The harmonic oscillator

Simon Greaves

Research Institute of Electrical Communication, Tohoku University, Sendai

QM11

1 / 27

The harmonic oscillator

A one-dimensional harmonic oscillator. A particle of mass m

oscillates about some point. x is the distance from the equilibrium

position.

The vibrational motion of atomic nuclei in molecules can be

approximated by a harmonic oscillator model.

2 / 27

The classical case

Classical harmonic oscillator

Classically, the potential

energy is given by V(x) = 12kx2.

The particle oscillates between

two turning points at x = ±x0.At x0 the energy is E =

1

2kx2

0,

the particle velocity is zero and

the energy is entirely potential

energy.

In the classical theory the

particle is not allowed outside

the range −x0 ≤ x ≤ x0.

The particle is most likely to be

found near x = ±x0 where itsvelocity is low.

3 / 27

Approximating a local minimum I

Parabolic potential well V(x) = 12kx2 V(x) approximates W(x) around x = a

4 / 27

Approximating a local minimum II

A local minimum at x = a in an arbitrary potential W(x) can beapproximated by a parabolic potential well of the form

V(x) = 12k(x − a)2.

Expand W(x) as a Taylor series about x = a:

W(x) = W(a) + (x − a)W ′(a) + 12(x − a)2W ′′(a) + ...

W(a) is constant everywhere and therefore exerts no force on theparticle.

W ′(a) = 0 at x = a and can be neglected for small deviations fromx = a.

So

W(x) ≈ 12(x − a)2W ′′(a) = 1

2k(x − a)2 = V(x) with k = W ′′(a)

5 / 27

The Schrödinger equation I

The classic equation of motion for a harmonic oscillator is

F = md2x

dt2= −kx

where k is the spring constant.

A solution to this equation is x(t) = A sin(ωt) + B cos(ωt), whichgives

d2x

dt2= −Aω2 sin(ωt)− Bω2 cos(ωt) = −ω2x

So −mω2x = −kx, i.e. k = mω2, where ω is the natural oscillationfrequency.

6 / 27

The Schrödinger equation II

If the potential minimum is located at x = 0 the potential energy ofthe system is V(x) = 1

2kx2 = 1

2mω2x2.

We want to solve the Schrödinger equation Eψ = Ĥψ with

Ĥ = − ~2

2m∇2 + 1

2mω2x2

In one dimension this becomes

− ~2

2m

d2ψ

dx2+

1

2mω2x2ψ = Eψ

7 / 27

Algebraic solution I

There are two approaches to solving the harmonic oscillator

problem. First, let’s look at the (simpler) algebraic method.

We rewrite the Schrödinger equation as

1

2m

[

(

~

i

)2 (

d

dx

)2

+ (mωx)2

]

ψ = Eψ

and try to factorise the term in the square brackets.

We define two operators a+ and a− as

a± −1√2m

(

~

i

d

dx± imωx

)

and we work out the product a−a+.

8 / 27

Algebraic solution II

We start by multiplying by f (x) to avoid mistakes and we have

(a−a+)f (x) =1

2m

(

~

i

d

dx− imωx

)(

~

i

d

dx+ imωx

)

f (x)

So

(a−a+)f (x) =1

2m

(

~

i

d

dx− imωx

)(

~

i

df

dx+ imωxf

)

This gives

(a−a+)f (x) =1

2m

[

−~2 d2f

dx2+ ~mω

d

dx(xf )− ~mωxdf

dx+ (mωx)2f

]

Or

(a−a+)f (x) =1

2m

[

−~2 d2f

dx2+ ~mωx

df

dx+ ~mωf − ~mωxdf

dx+ (mωx)2f

]

9 / 27

Algebraic solution III

Remove f (x) to get

(a−a+) =1

2m

[

−~2 d2

dx2+ ~mω + (mωx)2

]

We end up with

(a−a+) =1

2m

[

(

~

i

)2 (

d

dx

)2

+ (mωx)2

]

+~ω

2

This is almost the same as the Hamiltonian in our Schrödinger

equation. We can subtract the ~ω/2 term and write(

a−a+ −~ω

2

)

ψ = Eψ

10 / 27

Algebraic solution IV

The ordering of the operators a+ and a− makes a difference to the

sign of the ~ω/2 term. we can write a−a+ − a+a− = ~ω.

The Schrödinger equation reads(

a−a+ −~ω

2

)

ψ = Eψ or

(

a+a− +~ω

2

)

ψ = Eψ

We know that ψ is a solution to the Schrödinger equation withenergy E. We can apply the a+ operator to ψ and say that a+ψ isalso a solution to the Schrödinger equation with energy E + ~ω.

In other words(

a+a− +~ω

2

)

a+ψ = (E + ~ω)a+ψ

11 / 27

Algebraic solution V

To prove this, we write the left hand side as(

a+a−a+ +~ω

2a+

)

ψ = a+

(

a−a+ +~ω

2

)

ψ

So

a+

(

a−a+ +~ω

2

)

ψ = a+

((

a−a+ −~ω

2

)

+ ~ω

)

ψ

Finally

a+

((

a−a+ −~ω

2

)

+ ~ω

)

ψ = a+(E + ~ω)ψ = (E + ~ω)a+ψ

Similarly, a−ψ is a solution to the Schrödinger equation with energyE − ~ω.

12 / 27

Ladder of states

Ladder of stationary states

If we have a solution ψ to theSchrödinger equation, with energy E,

then we can obtain solutions for other

energy levels by using the a− and a+operators.

There must be a ground state ψ0 suchthat a−ψ0 = 0.

i.e.1√2m

(

~

i

dψ0dx

− imωxψ0)

= 0

13 / 27

Finding the ground state I

We want to solve

1√2m

(

~

i

dψ0dx

− imωxψ0)

= 0 i.e.dψ0dx

= −mωx~ψ0

Collecting ψ0 and x terms on either side and integrating gives

ln(ψ0) = −mωx2

2~+ c or ψ0 = A0e

−mω2~

x2

To normalise this we write

1 =

∫

∞

−∞

|ψ0|dx = A20∫

∞

−∞

e−mω~

x2 dx

14 / 27

Finding the ground state II

Using the identity

∫

∞

−∞

e−au2

du =

√

π

awith a =

mω

~we find A20

√

π~

mω= 1

The ground state wavefunction is therefore

ψ0(x) =(mω

π~

)1/4e−

mω2~

x2

To find the energy of the ground state we use the Schrödinger

equation(

a+a− +~ω

2

)

ψ0 = E0ψ0

By definition a−ψ0 = 0, so E0 = ~ω/2.

15 / 27

General solutions I

Starting with the ground state, we can write a general solution to

the Schrödinger equation as

ψn(x) = An(a+)ne−

mω2~

x2 with En =

(

n +1

2

)

~ω

The normalisation coefficients are given by

An =(mω

π~

)1/4 (−i)n√

n!(~ω)n

So

ψ1(x) = A1a+e−

mω2~

x2 = A11√2m

(

~

i

d

dx+ imωx

)

e−mω2~

x2

16 / 27

General solutions II

Thus

ψ1(x) =A1√2m

(

~

i

(

−mω~

x)

+ imωx

)

e−mω2~

x2

And

ψ1(x) = (iA1ω√

2m)xe−mω2~

x2 with A1 =(mω

π~

)1/4 −i√~ω

Put simply, ψ1(x) = f (xe−x2) and E1 = 3~ω/2.

Higher energy states follow a similar pattern, e.g. ψ2 = f (x2e−x

2

)etc.

17 / 27

Wavefunctions for the harmonic oscillator

18 / 27

Probability densities for the harmonic oscillator

19 / 27

Properties of the wavefunctions

There is a non-zero probability of finding the particle outside the

classically allowed range.

For odd states the probability of finding the particle at the centre of

the well is zero.

As n becomes large the wavefunction begins to approximate the

classical probability distribution.

20 / 27

Analytical solution I

We begin with the Schrd̈inger equation

− ~2

2m

d2ψ

dx2+

1

2mω2x2ψ = Eψ

Let

ξ2 =mω

~x2 , d2x =

~

mωd2ξ and λ =

2E

~ω

Substituting into the Schrödinger equation gives

−d2ψ

dξ2+ ξ2ψ = λψ or

d2ψ

dξ2= ψ(ξ2 − λ)

If ξ is very large i.e. x is very large then the term ψλ can beneglected and we have

d2ψ

dξ2≈ ξ2ψ

21 / 27

Analytical solution II

An approximate solution is

ψ(ξ) ≈ Ae−ξ2/2 + Beξ2/2

The B term cannot be normalised as it tends to ∞ as x → ∞, so weare left with ψ(ξ) = H(ξ)e−ξ

2/2, where H(ξ) are functions of ξ whichmust not affect the asymptotic behaviour of ψ.

Differentiating ψ(ξ) gives

dψ

dξ=

(

dH

dξ− Hξ

)

e−ξ2/2

Differentiating again gives

d2ψ

dξ2=

(

d2H

dξ2−

(

H + ξdH

dξ

))

e−ξ2/2 +

(

Hξ2 − ξ dHdξ

)

e−ξ2/2

22 / 27

Analytical solution III

Substitute into

d2ψ

dξ2= ψ(ξ2 − λ) to get H(ξ2 − λ) = d

2H

dξ2− 2ξ dH

dξ+ Hξ2 − H

Sod2H

dξ2− 2ξ dH

dξ+ H(λ− 1) = 0

which is called the Hermite equation.

To solve this equation we expand H(ξ) as a power series in ξ:

H(ξ) = c0 + c1ξ + c2ξ2 + ... =

∞∑

j=0

cjξj

23 / 27

Analytical solution IV

Differentiating once gives

dH

dξ= c1 + 2c2ξ + 3c3ξ

2 + ... =

∞∑

j=0

jcjξj−1

Differentiating again gives

d2H

dξ2= 2c2 + 2 · 3c3ξ + 3 · 4c4ξ2 + ... =

∞∑

j=0

(j + 1)(j + 2)cj+2ξj

The Schrödinger equation becomes

∞∑

j=0

[(j + 1)(j + 2)cj+2 − 2jcj + (λ− 1)cj] ξj = 0

24 / 27

Analytical solution V

We can see from the power expansion that the coefficient of each

power of ξ must vanish so we can write

(j + 1)(j + 2)cj+2 − 2jcj + (λ− 1)cj = 0

Therefore

cj+2 =2j + 1 − λ

(j + 1)(j + 2)cj

So if we know c0 and c1 we can generate all the other values of cj.

We can write H(ξ) as H(ξ) = Heven(ξ) + Hodd(ξ) with

Heven(ξ) = c0 + c2ξ2 + c4ξ

4 + ... and Hodd(ξ) = c1ξ+ c3ξ3 + c5ξ

5 + ...

It is obvious that unless the series terminates at some point then

H(ξ) will tend to infinity and we will not have a physically realisticsolution.

25 / 27

Analytical solution VI

To make the series terminate after a certain number of terms we

can set λ = 2n + 1 then, when j = n, the top part of the recursionformula will equal zero and the series will end.

The recursion formula becomes

cj+2 =2(j − n)

(j + 1)(j + 2)cj

If we set λ = 2n + 1 to terminate the power series then, sinceλ = 2E/~ω, we find that En = (n +

1

2)~ω, which is the same as the

algebraic solution.

We set the value of λ to terminate the power series. This willterminate either Heven or Hodd. If we use the Heven series then we

must set c1 = 0, so that the odd series is zero, and vice-versa.

26 / 27

Analytical solution VII

Suppose n = 0 and c1 = 0 (no odd series), then H0(ξ) = c0 and

ψ0(ξ) = c0e−ξ2/2.

For n = 1 and c0 = 0 (no even series), then H1(ξ) = c1 and

ψ1(ξ) = c1ξe−ξ2/2.

For n = 2 and c1 = 0 we have

c2 = −4

2c0 = −2c0 so ψ2(ξ) = c0(1 − 2ξ2)e−ξ

2/2

The power series of ξ are known as Hermite polynomials. Thenormalised, stationary states are given by

ψn(x) =(mω

π~

)1/4 1√2nn!

Hn(ξ)e−

mω2~

x2

27 / 27