∞

a

(Be−γx)2 dx = 2

∫ ∞

aB2e−2γx dx

where we used the fact that the probability density function is symmetrical so thatwe can write the total probability of being outside the well as being two times theprobability of being outside the well on the right hand side. This gives

Pout = 2B2[− 1

2γe−2γx

]∞a

=B2

γe−2γa

1

HencePin

Pout=

B2e−2γa

k cos2 ka

(ka + 1

2 sin 2ka)

B2

γ e−2γa=

γ

k

[ka + 1

2 sin 2ka

cos2 ka

]Inserting the values for k and γ into this formula gives Pin/Pout ∼ 170 for the loweststate and Pin/Pout ∼ 1.3 for the highest state. For states near the top of the well theprobability of the particle being in the classically forbidden region can be very large.

(e) Changing V and not a means only the solid line changes its radius. To have twostates just fit in the well the radius of the circle has to decrease until it just meetsthe curve for the second state. This would require the radius to have the value foundin part (b), i.e. √

2mV ′

h=

π

2aand so

V ′ =h2π2

8ma2=

(1.055× 10−34 × 3.142)2

8× 9.109× 10−31 × 0.7× 10−9 × 0.7× 10−9= 3.08×10−20 J = 0.19 eV

2. (a) The wavefunctions are

x < 0 : uL = A exp(ikLx) + B exp(−ikLx)x > 0 : uR = C exp(ikRx) + D exp(−ikRx)

where k2L = 2mE/h2 and k2

R = 2m(E − V0)/h2. The particles are incident from theleft so D = 0. Continuity of u and u′ at x = 0 leads to

A + B = C, A−B =kR

kLC

which give

A =kL + kR

2kLC, B =

kL − kR

2kLC =

kL − kR

kL + kRA

Including the flux factors of hk/m, we find

Reflection coeff :|B exp(−ikLx)|2hkL/m

|A exp(ikLx)|2hkL/m=

|B|2

|A|2=(

kL − kR

kL + kR

)2

Transmission coeff :|C exp(ikRx)|2hkR/m

|A exp(ikLx)|2hkL/m=

|C|2

|A|2kR

kL=

4kLkR

(kL + kR)2

Summing the above two expressions gives

R + T =(kL − kR)2

(kL + kR)2+

4kLkR

(kL + kR)2=

k2L + k2

R − 2kLkR + 4kLkR

(kL + kR)2

=k2

L + k2R + 2kLkR

(kL + kR)2=

(kL + kR)2

(kL + kR)2= 1

Let y = (E − V0)/V0 so E = V0(y + 1) and hence

kL =√

2mE

h=√

y + 1√

2mV0

h, kR =

√2m(E − V0)

h=√

y

√2mV0

h

In terms of y, the expressions for R and T become

R =(√

y + 1−√y)2

(√

y + 1 +√

y)2, T =

4√

y(y + 1)(√

y + 1 +√

y)2

These are shown below.

2

(b) Classically, for E > V0 all particles would pass through, which corresponds to thehigh energy limit in quantum mechanics.

(c) The only change for V (x) = −V0 is that now k2R = 2m(E + V0)/h2. Otherwise, the

calculation follows as above. In particular, the expressions for R and T in terms ofkL and kR are unaffected. Defining z = E/V0, then E = zV0 and E + V0 = (z + 1)V0

so

kL =√

2mE

h=√

z

√2mV0

h, kR =

√2m(E + V0)

h=√

z + 1√

2mV0

h

Hence

R =(√

z −√

z + 1)2

(√

z +√

z + 1)2T =

4√

z(z + 1)(√

z +√

z + 1)2

These are identical functions of z to the expressions before for y so the form ofthe result is the same in terms of the energy above the minimum for either case,independent of the sign of V0.Note that for very low energies, there is a large probability of reflection. Classically,all particles would pass through the barrier, and this again corresponds to the highenergy limit.

3. The potential is shown below.

The wavefunctions in the three regions are

uL = 0 since V = ∞uC = A exp(ikx) + B exp(−ikx)uR = C exp(−γx) + D exp(γx)

3

where k2 = 2mE/h2 and γ2 = 2m(W − E)/h2. We take γ as positive and put D = 0,otherwise uR diverges as x →∞.

Continuity of u at x = 0 : A + B = 0 (1)Continuity of u at x = a : A exp(ika) + B exp(−ika) = C exp(−γa) (2)Continuity of u′ at x = a : ik[A exp(ika)−B exp(−ika)] = −γC exp(−γa) (3)

Substituting eqn(1) into (2) and (3) and using the identities

eiθ − e−iθ = 2i sin θ, eiθ + e−iθ = 2 cos θ

with θ = ka gives

A · 2i sin ka = C exp(−γa), A · 2 cos ka =iγ

kC exp(−γa)

and so dividing gives γ = −k cot ka as required. [12 marks]

The energy levels are obtained by solving the equations:

γ = −k cot ka and k2 + γ2 =2mW

h2

Graphical solutions are shown in the figure below. The first equation gives the series ofcotangent curves, the second equation gives the circle. For a solution, then the radius of

the circle has to be bigger than π/2a so as to overlap the first cotangent curve. Hence√2mW

h2 >π

2a

which means2mW

h2 >π2

4a2

so

W >h2π2

8ma2

[4 marks]

If W is much larger then the energy level spacing then the circle radius will be large.Hence, it will come out almost horizontally from the γ axis, so the value of k1 will be veryclose to its limiting value of π/a. Hence

E1 =h2k2

1

2m≈ h2π2

2ma2

[4 marks]

4


Simple harmonic oscillators

Paul Dauncey, Released 4 Nov 2011

1. Recall the Hermitian condition∫ψ∗1Aψ2 dx =

∫(Aψ1)∗ψ2 dx

To prove that αx+ ip is not Hermitian, consider∫ψ∗1(αx+ ip)ψ2 dx = α

∫ψ∗1xψ2 dx+ i

∫ψ∗1 pψ2 dx

= α

∫(xψ1)∗ψ2 dx+ i

∫(pψ1)∗ψ2 dx

=∫

(αxψ1)∗ψ2 dx−∫

(ipψ1)∗ψ2 dx

which uses the fact that x and p are Hermitian. This can be written as∫ψ∗1(αx+ ip)ψ2 dx =

∫[(αx− ip)ψ1]∗ψ2 dx

Hence ∫ψ∗1(αx+ ip)ψ2 dx 6=

∫[(αx+ ip)ψ1]∗ψ2 dx

and we conclude that αx + ip is not Hermitian. Similarly, to prove that αx − ip is notHermitian∫

ψ∗1(αx− ip)ψ2 dx = α

∫ψ∗1xψ2 dx− i

∫ψ∗1 pψ2 dx

= α

∫(xψ1)∗ψ2 dx− i

∫(pψ1)∗ψ2 dx

=∫

(αxψ1)∗ψ2 dx+∫

(ipψ1)∗ψ2 dx =∫

[(αx+ ip)ψ1]∗ψ2 dx

Hence ∫ψ∗1(αx− ip)ψ2 dx 6=

∫[(αx− ip)ψ1]∗ψ2 dx

and we conclude that αx− ip is also not Hermitian.

2. (a) Taking a derivative of u0 gives

du0

dx= −mωx

hAe−mωx2/2h = −mωx

hu0

Hence using the product rule

d2u0

dx2= −mω

hu0 −

mωx

h

du0

dx= −mω

hu0 +

(mωx

h

)2

u0

Substituting into the TISE with the SHO potential V (x) = mω2x2/2

− h2

2md2u0

dx2+ V (x)u0 =

hω

2u0 −

mω2x2

2u0 +

mω2x2

2u0 =

hω

2u0 = E0u0

Hence, u0 is indeed an energy eigenstate and its eigenvalue is E0 = hω/2.

1

(b) Operating with mωx− ip gives

u1 = (mωx− ip)u0 = mωxu0 − hdu0

dx= mωxu0 +mωxu0 = 2mωxu0

Taking derivatives gives

du1

dx= 2mωu0 + 2mωx

du0

dx= 2mωu0 −

2m2ω2x2

hu0 = 2mω

(1− mωx2

h

)u0

and so

d2u1

dx2= −4m2ω2x

hu0 + 2mω

(1− mωx2

h

)du0

dx

= −4m2ω2x

hu0 − 2mω

(1− mωx2

h

)mωx

hu0

=

(−4m2ω2x

h− 2m2ω2x

h+

2m3ω3x3

h2

)u0

=

(−3mω

h+m2ω2x2

h2

)u1

Hence, substituting into the TISE gives

− h2

2md2u1

dx2+mω2x2

2u1 =

3hω2u1 −

mω2x2

2u1 +

mω2x2

2u1 =

3hω2u1 = E1u1

Therefore u1 is a solution of the TISE with eigenvalue E1 = 3hω/2. From the formulagiven in the lectures En = (n + 1/2)hω, then it is seen that this must be the firstexcited state.

(c) Operating with mωx+ ip on u1 gives

(mωx+ ip)u1 = mωxu1 + hdu1

dx= 2m2ω2x2u0 + 2mωh

(1− mωx2

h

)u0 = 2mωhu0

and so this is proportional to u0, the ground state.

(d) Operating with mωx+ ip on u0 gives

(mωx+ ip)u0 = mωxu0 + hdu0

dx= mωxu0 −mωxu0 = 0

Since u0 is the ground state, it in not possible to go any lower than this in energy.Hence, operating with mωx+ ip on this state gives zero, not another eigenstate.

This use of operators to jump between simple harmonic oscillator eigenstates will be dis-cussed in more detail later in the course.

3. (a) For a harmonic oscillator the energy is En = (n + 1/2)hω. For an ion in the n = 2state the energy is

E2 =5h2× 2πν = 2.5× 1.05× 10−34 × 2× 3.1415× 1× 106

= 1.65× 10−27 J = 1.03× 10−8 eV

(b) This gives a temperature around 2.4× 10−4 K.

2

(c) If another measurement is made immediately after the first, the same result is foundsince the system is no longer in a superposition. The wavefunction collapsed onto then = 2 eigenstate as a result of the first measurement.As the n = 2 eigenstate is an energy eigenstate, the system will stay in this energyas long as it is not disturbed. Hence, an isolated ion would always give the value forn = 2 for any future measurement.

(d) In reality the ion will be disturbed by the black-body radiation within the cavity.If the cavity is at a very low temperature, much smaller than the characteristictemperature found above, then the radiation will not have high enough energy todisturb the ion significantly and so the energy measurement would still always givethe value for n = 2. However, for higher temperatures, it will in general be in asuperposition of average energy around the appropriate kBT/2 and so the energymeasurement will tend to give a result around this value.

4. (i) The energies of a simple harmonic oscillator are E = (n + 1/2)hω. Hence, theE = 3hω/2 state has n = 1. The probability that this energy will be measured isgiven by the modulus squared of the overlap integral for the wavefunction with thisstate∫ ∞

−∞u∗1ψ dx =

(2mω1

πh

)1/4 ( 4π

)1/4 (mω1

h

)3/4 ∫ ∞

−∞x exp(−mω1x

2/2h) exp(−mω1x2/h) dx

=(

2mω1

πh

)1/4 ( 4π

)1/4 (mω1

h

)3/4 ∫ ∞

−∞x exp(−3mω1x

2/2h) dx

This is clearly an odd function and so the integral will give zero.[4 marks]

(ii) For the E = 5hω/2 state, then n = 2. Hence, the probability is given by the squareof the modulus of∫ ∞

−∞u∗2ψ dx =

(2mω1

πh

)1/4 (mω1

4πh

)1/4 ∫ ∞

−∞[2(mω1/h)x2 − 1] exp(−3mω1x

2/2h) dx

=

(m2ω2

1

2π2h2

)1/4 ∫ ∞

−∞

2mω1

hx2 exp(−3mω1x

2/2h)− exp(−3mω1x2/2h) dx

Using the given integrals, then∫ ∞

−∞x2 exp(−3mω1x

2/2h) dx =h

3mω1

√2hπ

3mω1

and ∫ ∞

−∞exp(−3mω1x

2/2h) dx =

√2hπ

3mω1

so the overlap is

=

(m2ω2

1

2π2h2

)1/4(2mω1

h

h

3mω1

√2hπ

3mω1−√

2hπ3mω1

)

=

(m2ω2

1

2π2h2

)1/4√2hπ

3mω1

(23− 1

)

= −13

(29

)1/4

= −0.229

3

Hence

P1 =19

(29

)1/2

=√

227

= 0.052

[6 marks]

(iii) The required overlap integral is now∫ ∞

−∞u′∗0 ψ dx =

(2mω1

πh

)1/4 (mω2

πh

)1/4 ∫ ∞

−∞exp(−mω1x

2/h) exp(−mω2x2/2h) dx

=

(2m2ω1ω2

π2h2

)1/4 ∫ ∞

−∞exp[−m(2ω1 + ω2)x2/2h] dx

=

(2m2ω1ω2

π2h2

)1/4√2hπ

m(2ω1 + ω2)

= (2ω1ω2)1/4

√2

(2ω1 + ω2)

and so

P2 = (2ω1ω2)1/2 2

(2ω1 + ω2)

=√

2ω1ω2

ω1 + ω2/2

[5 marks]

(iv) For P2 = 1, then√

2ω1ω2 = ω1 + ω2/22ω1ω2 = ω2

1 + ω22/4 + ω1ω2

0 = ω21 + ω2

2/4− ω1ω2

0 = (ω1 − ω2/2)2

and hence ω2 = 2ω1. For this case, then the ground state of the new potential wouldbe (

mω2

πh

)1/4

exp(−mω2x2/2h) =

(2mω1

πh

)1/4

exp(−mω1x2/h) = ψ

and hence for P2 = 1, then the wavefunction ψ is the ground state of the new potential.[5 marks]

4


Expectation values and time dependence


1. (a) The required overlap integral is

a0 =∫ ∞

−∞u∗0ψ dx =

√αβ

π

∫ ∞

−∞exp

[−(α2 + β2)x2

2

]dx =

√2αβ

α2 + β2

Hence, the probability of measuring E0 is the modulus squared of a0

P0 = |a0|2 =2αβ

α2 + β2

(b) The expectation value of kinetic energy is given by⟨p2

2m

⟩=

12m

∫ ∞

−∞ψ∗p2ψ dx = − h2

2m

∫ ∞

−∞ψ∗d2ψ

dx2dx

Since

d2ψ

dx2=

d

dx

(−β2xψ

)= −β2ψ + β4x2ψ

then ⟨p2

2m

⟩= − h2

2mβ√π

∫ ∞

−∞exp(−β2x2)(β4x2 − β2) dx

= − h2β2

4m+h2β2

2m=h2β2

4m

(c) For β = α, then

P0 =2α2

α2 + α2= 1

so the particle must be in the ground state; this is clear by putting β = α into theexpression for ψ. Also, the expectation value of the kinetic energy is⟨

p2

2m

⟩=h2α2

4m=

h2

4mmω0

h=hω

4

The total energy of the ground state is hω/2 so the ratio of the two is 1/2.For a classical SHO particle, then the momentum varies sinusoidally; e.g. p =p0 sin(ω0t), where p0 is the maximum momentum of the particle, which it has whenx = 0. The total energy is therefore the kinetic energy at this maximum, i.e.E = p2

0/2m and the average kinetic energy is

T =p20

2m

∫ 2π/ω0

0sin2(ωt) dt =

p20

4m=E

2

since sin2 averages to 1/2. Hence the ratio of the expectation value (or average) ofthe kinetic energy to the total energy is the same for classical and quantum SHOsystems.

1

2. (a) The standard method for evaluating the expectation value is to (i) check the normal-isation of ψ(x), then (ii) evaluate the integral of ψ∗(x)Hψ(x) over all space; this willalways work. But we note here that the wave function ψ(x) is written as a sum ofthe two energy eigenfunctions, u0(x) and u1(x), and this allows a short cut.

ψ(x) =(α√π

)1/2 (1√2

+ αx

)exp(−α2x2/2)

can be written as

ψ(x) =1√2

[u0(x) + u1(x)] = a0u0 + a1u1

with a0 = a1 = 1/√

2 and all other coefficients for other energy states are zero. Since∑n

|an|2 =12

+12

= 1

then the wavefunction is indeed normalised. Hence the probabilities of measuring theenergies are

P0 = |a0|2 =12, P1 = |a1|2 =

12

and so

〈E〉 =∑n

PnEn =12E0 +

12E1 =

12.12hω0 +

12.32hω0 = hω0

(b) The time dependent wave function ψ(x, t) is obtained from the wave function att = 0, by including the complex exponential time dependence on each of the energyeigenfunctions

ψ(x, t = 0) =1√2

[u0(x) + u1(x)]

ψ(x, t) =1√2

[u0(x) exp(−iE0t/h) + u1(x) exp(−iE1t/h)]

(c) To determine the time dependence of 〈x〉 we need to use the time dependent wavefunction ψ(x, t). Noting u0 and u1 are real, so u∗i = ui, then

〈x〉 =∫ ∞

−∞ψ∗(x, t)xψ(x, t) dx

=∫ ∞

−∞

12u2

0x+12u2

1x+ <{u0u1x exp[i(E1 − E0)t/h]} dx

=∫ ∞

−∞

12u2

0x+12u2

1x+ +u0u1x cos[(E1 − E0)t/h] dx

=12

∫ ∞

−∞u2

0x dx+12

∫ ∞

−∞u2

1x dx+ cos[(E1 − E0)t/h]∫ ∞

−∞u0u1xdx

The first two integrals are zero as they are odd functions and the third is a constantwith time, so

〈x〉 =∝ cos[(E1 − E0)t/h]

which is SHM with angular frequency

ω =E1 − E0

h=

3hω0/2− hω0/2h

= ω0

2

(d) The frequency of oscillation of a superposition of two energy states is always ∆E/h,so for the ground state and the second excited state, it would be 2ω0.

3. The eigenstates φn and eigenvalues an of A satisfy

Aφn = anφn

Hence

φ∗nAφn = anφ∗nφn∫ ∞

−∞φ∗nAφn dx = an

∫ ∞

−∞φ∗nφn dx

Using the Hermitian condition as given in the question for the particular case whereψ1 = ψ2 = φn, then the above leads to∫ ∞

−∞(Aφn)∗φn dx = an

∫ ∞

−∞φ∗nφn dx

The complex conjugate of the eigenvalue equation is

(Aφn)∗ = a∗nφ∗n

so that

(Aφn)∗φn = a∗nφ∗nφn∫ ∞

−∞(Aφn)∗φn dx = a∗n

∫ ∞

−∞φ∗nφn dx

Hence

an

∫ ∞

−∞φ∗nφn dx = a∗n

∫ ∞

−∞φ∗nφn dx

and so an = a∗n, which means an must be real.

[4 marks]

The expectation value of the operator in any state ψ is

〈A〉 =∫ ∞

−∞ψ∗Aψ dx

Hence

d

dt〈A〉 =

∫ ∞

−∞

∂ψ∗

∂tAψ + ψ∗A

∂ψ

∂tdx

The Schrodinger equation and its complex conjugate give

∂ψ

∂t= − i

hHψ,

∂ψ∗

∂t=i

h(Hψ)∗

so substituting in

d

dt〈A〉 =

i

h

∫ ∞

−∞(Hψ)∗Aψ − ψ∗AHψ dx =

i

h

∫ ∞

−∞ψ∗HAψ − ψ∗AHψ dx

=i

h

∫ ∞

−∞ψ∗[H, A]ψ dx =

i

h〈[H, A]〉

3

[7 marks]

If an operator commutes with the Hamiltonian operator, then the RHS of the above iszero. Hence, the expectation value does not change with time and so it is a conservedquantity. One example is conservation of momentum when there is no potential. In thiscase

H =p2

2m

so clearly

[H, p] =1

2m[p2, p] = 0

and so

d

dt〈p〉 =

i

h〈[H, p]〉 = 0

A second example is conservation of energy, which is always true for the conservativesystems considered here, i.e. where all the forces are due to a potential V = V (x). In thiscase then trivially

[H, H] = 0

so

d

dt〈H〉 =

i

h〈[H, H]〉 = 0

[3 marks]

Classically momentum is not conserved if there is a potential which gives a force. Thegeneral Hamiltonian operator with a potential is

H =p2

2m+ V (x)

for which

[H, p] = [V, p]

This commutator can be found by considering

[V, p]ψ = V pψ − pV ψ

= −ihV ∂ψ∂x

+ ih∂(V ψ)∂x

= −ihV ∂ψ∂x

+ ihV∂ψ

∂x+ ih

dV

dxψ

= ihdV

dxψ

Hence

d

dt〈p〉 =

i

h

⟨ihdV

dx

⟩= −

⟨dV

dx

⟩

4

This is a form of Newton’s law of motion which states that classically

dp

dt= F = −dV

dx

This is also the standard form for one of the two Hamilton equations resulting from theHamiltonian formalism of classical mechanics.

[6 marks]

5


Momentum space and the Uncertainty Principle


1. (a) The RMS uncertainty in x is given by

(∆x)2 = 〈x2〉 − 〈x〉2

Since |ψ|2 is an even function, then x|ψ|2 gives an odd function and so 〈x〉 = 0. Hence

(∆x)2 = 〈x2〉 =∫ ∞

−∞ψ∗x2ψ dx =

12a

∫ a

−ax2 dx =

12a

[13x3]a−a

=a2

3

so ∆x = a/√

3.(b) The amplitude for momentum is given by the overlap integral of ψ with a momentum

eigenstate, i.e. the Fourier transform of the spatial wavefunction

a(p) = A∗∫ ∞

−∞e−ipx/hψ(x) dx = A∗

∫ a

−a

1√2ae−ipx/h dx

=A∗√2a

−hip

[e−ipx/h

]a−a

=A∗√2a

(−hip

)[−2i sin(pa/h)] =

√2aA∗

sin(pa/h)pa/h

Hence the probability density is

|a(p)|2 = 2a|A|2 sin2(pa/h)(pa/h)2

(c) The relative probability for p = πh/2a to p = 0 is

1(π/2)2

:sin(0)

0=

4π2

: 1

(d) For a beam of particles of momentum pz incident on a slit of width x = 2a, thefunction |a(px)|2 determines the probability that a particle emerging from the slit hasmomentum px. The angle of the particle to the beam is thus θ ≈ tan θ = px/pz, so|a(px)|2 above determines the angular distribution, i.e. the diffraction pattern. Thisis shown below (with a(px) = Φ(px)).

The first diffraction minimum is when sin pxa/h = 0, i.e. px = hπ/a. This occurs atan angle θ given by

θ ≈ tan θ =px

pz=hπ

a

λ

2πh=

λ

2a

where λ is the de Broglie wavelength 2πh/p ≈ 2πh/pz.

1

2. Remembering that the expectation values 〈x〉 and 〈p〉 are just numbers, then the conditionQ′ψ = λR′ψ from Handout 5 amounts to

(p− 〈p〉)u = λ(x− 〈x〉)u

−ihdudx

= (λx+ 〈p〉 − λ〈x〉)u

du

u=

i

h(λx+ 〈p〉 − λ〈x〉) dx

ln(u) =i

h

(λx2

2+ [〈p〉 − λ〈x〉]x

)+ C

u = exp(C) exp(iλx2/2h) exp[i(〈p〉 − λ〈x〉)x/h]

The middle term is oscillatory for real λ and so is not normalisable unless λ has animaginary part to damp down the solution at large x. Any real part of λ results in anoscillatory term which cancels in the probability density, so for simplicity, let λ = iah.Hence

u = exp(C) exp(−ax2/2) exp(i〈p〉x/h+ a〈x〉x)= exp(C) exp[−a(x2/2− 〈x〉x)] exp(i〈p〉x/h)= exp(C + a〈x〉2/2) exp[−a(x2/2− 〈x〉x+ 〈x〉2/2)] exp(i〈p〉x/h)= exp(C + a〈x〉2/2) exp[−a(x− 〈x〉)2/2] exp(i〈p〉x/h)= A exp[−a(x− 〈x〉)2/2] exp(i〈p〉x/h)

where A is a constant and so

|u|2 = |A|2 exp[−a(x− 〈x〉)2]

as required.

To evaluate {x, p}, then since [x, p] = xp− px = ih

{x, p} = xp+ px = 2xp− ih = −2ih(xd

dx+

12

)Hence ⟨

12{x, p}

⟩= −ih

⟨xd

dx+

12

⟩Using the expression for u(x) above

xdu

dx= x

[−a(x− 〈x〉) +

i〈p〉h

]u

then ⟨12{x, p}

⟩= iah〈x2〉 − iah〈x〉2 + 〈x〉〈p〉 − ih

2

Explicitly, with y =√a(x− 〈x〉), then

〈x2〉 = |A|2∫x2 exp[−a(x− 〈x〉)2] dx =

|A|2√a

∫ (y√a

+ 〈x〉)2

exp(−y2) dy

=|A|2√a

∫ (y2

a+

2y〈x〉√a

+ 〈x〉2)

exp(−y2) dy

=|A|2

a√a

∫y2 exp(−y2) dy +

|A|2√a〈x〉2

∫exp(−y2) dy

2

since the middle integral is an odd function and so is zero. Using the given integrals, thenfor the second integral

|A|2√a

∫exp(−y2) dy =

|A|2√a

√π = |A|2

√π

a= 1

so |A|2 =√a/π and so for the second integral

|A|2

a√a

∫y2 exp(−y2) dy =

|A|2

a√a

√π

2= |A|2

√π

a

12a

=12a

Hence

〈x2〉 =12a

+ 〈x〉2

and so ⟨12{x, p}

⟩=ih

2+ iah〈x〉2 − iah〈x〉2 + 〈x〉〈p〉 − ih

2= 〈x〉〈p〉

and hence the anticommutator term is indeed zero.

3. (i) The expectation value of the observable associated with an operator Q is

〈Q〉 =∫ ∞

−∞ψ∗Qψ dx

[2 marks]

(ii) (a) To normalise the wavefunction then we require∫ ∞

−∞ψ∗ψ dx = 1 =

∫ ∞

−∞|A|2 exp(−ax2) dx

Setting y =√ax, so dx = dy/

√a then

1 =∫ ∞

−∞|A|2 exp(−y2)

dy√a

= |A|2√π

a

Hence, we need to set

A =(a

π

)1/4

[3 marks](b) The expectation value of x is

〈x〉 =√a

π

∫ ∞

−∞x exp(−ax2) dx = 0

as this is an odd function of x.[1 mark]

(c) In general the expectation value of the mean square of a variable is

∆Q2 = 〈(Q− 〈Q〉)2〉 = 〈Q2 − 2Q〈Q〉+ 〈Q〉2〉= 〈Q2〉 − 2〈Q〉〈Q〉+ 〈Q〉2〈〉

3

The last expression here is 〈〉 =∫ψ∗ψ dx = 1 so

∆Q2 = 〈Q2〉 − 2〈Q〉2 + 〈Q〉2 = 〈Q2〉 − 〈Q〉2

Since 〈x〉 = 0, then the mean square of x is

∆x2 = 〈x2〉 =√a

π

∫ ∞

−∞x2 exp(−ax2) dx

Again with y =√ax, so dx = dy/

√a then this is

∆x2 = 〈x2〉 =√a

π

∫ ∞

−∞

y2

aexp(−ax2)

dy√a

=1

a√π

√π

2=

12a

Hence, the rms is ∆x = 1/√

2a.[2 marks]

(d) For momentum, we will need

pψ = −ih ddx

[√a

πexp(ikx) exp

(−ax2

2

)]

= −ih[ik

√a

πexp(ikx) exp

(−ax2

2

)− ax

√a

πexp(ikx) exp

(−ax2

2

)]= −ih(ik − ax)ψ= h(k + iax)ψ

Hence

〈p〉 =∫ ∞

−∞ψ∗pψ dx = hk

(∫ ∞

−∞ψ∗ψ dx

)+ iha〈x〉 = hk

[4 marks](e) As above, the mean square of momentum is given by

∆p2 = 〈p2〉 − 〈p〉2

We need

p2ψ = p [h(k + iax)ψ]= h2(k + iax)2ψ + (−ih)(hia)ψ= h2(k2 − a2x2 + 2ikax)ψ + h2aψ

= h2(k2 + a− a2x2 + 2ikax)ψ

Hence

〈p2〉 = h2(k2 + a)(∫ ∞

−∞ψ∗ψ dx

)− h2a2〈x2〉+ 2ih2ka〈x〉

= h2(k2 + a)− h2a

2

= h2(k2 +

a

2

)and so the mean square is

∆p2 = 〈p2〉 − 〈p〉2 = h2(k2 +

a

2

)− h2k2 =

h2a

2

and so the rms is h√a/2.

[5 marks]

4

(f) Hence the product of the two rms values is

∆x∆p =1√2a

× h

√a

2=h

2

[1 mark]

(iii) The above product is the minimum possible value allowed by the Uncertainty Prin-ciple. This arises from the fact that the momentum space distribution is the Fouriertransform of the position space distribution, so a narrow distribution in one vari-able results in a wide spread in the other. The wavefunction here is the minimumuncertainty product because it is of the form of a Gaussian, i.e. |ψ|2 ∼ exp(−ax2).

[2 marks]

5


SHO ladder operators


1. This problem covers the calculation of the uncertainty product for any eigenstate of theharmonic oscillator.

(a) Subtracting and adding the equations gives

x =1√2

√h

mω(A + A†)

p = − i√2

√hmω(A− A†)

(b) Hence

x2 = xx =h

2mω(A + A†)(A + A†)

=h

2mω(A2 + AA† + A†A + A†2)

p2 = pp = − hmω

2(A− A†)(A− A†)

= − hmω

2(A2 − AA† − A†A + A†2)

(c) The expectation values are evaluated by sandwiching the operators between u∗n(x)and un(x) and integrating over all space. Since A is the lowering operator, then

〈A〉 =∫

u∗nAun dx ∝∫

u∗nun−1 dx = 0

as the eigenstates are orthogonal. A similar calculation shows 〈A†〉 = 0. Hence

〈x〉 = 0, 〈p〉 = 0

For x2 and p2, the first term from each of the above expressions for these operatorsgives

〈A2〉 =∫

u∗nA2un dx ∝∫

u∗nAun−1 dx ∝∫

u∗nun−2 dx = 0

again since these two eigenstates are orthogonal. Similarly, the A†2 terms also givezero. However, the second term for each gives

〈AA†〉 =∫

u∗nAA†un dx ∝∫

u∗nAun+1 dx ∝∫

u∗nun dx 6= 0

and so does not go to zero. Hence, the only terms left are the middle two, so

〈x2〉 =h

2mω〈AA† + A†A〉

〈p2〉 =hmω

2

⟨AA† + A†A

⟩

1

(d) Using the equation for the Hamiltonian operator

AA† + A†A =2H

hω

so

〈x2〉 =〈H〉mω2

and 〈p2〉 = m〈H〉 .

But 〈H〉 for the state un is just En = (n + 1/2)hω so

∆x =√〈x2〉 − 〈x〉2 =

√〈x2〉 =

√(n + 1/2)h/mω

and ∆p =√〈p2〉 − 〈p〉2 =

√〈p2〉 =

√(n + 1/2)mhω

Hence ∆x∆p = (n + 1/2)h.

2. (a) The ground state by definition is the lowest energy state. However, A generallychanges a state into the state with an energy lower by hω. This is not possible forthe ground state so A must give zero in this case.

(b) As given in Qu. 1, in terms of A and A†, the Hamiltonian can be written as

H = hω

(AA† − 1

2

)= hω

(A†A +

12

)=

hω

2

(AA† + A†A

)Using the second of these forms, then

Humin = hω

(A†A +

12

)umin =

12hωumin

Hence, the energy of the ground state is hω/2.

(c) Using the explicit operators x = x and p = −ih d/dx, then the equation is

Aumin = 0 =1√2

√mω

hx +

√h

mω

d

dx

umin

so

dumin

dx= −mω

hxumin

dumin

umin= −mω

hx dx

ln(umin) = −mω

2hx2 + C

umin = exp(C) exp(−mωx2/2h)

is the ground state of the SHO, where C is a constant of integration which could befixed by normalisation.

(d) The case for

A†umax = 0

is very similar. Using the first form for H above, the energy is given by

Humax = hω

(AA† − 1

2

)umax = −1

2hωumax

2

so the energy of such a state would be −hω/2, which is clearly below the ground statewe found above. The state itself would be given by

A†umax = 0 =1√2

√mω

hx−

√h

mω

d

dx

umax

so

dumax

dx=

mω

hxumax

dumax

umax=

mω

hx dx

ln(umax) =mω

2hx2 + C

umax = exp(C) exp(mωx2/2h)

This would go to infinity as x → ±∞ and so is not normalisable. Hence, while it isa mathematical solution of the TISE, it does not correspond to a physical state.

3. (i) Consider

R−R+ = (P − iX)(P + iX) = P 2 + X2 + i[P , X] = P 2 + X2 + 1

Hence

H =hω

2(P 2 + X2) =

hω

2(R−R+ − 1)

so

R−R+ =2H

hω+ 1

[4 marks]

(ii) Using the above and the similar expression given in the question, then

H

hω=

12(R−R+ − 1) =

12(R+R− + 1)

Hence, using both the above, the commutator can be written as[R+,

H

hω

]= R+

H

hω− H

hωR+ = R+

12(R−R+ − 1)− 1

2(R+R− + 1)R+

=12(R+R−R+ − R+ − R+R−R+ − R+) =

12(−2R+) = −R+

[4 marks]

(iii) Consider

H(R+un) = HR+un − R+Hun + R+Hun = [H, R+]un + R+Enun

= hωR+un + EnR+un = (En + hω)(R+un)

so R+un is an eigenstate of H with an eigenvalue En + hω.[5 marks]

3

(iv) Since R+ is the adjoint of R−, then∫u∗nR−R+un dx =

∫(R+un)∗(R+un) dx =

∫c∗n+1u

∗n+1cn+1un+1 dx

= |cn+1|2∫

u∗n+1un+1 dx = |cn+1|2

[3 marks]

(v) Since En = (2n + 1)hω/2, then

∫u∗n

2H

hωun dx =

∫u∗n

2En

hωun dx = (2n + 1)

∫u∗nun dx = 2n + 1

and so

2n + 1 =∫

u∗n(R−R+ − 1)un dx =∫

u∗nR−R+un dx−∫

u∗nun dx = |cn+1|2 − 1

With cn+1 real, then

c2n+1 = 2n + 2 = 2(n + 1)

cn+1 =√

2(n + 1)

and so

R+un = cn+1un+1 =√

2(n + 1)un+1

[4 marks]

4


Perturbation solutions

Paul Dauncey, Released 2 Dec 2011

1. (a) The perturbed potential is shown below and is divided into a sum of the originalpotential plus a perturbing potential of height δ.

(b) The first order correction to the ground state energy is

E(1)1 =

∫ 0

−au∗1(x)δu1(x) dx =

δ

a

∫ 0

−acos

πx

2acos

πx

2adx

=2δ

π

∫ 0

−π/2cos2 θ dθ where θ = πx/2a

=2δ

π

[θ

2+

14

sin 2θ

]0−π/2

=2δ

π

π

4=

δ

2

i.e. half the wavefunction is in a potential of δ and the other half is in a potentialof 0, so ignoring the change to the eigenstate itself (as is the case for this first ordercalculation), then the energy is increased by δ/2.

(c) The first order correction to the ground state wavefunction from state um is

u(1)1 =

∑m6=1

a1mum =∑m6=1

1E1 − Em

(∫ 0

−au∗m δ u1 dx

)um

For the m = 2 contribution, then the integral is

a12 =1

E1 − E2

δ

a

∫ 0

−asin

2πx

2aδ cos

πx

2adx =

1E1 − E2

2δ

π

∫ 0

−π/2sin 2θ cos θ dθ

=1

E1 − E2

2δ

π

[−1

2cos θ − 1

6cos 3θ

]0−π/2

=1

E1 − E2

2δ

π

(−1

2− 1

6

)= − 1

E1 − E2

4δ

3π

Now E2 = π222/2ma2 = 4E1 so

a12 =4δ

9πE1≈ 0.14

δ

E1

If we limit ourselves to the admixture of the u2 state we have an approximation tothe ground state of the perturbed well given by U1 = u1 +a12u2. Since a12 is positive,then the sine function of u2 adds to u1 for x > 0 and subtracts for x < 0. Hence, thereis a greater probability of finding the electron on the right hand side of the well thanfor the unperturbed ground state wavefunction and conversely a lower probability onthe left hand side. This is to be expected since the the energy is higher on the left,i.e. the perturbation “repels” the electron weakly from the left hand side of the well.

1

(d) If δ becomes very large so the well can be considered to be an infinite square well inthe range 0 < x < a, then the width is half that of the original well. Hence, sincethe energies are proportional to 1/a2, then the new ground state energy E′

1 will beE′

1 = 4E1.Another way to see this is that the new ground state solution which goes to zero atx = 0 and x = a will have an identical shape to the right hand half of the originalfirst excited state. Hence, E′

1 = E2 = 4E1.The first order perturbation correction for the energy is proportional to δ so if thiswas valid for large δ, it would predict a very large ground state energy. This is seento be completely incorrect, showing the perturbation results are only valid for smallδ.

2. (i) The ground state correction is given by

u(1)0 =

∑m6=0

∫∞−∞ u∗mεxu0dx

E0 − Emum =

∑m6=0

C

∫ ∞

−∞

u∗mu1dx

E0 − Emum

=∑m6=0

Cδm1

E0 − Emum =

Cu1

E0 − E1

[4 marks]

(ii) Assuming the given relation, then

εxu0 = Cu1

εx

(α

π

)1/4

e−αx2/2 = C

(4π

)1/4

α3/4xe−αx2/2

C = ε

(12α

)1/2

= ε

(h

2mω

)1/2

Hence

U0 = u0 + u(1)0 = u0 + ε

(h

2mω

)1/2

×(

1−hω

)u1 = u0 − ε

(1

2mhω3

)1/2

u1

[3 marks]

(iii) The expectation value of x is

〈x〉 =∫ ∞

−∞u0xu0dx =

∫ ∞

−∞

(α

π

)1/4

e−αx2/2x

(α

π

)1/4

e−αx2/2dx

=(

α

π

)1/2 ∫ ∞

−∞xe−αx2

dx = 0 odd integrand.

[2 marks]

(iv) Including the perturbation, this changes to

〈x〉 =∫ ∞

−∞(u∗0 −

C∗

hωu∗1)x(u0 −

C

hωu1)dx

=∫ ∞

−∞u0xu0dx +

C2

h2ω2

∫ ∞

−∞u1xu1dx− 2C

hω

∫ ∞

−∞u0xu1dx

The first two integrands are odd leaving only the third to evaluate.

〈x〉 = −2C

hω

∫ ∞

−∞

(α

π

)1/4

e−αx2/2 x

(4π

)1/4

α3/4 x e−αx2/2dx

2

= −2C

hω

(α

π

)1/4 ( 4π

)1/4

α3/4∫ ∞

−∞x2 e−αx2

dx

= −2C

hω

41/4α

π1/2

1α3/2

∫ ∞

−∞y2 e−αy2

dy

where αx2 = y2 so that dx = dy/√

α.

〈x〉 = −2C

hω

41/4

π1/2

1α1/2

π1/2

2

= − C

hω

(2α

)1/2

= − ε

hω

(h

2mω

)1/2 ( 2h

mω

)1/2

= − ε

hω

(h2

m2ω2

)1/2

= −ε/mω2

Or alternatively, avoiding using the standard integral:

〈x〉 =∫ ∞

−∞(u∗0 −

C∗

hωu∗1)x(u0 −

C

hωu1)dx

=∫ ∞

−∞u0xu0dx +

C2

h2ω2

∫ ∞

−∞u1xu1dx− 2C

hω

∫ ∞

−∞u0xu1dx

The first two integrands are odd leaving only the third to evaluate. Now note thatu0x = Cu1/ε so that

〈x〉 = −2C

hω

C

ε

∫ ∞

−∞u∗1u1 = −2C2

hωε= − 2ε2

hωε2α= − εh

hωmω= − ε

mω2

The negative sign means the particle shifts slightly to the left (i.e. negative x) wherethe potential is lowered by the perturbation.

[8 marks]

(v) The first order energy shift is

E(1)0 =

∫ ∞

−∞u0εxu0dx =

∫ ∞

−∞

(α

π

)1/4

e−αx2/2εx

(α

π

)1/4

e−αx2/2dx

=(

α

π

)1/2

ε

∫ ∞

−∞xe−αx2

dx = 0

since it is an odd integral, the same as for part (iii).Or alternatively by noting εxu0 = Cu1

∆E =∫ ∞

−∞u0εxu0dx = C

∫ ∞

−∞u0u1dx = 0

since u0 and u1 are orthogonal.[3 marks]

3


Angular momentum


1. (a) i. Firstly, since r = (x2 + y2 + z2)1/2, then

∂r

∂x=

12(x2 + y2 + z2)−1/2 × (2x) =

x

r

Hence the required derivative of the function f(r) is

∂

∂xf(r) =

df

dr

∂r

∂x= f ′(r)

x

r

and similarly for ∂f/∂y and ∂f/∂z.ii. Consider

Lxf(r) = −ih

(y

∂

∂z− z

∂

∂y

)f(r) = −ih

(yf ′

z

r− zf ′

y

r

)= 0× f(r)

Lyf(r) = −ih

(z

∂

∂x− x

∂

∂z

)f(r) = −ih

(zf ′

x

r− xf ′

z

r

)= 0× f(r)

Lzf(r) = −ih

(x

∂

∂y− y

∂

∂x

)f(r) = −ih

(xf ′

y

r− yf ′

x

r

)= 0× f(r)

Hence

L2f(r) = LxLxf(r) + LyLyf(r) + LzLzf(r) = 0× f(r)

Hence any spherically symmetric function f(r) is unusual and unique, in thatit is an eigenstate simultaneously of all components of angular momentum. Itcorresponds to the case of all eigenvalues being zero, i.e. no angular momentum.This is possible as this state has no ml degeneracy, since the only allowed valueis ml = 0.

(b) i. Using the above results

Lxgx = −ih

(y

∂

∂z− z

∂

∂y

)[xf(r)] = −ih

(yxf ′

z

r− zxf ′

y

r

)= 0× gx

Lxgy = −ih

(y

∂

∂z− z

∂

∂y

)[yf(r)] = −ih

(yyf ′

z

r− zyf ′

y

r− zf

)= ihgz

Lxgz = −ih

(y

∂

∂z− z

∂

∂y

)[zf(r)] = −ih

(yzf ′

z

r− zzf ′

y

r+ yf

)= −ihgy

Hence

LxX0 = 0×X0

and

LxX± =1√2Lx(gy ± igz) =

1√2(ihgz ± hgy) = ± h√

2(gy ± igz) = ±hX±

So X0 is an eigenstate of Lx with eigenvalue zero, while X± are eigenstateswith eigenvalues ±h. To check these functions are also eigenstates of L2 we use

1

L2 = LxLx + LyLy + LzLz. Making use of the Lx results above, and noting thecyclic behaviour:

Lxgx = 0,Lxgy = ihgz,Lxgz = −ihgy,

Lygx = −ihgz,Lygy = 0,Lygz = ihgx,

Lzgx = ihgy

Lzgy = −ihgx

Lzgz = 0

Therefore,

LxLxgx = 0,LxLxgy = ih(−ih)gy = h2gy,LxLxgz = −ih(ih)gz = h2gz,

LyLygx = h2gx,LyLygy = 0,LyLygz = h2gz,

LzLzgx = h2gx

LzLzgy = h2gy

LzLzgz = 0

So gx, gy and gz, and hence X± and X0, are all eigenstates of L2 with eigenvalue2h2, i.e. eigenvalues l(l + 1)h2 where l = 1.

ii. We need to express X− as a sum of the eigenstates of Lz.

X− =∑n

anZn = a+Z+ + a0Z0 + a−Z−

= a+(gx + igy)√

2+ a0gz + a−

(gx − igy)√2

=(a+ + a−)√

2gx + i

(a+ − a−)√2

gy + a0gz

Comparing this with X−, we see (by inspection) that

a+ + a− = 0,i√2(a+ − a−) =

1√2, a0 =

−i√2

.

Hencea+ =

−i

2, a− =

i

2, a0 =

−i√2

andX− =

−i

2Z+ − i√

2Z0 +

i

2Z−

iii. The particle is in the state X−. This has been expressed above as an expansionin eigenstates of Lz. The expansion contains Z± and Z0, corresponding to eigen-values L2 = 2h2, and Lz = h, 0 and −h. Hence, the only possible result of ameasurement of L2 is 2h2. The possible results of a measurement of Lz are h, 0and −h.The probabilities of the possible Lz measurements are given by

P1 = |a+|2 = 1/4, P0 = |a0|2 = 1/2, P−1 = |a−|2 = 1/4

respectively.The expectation value of Lz is:

〈Lz〉 =∑

Pmmh =14(h) +

12(0) +

14(−h) =

h

4+

02− h

4= 0

One gets the same result using the integral definition of 〈Lz〉, and exploitingorthogonality:

〈Lz〉 =∫ ∫ ∫

(X−)∗LzX− dx dy dz =∫ ∫ ∫ (

h

4(Z+)∗Z+ − h

4(Z−)∗Z−

)dx dy dz = 0

The result zero was probably obvious; why should a state of definite Lx show apreference for positive or negative values in the z direction?

2

(c) By inspection, the combination

gxx + gyy + gzz = (x2 + y2 + z2)f(r) = r2f(r)

and so is purely a function of r. In Qu 1(a), this has been shown to be an eigenstatewith l = 0. Hence, this combination is not l = 2.For reference, the l = 2 states with definite Lz values constructed from the other fiveindependent combinations are

Z0 = gzz, Z±1 =gxz ± igyz√

2, Z±2 =

gxx − gyy ± 2igxy√6

as can be verified by explicit operation with Lz.

2. (i) The commutator is

[L2x, Lz] = LxLxLz − LzLxLx = Lx(LxLz)− (LzLx)Lx

But LzLx − LxLz = ihLy so that LxLz = −ihLy + LzLx giving

[L2x, Lz] = Lx(−ihLy + LzLx)− (ihLy + LxLz)Lx

= −ih(LxLy + LyLx)

Also[L2

z, Lz] = LzLzLz − LzLzLz = 0

We can put these together with the expression for [L2y, Lz] given in the question so

that

[L2, Lz] = [L2x, Lz] + [L2

y, Lz] + [L2z, Lz] = −ih(LxLy + LyLx) + ih(LxLy + LyLx) = 0

The fact that L2 commutes with Lx, Ly and Lz but that Lx, Ly and Lz do notmutually commute amongst themselves suggests that we may simultaneously knowthe magnitude of the angular momentum and one of its components, but that wecannot know all three of its components simultaneously. There exist generalisedHeisenberg uncertainty relationships between the individual components of angularmomentum.

[7 marks]

(ii) The commutator is

[Lz, L−] = LzL− − L−Lz = Lz(Lx − iLy)− (Lx − iLy)Lz

= LzLx − LxLz + iLyLz − iLzLy = [Lz, Lx] + i[Ly, Lz]= ihLy + i2hLx = −h(Lx − iLy) = −hL−

[3 marks]

(iii) Applying the above, then

LzL−φm − L−Lzφm = −hL−φm

LzL−φm −mhL−φm = −hL−φm

Lz(L−φm) = (−h + mh)(L−φm)Lz(L−φm) = (m− 1)h(L−φm)

This is an eigenvalue equation for the state L−φm with eigenvalue (m− 1)h.[3 marks]

3

(iv) Operating with Lz gives, for φ−1,

Lzφ−1 = (xpy − ypx)(x− iy)√

2

=[x

(−ih

∂

∂y

)− y

(−ih

∂

∂x

)](x− iy)√

2

=−ih√

2

[x

∂

∂yx− ix

∂

∂yy − y

∂

∂xx + iy

∂

∂xy

]=

−ih√2

(0− ix− y − 0) = −h(x− iy)√

2= −hφ−1

and for φ0,

Lzφ0 = (xpy − ypx)z

= x

(−ih

∂

∂yz

)− y

(−ih

∂

∂xz

)= 0

(Note, these functions correspond to those in Qu 1(b) above with f(r) = 1.)[3 marks]

(v) Operating with L− gives

L−φ0 = (Lx − iLy)z= [(ypz − zpy)− i(zpx − xpz)] z= ypzz + ixpzz = (y + ix)(−ih) = h(x− iy)=

√2hφ−1

It only remains to show that this state is an eigenstate of Lz with eigenvalue −h. Wealready know φ−1 is such a state, so

Lz

√2hφ−1 =

√2hLzφ−1 =

√2h(−h)φ−1 = −h(

√2hφ−1)

Hence√

2hφ−1 is an eigenstate of Lz with eigenvalue −h.[4 marks]

4


Angular momentum and central potentials


1. (a) The potential for the 3D SHO has the form

V (r) =12mω2

xx2 +

12mω2

yy2 +

12mω2

zz2

For the isotropic case we have ωx = ωy = ωz; let this equal ω0 which gives

V (r) =12mω2

0r2 = V (r)

i.e. this is a central potential.(b) The separations of variables (see lecture 21) results in energy eigenstates of the form

unxnynz = unx(x)uny(y)unz(z)

where the un are the 1D solutions, and the energy is

E = Ex + Ey + Ez =(nx + ny + nz +

32

)hω0

For the ground state nx = ny = nz = 0. The required 1D ground state energyeigenstate is

u0(x) = Ae−ax2/2

Hence, the 3D ground state energy eigenstate is therefore

u000 = A3e−ax2/2e−ay2/2e−az2/2 = Be−a(x2+y2+z2)/2 = Be−ar2/2

where B = A3. The energy is

E0 =32hω0

(c) To normalise this eigenstate, we need to integrate over all space. As the state is afunction of r only, this is most easily done in spherical polars, so∫ ∞

0

∫ π

0

∫ 2π

0|B|2e−ar2

r2 sin θ dθ dφ = 1

Doing the angular integration gives

4π|B|2∫ ∞0

r2e−ar2dr = 1

Using the standard integral, then this is

4π|B|2 14

√π

a3= 1

so

|B|2 =

√a3

π3=

(a

π

)3/2

Taking B to be real, we can write our normalised wavefunction as

u000 =(a

π

)3/4

e−ar2/2

As a cross-check, the 1D ground state normalisation as given in the lectures is A =(a/π)1/4, which is clearly consistent with the above.

1

(d) As shown in the lectures, when expressed in spherical polars, the operator L2 containsonly derivatives of the angles θ and φ. However, u000 is a function of r only. Hence,L2u000 = 0, so this state is an eigenstate of L2 with eigenvalue 0, i.e. l = 0.

(e) Given the equation for the energy above, the first excited states clearly correspondto nx + ny + nz = 1 so

E1 =52hω0

There are three states with this energy, namely u100, u010 and u001. Hence, thedegeneracy is three. There are generally 2l+ 1 states for a given l so with 2l+ 1 = 3,then l = 1 for these three states.

2. (a) The wave function here is expressed as a sum of normalised energy eigenstates. Thesquared modulus of the coefficients in this expansion correspond to the probabilityto find the system in the corresponding eigenstate, so their sum must equal unity fora normalised wave function. Hence, writing

ψ =∑nlml

anlmlunlml

= A[3u100 − 2u211 + 3u210 +

√3u21−1

]then we need ∑

|anlml|2 = |A|2(32 + 22 + 32 +

√32) = 25|A|2 = 1

We can take A to be real which means it has a value A = 1/5. Hence

ψ =15

[3u100 − 2u211 + 3u210 +

√3u21−1

](b) Similarly, normalising U2 gives

|B|2(22 + 32 +√

32) = 16|B|2 = 1

so B = 1/4 and

U2 = −12u211 +

34u210 +

√3

4u21−1

Henceψ =

35u100 +

45U2

(c) The state is a combination of n = 1 and n = 2 states, so only these two energies arepossible measurement outcomes. These have values of

E1 = −13.6 eV12

= −13.6 eV, E2 = −13.6 eV22

= −3.4 eV

The probability of E1 is given by the square of the first coefficient

Pn=1 =∣∣∣∣35

∣∣∣∣2 =925

while the probability of E2 is the square of the second

Pn=2 =∣∣∣∣45

∣∣∣∣2 =1625

Clearly Pn=1 + Pn=2 = 1 as expected.

2

(d) There are three states which could give E2, so with the total probability of thatoutcome as the sum of the separate probabilities

Pn=2 =∣∣∣∣−2

5

∣∣∣∣2 +∣∣∣∣35

∣∣∣∣2 +

∣∣∣∣∣√

35

∣∣∣∣∣2

=425

+925

+325

=1625

as above.

(e) Similarly, there are two possible measurement outcomes for L2 corresponding to thel = 0 and l = 1 quantum numbers. The measurement outcomes for L2 are l(l + 1)h2

so these are 0 and 2h2, respectively. The probabilities happen to be the same as forthe energy states above, as the n = 1 state is also l = 0 and the n = 2 states are alll = 1. Hence

Pl=0 =925, Pl=1 =

1625

(f) For Lz, there are values of ml = 0 and ±1, so there are three possible measure-ment outcomes with Lz values of 0 and ±h, respectively. Evaluating the separateprobabilities for convenience, then for ml = 0

Pml=0 =∣∣∣∣35

∣∣∣∣2 +∣∣∣∣35

∣∣∣∣2 =925

+925

=1825

while for ml = 1 the probability is

Pml=1 =∣∣∣∣−2

5

∣∣∣∣2 =425

and finally for ml = −1 it is

Pml=−1 =

∣∣∣∣∣√

35

∣∣∣∣∣2

=325

Again Pml=0 + Pml=1 + Pml=−1 = 1 as expected.

(g) If a measurement of L2 is made with an outcome of 0, then the wavefunction musthave collapsed to be purely the state u100. As this is also an energy eigenstate, thenit will stay in that state indefinitely, if not disturbed. This means the only possibleoutcome of a subsequent Lz measurement has ml = 0 and so gives Lz = 0. This is adefinite outcome and so has a probability of 1.

3. (i) Writing

ψ =∑n

anun

then

[Q, R]ψ =∑n

an(QRun − RQun) =∑n

an(Qrnun − Rqnun)

=∑n

an(rnqnun − qnrnun) = 0

[3 marks]

3

(ii) Applying L2

L2Y11 = h2

√38π

{1

sin2 θ

∂2

∂φ2(sin θeiφ) +

1sin θ

∂

∂θ

[sin θ

∂

∂θ(sin θeiφ)

]}

= h2

√38π

[− eiφ

sin θ+

eiφ

sin θd

dθ(sin θ cos θ)

]

= h2

√38π

eiφ

sin θ

[cos2 θ − sin2 θ − 1

]= h2

√38π

eiφ

sin θ

[− sin2 θ − (1− cos2 θ)

]= h2

√38π

eiφ

sin θ(−2 sin2 θ) = −2h2

√38π

sin θeiφ = 2h2Y11

Hence, Y11 is an eigenstate of L2. We know that L2Ylm = l(l + 1)h2Ylm so we musthave l(l + 1) = 2 giving l = 1. For Lz

LzY11 = ih

√38π

∂

∂φ(sin θeiφ) = ih

√38π

sin θ∂

∂φ(eiφ)

= ih

√38π

sin θ(ieiφ) = −h√

38π

sin θeiφ = hY11

and so Y11 is also an eigenstate of Lz. We know that LzYlm = mlhYlm so we musthave ml = 1.

[8 marks]

(iii) By inspection

ψ(r) = R(r)[

12√

2πsin θe−iφ +

1√2π

cos θ]

= R(r)

[1

2√

2π

√8π3Y1−1 +

1√2π

√4π3Y10

]

= R(r)

[1√3Y1−1 +

√23Y10

]

[3 marks]

(iv) (a) Since only spherical harmonics with l = 1 are in the superposition, a measurementof L2 will yield the result 2h2.

(b) The probability that a measurement of Lz yields the result −h is (1/√

3)2 = 1/3.(c) The expectation value of Lz is

∣∣∣∣ 1√3

∣∣∣∣2 × (−h) +

∣∣∣∣∣√

23

∣∣∣∣∣2

× (0) = −h/3

[6 marks]

4


Spin

Paul Dauncey, Released 13 Jan 2012

1. (a) For each of the matrices in turn

σ2x =

(0 11 0

) (0 11 0

)=

(1 00 1

)= I

σ2y =

(0 −ii 0

) (0 −ii 0

)=

(1 00 1

)= I

σ2z =

(1 00 −1

) (1 00 −1

)=

(1 00 1

)= I

For any of the matrices, the eigenvalue equation is

σiχ = λχ

so reapplying the matrix gives

σ2i χ = λσiχ so Iχ = λ2χ

Hence

λ2 = 1 so λ = ±1

Since Si = (h/2)σi, then acting on an eigenvector χ gives

Siχ =h

2σiχ = ± h

2χ

so the Si eigenvalues are ±h/2.(b) The eigenvalue equation for Sy is

h

2

(0 −ii 0

) (ab

)= λ

(ab

)For the two eigenvalues

λ = +h/2 : b = ia, ηy+ = a

(1i

)λ = −h/2 : b = −ia, ηy− = a

(1−i

)

To normalise them, then

1 = η†y+ηy+ = |a|2 ( 1 −i )(

1i

)= 2|a|2

so we can take a = 1/√

2. Hence

ηy+ =1√2

(1i

), ηy− =

1√2

(1−i

)For the orthogonality

η†y+ηy− =12

( 1 −i )(

1−i

)=

12(1− 1) = 0

so they are indeed orthogonal.

1

(c) The expectation values are

〈Sz〉 = χ†Szχ =h

2× 25( 3 4 )

(1 00 −1

) (34

)=

h

50( 3 4 )

(3−4

)= −7h

50

〈Sy〉 = χ†Syχ =h

2× 25( 3 4 )

(0 −ii 0

) (34

)=

h

50( 3 4 )

(−i4i3

)= 0

(d) Writingχ = a+ηy+ + a−ηy−

then using orthonormalitya± = η†y±χ

Explicitly

a+ =1

5√

2( 1 −i )

(34

)=

15√

2(3− 4i)

a− =1

5√

2( 1 i )

(34

)=

15√

2(3 + 4i)

Hence, the probability of measuring Sy = −h/2 is

|a−|2 =3 + 4i

5√

23− 4i

5√

2=

32 + 42

25× 2=

12

(e) As the Sy measurement yields result −h/2, then the state collapses to χ =1√2

(1−i

).

Hence the expectation value of Sz is

〈Sz〉 =h

2· 1√

2( 1 i )

(1 00 −1

)1√2

(1−i

)=

h

4( 1 i )

(1i

)=

h

4(1− 1) = 0

This gives a different result as the Sy measurement has collapsed the wavefunctionand so changed 〈Sz〉. Note, the eigenstates of Sy should not have any preference for±Sz so the expectation value would be expected to be zero, as found above.

2. (a) The relevant term in the Hamiltonian is

Hspin =2µBB

hSz

so the energy eigenstates are the Sz eigenstates, i.e.(10

),

(01

)The energy eigenvalues E± are (2µBB/h)× (eigenvalue of Sz = ±h/2) = ±µBB. Att = 0

χ(t = 0) =15

(34

)=

35

(10

)+

45

(01

)so at a later time

χ(t) =35

(10

)exp(−iE+t/h) +

45

(01

)exp(−iE−t/h)

=35

(10

)exp(−iωt) +

45

(01

)exp(iωt)

=15

[3 exp(−iωt)4 exp(iωt)

]where ω = µBB/h.

2

(b) The expectation value of Sz is given by

〈Sz〉 =h

2χ†(t)

(1 00 −1

)χ(t) =

h

50[ 3 exp(iωt) 4 exp(−iωt) ]

[3 exp(−iωt)−4 exp(iωt)

]=

h

50(9− 16) = −7h

50

and hence is not a function of time.

(c) Similarly, the expectation value of Sx is given by

〈Sx〉 =h

2χ†(t)

(0 11 0

)χ(t) =

h


[4 exp(iωt)

3 exp(−iωt)

]=

h

50[12 exp(i2ωt) + 12 exp(−i2ωt)] =

24h

50cos(2ωt)

and the expectation value of Sy is given by

〈Sy〉 =h

2χ†(t)

(0 −ii 0

)χ(t) =

h


[−4i exp(iωt)3i exp(−iωt)

]=

h

50[−12i exp(i2ωt) + 12i exp(−i2ωt)] =

24h

50sin(2ωt)

Hence, these both undergo harmonic motion with an angular frequency of 2ω =2µBB/h.

(d) The sum of the squared expectation values is

〈Sx〉2 + 〈Sy〉2 + 〈Sz〉2 =576h2

2500cos2(2ωt) +

576h2

2500sin2(2ωt) +

49h2

2500

=576h2

2500+

49h2

2500=

625h2

2500=

h2

4

Hence, the sum of the squared expectation values is constant and equal to (h/2)2.

3. (i) Calculating explicitly

[Sx, Sy] =h2

4

[(0 11 0

) (0 −ii 0

)−

(0 −ii 0

) (0 11 0

)]=

h2

4

[(i 00 −i

)−

(−i 00 i

)]=

h2

2

(i 00 −i

)= ihSz

[5 marks]

(ii) The magnetic interaction energy operator for B = (B/√

2)(0, 1, 1) is

H =e

mB.S =

e

m

B√2(Sy + Sz)

=e

m

B√2

h

2

[(0 −ii 0

)+

(1 00 1

)]=

eh

2m

B√2

(1 −ii −1

)[3 marks]

(iii) Using

u1,2 = N1,2

(1

±i(√

2∓ 1)

)

3

then

Hu1,2 =eh

2m

B√2N1,2

(1 −ii −1

) (1

±i(√

2∓ 1)

)=

eh

2m

B√2N1,2

(1± (

√2∓ 1)

i∓ i(√

2∓ 1)

)=

eh

2m

B√2N1,2

(±√

2i(2∓

√2)

)= ±eBh

2mN1,2

(1

±i(√

2∓ 1)

)= ±eBh

2mu1,2

Hence u1 is an eigenstate with E1 = eBh/2m and u2 is an eigenstate with E2 =−eBh/2m.

[4 marks]

(iv) By explicit calculation

u†1u2 = N∗1 N2 ( 1 −i(

√2− 1) )

(1

−i(√

2 + 1)

)= N∗

1 N2

(1− (

√2 + 1)(

√2− 1)

)= N∗

1 N2 (1− 2 + 1) = 0

and hence u1 and u2 are orthogonal.[1 mark]

(v) The energy expectation value of the spin state χ = 1√2

(1i

)is given by

〈H〉 = χ†Hχ =eh

2m

B√2

12

( 1 −i )(

1 −ii −1

) (1i

)=

eh

4m

B√2

( 1 −i )(

20

)=

eh

2m

B√2

[3 marks]

(vi) Since there are only two possible energy measurement values

Prob(E1) + Prob(E2) = 1

The expectation value is the average value so

Prob(E1)E1 + Prob(E2)E2 =eh

2m

B√2

Prob(E1)(

eBh

2m

)+ Prob(E2)

(−eBh

2m

)=

eh

2m

B√2

Prob(E1)− Prob(E2) =1√2

Adding these two equations gives

2Prob(E1) = 1 +1√2

=√

2 + 1√2

while subtracting gives

2Prob(E2) = 1− 1√2

=√

2− 1√2

Hence

Prob(E1)Prob(E2)

=√

2 + 1√2− 1

4

Alternatively by explicitly calculating overlaps

c1,2 = u†1,2χ =N∗

1,2√2

( 1 ∓i(√

2∓ 1) )(

1i

)=

N∗1,2√2

[1± (√

2∓ 1)] =N∗

1,2√2

(±√

2) = ±N1,2

Hence, the probabilities of measuring E1 and E2 are in the ratio:

Prob(E1)Prob(E2)

=|c1|2

|c2|2=

N21

N22

=4 + 2

√2

4− 2√

2=√

2 + 1√2− 1

[4 marks]

5

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