10 The 1D Harmonic Oscillator

(Notes adapted from lectures of Prof. Sunil Mukhi)

Let us now consider a problem which has widespread applicability; a particle of mass m movingin one dimension in a harmonic oscillator potential. It is actually possible to make systems wherethe motion is confined to one dimension and the harmonic oscillator quadratic potential V (x) ∼ x2

approximates quite well any generic small oscillations problem (for any potential). Some examplesof this are vibrating nuclei in diatomic molecules, or the torsional oscillations of molecule groupsin large molecules.

The potential for the harmonic oscillator is V (x) = 12kx2 = 1

2mω2x2, where ω =�

k/m is the(classical) angular frequency of the oscillator. For this system the TISE is

�− h2

2m

d2

dx2 + 12mω2x2

�ψ(x) = Eψ(x) (50)

This is an ordinary differential equation, and we can just look up the solutions to this in a book.However it is very instructive to first observe some general features of the problem.

In classical physics, the total energy of a 1d harmonic oscillator is

E = p2

2m+ 1

2mω2x2

Clearly E ≥ 0 since it is a sum of two terms, that are each ≥ 0. There also exists a solution withE = 0 which is quite trivial: the oscillator starts at the origin with zero initial velocity. Then itstays at rest and has p = 0 and x = 0 for all time. From our understanding so far, it seems unlikelythat there is a quantum-mechanical analogue of this classically allowed behavior, since we knowthat ΔpΔx ≥ h

2 . This would be inconsistent with having p and x vanishing. This implies that thesystem has some positive minimum energy (called the zero-point energy).

Let us now try to estimate this minimum energy using the same technique that we used for thehydrogen atom. We begin by scaling x = λx�, where λ is some positive constant. In equation 50,the Hamiltonian H is then scaled to

H � =�

− 1λ2

h2

2m

d2

dx�2 + 12mω2λ2x�2

�(51)

Now we choose λ such that the coefficient of the derivative term becomes equal to the coefficientof the x�2 term

h2

2mλ2 = 12mω2λ2

giving

λ =�

h

mω(52)

The Hamiltonian is thus

H � = 12 hω

�− d2

dx�2 + x�2�

(53)

Sourabh Dube - PHY202 31

and the TISE can be written as12 hω

�− d2

dx�2 + x�2�

ψ(x�) = Eψ(x�) (54)

Let us set E � = 2E/(hω) and get our TISE�

− d2

dx�2 + x�2�

ψ(x�) = E �ψ(x�) (55)

We make the same argument as before, that since this equation has no parameters to change thevalue of E � dramatically, thus E �

min is of the order 1. This gives us Emin = (1/2)hω.

In the case of the harmonic oscillator, this estimate turns out to be exact. However at the moment,let us assume that there is an eigenvalue E = (1/2)hω, and try to find the appropriate eigenfunc-tion. Thus we are trying to solve

H �ψ = (1/2)hωψ (56)

Using the form of H �,

12 hω

�− d2

dx�2 + x�2�

ψ(x�) = (1/2)hωψ(x�)

Dividing both sides by (1/2)hω and bringing all terms to one side, we get�

− d2

dx�2 + x�2 − 1�

ψ(x�) = 0

This can be written as �− d

dx� + x�� �

d

dx� + x��

ψ(x�) = 0 (57)

(Please check yourself that expanding this equation leads to the previous one, recall that a differ-ential operator like d

dxacts on everything to its right).

For ψ(x�) to solve this equation, it is sufficient for us to solve�

d

dx� + x��

ψ(x�) = 0 (58)

which is a first order ordinary differential equation and can easily be integrated to give

ψ(x�) = e− x�22 (59)

which is a gaussian function. Finally returning to the original variable x = λx� =�

hmω

x� we get

ψ(x) =�

mω

hπ

� 14

e− mωx22h (60)

where we have also put in the normalisation factor.

We can check that this is a solution by inserting this wave function into the above equation 56 andconfirming that it satisfies the equation with energy eigenvalue E = (1/2)hω.

As usual, the full wavefunction is obtained by putting back the time-dependence

ψ(x, t) = ψ(x)e− ih

Et =�

mω

hπ

� 14

e− mωx22h

− i2 ωt (61)

Sourabh Dube - PHY202 32

10.1 Properties of the ground state

The wave function that just found in the preceding section is called the ground state wave functionbecause it describes the closest the system can come to settling down to a state of zero energy. Weshall demonstrate this conclusively in the next sections. For now, let us take it as fact.

It is easy to calculate Δp and Δx for this wave function.

�p� =�

dxψ∗(�x)�

−ihd

dx

�ψ(�x)

�p2� =�

dxψ∗(�x)�

−h2 d2

dx2

�ψ(�x)

�x� =�

dxψ∗(�x) (x) ψ(�x)

�x2� =�

dxψ∗(�x)�x2

�ψ(�x)

We find that �x� and �p� are zero due to the symmetry of the Gaussian about the origin (both inposition and momentum space). We find that

�x2� = h

2mω�p2� = mωh

2

giving us

Δx =�

�x2� − �x�2 =�

h

2mω

Δp =�

�p2� − �p�2 =�

mωh

2and thus we get ΔxΔp = h

2 . This was expected given that the ground state was a Gaussian.

The probability density for the ground state is given as

ψ∗(�x)ψ(�x) =�

mω

hπe− mωx2

h

This is also a Gaussian. Let us now plot the probability density and the potential V (x) = (1/2)mω2x2

together, for the ground state energy E = (1/2)hω.

Classically, the particle is confined to the region inside the potential. But in Fig. 7 we see thatalthough the probability is concentrated mainly in the classically allowed region, there is somefinite non-zero probability extending to infinity in both directions. This means there is a finiteprobability of finding the particle in this classically forbidden region. Note that this probabilitydies off very rapidly as we go deeper in the forbidden region. This highlights a crucial differencebetween quantum and classical physics, and this property of nature at the quantum scale also hasreal world applications. We shall return to this later in the course.

Sourabh Dube - PHY202 33

x-4 -3 -2 -1 0 1 2 3 4-1

0

1

2

3

4

5

V(x)

2|0

ψ|

Figure 7: The shapes of the harmonic oscillator potential V (x), and the ground state wavefunctionψ0(x).

10.2 Excited states

Let us now get the other eigenfunctions for the harmonic oscillator. Let us start from equation 54.

12 hω

�− d2

dx�2 + x�2�

ψ(x�) = Eψ(x�) (62)

Let us divide both sides by hω, and define �n = En/(hω)

12

�− d2

dx�2 + x�2�

ψn(x�) = �nψn(x�) (63)

We can think of the operator on the LHS as the “reduced Hamiltonian” and �n as the “reducedenergy”. We factorize the reduced Hamiltonian and can write

12

�− d2

dx�2 + x�2�

= 1√2

�− d

dx� + x��

1√2

�d

dx� + x��

+ 12 (64)

Let us define

a = 1√2

�d

dx� + x��

a† = 1√2

�− d

dx� + x��

(65)

and rewrite the reduced Hamiltonian as

12

�− d2

dx�2 + x�2�

= a†a + 12 (66)

and the reduced TISE as �a†a + (1/2)

�ψn(x�) = �nψn(x�) (67)

An interesting property that a and a† have is

[a, a†] ≡ aa† − a†a = 1

Sourabh Dube - PHY202 34

Here we have defined the commutator of a and a†. The commutator of two operators A and Bis [A, B] = AB − BA, and since these are operators, can be non-zero. Moreover, we have alsolabelled one operator as the “dagger” of the other. For a matrix, this symbol denotes Hermitianconjugation. As we shall see in more detail later, two operators A and B obeying the relation

�ψ∗

1(Bψ2) =�

(Aψ1)∗ψ2

for any two wavefunctions ψ1 and ψ2, are said to be Hermitian conjugates of each other, and wewrite B = A†.

Now we had found one solution to the TISE by requiring

aψ0 = 1√2

�d

dx� + x��

ψ0(x�) = 0 (68)

Now let us consider ψ1 = a†ψ0 and solve the TISE for ψ1

H �ψ1 = (a†a + (1/2))ψ1 = (a†a + (1/2))(a†ψ0)= a†aa†ψ0 + (1/2)ψ0

= a†(aa†)ψ0 + (1/2)ψ0

= a†(1 + a†a)ψ0 + (1/2)ψ0

= a†ψ0 + (a†a†)aψ0 + (1/2)ψ0

= (3/2)a†ψ0

= (3/2)ψ1

Here we used [a, a†] = 1, and aψ0 = 0. We thus see that ψ1(= a†ψ0) is an eigenfunction ofthe reduced Hamiltonian with eigenvalue 3

2(= 1 + 12). Explicitly, the wave function ψ1 without

normalisation prefactors is

ψ1 = a†ψ0 ∼�

− d

dx� + x��

e− x�22 ∼ x�e− x�2

2

Going back to the original variables, the energy of this eigenfunction is E1 = 32 hω. We can now

similarly show that ψn will be an eigenfunction with eigenvalue �n = n + 12 , where

ψn(x�) =�a†

�nψ0(x�) (69)

and which satisfies �a†a + 1

2

�ψn = �nψn (70)

Thus the 1D harmonic oscillator has a tower of energy eigenfunctions ψn, n = 0, 1, 2, ... withenergies

En = (n + 12)hω (71)

The explicit form of ψn(x�) can be worked out using the above formula, and it always turns out tobe a specific polynomial of degree n (called a Hermite polynomial) multiplying the same Gaussian.

Sourabh Dube - PHY202 35

It is interesting to note the process that we used above. A quick summary is that for any solutionof the equation

a†aψ = λψ

we can show that the functions aψ and a†ψ obey�a†a

�a†ψ = (λ + 1) a†ψ

�a†a

�aψ = (λ − 1) aψ

This shows us that a† is a “raising operator”, which takes an eigenfunction to another eigenfunctionwith the eigenvalue raised by 1, and a is similarly a “lowering operator”. The state ψ0 is annihilatedby the lowering operator because there is no state for it to be lowered to, and is thus the groundstate.

10.3 Properties of the solutions

We saw in the previous subsection that the reduced TISE can be written as�

a†a + 12

�ψn = �nψn

Using the commutation relation [a, a†] = 1, we can rewrite this equation as�

aa† − 12

�ψn = �nψn

and thus find a wavefunction ψ0 obeying

a†ψ0 ∼�

− d

dx� + x��

ψ0(x�) = 0

with energy � = −(1/2). This seems surprising since we have already concluded that the groundstate energy is � = 1/2 (and there is no lower energy).

The resolution of this puzzle carries as important lesson. The wave function obtained by solvingthe above equation (for � = −(1/2)) is

ψ0(x�) = ex�2

2

This differs from the ground-state wave function ψ0(x�) only by the sign of the exponent. But thisis a crucial change of sign. The wavefunction ψ0 actually cannot be normalized as it grows withoutlimit as we move towards x� → ±∞. Such solutions must be rejected on physical grounds. Tosummarise, the harmonic oscillator has infinitely many solutions of the form

(a†)nψ0(x�) where aψ0(x�) = 0

which fall of at infinity, which can be normalised, and thus are physically meaningful and anotherinfinite set

(a†)nψ0(x�) where aψ0(x�) = 0

Sourabh Dube - PHY202 36

which grow at infinity and therefore cannot be normalised and are not physically meaningful.

Now let us observe another interesting property of the harmonic oscillator stationary states

I =�

dx�ψ∗1(x�)ψ0(x�) =

�dx�

�a†ψ0(x�)

�∗ψ0(x�)

=�

dx�ψ∗0(x�) (aψ0(x�))

= 0

This shows us that the wave functions ψ0 and ψ1 are orthogonal. More generally any two eigen-functions of the 1d harmonic oscillator are mutually orthogonal. In addition, the wavefunctionscan be all normalised. We write these properties together as

�dx�ψ∗

m(x�)ψn(x�) = δmn

where δmn is the Kronecker-delta function. The collection of ψn are said to be a set of orthonormalset of wave functions, similar to the basis vectors in three-dimensional space (i, j, k).

There is another property of the ψn which is less obvious: they form a complete basis. Thismeans that all functions of x� that vanish sufficiently fast at infinity can be expressed as a linearcombination of these stationary states.

f(x�) =∞�

n=0cnψn(x�)

for some choice of coefficients cn. Recall that this property was also true of the solutions to theinfinite potential well.

Sourabh Dube - PHY202 37