EPGY Summer Institute, Quantum Mechanics QM Lecture 7

EPGY Summer Institute, Quantum Mechanics

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QM Lecture 7

In the previous lecture we gave the Schrödinger equation (1926) for a particle of mass m,

ih t(x, t) = h2

2m2

x2(x, t)+V(x)(x, t) .

We arrived at the form by considering the non-relativistic energy, E = _ mv2 + V = p2/2m + V,and replacing E and p by their operator forms acting on the wave function,

E E = ih t

& p p = ih x

E =p2

2m +V

We also examined some simple cases of typical problems.

1) Free particle: V = 0

ih t(x, t) = h2

2m2

x2(x, t) , (x, t) = Aei(kxt)

2) Schrödinger equation in three dimensions:

ih t(rr, t) = h2

2m2

x2+2

y2+2

z2

(rr, t)+V(| rr |)(rr , t) h2

2m2(rr, t)+V(| rr |)(rr, t)

3) Particle in a one-dimensional box (infinite well): (We will solve this problem in this lecture)V(x) = 0 for 0 < x < LV(x) = for L < x < 0.

ih t(x, t) = h2

2m2

x2(x, t) . for 0 < x < L

4) Hydrogen Atom: (We will not solve this one).

ih t(rr, t) = h2

2m2(rr, t) ke2

r(rr, t)

Before proceeding some general remarks and definitions are in order. The Schrödinger equation isoften written in a simpler form,

ih t = H , H = h2

2m2

x2+V This operator is called the Hamiltonian, its name

derives from the classical counterpart, (H = KE + PE), which is closely related to the energy.

Often we will consider systems in which the energy of an object remains constant. As anexample, the particle in a box will have one energy and since it is not interacting with anything itcan not change. (We will find the allowable energies that it could have). If this is the case, theSchrödinger equation simplifies to,

ih t = E =

h2

2m2

x2 +V . where now E is simply a constant.


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In this equation we can limit to be a function of position alone,

E(x) = h2

2md2

dx2(x)+V(x)(x) , or E = H

This equation is called the time-independent Schrödinger equation.*For those fans of calculus, note that since our function is only a function of x now, our partial derivatives convert toordinary derivatives.

An aside on operators: As we mentioned, the quantities E, P, H are called operators. They are notfunctions but are objects which act on functions (a derivative is a simple operator, it must act on something to becomplete, but you can still consider it as an separate quantity). When we introduced the operators, they acted onwavefunctions, (x), which are arguments of position (and time). It is said that we were in the position basis. In thisbasis, the position variable is not an operator (or you can consider it as x times the unit operator

ˆ1 which does not do

much), but the momentum is.Likewise, we could consider a wavefunction which describes the possibility to obtain values of momentum

p, (p). Just as with the position basis wavefunction, the modulus square of (p) gives the probability distribution,P(p), to find the object with momentum p. This is the momentum basis. In this basis, the momentum is represented asan ordinary number (or again as a variable multiplied by the unit operator), and the position is represented by an

operator,ˆX

., (such that X(p) = x(p) and in case you are wondering its form is given by X = ih

p). We can

also examine a problem in the energy basis. In this basis E is an ordinary number (called a c-number) and positionand momentum are represented as operators. This basis is useful in solving the harmonic oscillator problem.

Before proceeding with the one-dimensional box, lets discuss some of the generalproperties of wavefunctions. They are mathematical functions which,

1) are continuous. (e.g. the wavefunction must match across all boundaries).2) has a derivative that is continuous (slope is continuous, no kinks) except where the

potential may be infinite.3) tend to zero as you go an infinite distance away. We do not want the probability to

blow up very far away.4) The wavefunctions are normalized. This requirement just says that the total

probability to find a particle in all space is 1.

Prob( ) =1= dxP(x) =

dx |(x) |2

.

Example 1: Free Particle in a One Dimensional Box

This is an unrealistic situation but it does help to understand the nature of quantizedenergy levels. The potential is infinite outside of the box and 0 within (for 0 < x < L).

To understand how to attack this first we will consider the case of a wavefunction in aregion where the potential is a constant, V. Then the SE takes the following form,

E =

h2

2md2dx2

+V or d2

dx2=

2mh2

(V E) .

Note that this is a simple form and that an exponential will solve this equation.

(Recall that

d2

dx2eax = a2eax ). Let’s define

k = ± 2m

h2(V E)


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and then we see that a solution is of the following form, =Aekx +Bekx , where A is anormalization factor. Observe,

A d2ekx

dx2 = k2 = 2mh2 (VE)

There are 3 scenarios we want to examine.

1) V = 0In this case we get,

k = ±

2mEh2

= ±i 2mEh2

and we get our harmonic solution

= Aexp +i 2mE

h2x

+ Bexp i 2mE

h2x

2) V < ENow we have,

k = ± 2m(V E)

h2= ±

2m(E V)h2

= ±i 2m(E V)h2

and we have the same form of the solution as in the first case except for E-V in place of E.

3) V > EIn this case we have,

k = ± 2m(V E)

h2,

and the form of our solution is,

= Aexp +

2m(V E)h2

x

+ Bexp

2m(V E)h2

x

.

Returning to the infinite well we will have to consider three regions,Region I x < 0, V = . Solution is of type 3 above.Region II 0 < x < L, V = 0. Solution is of type 1 above.Region III x > L, V = . Solution is of type 3 above.

Region I:

I = AI exp +

2m((V =) E)h2

x

+ BI exp

2m((V =) E)h2

x

Within this region x is always negative. Thus the second term can not be included in ourwavefunction (or, that BI = 0). Why? Well, the wavefunction is infinite, e. The wave functioneventually becomes infinite which is unphysical.

The first term is e- = 0. Conclusion: The wave function is 0 in region I.

Region III:


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This is of the same form as region I. Restricting to physical solutions requires setting AIII= 0 and the second term vanishes. The wavefunction is zero in regions I and III.

Region II:Our solution is,

II = AIIe

ikx + BIIeikx , k = 2mE

h2

Once we have established the form of our solutions we must insure that it is continuous.Thus there are boundary conditions which must be satisfied (and in most problems this is wherethe solutions take shape). The boundary conditions we need to consider are,

I(0) = 0 = II(0)II(L) = III(L) = 0.

(We will not need the other condition that the derivative is continuous). These conditions simplystate that our solution inside of the box must vanish at the walls.

We can explain this physically since we must have standing waves in region II. To havestanding waves, the ends must be nodes. This limits the possible solution.

With these constraints we can find the allowable energies for a particle within this box. (It can nottake on any value). The constraint is that an integer (or half integer) number of wavelengths mustfit within the box.

Examine the left boundary, II(0) = 0, this limits the form of our solution to a sin function.We can rewrite our solution above as a collection of sines and cosines (with the Euler relation),but a cosine function will not vanish at x = 0. Thus we rewrite our solution as,

II = Asinkxexamining our boundary conditions we have,

II (0) = 0 = Asin(0) II (L) = 0 = Asin(kL)

Thus if kL = n where n = 1, 2, 3, … then the boundary condition is satisfied. We do not includen = 0 as this would simply give the wavefunction to be zero everywhere which does not describethis case. (The probability would be 0 to find it anywhere). Writing k out in full we find theallowable energy levels,

kL = n 2mE

h= n E =

n2h22

2mL2.

Particle trapped in a semi-finite well in one dimensionIn this case we simply modify the previous example to have the potential to the right of

the box to be zero and that within the well to be some negative constant, -V. (This is done becausewe usually like to write bound states with negative energy). We are interested in the case wherethe particle initially has energy less than E. The boundary will be at x = +L. Thus repeating theprocess above we have (remember that E < 0).,


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V(x) region Schrodinger Equation DiffiQ to solve,

, I: x < 0 (E+)I = h2

2md2I

dx2

-V, II: 0 < x < L (E+V)II = h2

2md2II

dx2

0, III: x > L EIII = h2

2md2III

dx2

d2II

dx2= k2II , k 2m(E+V)

h

d2III

dx2= 2III ,

2mEh

Now that we have the equations to solve we note that in region III we have V > E and we expect adecaying exponential and in II E > V so that we expect an oscillatory form. Indeed, we can writedown the general form of the solutions in the three regions as,

Condition 3 limits these to:I : I = 0

II : II = AII cos(kx)+ BII sin(kx)

III : III = AIIIex + BIIIe

x

0

II = AII cos(kx)+ BII sin(kx)III = BIIIe

x

where in the last column we have eliminated exponentially increasing terms.What is left is to utilize the boundary conditions, these are the conditions that enforce condition 1and 2 (continuity and smoothness) – this is generally where the “meat” of problems lie. The twoboundaries are at x = 0 and x = L, we have,

continuity (1(x0 ) = 2 (x0 )) smoothness ( '1(x0 ) = '2 (x0 ))

b.c. @ x = 0: 0= AII cos(k0) BII sin(k0) = AII *doesn't apply since V = *

b.c. @ x = L: BII sin(kL) = BIIIeL kBII cos(kL) = BIIIe

L

The first boundary condition gives us that AII = 0, this is what we found for the infinite well. Wehave taken this into account in the second line. We can sub the first condition into the second tofind the condition on and k,

BII sin(kL) = BIIIeL kBII cos(kL) = (BIIIe

L ) = BII sin(kL)

k cos(kL) = sin(kL) or

k= tan(kL) expressed in terms of their original forms, V

E1 = tan L

h2m(V+ E)

This is a difficult equation to solve, but can be performed by a graphical method. We find thatthere are only specific values of E for which this expression is solved: energy is quantized withinthe well.

Particle trapped in a finite well in one dimension: tunneling We will now examine one more example that will demonstrate and important property of

quantum objects that is completely different than what classical physics allows.This is case of a particle trapped in a finite well (that is, in region II below). Classically if

a particle is placed in such a well such that its energy is less than the potential energy to surmountit, it will never escape. (Think of gravitational potential energy and the potential barrier as a ‘wall’that the object must pass over, if its KE is less than PE at that point, it will never reach into regionIII).

I V = , x < 0II V = 0, 0 < x < LIII V = V0, L < x < L + dIV V = 0, x > L + d


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Region I is the same as before, the wavefunction vanishes and the solution in region II will needto vanish at x = 0. And consider the case where the energy of the particle is less than the barrier’spotential energy, 0 < E < V.

We want to find the form of the solution in all 4 regions.

I V = , x < 0 : I = 0II V = 0, 0 < x < L : II is oscillatory ~ AIIcos(kIIx) + BIIsin(kIIx)III V = V0, L < x < L + d : III is exponential ~ AIIIexp(kIIIx) + BIIIexp(-kIIIx) (E<V)IV V = 0, x > L + d : IV is oscillatory ~ AIVcos(kIVx) + BIVsin(kIVx)

And we have for the wavenumbers, k,

II kII =

2mEh

III kIII =

2m(V0 E)h

IV kIV =

2mEh

Now we apply the boundary conditions:

1) I (x = 0) = II (x = 0)

2) II (x = L) = III (x = L)3) III (x = L+ d) = IV (x = L+ d)

Condition 1 tells us that our solution in region II must be a sine. We can have cos(0) = 0.Conditions 2 and 3 will give us relations between kII, kIII, and kIV. So the forms of our solution is,

I V = , x < 0 : I = 0II V = 0, 0 < x < L : II(x) = BIIsin(kIIx)III V = V0, L < x < L + d : III(x)= BIIIexp(-kIIIx) (E<V)IV V = 0, x > L + d : IV(x)=AIVcos(kIVx) + BIVsin(kIVx)

In region III we only choose the negative exponent since the positive one would tell usthat the chance of getting through the barrier is greater the thicker it is, that is physicallyunacceptable.

A plot of the wave function might look something like,

The important point here is that fact the wavefunction does not vanish in region IV.It is non-zero within and past the boundary. Thus the square of the wavefunction is non-zero,meaning that there is a finite probability of discovering this particle outside of the wall! This isabsolutely not allowed classically, but here we see that the particle can “tunnel” through thebarrier to the outside. This phenomenon is called quantum tunneling and is the basis of allsemiconductor technology.

Notice that the wider the barrier the smaller the probability to escape is. An estimate of theprobability of transmission can be found by examining the probability to find the particle inregion IV. It turns out to be something like,

Ptunneling ~| III (d) |2 ~ e

2dh

2m( V0E )


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Appendix 7.A: More Mathematically Formal Statement of Quantum Mechanics

1) State of a particle is represented by a vector in a Hilbert space.2) Independent variables, x and p are represented by operators

ˆˆ X and P. All other

observables are given as functions of these ( (X, P) ).3) Particle in state will yield one of the eigenvalues of an observable with

probability P=||2.4) The state vector, , obeys the Schrödinger equation.

Some comments on the terminology used in the above statement.Hilbert space. This is the space of all states . It is the analog of the state space in

classical mechanics (called phase space of x and p). A rather esoteric idea: essentially think of thefirst comment as saying that the proper way to represent the state of a particle in quantummechanics is with the wavefunction.

Eigenvalue. This is a mathematical term that would take a bit to explain. However, we didexamine some eigenvalues in the last lecture. Recall we said that we could define the energyoperator as,

E = ih

t

= E .

The operator acts on a function (formally called an eigenfunction) yield the energy (a simplenumber) times the eigenfunction . The number, E, is an eigenvalue of the energy operator.Whenever an operator acts on a function and yields a simple number times the original function,that number is called an eigenvalue.

EPGY Summer Institute, Quantum Mechanics QM Lecture 7

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