# 9. The Simple Harmonic Oscillator

*Michael Fowler *

### Einstein’s Solution of the Specific Heat Puzzle

The simple harmonic oscillator, a nonrelativistic particle
in a potential ${\scriptscriptstyle \frac{1}{2}}k{x}^{2},$ is an excellent model for a wide range of
systems in nature. In fact, not long
after Planck’s discovery that the black body radiation spectrum could be
explained by assuming energy to be exchanged in quanta, Einstein applied the
same principle to the simple harmonic oscillator, thereby solving a
long-standing puzzle in solid state physics$\u2014$the mysterious *drop in specific heat* of all solids at
low temperatures. Classical thermodynamics, a very successful theory in many
ways, predicted no such drop—with the standard equipartition of energy, $kT$ in each mode (potential plus kinetic), the
specific heat should remain more or less constant as the temperature was
lowered (assuming no phase change).

To explain the anomalous low temperature behavior, Einstein
assumed each atom to be an independent (quantum) simple harmonic oscillator,
and, just as for black body radiation, he assumed the oscillators could only
absorb or emit energy in *quanta*.
Consequently, at low enough temperatures there is rarely sufficient
energy in the ambient thermal excitations to excite the oscillators, and they
freeze out, just as blue oscillators do in low temperature black body
radiation. Einstein’s picture was later
somewhat refined—the basic set of oscillators was taken to be standing sound
wave oscillations in the solid rather than individual atoms (making the picture
even more like black body radiation in a cavity) but the main conclusion—the
drop off in specific heat at low temperatures—was not affected.

### The Classical Simple Harmonic Oscillator

The classical equation of motion for a one-dimensional simple harmonic oscillator with a particle of mass $m$ attached to a spring having spring constant $k$ is

$$m\frac{{d}^{2}x}{d{t}^{2}}=-kx.$$

The solution is

$$x={x}_{0}\mathrm{sin}(\omega t+\delta ),\text{}\omega =\sqrt{\frac{k}{m}}\text{\hspace{0.17em}},$$

and the momentum $p=mv$ has time dependence

$p=m{x}_{0}\omega \mathrm{cos}(\omega t+\delta ).$

The total energy

$(1/2m)({p}^{2}+{m}^{2}{\omega}^{2}{x}^{2})=E$

is clearly constant in time.

It is often useful to picture the time-development of a
system in ** phase space**, in this case a two-dimensional plot with
position on the

*$x$*-axis, momentum on the $y$ -axis. Actually, to have $\left(x,y\right)$ coordinates with the same dimensions, we use $\left(m\omega x,p\right).$

It is evident from the above expression for the total energy that in these variables the point representing the system in phase space moves clockwise around a circle of radius $\sqrt{2mE}$ centered at the origin.

Note that in the *classical* problem we could choose any point $\left(m\omega x,p\right),$ place the system there and it would then move
in a circle about the origin. In the *quantum*
problem, on the other hand, we cannot specify the initial coordinates $\left(m\omega x,p\right)$ precisely, because of the uncertainly
principle. The best we can do is to place the system initially in a small cell
in phase space, of size $\Delta x\cdot \Delta p=\hslash /2$. In fact, we shall find that in quantum
mechanics phase space is always divided into cells of essentially this size for
each pair of variables.

### Schrödinger’s Equation and the Ground State Wave Function

From the classical expression for total energy given above, the Schrödinger equation for the quantum oscillator follows in standard fashion:

$$-\frac{{\hslash}^{2}}{2m}\frac{{d}^{2}\psi (x)}{d{x}^{2}}+\frac{1}{2}m{\omega}^{2}{x}^{2}\psi (x)=E\psi (x).$$

What will the solutions to this Schrödinger equation look
like? Since the potential ${\scriptscriptstyle \frac{1}{2}}m{\omega}^{2}{x}^{2}$^{ }increases without limit on going away
from $x=0,$ it follows that no matter how much kinetic
energy the particle has, for sufficiently large $x$ the potential energy dominates, and the (bound
state) wavefunction decays with increasing rapidity for further increase in $x.$ (Obviously, for a real physical oscillator
there is a limit on the height of the potential$\u2014$we will assume
that limit is much greater than the energies of interest in our problem.)

We know that when a particle penetrates a barrier of
constant height ${V}_{0}$ (greater than the particle’s kinetic energy)
the wave function decreases exponentially into the barrier, as ${e}^{-\alpha x}$,
where $\alpha =\sqrt{2m\left({V}_{0}-E\right)/{\hslash}^{2}}.$ But, in contrast to this constant height
barrier, the “height” of the simple harmonic oscillator potential *continues to increase* as the particle
penetrates to larger $x.$ Obviously, in this situation the decay will be
faster than exponential. If we (rather
naïvely) assume it is more or less *locally* exponential, but with a local
$\alpha $ varying with ${V}_{0},$ neglecting $E$ relative to ${V}_{0}$ in the
expression for $\alpha $ suggests that $\alpha $ itself is proportional to $x$ (since
the potential is proportional to ${x}^{2},$ and $\alpha \propto \sqrt{V}$ ) so maybe the wavefunction decays as ${e}^{-\left(\text{constant}\right){x}^{2}}$?

To check this idea, we insert $\psi (x)={e}^{-\text{\hspace{0.05em}}\text{\hspace{0.17em}}{x}^{2}/2{b}^{2}}$ in the Schrödinger equation, using

$$\frac{{d}^{2}\psi}{d{x}^{2}}=-\frac{1}{{b}^{2}}\psi +\frac{{x}^{2}}{{b}^{4}}\psi $$

to find

$$$$$-\frac{{\hslash}^{2}}{2m}\left(-\frac{1}{{b}^{2}}+\frac{{x}^{2}}{{b}^{4}}\right)\psi (x)+\frac{1}{2}m{\omega}^{2}{x}^{2}\psi (x)=E\psi (x)$.

The $\psi \left(x\right)$ is just a factor here, and it is never zero, so can be cancelled out. This leaves a quadratic expression which must have the same coefficients of ${x}^{0},{x}^{2}$ on the two sides, that is, the coefficient of ${x}^{2}$ on the left hand side must be zero:

$$\frac{{\hslash}^{2}}{2m{b}^{4}}=\frac{m{\omega}^{2}}{2},\text{so}b=\sqrt{\frac{\hslash}{m\omega}}.$$

This fixes the wave function. Equating the constant terms fixes the energy:

$$E=\frac{{\hslash}^{2}}{2m{b}^{2}}=\frac{1}{2}\hslash \omega .$$

So the conjectured form for the wave function is in fact the
*exact* solution for the lowest energy
state! (It’s the lowest state because it
has no nodes.)

Also note that even in this ground state the energy is *nonzero*,
just as it was for the square well. The
central part of the wave function must have some curvature to join together the
decreasing wave function on the left to that on the right. This “zero point energy” is sufficient in one
physical case to melt the lattice—helium is liquid even down to absolute
zero temperature (checked down to microkelvins!) because the wave function
spread destabilizes the solid lattice that will form with sufficient external
pressure.

### Higher Energy States

It is clear from the above discussion of the ground state that $b=\sqrt{\frac{\hslash}{m\omega}}$ is the natural unit of length in this problem, and $\hslash \omega $ that of energy, so to investigate higher energy states we reformulate in dimensionless variables,

$$\xi =\frac{x}{b}=x\sqrt{\frac{m\omega}{\hslash}},\text{}\epsilon =\frac{E}{\hslash \omega}.$$

Schrödinger’s equation becomes

$$\frac{{d}^{2}\psi (\xi )}{d{\xi}^{2}}=({\xi}^{2}-2\epsilon )\psi (\xi ).$$

Deep in the barrier, the $\epsilon $ term will become negligible, and just as for the ground state wave function, higher bound state wave functions will have ${e}^{-{\xi}^{2}/2}$ behavior, multiplied by some more slowly varying factor (it turns out to be a polynomial).

*Exercise*: find the relative contributions to the
second derivative from the two terms in ${x}^{n}{e}^{-{x}^{2}/2}.$ For given $n,$ when do the contributions involving the first
term become small? Define “small”.

The standard approach to solving the general problem is to factor out the ${e}^{-{\xi}^{2}/2}$ term,

$\psi (\xi )=h(\xi ){e}^{-{\xi}^{2}/2}$

giving a differential equation for $h\left(\xi \right)$:

$$\frac{{d}^{2}h}{d{\xi}^{2}}-2\xi \frac{dh}{d\xi}+(2\epsilon -1)h=0$$

We try solving this with a power series in $\xi :$

$h(\xi )={h}_{0}+{h}_{1}\xi +{h}_{2}{\xi}^{2}=\mathrm{...}$ .

Inserting this in the differential equation, and requiring that the coefficient of each power ${\xi}^{n}$ vanish identically, leads to a recurrence formula for the coefficients ${h}_{n}:$

$${h}_{n+2}=\frac{(2n+1-2\epsilon )}{(n+1)(n+2)}{h}_{n}.$$

Evidently, the series of odd powers and that of even powers are independent solutions to Schrödinger’s equation. (Actually this isn’t surprising: the potential is even in $x,$ so the parity operator $P$ commutes with the Hamiltonian. Therefore, unless states are degenerate in energy, the wave functions will be even or odd in $x.$ ) For large $n,$ the recurrence relation simplifies to

$${h}_{n+2}\approx \frac{2}{n}{h}_{n},\text{\hspace{1em}}n\gg \epsilon .$$

The series therefore tends to

$\sum \frac{{2}^{n}{\xi}^{2n}}{(2n-2)(2n-4)\mathrm{...2}}}=2{\xi}^{2}{\displaystyle \sum \frac{{\xi}^{2\left(n-1\right)}}{\left(n-1\right)!}=}\text{\hspace{0.17em}}{e}^{{\xi}^{2}}.$

Multiply this by the ${e}^{-{\xi}^{2}/2}$ factor to recover the full wavefunction, we find $\psi $ diverges for large $\xi $ as ${e}^{+{\xi}^{2}/2}$.

Actually we should have expected this—for a general value of the energy, the Schrödinger equation has the solution $\approx A{e}^{+{\xi}^{2}/2}+B{e}^{-{\xi}^{2}/2}$ at large distances, and only at certain energies does the coefficient $A$ vanish to give a normalizable bound state wavefunction.

So how do we find the *nondiverging* solutions? It is clear that the infinite power series
must be stopped! The key is in the
recurrence relation.

If the energy satisfies

$2\epsilon =2n+1,\text{\hspace{1em}}n\text{aninteger,}$

*then* ${h}_{n+2}$ *and all
higher coefficients vanish*.

This requirement in fact *completely
determines the polynomial* (except for an overall constant) because with $2\epsilon =2n+1$ the coefficients ${h}_{m}$ for $mn$ are determined by

$${h}_{m+2}=\frac{(2m+1-2\epsilon )}{(m+1)(m+2)}{h}_{m}=\frac{(2m+1-\left(2n+1\right))}{(m+1)(m+2)}{h}_{m}.$$

This ${n}^{\text{th}}$ order polynomial is called a *Hermite
polynomial* and written ${H}_{n}\left(\xi \right).$ The
standard normalization of the Hermite polynomials ${H}_{n}\left(\xi \right)$ is to take the coefficient of the highest
power ${\xi}^{n}$ to be ${2}^{n}.$ The
other coefficients then follow using the recurrence relation above, giving:

${H}_{0}(\xi )=1,\text{}{H}_{1}(\xi )=2\xi ,\text{}{H}_{2}(\xi )=4{\xi}^{2}-2,\text{}{H}_{3}(\xi )=8{\xi}^{3}-12\xi ,\text{etc}\text{.}$

So the bottom line is that the wavefunction for the ${n}^{\text{th}}$ excited state, having energy $\epsilon =n+{\scriptscriptstyle \frac{1}{2}}$, is ${\psi}_{n}(\xi )={C}_{n}{H}_{n}(\xi ){e}^{-{\xi}^{2}/2}$, where ${C}_{n}$ is a normalization constant to be determined in the next section.

It can be shown (see exercises at the end of this lecture) that ${H}_{n}{}^{\prime}\left(\xi \right)=2n{H}_{n-1}\left(\xi \right)$. Using this, beginning with the ground state, one can easily convince oneself that the successive energy eigenstates each have one more node—the ${n}^{\text{th}}$ state has $n$ nodes. This is also evident from numerical solution using the spreadsheet, watching how the wave function behaves at large $x$ as the energy is cranked up.

The spreadsheet can also be used to plot the wave function
for large $n,$ say $n=200.$ It is instructive to compare the probability
distribution with that for a *classical*
pendulum, one oscillating with fixed amplitude and observed many times at
random intervals. For the pendulum, the probability peaks at the end of the
swing, where the pendulum is slowest and therefore spends most time. The $n=200$ distribution amplitude follows this pattern,
but of course oscillates. However, in
the large $n$ limit these oscillations take place over
undetectably small intervals.

The *classical*
pendulum when not at rest clearly has a time-dependent probability distribution—it swings backwards and forwards.
This means it *cannot* be in an
eigenstate of the energy. In fact, the
quantum state most like the classical is a *coherent
state* built up of neighboring energy eigenstates. We shall discuss coherent states later in the
course.

### Operator Approach to the Simple Harmonic Oscillator

Having scaled the position coordinate $x$ to the dimensionless $\xi \text{by}\xi =x/b=x\sqrt{m\omega /\hslash},$ let us also scale the momentum from $p$ to $\pi =-i\text{\hspace{0.17em}}d/d\xi $ (so $\pi =bp/\hslash =p/\sqrt{\hslash m\omega}$ ).

The Hamiltonian is

$$H=\frac{{p}^{2}+{m}^{2}{\omega}^{2}{x}^{2}}{2m}=\frac{\hslash \omega}{2}\left({\pi}^{2}+{\xi}^{2}\right).$$

Dirac had the brilliant idea of factorizing this expression: the obvious thought $\left({\xi}^{2}+{\pi}^{2}\right)=\left(\xi +i\pi \right)\left(\xi -i\pi \right)$ isn’t quite right, because it fails to take account of the noncommutativity of the operators, but the symmetrical version

$$H=\frac{\hslash \omega}{4}\left[\left(\xi +i\pi \right)\left(\xi -i\pi \right)+\left(\xi -i\pi \right)\left(\xi +i\pi \right)\right]$$

is fine, and we shall soon see that it leads to a very easy way of finding the eigenvalues and operator matrix elements for the oscillator, far simpler than using the wave functions we found above. Interestingly, Dirac’s factorization here of a second-order differential operator into a product of first-order operators is close to the idea that led to his most famous achievement, the Dirac equation, the basis of the relativistic theory of electrons, protons, etc.

To continue, we define new operators $a,\text{\hspace{0.17em}}{a}^{\u2020}$ by

$$a=\frac{\xi +i\pi}{\sqrt{2}}=\frac{1}{\sqrt{2\hslash m\omega}}\left(m\omega x+ip\right),\text{\hspace{1em}}{a}^{\u2020}=\frac{\xi -i\pi}{\sqrt{2}}=\frac{1}{\sqrt{2\hslash m\omega}}\left(m\omega x-ip\right).$$

(We’ve expressed $a$ in terms of the original variables $x,p$ for later use.)

From the commutation relation $\left[i\pi ,\xi \right]=1$ it follows that

$[a,{a}^{\u2020}]=1.$

Therefore the Hamiltonian can be written:

$H=\hslash \omega \left({a}^{\u2020}a+\frac{1}{2}\right)=\hslash \omega \left(N+\frac{1}{2}\right),\text{where}N={a}^{\u2020}a.$

Note that the operator $N$ can only have *non-negative* eigenvalues,
since

$\langle \psi |N|\psi \rangle =\langle \psi |{a}^{\u2020}a|\psi \rangle =\langle {\psi}_{a}|{\psi}_{a}\rangle \text{\hspace{0.17em}}\text{\hspace{0.17em}}\ge \text{\hspace{0.17em}}\text{\hspace{0.17em}}0.$

Now

$\left[N,{a}^{\u2020}\right]={a}^{\u2020}a{a}^{\u2020}-{a}^{\u2020}{a}^{\u2020}a={a}^{\u2020}\left[a,{a}^{\u2020}\right]={a}^{\u2020}$

Suppose $N$ has an eigenfunction $|\nu \rangle $ with eigenvalue $\nu $,

$N|\nu \rangle \text{\hspace{0.17em}}=\nu |\nu \rangle .$

From the two equations above

$N{a}^{\u2020}|\nu \rangle \text{\hspace{0.17em}}=\text{\hspace{0.17em}}{a}^{\u2020}N|\nu \rangle +{a}^{\u2020}|\nu \rangle \text{\hspace{0.17em}}=\left(\nu +1\right){a}^{\u2020}|\nu \rangle $

so ${a}^{\u2020}|\nu \rangle $ is an eigenfunction of $N$ with eigenvalue $\nu +1.$ Operating with ${a}^{\u2020}$ again and again, we climb an infinite ladder of eigenstates equally spaced in energy.

${a}^{\u2020}$ is often termed a ** creation operator**,
since the quantum of energy $\hslash \omega $ added each time it operates is equivalent to
an added photon in black body radiation (electromagnetic oscillations in a
cavity).

It is easy to check that the state $a|\nu \rangle $_{ }is an eigenstate with eigenvalue $\nu -1,$* *provided it is nonzero, so the operator $a$ takes us *down* the ladder. However, this
cannot go on indefinitely—we have established that $N$ cannot have negative eigenvalues. We must
eventually reach a state $|\nu \rangle \text{\hspace{0.17em}}\text{forwhich}a|\nu \rangle \text{\hspace{0.17em}}=0,$ the operator $a$ *annihilates*
the state. (At each step down, $a$ annihilates one quantum of energy—so $a$ is often called an *annihilation* or *destruction
*operator.)

Since the norm squared of $a|\nu \rangle ,$ ${\left|a|\nu \rangle \right|}^{2}=\langle \nu |{a}^{\u2020}a|\nu \rangle \text{\hspace{0.17em}}=\langle \nu |N|\nu \rangle \text{\hspace{0.17em}}=\nu \langle \nu |\nu \rangle ,$ and since $\langle \nu |\nu \rangle \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0\text{\hspace{0.17em}}$ for any nonvanishing state, it must be that *the lowest eigenstate* (the $|\nu \rangle \text{\hspace{0.17em}}\text{forwhich}a|\nu \rangle \text{\hspace{0.17em}}=0$ ) *has $\nu =0.$ * It
follows that the $\nu $ ’s on the ladder are *the positive integers*, so from this point on we relabel the
eigenstates with $n$ in place of $\nu .$

That is to say, we have proved that the only possible eigenvalues of $N$ are zero and the positive integers: 0, 1, 2, 3… .

$N$ is called the** number operator**: it measures
the number of quanta of energy in the oscillator above the irreducible ground
state energy (that is, above the “zero-point energy” arising from the wave-like
nature of the particle).

Since from above the Hamiltonian

$H=\hslash \omega \left({a}^{\u2020}a+\frac{1}{2}\right)=\hslash \omega \left(N+\frac{1}{2}\right)$

the energy eigenvalues are

$H|n\rangle =\left(n+{\scriptscriptstyle \frac{1}{2}}\right)\hslash \omega |n\rangle .$

It is important to appreciate that Dirac’s factorization
trick and very little effort has given us *all* the eigenvalues of the
Hamiltonian

$$H=\frac{\hslash \omega}{2}\left({\pi}^{2}+{\xi}^{2}\right).$$

Contrast the work needed in this section with that in the standard Schrödinger approach. We have also established that the lowest energy state $|0\rangle $, having energy ${\scriptscriptstyle \frac{1}{2}}\hslash \omega ,$ must satisfy the first-order differential equation $a|0\rangle =0,$ that is,

$$(\xi +i\pi )|0\text{\hspace{0.17em}}=\left(\xi +\frac{d}{d\xi}\right){\psi}_{0}(\xi )=0.$$

The solution, unnormalized, is

${\psi}_{0}(\xi )=C{e}^{-{\xi}^{2}/2}.$

(In fact, we’ve seen this equation and its solution before: this was the condition for the “least uncertain” wave function in the discussion of the Generalized Uncertainty Principle.)

We denote the *normalized* set of eigenstates $|0\rangle ,|1\rangle ,|2\rangle ,\dots |n\rangle \dots \text{with}\langle n|n\rangle \text{\hspace{0.17em}}=1.$ Now ${a}^{\u2020}|n\rangle \text{\hspace{0.17em}}={C}_{n}|n+1\rangle ,$ and *C _{n}* is easily found:

$|{C}_{n}{|}^{2}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}|{C}_{n}{|}^{2}\langle n+1|n+1\rangle \text{\hspace{0.17em}}=\text{\hspace{0.17em}}\langle n|a{a}^{\u2020}|n\rangle \text{\hspace{0.17em}}=\left(n+1\right),$

and

${a}^{\u2020}|n\rangle =\sqrt{n+1}|n+1\rangle .$

Therefore, if we take the set of orthonormal states $|0\rangle ,|1\rangle ,|2\rangle ,\dots |n\rangle \dots $ as the basis in the Hilbert space, the *only*
nonzero matrix elements of ${a}^{\u2020}$ are $\langle n+1|{a}^{\u2020}|n\rangle \text{\hspace{0.17em}}=\sqrt{n+1}.$ That is to say,

${a}^{\u2020}=\left(\begin{array}{ccccc}0& 0& 0& 0& \dots \\ \sqrt{1}& 0& 0& 0& \dots \\ 0& \sqrt{2}& 0& 0& \dots \\ 0& 0& \sqrt{3}& 0& \dots \\ \vdots & \vdots & \vdots & \vdots & \ddots \end{array}\right).$

(The column vectors in the space this matrix operates on
have an infinite number of elements: the lowest energy, the ground state
component, is the entry at the *top *of the infinite vector—so up the
energy ladder is down the vector!)

The adjoint

$a=\left(\begin{array}{ccccc}0& \sqrt{1}& 0& 0& \dots \\ 0& 0& \sqrt{2}& 0& \dots \\ 0& 0& 0& \sqrt{3}& \dots \\ 0& 0& 0& 0& \dots \\ \vdots & \vdots & \vdots & \vdots & \ddots \end{array}\right).$

So

$a|n\rangle \text{\hspace{0.17em}}=\sqrt{n}|n-1\rangle .$

For practical computations, we need to find the matrix elements of the position and momentum variables between the normalized eigenstates. Now

$x=\sqrt{\hslash /2m\omega}\text{\hspace{0.17em}}\left({a}^{\u2020}+a\right),\text{\hspace{0.17em}}\text{\hspace{1em}}p=i\sqrt{m\omega \hslash /2}\text{\hspace{0.17em}}\left({a}^{\u2020}-a\right)$

so

$x=\sqrt{\hslash /2m\omega}\left(\begin{array}{ccccc}0& \sqrt{1}& 0& 0& \dots \\ \sqrt{1}& 0& \sqrt{2}& 0& \dots \\ 0& \sqrt{2}& 0& \sqrt{3}& \dots \\ 0& 0& \sqrt{3}& 0& \dots \\ \vdots & \vdots & \vdots & \vdots & \ddots \end{array}\right),\text{\hspace{1em}}p=i\sqrt{m\omega \hslash /2}\left(\begin{array}{ccccc}0& -\sqrt{1}& 0& 0& \dots \\ \sqrt{1}& 0& -\sqrt{2}& 0& \dots \\ 0& \sqrt{2}& 0& -\sqrt{3}& \dots \\ 0& 0& \sqrt{3}& 0& \dots \\ \vdots & \vdots & \vdots & \vdots & \ddots \end{array}\right).$

These matrices are, of course, Hermitian (not forgetting the $i$ factor in $p.$ )

To find the matrix elements between eigenstates of any product of $x$ ’s and $p$ ’s, express all the $x$ ’s and $p$ ’s in terms of $a$ ’s and ${a}^{\u2020}$ ’s, to give a sum of products of $a$ ’s and ${a}^{\u2020}$ ’s. Each product in this sum can be evaluated sequentially from the right, because each $a$ or ${a}^{\u2020}$ has only one nonzero matrix element when the product operates on one eigenstate.

### Normalizing the Eigenstates in *x*-space

The normalized ground state wave function is

$${\psi}_{0}(\xi )=C{e}^{-{\xi}^{2}/2}={\left(\frac{m\omega}{\pi \hslash}\right)}^{{\scriptscriptstyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}}{e}^{-m\omega {x}^{2}/2\hslash},$$

where we have gone back to the $x$ variable, and normalized using $\int}_{-\infty}^{\infty}{e}^{-a{x}^{2}}dx=\sqrt{\pi /a$.

To find the normalized wave functions for the higher states, they are first constructed formally by applying the creation operator ${a}^{\u2020}$ repeatedly on the ground state $|0\rangle .$ Next, the result is translated into $x$ -space (actually $\xi =x/b$ ) by writing ${a}^{\u2020}$ as a differential operator, acting on ${\psi}_{0}\left(\xi \right).$

Using $\langle n|{a}^{\u2020}|n-1\rangle \text{\hspace{0.17em}}=\sqrt{n},$

$$|n\rangle \text{\hspace{0.17em}}=\frac{{a}^{\u2020}}{\sqrt{n}}|n-1\rangle \text{\hspace{0.17em}}=\dots =\frac{{\left({a}^{\u2020}\right)}^{n}}{\sqrt{n!}}|0\rangle .$$

Now

${a}^{\u2020}=\left(1/\sqrt{2}\right)\left(\xi -i\pi \right)=\left(1/\sqrt{2}\right)\left(\xi -d/d\xi \right),$

so

$${\psi}_{n}(\xi )=\frac{{\left({a}^{\u2020}\right)}^{n}}{\sqrt{n!}}|0\rangle =\frac{1}{\sqrt{n!}}{\left(\frac{1}{\sqrt{2}}\left(\xi -\frac{d}{d\xi}\right)\right)}^{n}{\left(\frac{m\omega}{\pi \hslash}\right)}^{{\scriptscriptstyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}}{e}^{-{\xi}^{2}/2}.$$

We need to check that this expression is indeed the same as the Hermite polynomial wave function derived earlier, and to do that we need some further properties of the Hermite polynomials.

### Some Properties of Hermite Polynomials

The mathematicians *define*
the Hermite polynomials by:

$${H}_{n}(\xi )={(-)}^{n}{e}^{{\xi}^{2}}\frac{{d}^{n}}{d{\xi}^{n}}{e}^{-{\xi}^{2}}$$

so

${H}_{0}(\xi )=1,\text{}{H}_{1}(\xi )=2\xi ,\text{}{H}_{2}(\xi )=4{\xi}^{2}-2,\text{}{H}_{3}(\xi )=8{\xi}^{3}-12\xi ,\text{etc}\text{.}$

It follows immediately from the definition that the coefficient of the leading power is ${2}^{n}.$

It is a straightforward exercise to check that ${H}_{n}$ is a solution of the differential equation

$$\left(\frac{{d}^{2}}{d{\xi}^{2}}-2\xi \frac{d}{d\xi}+2n\right){H}_{n}(\xi )=0,$$

so these are indeed the same polynomials we found by the series solution of Schrödinger’s equation earlier (recall the equation for the polynomial component of the wave function was

$$\frac{{d}^{2}h}{d{\xi}^{2}}-2\xi \frac{dh}{d\xi}+(2\epsilon -1)h=0$$

with $2\epsilon =2n+1$.)

We have found ${\psi}_{n}(\xi )$ in the form

$${\psi}_{n}(\xi )=\frac{1}{\sqrt{n!}}{\left(\frac{1}{\sqrt{2}}\left(\xi -\frac{d}{d\xi}\right)\right)}^{n}{\left(\frac{m\omega}{\pi \hslash}\right)}^{{\scriptscriptstyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}}{e}^{-{\xi}^{2}/2}.$$

We shall now prove that the polynomial component is exactly equivalent to the Hermite polynomial as defined at the beginning of this section.

We begin with the operator identity:

$$\left(\xi -\frac{d}{d\xi}\right)=-{e}^{{\xi}^{2}/2}\frac{d}{d\xi}{e}^{-{\xi}^{2}/2}$$

Both sides of this expression are to be regarded as *operators*, that is, it is assumed that
both are operating on some function $f\left(\xi \right)$.

Now take the ${n}^{\text{th}}$ power of both sides: on the right, we find, for example,

$$\begin{array}{c}{\left(-{e}^{{\xi}^{2}/2}\frac{d}{d\xi}{e}^{-{\xi}^{2}/2}\right)}^{3}={(-)}^{3}{e}^{{\xi}^{2}/2}\frac{d}{d\xi}{e}^{-{\xi}^{2}/2}{e}^{{\xi}^{2}/2}\frac{d}{d\xi}{e}^{-{\xi}^{2}/2}{e}^{{\xi}^{2}/2}\frac{d}{d\xi}{e}^{-{\xi}^{2}/2}\\ ={(-)}^{3}{e}^{{\xi}^{2}/2}\frac{{d}^{3}}{d{\xi}^{3}}{e}^{-{\xi}^{2}/2}\end{array}$$

since the intermediate exponential terms cancel against each other.

So:

$${\left(\xi -\frac{d}{d\xi}\right)}^{n}={(-)}^{n}{e}^{{\xi}^{2}/2}\frac{{d}^{n}}{d{\xi}^{n}}{e}^{-{\xi}^{2}/2}$$

and substituting this into the expression for ${\psi}_{n}(\xi )$ above,

$$\begin{array}{c}{\psi}_{n}\left(\xi \right)=\frac{1}{\sqrt{{2}^{n}n!}}{(-)}^{n}\left({e}^{{\xi}^{2}/2}\frac{{d}^{n}}{d{\xi}^{n}}{e}^{-{\xi}^{2}/2}\right){\left(\frac{m\omega}{\pi \hslash}\right)}^{1/4}{e}^{-{\xi}^{2}/2}\\ =\frac{1}{\sqrt{{2}^{n}n!}}{(-)}^{n}{\left(\frac{m\omega}{\pi \hslash}\right)}^{1/4}{e}^{-{\xi}^{2}/2}\left({e}^{{\xi}^{2}}\frac{{d}^{n}}{d{\xi}^{n}}{e}^{-{\xi}^{2}}\right)\\ =\frac{1}{\sqrt{{2}^{n}n!}}{\left(\frac{m\omega}{\pi \hslash}\right)}^{1/4}{H}_{n}\left(\xi \right){e}^{-{\xi}^{2}/2},\text{with}\xi =\sqrt{\frac{m\omega}{\hslash}}x.\end{array}$$

This established the equivalence of the two approaches to Schrödinger’s equation for the simple harmonic oscillator, and provides us with the overall normalization constants without doing integrals. (The expression for ${\psi}_{n}(\xi )$ above satisfies $\int |{\psi}_{n}{|}^{2}dx=1$.)

*Exercises: *

Use ${H}_{n}\left(\xi \right)={(-)}^{n}{e}^{{\xi}^{2}}\frac{{d}^{n}}{d{\xi}^{n}}{e}^{-{\xi}^{2}}$ to prove:

(a) the coefficient of ${\xi}^{n}$ is ${2}^{n}$.

(b) ${H}_{n}{}^{\prime}\left(\xi \right)=2n{H}_{n-1}\left(\xi \right)$

(c) ${H}_{n+1}\left(\xi \right)=2\xi {H}_{n}\left(\xi \right)-2n{H}_{n-1}\left(\xi \right)$

(d) ${\int}_{-\infty}^{\infty}{e}^{-{\xi}^{2}}{H}_{n}^{2}\left(\xi \right)}\text{\hspace{0.05em}}d\xi ={2}^{n}n!\sqrt{\pi$

(Hint: rewrite as ${\int}_{-\infty}^{\infty}{H}_{n}\left(\xi \right){(-)}^{n}\frac{{d}^{n}}{d{\xi}^{n}}{e}^{-{\xi}^{2}}}\text{\hspace{0.05em}}d\xi $,
then integrate by parts *n* times, and
use (a).)

(e) ${\int}_{-\infty}^{\infty}{e}^{-{\xi}^{2}}{H}_{n}\left(\xi \right)}\text{\hspace{0.05em}}{H}_{m}\left(\xi \right)d\xi =0,\text{for}m\ne n.$

It’s worth doing these exercises to become more familiar with the Hermite polynomials, but in evaluating matrix elements (and indeed in establishing some of these results) it is almost always far simpler to work with the creation and annihilation operators.

*Exercise*: use the creation and annihilation
operators to find $\langle n|{x}^{4}|n\rangle $. This matrix element is useful in estimating
the energy change arising on adding a small nonharmonic potential energy term
to a harmonic oscillator.

### Time-Dependent Wave Functions

The set of normalized eigenstates $|0\rangle ,|1\rangle ,|2\rangle ,\dots |n\rangle \dots \text{}$ discussed above are of course solutions to the
time-*independent* Schrödinger
equation, or in ket notation eigenstates of the Hamiltonian $H|n\rangle =\left(n+{\scriptscriptstyle \frac{1}{2}}\right)\hslash \omega |n\rangle .$ Putting in the time-dependence explicitly, $|n,t\rangle ={e}^{-iHt/\hslash}|n,t=0\rangle ={e}^{-i\left(n+{\scriptscriptstyle \frac{1}{2}}\right)\omega t}|n\rangle $. It is necessary to include the time
dependence when dealing with a state which is a superposition of states of
different energies, such as $\left(1/\sqrt{2}\right)\left(|0\rangle +|1\rangle \right),$ which then becomes $\left(1/\sqrt{2}\right)\left({e}^{-i\omega t/2}|0\rangle +{e}^{-3i\omega t/2}|1\rangle \right).$ Expectation values of combinations of position
and/or momentum operators in such states are best evaluated by expressing everything
in terms of annihilation and creation operators.

### Solving Schrödinger’s Equation in Momentum Space

In the lecture on Function Spaces, we established that the basis of $|x\rangle $ states (eigenstates of the position operator) and that of $|k\rangle $ states (eigenstates of the momentum operator) were both complete bases in Hilbert space (physicist’s definition) so we could work equally well with either from a formal point of view. Why then do we almost always work in $x$-space? Well, probably because we live in $x$-space, but there’s another reason. The momentum operator in the $x$-space representation is $p=-i\hslash d/dx$, so Schrödinger’s equation, written $\left({p}^{2}/2m+V\left(x\right)\right)\psi \left(x\right)=E\psi \left(x\right)$, with $p$ in operator form, is a second-order differential equation. Now consider what happens to Schrödinger’s equation if we work in $p$-space. Since the operator identity $\left[x,p\right]=i\hslash $ is true regardless of representation, we must have $x=i\hslash d/dp$. So for a particle in a potential $V\left(x\right)$, writing Schrödinger’s equation in $p$-space we are confronted with the nasty looking operator $V\left(i\hslash d/dp\right)$! This will produce a differential equation in general a lot harder to solve than the standard $x$-space equation—so we stay in $x$-space.

*But* there are two
potentials that can be handled in momentum space: first, for a *linear *potential $V\left(x\right)=-Fx$,
the momentum space analysis is actually easier—it’s just a first-order
equation. Second, for a particle in a *quadratic* potential—a simple harmonic
oscillator—the two approaches yield the *same*
differential equation. That means that the eigenfunctions in momentum space
(scaled appropriately) must be *identical*
to those in position space—the simple harmonic eigenfunctions are their own
Fourier transforms!