# Harmonic oscillator in 2-d and 3-d, and in polar and spherical coordinates

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 10, Exercises 10.2.2 – 10.2.3.

We’ve seen that the 3-d isotropic harmonic oscillator can be solved in rectangular coordinates using separation of variables. The Hamiltonian is

$\displaystyle H=\frac{p_{x}^{2}+p_{y}^{2}+p_{z}^{2}}{2m}+\frac{m\omega^{2}}{2}\left(x^{2}+y^{2}+z^{2}\right) \ \ \ \ \ (1)$

The solution to the Schrödinger equation is just the product of three one-dimensional oscillator eigenfunctions, one for each coordinate. That is

$\displaystyle \psi_{n}\left(x,y,z\right)=\psi_{n_{x}}\left(x\right)\psi_{n_{y}}\left(y\right)\psi_{n_{z}}\left(z\right) \ \ \ \ \ (2)$

Each one-dimensional eigenfunction can be expressed in terms of Hermite polynomials as

$\displaystyle \psi_{n_{x}}\left(x\right)=\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\frac{1}{\sqrt{2^{n_{x}}n_{x}!}}H_{n_{x}}\left(\sqrt{\frac{m\omega}{\hbar}}x\right)e^{-m\omega x^{2}/2\hbar} \ \ \ \ \ (3)$

with the functions for ${y}$ and ${z}$ obtained by replacing ${x}$ by ${y}$ or ${z}$ and ${n_{x}}$ by ${n_{y}}$ or ${n_{z}}$. We also saw earlier that in the 3-d oscillator, the total energy for state ${\psi_{n}\left(x,y,z\right)}$ is given in terms of the quantum numbers of the three 1-d oscillators as

$\displaystyle E_{n}=\hbar\omega\left(n+\frac{3}{2}\right)=\hbar\omega\left(n_{x}+n_{y}+n_{z}+\frac{3}{2}\right) \ \ \ \ \ (4)$

and that the degeneracy of level ${n}$ is ${\frac{1}{2}\left(n+1\right)\left(n+2\right)}$.

Since the Hermite polynomial ${H_{n_{x}}}$ has parity ${\left(-1\right)^{n_{x}}}$ (that is, odd (even) polynomials are odd (even) functions), the 3-d wave function ${\psi_{n}}$ has parity ${\left(-1\right)^{n_{x}}\left(-1\right)^{n_{y}}\left(-1\right)^{n_{z}}=\left(-1\right)^{n}}$.

We can write the one ${n=0}$ state and three ${n=1}$ states in spherical coordinates using the standard transformation

 $\displaystyle x$ $\displaystyle =$ $\displaystyle r\sin\theta\cos\phi\ \ \ \ \ (5)$ $\displaystyle y$ $\displaystyle =$ $\displaystyle r\sin\theta\sin\phi\ \ \ \ \ (6)$ $\displaystyle z$ $\displaystyle =$ $\displaystyle r\cos\phi \ \ \ \ \ (7)$

Using the notation ${\psi_{n}=\psi_{n_{x}n_{y}n_{z}}=\psi_{n_{x}}\psi_{n_{y}}\psi_{n_{z}}}$, we have, using ${H_{0}\left(y\right)=1}$ and ${H_{1}\left(y\right)=2y}$:

 $\displaystyle \psi_{000}$ $\displaystyle =$ $\displaystyle \left(\frac{m\omega}{\pi\hbar}\right)^{3/4}e^{-m\omega r^{2}/2\hbar}\ \ \ \ \ (8)$ $\displaystyle \psi_{100}$ $\displaystyle =$ $\displaystyle \sqrt{\frac{2m\omega}{\hbar}}\left(\frac{m\omega}{\pi\hbar}\right)^{3/4}e^{-m\omega r^{2}/2\hbar}r\sin\theta\cos\phi\ \ \ \ \ (9)$ $\displaystyle \psi_{010}$ $\displaystyle =$ $\displaystyle \sqrt{\frac{2m\omega}{\hbar}}\left(\frac{m\omega}{\pi\hbar}\right)^{3/4}e^{-m\omega r^{2}/2\hbar}r\sin\theta\sin\phi\ \ \ \ \ (10)$ $\displaystyle \psi_{001}$ $\displaystyle =$ $\displaystyle \sqrt{\frac{2m\omega}{\hbar}}\left(\frac{m\omega}{\pi\hbar}\right)^{3/4}e^{-m\omega r^{2}/2\hbar}r\cos\theta \ \ \ \ \ (11)$

We can check that these are the correct spherical versions of the eigenfunctions by using the Schrödinger equation in spherical coordinates, which is

$\displaystyle H\psi=\left[-\frac{\hbar^{2}\nabla^{2}}{2m}+\frac{m\omega^{2}}{2}r^{2}\right]\psi=E\psi \ \ \ \ \ (12)$

The spherical laplacian operator is

$\displaystyle \nabla^{2}\psi=\frac{1}{r^{2}}\frac{\partial}{\partial r}\left(r^{2}\frac{\partial\psi}{\partial r}\right)+\frac{1}{r^{2}\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial\psi}{\partial\theta}\right)+\frac{1}{r^{2}\sin^{2}\theta}\frac{\partial^{2}\psi}{\partial\phi^{2}} \ \ \ \ \ (13)$

You can grind through the derivatives by hand if you like, but I just used Maple to do it, giving the results

 $\displaystyle H\psi_{000}$ $\displaystyle =$ $\displaystyle \frac{3}{2}\hbar\omega\psi_{000}\ \ \ \ \ (14)$ $\displaystyle H\psi_{100}$ $\displaystyle =$ $\displaystyle \frac{5}{2}\hbar\omega\psi_{100}\ \ \ \ \ (15)$ $\displaystyle H\psi_{010}$ $\displaystyle =$ $\displaystyle \frac{5}{2}\hbar\omega\psi_{010}\ \ \ \ \ (16)$ $\displaystyle H\psi_{001}$ $\displaystyle =$ $\displaystyle \frac{5}{2}\hbar\omega\psi_{001} \ \ \ \ \ (17)$

In two dimensions, the analysis is pretty much the same. In the more general case where the masses are equal, but ${\omega_{x}\ne\omega_{y}}$, the Hamiltonian is

$\displaystyle H=\frac{p_{x}^{2}+p_{y}^{2}}{2m}+\frac{m}{2}\left(\omega_{x}^{2}x^{2}+\omega_{y}^{2}y^{2}\right) \ \ \ \ \ (18)$

A solution by separation of variables still works, with the result

$\displaystyle \psi_{n}\left(x,y\right)=\psi_{n_{x}}\left(x\right)\psi_{n_{y}}\left(y\right) \ \ \ \ \ (19)$

The total energy is

$\displaystyle E_{n}=E_{n_{x}}+E_{n_{y}}=\hbar\omega\left(n_{x}+\frac{1}{2}+n_{y}+\frac{1}{2}\right)=\hbar\omega\left(n+1\right) \ \ \ \ \ (20)$

For a given energy level ${n=n_{x}+n_{y}}$, there are ${n+1}$ ways of forming ${n}$ out of a sum of 2 non-negative integers, so the degeneracy of level ${n}$ is ${n+1}$.

The one ${n=0}$ state and two ${n=1}$ states are

 $\displaystyle \psi_{00}$ $\displaystyle =$ $\displaystyle \left(\frac{m\omega}{\pi\hbar}\right)^{1/2}e^{-m\omega\left(x^{2}+y^{2}\right)/2\hbar}\ \ \ \ \ (21)$ $\displaystyle \psi_{10}$ $\displaystyle =$ $\displaystyle \sqrt{\frac{2m\omega}{\hbar}}\left(\frac{m\omega}{\pi\hbar}\right)^{1/2}e^{-m\omega\left(x^{2}+y^{2}\right)/2\hbar}x\ \ \ \ \ (22)$ $\displaystyle \psi_{01}$ $\displaystyle =$ $\displaystyle \sqrt{\frac{2m\omega}{\hbar}}\left(\frac{m\omega}{\pi\hbar}\right)^{1/2}e^{-m\omega\left(x^{2}+y^{2}\right)/2\hbar}y \ \ \ \ \ (23)$

To translate to polar coordinates, we use the transformations

 $\displaystyle x$ $\displaystyle =$ $\displaystyle \rho\cos\phi\ \ \ \ \ (24)$ $\displaystyle y$ $\displaystyle =$ $\displaystyle \rho\sin\phi \ \ \ \ \ (25)$

so we have

 $\displaystyle \psi_{00}$ $\displaystyle =$ $\displaystyle \left(\frac{m\omega}{\pi\hbar}\right)^{1/2}e^{-m\omega\rho^{2}/2\hbar}\ \ \ \ \ (26)$ $\displaystyle \psi_{10}$ $\displaystyle =$ $\displaystyle \sqrt{\frac{2m\omega}{\hbar}}\left(\frac{m\omega}{\pi\hbar}\right)^{1/2}e^{-m\omega\rho^{2}/2\hbar}\rho\cos\phi\ \ \ \ \ (27)$ $\displaystyle \psi_{01}$ $\displaystyle =$ $\displaystyle \sqrt{\frac{2m\omega}{\hbar}}\left(\frac{m\omega}{\pi\hbar}\right)^{1/2}e^{-m\omega\rho^{2}/2\hbar}\rho\sin\phi \ \ \ \ \ (28)$

Again, we can check this by plugging these polar formulas into the polar Schrödinger equation, where the 2-d Laplacian is

$\displaystyle \nabla^{2}=\frac{\partial^{2}}{\partial\rho^{2}}+\frac{1}{\rho}\frac{\partial}{\partial\rho}+\frac{1}{\rho^{2}}\frac{\partial^{2}}{\partial\phi^{2}} \ \ \ \ \ (29)$

The results are

 $\displaystyle H\psi_{00}$ $\displaystyle =$ $\displaystyle \hbar\omega\psi_{00}\ \ \ \ \ (30)$ $\displaystyle H\psi_{10}$ $\displaystyle =$ $\displaystyle 2\hbar\omega\psi_{10}\ \ \ \ \ (31)$ $\displaystyle H\psi_{01}$ $\displaystyle =$ $\displaystyle 2\hbar\omega\psi_{01} \ \ \ \ \ (32)$

# Uncertainties in the harmonic oscillator and hydrogen atom

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 9, Exercises 9.4.1 – 9.4.2.

Here we’ll look at a couple of calculations relevant to the application of the uncertainty principle to the hydrogen atom. When calculating uncertainties, we need to find the average values of various quantities. First, we’ll look at an average in the case of the harmonic oscillator.

$\displaystyle \psi_{n}(x)=\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\frac{1}{\sqrt{2^{n}n!}}H_{n}\left(\sqrt{\frac{m\omega}{\hbar}}x\right)e^{-m\omega x^{2}/2\hbar} \ \ \ \ \ (1)$

where ${H_{n}}$ is the ${n}$th Hermite polynomial. For ${n=1,}$ we have

$\displaystyle H_{1}\left(\sqrt{\frac{m\omega}{\hbar}}x\right)=2\sqrt{\frac{m\omega}{\hbar}}x \ \ \ \ \ (2)$

so

$\displaystyle \psi_{1}(x)=\frac{\sqrt{2}}{\pi^{1/4}}\left(\frac{m\omega}{\hbar}\right)^{3/4}x\;e^{-m\omega x^{2}/2\hbar} \ \ \ \ \ (3)$

For this state, we can calculate the average

 $\displaystyle \left\langle \frac{1}{X^{2}}\right\rangle$ $\displaystyle =$ $\displaystyle \int_{-\infty}^{\infty}\psi_{1}^{2}(x)\frac{1}{x^{2}}dx\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2}{\sqrt{\pi}}\left(\frac{m\omega}{\hbar}\right)^{3/2}\int_{-\infty}^{\infty}e^{-m\omega x^{2}/\hbar}dx\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2}{\sqrt{\pi}}\left(\frac{m\omega}{\hbar}\right)^{3/2}\sqrt{\frac{\pi\hbar}{m\omega}}\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2m\omega}{\hbar} \ \ \ \ \ (7)$

where we evaluated the Gaussian integral in the second line.

We can compare this to ${1/\left\langle X^{2}\right\rangle }$ as follows:

 $\displaystyle \left\langle X^{2}\right\rangle$ $\displaystyle =$ $\displaystyle \int_{-\infty}^{\infty}\psi_{1}^{2}(x)x^{2}dx\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2}{\sqrt{\pi}}\left(\frac{m\omega}{\hbar}\right)^{3/2}\int_{-\infty}^{\infty}e^{-m\omega x^{2}/\hbar}x^{4}dx\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2}{\sqrt{\pi}}\left(\frac{m\omega}{\hbar}\right)^{3/2}\frac{3\sqrt{\pi}}{4}\left(\frac{\hbar}{m\omega}\right)^{5/2}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{3}{2}\frac{\hbar}{m\omega}\ \ \ \ \ (11)$ $\displaystyle \frac{1}{\left\langle X^{2}\right\rangle }$ $\displaystyle =$ $\displaystyle \frac{2}{3}\frac{m\omega}{\hbar} \ \ \ \ \ (12)$

Thus ${\left\langle \frac{1}{X^{2}}\right\rangle }$ and ${\frac{1}{\left\langle X^{2}\right\rangle }}$ have the same order of magnitude, although they are not equal.

In three dimensions, we consider the ground state of hydrogen

$\displaystyle \psi_{100}\left(r\right)=\frac{1}{\sqrt{\pi}a_{0}^{3/2}}e^{-r/a_{0}} \ \ \ \ \ (13)$

where ${a_{0}}$ is the Bohr radius

$\displaystyle a_{0}\equiv\frac{\hbar^{2}}{me^{2}} \ \ \ \ \ (14)$

with ${m}$ and ${e}$ being the mass and charge of the electron. The wave function is normalized as we can see by doing the integral (in 3 dimensions):

 $\displaystyle \int\psi_{100}^{2}(r)d^{3}\mathbf{r}$ $\displaystyle =$ $\displaystyle \frac{4\pi}{\pi a_{0}^{3}}\int_{0}^{\infty}e^{-2r/a_{0}}r^{2}dr \ \ \ \ \ (15)$

We can use the formula (given in Shankar’s Appendix 2)

$\displaystyle \int_{0}^{\infty}e^{-r/\alpha}r^{n}dr=\frac{n!}{\alpha^{n+1}} \ \ \ \ \ (16)$

We get

$\displaystyle \int\psi_{100}^{2}(r)d^{3}\mathbf{r}=\frac{4\pi}{\pi a_{0}^{3}}\frac{2!}{2^{3}}a_{0}^{3}=1 \ \ \ \ \ (17)$

as required.

For a spherically symmetric wave function centred at ${r=0}$,

$\displaystyle \left(\Delta X\right)^{2}=\left\langle X^{2}\right\rangle -\left\langle X\right\rangle ^{2}=\left\langle X^{2}\right\rangle \ \ \ \ \ (18)$

with identical relations for ${Y}$ and ${Z}$. Since

 $\displaystyle r^{2}$ $\displaystyle =$ $\displaystyle x^{2}+y^{2}+z^{2}\ \ \ \ \ (19)$ $\displaystyle \left\langle r^{2}\right\rangle$ $\displaystyle =$ $\displaystyle \left\langle x^{2}\right\rangle +\left\langle y^{2}\right\rangle +\left\langle z^{2}\right\rangle =3\left\langle X^{2}\right\rangle \ \ \ \ \ (20)$ $\displaystyle \left\langle X^{2}\right\rangle$ $\displaystyle =$ $\displaystyle \frac{1}{3}\left\langle r^{2}\right\rangle \ \ \ \ \ (21)$

Thus

 $\displaystyle \left\langle X^{2}\right\rangle$ $\displaystyle =$ $\displaystyle \frac{1}{3}\int\psi_{100}^{2}(r)r^{2}d^{3}\mathbf{r}\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{4\pi}{3\pi a_{0}^{3}}\int_{0}^{\infty}e^{-2r/a_{0}}r^{4}dr\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{4}{3a_{0}^{3}}\frac{4!}{2^{5}}a_{0}^{5}\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle a_{0}^{2}\ \ \ \ \ (25)$ $\displaystyle \Delta X$ $\displaystyle =$ $\displaystyle a_{0}=\frac{\hbar^{2}}{me^{2}} \ \ \ \ \ (26)$

We can also find

 $\displaystyle \left\langle \frac{1}{r}\right\rangle$ $\displaystyle =$ $\displaystyle \int\psi_{100}^{2}(r)\frac{1}{r}d^{3}\mathbf{r}\ \ \ \ \ (27)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{4\pi}{\pi a_{0}^{3}}\int_{0}^{\infty}e^{-2r/a_{0}}r\;dr\ \ \ \ \ (28)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{4}{a_{0}^{3}}\frac{a_{0}^{2}}{4}\ \ \ \ \ (29)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{a_{0}}\ \ \ \ \ (30)$ $\displaystyle \left\langle r\right\rangle$ $\displaystyle =$ $\displaystyle \int\psi_{100}^{2}(r)r\;d^{3}\mathbf{r}\ \ \ \ \ (31)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{4\pi}{\pi a_{0}^{3}}\int_{0}^{\infty}e^{-2r/a_{0}}r^{3}dr\ \ \ \ \ (32)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{4}{a_{0}^{3}}\frac{6a_{0}^{4}}{16}\ \ \ \ \ (33)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{3}{2}a_{0} \ \ \ \ \ (34)$

Thus both ${\left\langle \frac{1}{r}\right\rangle }$ and ${\frac{1}{\left\langle r\right\rangle }}$ are of the same order of magnitude as ${1/a_{0}=me^{2}/\hbar^{2}}$.

# Harmonic oscillator energies and eigenfunctions derived from the propagator

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 8. Section 8.6, Exercise 8.6.3.

Given the propagator for the harmonic oscillator, it is possible to work backwards and deduce the eigenvalues and eigenfunctions of the Hamiltonian, although this isn’t the easiest way to find them. We’ve seen that the propagator for the oscillator is

$\displaystyle U\left(x,t;x^{\prime}\right)=A\left(t\right)\exp\left[\frac{im\omega}{2\hbar\sin\omega t}\left(\left(x^{\prime2}+x^{2}\right)\cos\omega t-2x^{\prime}x\right)\right] \ \ \ \ \ (1)$

where ${A\left(t\right)}$ is some function of time which is found by doing a path integral. Shankar cheats a bit by just telling us what ${A}$ is:

$\displaystyle A\left(t\right)=\sqrt{\frac{m\omega}{2\pi i\hbar\sin\omega t}} \ \ \ \ \ (2)$

To deduce (some of) the energy levels, we can compare the propagator with its more traditional form

$\displaystyle U\left(t\right)=\sum_{n}e^{-iE_{n}t/\hbar}\left|E_{n}\right\rangle \left\langle E_{n}\right| \ \ \ \ \ (3)$

where ${E_{n}}$ is the ${n}$th energy level. In position space this is

$\displaystyle U\left(t\right)=\sum_{n}\psi_{n}^*\left(x\right)\psi_{n}\left(x\right)e^{-iE_{n}t/\hbar} \ \ \ \ \ (4)$

We can try finding the energy levels as follows. We take ${x=x^{\prime}=t^{\prime}=0}$, which is equivalent to taking the end time ${t}$ to be a multiple of a complete period of the oscillator, so that the particle has returned to its starting point. In that case, 1 becomes

$\displaystyle U\left(x,t;x^{\prime}\right)=A\left(t\right)=\sqrt{\frac{m\omega}{2\pi i\hbar\sin\omega t}} \ \ \ \ \ (5)$

If we can expand this quantity in powers of ${e^{-i\omega t}}$, we can compare it with the series 4 and read off the energies from the exponents in the series. To do this, we write

 $\displaystyle A\left(t\right)$ $\displaystyle =$ $\displaystyle \sqrt{\frac{m\omega}{\pi\hbar\left(e^{i\omega t}-e^{-i\omega t}\right)}}\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{m\omega}{\pi\hbar}}e^{-i\omega t/2}\frac{1}{\sqrt{1-e^{-2i\omega t}}} \ \ \ \ \ (7)$

To save writing, we’ll define the symbol

$\displaystyle \eta\equiv e^{-i\omega t} \ \ \ \ \ (8)$

so that

$\displaystyle A\left(t\right)=\sqrt{\frac{m\omega}{\pi\hbar}}\eta^{1/2}\frac{1}{\sqrt{1-\eta^{2}}} \ \ \ \ \ (9)$

We can now expand the last factor using the binomial expansion to get

$\displaystyle A\left(t\right)=\sqrt{\frac{m\omega}{\pi\hbar}}\eta^{1/2}\left[1+\frac{1}{2}\eta^{2}+\frac{3}{8}\eta^{4}+\ldots\right] \ \ \ \ \ (10)$

In terms of the original variables, we get

$\displaystyle A\left(t\right)=\sqrt{\frac{m\omega}{\pi\hbar}}\left[e^{-i\omega t/2}+\frac{1}{2}e^{-5i\omega t/2}+\frac{3}{8}e^{-9i\omega t/2}+\ldots\right] \ \ \ \ \ (11)$

Comparing with 4, we find energy levels of

$\displaystyle E=\frac{\hbar\omega}{2},\frac{5\hbar\omega}{2},\frac{9\hbar\omega}{2},\ldots \ \ \ \ \ (12)$

These correspond to ${E_{0},E_{2},E_{4},\ldots}$. The odd energy levels ${\left(\frac{3\hbar\omega}{2},\frac{7\hbar\omega}{2},\ldots\right)}$ are missing because the corresponding wave functions ${\psi_{n}\left(x\right)}$ are odd functions of ${x}$ and are therefore zero at ${x=0}$, so the corresponding terms in 4 vanish. The numerical coefficients in 11 give us ${\left|\psi_{n}\left(0\right)\right|^{2}}$ for ${n=0,2,4,\ldots}$.

To get the other energies, as well as the eigenfunctions, from a comparison of 1 and 4 is possible, but quite messy, even for the lower energies. To do it, we take ${t^{\prime}=0}$ as before, but now we take ${x=x^{\prime}\ne0}$. That is, we start the oscillator off at some location ${x^{\prime}\ne0}$ and then look at it exactly one period later, when it has returned to the same position. The propagator 1 now becomes

 $\displaystyle U\left(x,t;x^{\prime}\right)$ $\displaystyle =$ $\displaystyle \sqrt{\frac{m\omega}{2\pi i\hbar\sin\omega t}}\exp\left[\frac{im\omega}{2\hbar\sin\omega t}\left(2x^{2}\left(\cos\omega t-1\right)\right)\right]\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{m\omega}{\pi\hbar\left(e^{i\omega t}-e^{-i\omega t}\right)}}\exp\left[-\frac{m\omega}{\hbar\left(e^{i\omega t}-e^{-i\omega t}\right)}\left(x^{2}\left(\left(e^{i\omega t}+e^{-i\omega t}\right)-2\right)\right)\right]\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{m\omega}{\pi\hbar}}\eta^{1/2}\frac{1}{\sqrt{1-\eta^{2}}}\exp\left[-\frac{m\omega x^{2}}{\hbar}\left(\frac{\frac{1}{\eta}+\eta-2}{\frac{1}{\eta}-\eta}\right)\right]\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{m\omega}{\pi\hbar}}\eta^{1/2}\frac{1}{\sqrt{1-\eta^{2}}}\exp\left[-\frac{m\omega x^{2}}{\hbar}\left(\frac{1+\eta^{2}-2\eta}{1-\eta^{2}}\right)\right] \ \ \ \ \ (16)$

We now need to expand this in a power series in ${\eta}$, which gets very messy so is best handled with software like Maple. Shankar asks only for the first two terms in the series (the terms corresponding to ${\eta^{1/2}}$ and ${\eta^{3/2}}$) but even doing this by hand can get very tedious. The result from Maple is, for the first two terms:

 $\displaystyle \eta^{1/2}$ $\displaystyle \rightarrow$ $\displaystyle \sqrt{\frac{m\omega}{\pi\hbar}}e^{-m\omega x^{2}/\hbar}\eta^{1/2}=\sqrt{\frac{m\omega}{\pi\hbar}}e^{-m\omega x^{2}/\hbar}e^{-i\omega t/2\hbar}\ \ \ \ \ (17)$ $\displaystyle \eta^{3/2}$ $\displaystyle \rightarrow$ $\displaystyle \sqrt{\frac{m\omega}{\pi\hbar}}\frac{2m\omega}{\hbar}e^{-m\omega x^{2}/\hbar}x^{2}\eta^{3/2}=\sqrt{\frac{m\omega}{\pi\hbar}}\frac{2m\omega}{\hbar}e^{-m\omega x^{2}/\hbar}x^{2}e^{-3i\omega t/2\hbar} \ \ \ \ \ (18)$

Comparing this with 4, we can read off:

 $\displaystyle E_{0}$ $\displaystyle =$ $\displaystyle \frac{\hbar\omega}{2}\ \ \ \ \ (19)$ $\displaystyle \left|\psi_{0}\left(x\right)\right|^{2}$ $\displaystyle =$ $\displaystyle \sqrt{\frac{m\omega}{\pi\hbar}}e^{-m\omega x^{2}/\hbar}\ \ \ \ \ (20)$ $\displaystyle E_{1}$ $\displaystyle =$ $\displaystyle \frac{3\hbar\omega}{2}\ \ \ \ \ (21)$ $\displaystyle \left|\psi_{1}\left(x\right)\right|^{2}$ $\displaystyle =$ $\displaystyle \sqrt{\frac{m\omega}{\pi\hbar}}\frac{2m\omega}{\hbar}e^{-m\omega x^{2}/\hbar}x^{2} \ \ \ \ \ (22)$

To check this, we recall the eigenfunctions we worked out earlier, using Hermite polynomials

$\displaystyle \psi_{n}(x)=\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\frac{1}{\sqrt{2^{n}n!}}H_{n}\left(\sqrt{\frac{m\omega}{\hbar}}x\right)e^{-m\omega x^{2}/2\hbar} \ \ \ \ \ (23)$

The first two Hermite polynomials are

 $\displaystyle H_{0}\left(\sqrt{\frac{m\omega}{\hbar}}x\right)$ $\displaystyle =$ $\displaystyle 1\ \ \ \ \ (24)$ $\displaystyle H_{1}\left(\sqrt{\frac{m\omega}{\hbar}}x\right)$ $\displaystyle =$ $\displaystyle 2\sqrt{\frac{m\omega}{\hbar}}x \ \ \ \ \ (25)$

Plugging these into 23 and comparing with 20 and 22 shows we got the right answer.

# Harmonic oscillator: momentum space functions and Hermite polynomial recursion relations from raising and lowering operators

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 7.5, Exercises 7.5.1 – 7.5.3.

Earlier, we found the position space energy eigenfunctions of the harmonic oscillator to be

 $\displaystyle \psi_{n}(y)$ $\displaystyle =$ $\displaystyle \left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\frac{1}{\sqrt{2^{n}n!}}H_{n}(y)e^{-y^{2}/2}\ \ \ \ \ (1)$ $\displaystyle \psi_{n}(x)$ $\displaystyle =$ $\displaystyle \left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\frac{1}{\sqrt{2^{n}n!}}H_{n}\left(\sqrt{\frac{m\omega}{\hbar}}x\right)e^{-m\omega x^{2}/2\hbar} \ \ \ \ \ (2)$

where ${y}$ in the first equation is shorthand for

$\displaystyle y=\sqrt{\frac{m\omega}{\hbar}}x \ \ \ \ \ (3)$

It turns out that an alternative method for deriving these functions uses the lowering operator ${a}$. Shankar gives the derivation of ${\psi_{n}\left(x\right)}$ in his section 7.5, but we can use the same technique to derive the momentum space functions. We start with the ground state and use

$\displaystyle a\left|0\right\rangle =0 \ \ \ \ \ (4)$

In terms of ${X}$ and ${P}$, we have

$\displaystyle a=\sqrt{\frac{m\omega}{2\hbar}}X+i\frac{1}{\sqrt{2m\omega\hbar}}P \ \ \ \ \ (5)$

To find the momentum space functions, we need to express ${X}$ and ${P}$ in terms of ${p}$:

 $\displaystyle X$ $\displaystyle =$ $\displaystyle i\hbar\frac{d}{dp}\ \ \ \ \ (6)$ $\displaystyle P$ $\displaystyle =$ $\displaystyle p \ \ \ \ \ (7)$

We thus have

$\displaystyle \left[i\hbar\sqrt{\frac{m\omega}{2\hbar}}\frac{d}{dp}+i\frac{1}{\sqrt{2m\omega\hbar}}p\right]\psi_{0}\left(p\right)=0 \ \ \ \ \ (8)$

If we define the auxiliary variable

$\displaystyle z\equiv\frac{p}{\sqrt{\hbar m\omega}} \ \ \ \ \ (9)$

we get

$\displaystyle \left(\frac{d}{dz}+z\right)\psi_{0}\left(z\right)=0 \ \ \ \ \ (10)$

This has the solution

$\displaystyle \psi_{0}\left(z\right)=Ae^{-z^{2}/2} \ \ \ \ \ (11)$

for some normalization constant ${A}$. Thus in terms of ${p}$ we have

$\displaystyle \psi_{0}\left(p\right)=Ae^{-p^{2}/2\hbar m\omega} \ \ \ \ \ (12)$

Normalizing in the usual way, making use of the Gaussian integral, we have

 $\displaystyle \int_{-\infty}^{\infty}\psi_{0}^{2}\left(p\right)dp$ $\displaystyle =$ $\displaystyle A^{2}\int_{-\infty}^{\infty}e^{-p^{2}/\hbar m\omega}dp=1\ \ \ \ \ (13)$ $\displaystyle A$ $\displaystyle =$ $\displaystyle \frac{1}{\left(\pi\hbar m\omega\right)^{1/4}} \ \ \ \ \ (14)$

This agrees with the earlier result which was obtained by solving a second-order differential equation.

We can also use ${a}$ and ${a^{\dagger}}$ to verify a couple of recursion relations for Hermite polynomials. Reverting back to position space we have

 $\displaystyle X$ $\displaystyle =$ $\displaystyle x\ \ \ \ \ (15)$ $\displaystyle P$ $\displaystyle =$ $\displaystyle -i\hbar\frac{d}{dx} \ \ \ \ \ (16)$

so 5 becomes

$\displaystyle a=\sqrt{\frac{m\omega}{2\hbar}}x+\frac{\hbar}{\sqrt{2m\omega\hbar}}\frac{d}{dx} \ \ \ \ \ (17)$

Also from 5 we have, since ${X}$ and ${P}$ are both hermitian operators

 $\displaystyle a^{\dagger}$ $\displaystyle =$ $\displaystyle \sqrt{\frac{m\omega}{2\hbar}}X-i\frac{1}{\sqrt{2m\omega\hbar}}P\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{m\omega}{2\hbar}}x-\frac{\hbar}{\sqrt{2m\omega\hbar}}\frac{d}{dx} \ \ \ \ \ (19)$

Defining

$\displaystyle y\equiv\sqrt{\frac{m\omega}{\hbar}}x \ \ \ \ \ (20)$

we have

 $\displaystyle a$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2}}\left(y+\frac{d}{dy}\right)\ \ \ \ \ (21)$ $\displaystyle a^{\dagger}$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2}}\left(y-\frac{d}{dy}\right) \ \ \ \ \ (22)$

We also recall the normalization conditions on the raising and lowering operators:

 $\displaystyle a\left|n\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{n}\left|n-1\right\rangle \ \ \ \ \ (23)$ $\displaystyle a^{\dagger}\left|n\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{n+1}\left|n+1\right\rangle \ \ \ \ \ (24)$

Applying 23 to 1 we have, after cancelling common factors from each side:

 $\displaystyle \frac{1}{\sqrt{2}}\frac{1}{\sqrt{2^{n}n!}}\left(y+\frac{d}{dy}\right)\left[H_{n}(y)e^{-y^{2}/2}\right]$ $\displaystyle =$ $\displaystyle \frac{\sqrt{n}}{\sqrt{2^{n-1}\left(n-1\right)!}}H_{n-1}\left(y\right)e^{-y^{2}/2}\ \ \ \ \ (25)$ $\displaystyle \frac{1}{2\sqrt{n}}\frac{1}{\sqrt{2^{n-1}\left(n-1\right)!}}e^{-y^{2}/2}\left[yH_{n}\left(y\right)-yH_{n}\left(y\right)+\frac{dH_{n}}{dy}\right]$ $\displaystyle =$ $\displaystyle \frac{\sqrt{n}}{\sqrt{2^{n-1}\left(n-1\right)!}}H_{n-1}\left(y\right)e^{-y^{2}/2}\ \ \ \ \ (26)$ $\displaystyle yH_{n}\left(y\right)-yH_{n}\left(y\right)+\frac{dH_{n}}{dy}$ $\displaystyle =$ $\displaystyle 2nH_{n-1}\left(y\right)\ \ \ \ \ (27)$ $\displaystyle H_{n}^{\prime}\left(y\right)$ $\displaystyle =$ $\displaystyle 2nH_{n-1}\left(y\right) \ \ \ \ \ (28)$

Another recursion relation for Hermite polynomials can be found as follows. We start with 22 to get

$\displaystyle a+a^{\dagger}=\sqrt{2}y \ \ \ \ \ (29)$

We now apply 23 and 24 to 1. We can cancel common factors, including ${e^{-y^{2}/2}}$, from both sides to get

 $\displaystyle \left(a+a^{\dagger}\right)\psi_{n}$ $\displaystyle =$ $\displaystyle \sqrt{2}y\psi_{n}\ \ \ \ \ (30)$ $\displaystyle \frac{\sqrt{2}y}{\sqrt{2^{n}n!}}H_{n}(y)$ $\displaystyle =$ $\displaystyle \frac{\sqrt{n}}{\sqrt{2^{n-1}\left(n-1\right)!}}H_{n-1}(y)+\frac{\sqrt{n+1}}{\sqrt{2^{n+1}\left(n+1\right)!}}H_{n+1}(y)\ \ \ \ \ (31)$ $\displaystyle \frac{y}{\sqrt{2^{n-1}n\left(n-1\right)!}}H_{n}(y)$ $\displaystyle =$ $\displaystyle \frac{\sqrt{n}}{\sqrt{2^{n-1}\left(n-1\right)!}}H_{n-1}(y)+\frac{1}{2\sqrt{2^{n-1}n\left(n-1\right)!}}H_{n+1}(y)\ \ \ \ \ (32)$ $\displaystyle yH_{n}\left(y\right)$ $\displaystyle =$ $\displaystyle nH_{n-1}\left(y\right)+\frac{1}{2}H_{n+1}\left(y\right)\ \ \ \ \ (33)$ $\displaystyle H_{n+1}\left(y\right)$ $\displaystyle =$ $\displaystyle 2yH_{n}\left(y\right)-2nH_{n-1}\left(y\right) \ \ \ \ \ (34)$

# Harmonic oscillator – raising and lowering operators as functions of time

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 7.4, Exercise 7.4.6.

We’ll consider here the problem of finding the averages of the raising and lowering operators (from the harmonic oscillator) as functions of time, that is, we want to find ${\left\langle a\left(t\right)\right\rangle }$ and ${\left\langle a^{\dagger}\left(t\right)\right\rangle }$. At first glance we might think they are both zero, since they are defined in terms of position and momentum as

 $\displaystyle a^{\dagger}$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2\hbar m\omega}}\left[-iP+m\omega X\right]\ \ \ \ \ (1)$ $\displaystyle a$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2\hbar m\omega}}\left[iP+m\omega X\right] \ \ \ \ \ (2)$

and the averages of ${P}$ and ${X}$ in any of the energy eigenstates of the harmonic oscillator are all zero. However, suppose we have a mixed state ${\left|\psi\right\rangle }$ which can be written as a sum over the eigenstates as

 $\displaystyle \psi\left(t\right)$ $\displaystyle =$ $\displaystyle \sum_{n=0}^{\infty}c_{n}e^{-iE_{n}t/\hbar}\left|n\right\rangle \ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{n=0}^{\infty}c_{n}e^{-i\left(2n+1\right)\omega t/2}\left|n\right\rangle \ \ \ \ \ (4)$

where in the second line we used the energies of the oscillator as

$\displaystyle E_{n}=\hbar\omega\left(n+\frac{1}{2}\right) \ \ \ \ \ (5)$

We now have

 $\displaystyle \left\langle a\left(t\right)\right\rangle$ $\displaystyle =$ $\displaystyle \left\langle \psi\left|a\right|\psi\right\rangle \ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{n=0}^{\infty}\sum_{m=0}^{\infty}c_{m}^*e^{i\left(2m+1\right)\omega t/2}c_{n}e^{-i\left(2n+1\right)\omega t/2}\left\langle m\left|a\right|n\right\rangle \ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{n=0}^{\infty}\sum_{m=0}^{\infty}c_{m}^*c_{n}e^{i\left(m-n\right)\omega t}\left\langle m\left|a\right|n\right\rangle \ \ \ \ \ (8)$

We can now use the formula

$\displaystyle a\left|n\right\rangle =\sqrt{n}\left|n-1\right\rangle \ \ \ \ \ (9)$

This gives

 $\displaystyle \left\langle a\left(t\right)\right\rangle$ $\displaystyle =$ $\displaystyle \sum_{n=0}^{\infty}\sum_{m=0}^{\infty}c_{m}^*c_{n}e^{i\left(m-n\right)\omega t}\sqrt{n}\left\langle m\left|n-1\right.\right\rangle \ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{n=0}^{\infty}\sum_{m=0}^{\infty}c_{m}^*c_{n}e^{i\left(m-n\right)\omega t}\sqrt{n}\delta_{m,n-1}\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle e^{-i\omega t}\sum_{n=0}^{\infty}c_{n-1}^*c_{n}\sqrt{n}\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle e^{-i\omega t}\left\langle a\left(0\right)\right\rangle \ \ \ \ \ (13)$

Note that if ${\left|\psi\right\rangle }$ is an eigenstate, then only one of the coefficients ${c_{n}}$ is non-zero, so ${\left\langle a\left(0\right)\right\rangle =0}$ as we’d expect.

The derivation for ${\left\langle a^{\dagger}\left(t\right)\right\rangle }$ is similar:

 $\displaystyle \left\langle a^{\dagger}\left(t\right)\right\rangle$ $\displaystyle =$ $\displaystyle \left\langle \psi\left|a^{\dagger}\right|\psi\right\rangle \ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{n=0}^{\infty}\sum_{m=0}^{\infty}c_{m}^*e^{i\left(2m+1\right)\omega t/2}c_{n}e^{-i\left(2n+1\right)\omega t/2}\left\langle m\left|a^{\dagger}\right|n\right\rangle \ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{n=0}^{\infty}\sum_{m=0}^{\infty}c_{m}^*c_{n}e^{i\left(m-n\right)\omega t}\left\langle m\left|a^{\dagger}\right|n\right\rangle \ \ \ \ \ (16)$

We can now use the formula

$\displaystyle a^{\dagger}\left|n\right\rangle =\sqrt{n+1}\left|n+1\right\rangle \ \ \ \ \ (17)$

This gives

 $\displaystyle \left\langle a^{\dagger}\left(t\right)\right\rangle$ $\displaystyle =$ $\displaystyle \sum_{n=0}^{\infty}\sum_{m=0}^{\infty}c_{m}^*c_{n}e^{i\left(m-n\right)\omega t}\sqrt{n+1}\left\langle m\left|n+1\right.\right\rangle \ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{n=0}^{\infty}\sum_{m=0}^{\infty}c_{m}^*c_{n}e^{i\left(m-n\right)\omega t}\sqrt{n+1}\delta_{m,n+1}\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle e^{i\omega t}\sum_{n=0}^{\infty}c_{n+1}^*c_{n}\sqrt{n+1}\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle e^{i\omega t}\left\langle a^{\dagger}\left(0\right)\right\rangle \ \ \ \ \ (21)$

# Harmonic oscillator – mixed initial state and Ehrenfest’s theorem

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 7.4, Exercise 7.4.5.

We’ve already done an example of a harmonic oscillator in a mixed initial state, but it’s useful to do this other example from Shankar so we can see how the modified Ehrenfest’s theorem fits in. In this case, we start with a particle in the mixed initial state

$\displaystyle \left|\psi\left(0\right)\right\rangle =\frac{1}{\sqrt{2}}\left[\left|0\right\rangle +\left|1\right\rangle \right] \ \ \ \ \ (1)$

The time-dependent solution is therefore

 $\displaystyle \left|\psi\left(t\right)\right\rangle$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2}}\left[e^{-iE_{0}\hbar t}\left|0\right\rangle +e^{-iE_{1}t}\left|1\right\rangle \right]\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2}}\left[e^{-i\omega t/2}\left|0\right\rangle +e^{-3i\omega t/2}\left|1\right\rangle \right] \ \ \ \ \ (3)$

since the first two energies are ${E_{0}=\hbar\omega/2}$ and ${E_{1}=3\hbar\omega/2}$.

The position and momentum operators can be written in terms of the raising and lowering operators

 $\displaystyle X$ $\displaystyle =$ $\displaystyle \sqrt{\frac{\hbar}{2m\omega}}\left(a^{\dagger}+a\right)\ \ \ \ \ (4)$ $\displaystyle P$ $\displaystyle =$ $\displaystyle i\sqrt{\frac{\hbar m\omega}{2}}\left(a^{\dagger}-a\right) \ \ \ \ \ (5)$

To find the mean position and momentum, we can use these equations:

 $\displaystyle \left\langle X\left(0\right)\right\rangle$ $\displaystyle =$ $\displaystyle \left\langle \psi\left(0\right)\left|X\right|\psi\left(0\right)\right\rangle \ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\sqrt{\frac{\hbar}{2m\omega}}\left[\left\langle 0\right|+\left\langle 1\right|\right]\left(a^{\dagger}+a\right)\left[\left|0\right\rangle +\left|1\right\rangle \right] \ \ \ \ \ (7)$

To work out the last line, remember that the stationary states are orthogonal so that ${\left\langle 0\left|1\right.\right\rangle =0}$, and that

 $\displaystyle a^{\dagger}\left|n\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{n+1}\left|n+1\right\rangle \ \ \ \ \ (8)$ $\displaystyle a\left|n\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{n}\left|n-1\right\rangle \ \ \ \ \ (9)$

We therefore get

$\displaystyle \left\langle X\left(0\right)\right\rangle =\frac{1}{2}\sqrt{\frac{\hbar}{2m\omega}}\left(1+1\right)=\sqrt{\frac{\hbar}{2m\omega}} \ \ \ \ \ (10)$

Doing a similar analysis for the momentum, we have

 $\displaystyle \left\langle P\left(0\right)\right\rangle$ $\displaystyle =$ $\displaystyle \left\langle \psi\left(0\right)\left|P\right|\psi\left(0\right)\right\rangle \ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{i}{2}\sqrt{\frac{\hbar m\omega}{2}}\left[\left\langle 0\right|+\left\langle 1\right|\right]\left(a^{\dagger}-a\right)\left[\left|0\right\rangle +\left|1\right\rangle \right]\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{\hbar m\omega}{2}}\frac{1}{2i}\left[\left\langle 0\right|+\left\langle 1\right|\right]\left(a-a^{\dagger}\right)\left[\left|0\right\rangle +\left|1\right\rangle \right]\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{\hbar m\omega}{2}}\frac{1}{2i}\left(1-1\right)\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (15)$

We can expand these equations to give the averages of position and momentum at all times by plugging in 3:

 $\displaystyle \left\langle X\left(t\right)\right\rangle$ $\displaystyle =$ $\displaystyle \left\langle \psi\left(t\right)\left|X\right|\psi\left(t\right)\right\rangle \ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\sqrt{\frac{\hbar}{2m\omega}}\left[\left\langle 0\right|e^{i\omega t/2}+\left\langle 1\right|e^{3i\omega t/2}\right]\left(a^{\dagger}+a\right)\left[e^{-i\omega t/2}\left|0\right\rangle +e^{-3i\omega t/2}\left|1\right\rangle \right]\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\sqrt{\frac{\hbar}{2m\omega}}\left(e^{-i\omega t}+e^{i\omega t}\right)\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{\hbar}{2m\omega}}\cos\omega t \ \ \ \ \ (19)$
 $\displaystyle \left\langle P\left(t\right)\right\rangle$ $\displaystyle =$ $\displaystyle \left\langle \psi\left(t\right)\left|P\right|\psi\left(t\right)\right\rangle \ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{i}{2}\sqrt{\frac{\hbar m\omega}{2}}\left[\left\langle 0\right|e^{i\omega t/2}+\left\langle 1\right|e^{3i\omega t/2}\right]\left(a^{\dagger}-a\right)\left[e^{-i\omega t/2}\left|0\right\rangle +e^{-3i\omega t/2}\left|1\right\rangle \right]\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{\hbar m\omega}{2}}\frac{1}{2i}\left[\left\langle 0\right|e^{i\omega t/2}+\left\langle 1\right|e^{3i\omega t/2}\right]\left(a-a^{\dagger}\right)\left[e^{-i\omega t/2}\left|0\right\rangle +e^{-3i\omega t/2}\left|1\right\rangle \right]\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{\hbar m\omega}{2}}\frac{1}{2i}\left(e^{-i\omega t}-e^{i\omega t}\right)\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\sqrt{\frac{\hbar m\omega}{2}}\sin\omega t \ \ \ \ \ (24)$

Although we can calculate ${\left\langle \dot{X}\left(t\right)\right\rangle }$ and ${\left\langle \dot{P}\left(t\right)\right\rangle }$ directly by taking the time derivative, we can also do it by using Ehrenfest’s theorem in the form

$\displaystyle \frac{d\left\langle \Omega\right\rangle }{dt}=-\frac{i}{\hbar}\left\langle \left[\Omega,H\right]\right\rangle \ \ \ \ \ (25)$

for some operator ${\Omega}$.

Since the energy of the oscillator in state ${\left|n\right\rangle }$ is ${\left(n+\frac{1}{2}\right)\hbar\omega}$, we can write the hamiltonian as

$\displaystyle H=\hbar\omega\left(a^{\dagger}a+\frac{1}{2}\right) \ \ \ \ \ (26)$

We also have the commutator

$\displaystyle \left[a,a^{\dagger}\right]=1 \ \ \ \ \ (27)$

To use this for ${X}$ and ${P}$ we need the commutators ${\left[a,H\right]}$ and ${\left[a^{\dagger},H\right]}$, which amounts to finding

 $\displaystyle \left[a,a^{\dagger}a\right]$ $\displaystyle =$ $\displaystyle aa^{\dagger}a-a^{\dagger}aa\ \ \ \ \ (28)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(1+a^{\dagger}a\right)a-a^{\dagger}aa\ \ \ \ \ (29)$ $\displaystyle$ $\displaystyle =$ $\displaystyle a\ \ \ \ \ (30)$ $\displaystyle \left[a^{\dagger},a^{\dagger}a\right]$ $\displaystyle =$ $\displaystyle a^{\dagger}a^{\dagger}a-a^{\dagger}aa^{\dagger}\ \ \ \ \ (31)$ $\displaystyle$ $\displaystyle =$ $\displaystyle a^{\dagger}a^{\dagger}a-a^{\dagger}\left(1+a^{\dagger}a^{\dagger}a\right)\ \ \ \ \ (32)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -a^{\dagger} \ \ \ \ \ (33)$

Therefore we have

 $\displaystyle \left[a,H\right]$ $\displaystyle =$ $\displaystyle \hbar\omega a\ \ \ \ \ (34)$ $\displaystyle \left[a^{\dagger},H\right]$ $\displaystyle =$ $\displaystyle -\hbar\omega a^{\dagger} \ \ \ \ \ (35)$

Finally we get

 $\displaystyle \left\langle \dot{X}\left(t\right)\right\rangle$ $\displaystyle =$ $\displaystyle -\frac{i}{\hbar}\left\langle \left[X,H\right]\right\rangle \ \ \ \ \ (36)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{i}{\hbar}\left\langle \left[a+a^{\dagger},H\right]\right\rangle \ \ \ \ \ (37)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{i}{\hbar}\hbar\omega\sqrt{\frac{\hbar}{2m\omega}}\left\langle a-a^{\dagger}\right\rangle \ \ \ \ \ (38)$ $\displaystyle$ $\displaystyle =$ $\displaystyle i\omega\sqrt{\frac{\hbar}{2m\omega}}\sqrt{\frac{2}{\hbar m\omega}}\frac{1}{i}\left\langle P\left(t\right)\right\rangle \ \ \ \ \ (39)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\omega\sqrt{\frac{\hbar}{2m\omega}}\sin\omega t \ \ \ \ \ (40)$

where we used 5 in the fourth line and 24 in the last line. The last line is indeed the time derivative of 19, so fortunately Ehrenfest’s theorem gives the correct answer.

For the momentum, we have

 $\displaystyle \left\langle \dot{P}\left(t\right)\right\rangle$ $\displaystyle =$ $\displaystyle -\frac{i}{\hbar}\left\langle \left[P,H\right]\right\rangle \ \ \ \ \ (41)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{i}{\hbar}\left\langle \left[a^{\dagger}-a,H\right]\right\rangle \ \ \ \ \ (42)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{i}{\hbar}\hbar\omega\sqrt{\frac{\hbar m\omega}{2}}i\left\langle -a^{\dagger}-a\right\rangle \ \ \ \ \ (43)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \omega\sqrt{\frac{\hbar m\omega}{2}}\sqrt{\frac{\hbar}{2m\omega}}\sqrt{\frac{2m\omega}{\hbar}}\left\langle -X\left(t\right)\right\rangle \ \ \ \ \ (44)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\omega\sqrt{\frac{\hbar m\omega}{2}}\cos\omega t \ \ \ \ \ (45)$

which is the correct derivative of 24.

# Virial theorem in classical mechanics; application to harmonic oscillator

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 7.4, Exercise 7.4.3.

We’ve seen the virial theorem in quantum mechanics, but this theorem was originally devised in classical mechanics. For a single particle, we consider the quantity

$\displaystyle G=\mathbf{r}\cdot\mathbf{p} \ \ \ \ \ (1)$

that is, the product of position and momentum. Taking the time derivative, we have

 $\displaystyle \frac{dG}{dt}$ $\displaystyle =$ $\displaystyle \mathbf{p}\cdot\frac{d\mathbf{r}}{dt}+\mathbf{r}\cdot\frac{d\mathbf{p}}{dt}\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle mv^{2}+\mathbf{r}\cdot\mathbf{F} \ \ \ \ \ (3)$

where ${\mathbf{v}=d\mathbf{r}/dt}$ is the velocity of the particle, ${\mathbf{p}=m\mathbf{v}}$ and ${\mathbf{F}=d\mathbf{p}/dt}$ is the force acting on the particle. If the force is a central force (that is, it depends only on the particle’s distance from some centre point ${\mathbf{r}=0}$, then the force can be written as the negative gradient of a potential ${V}$ that depends only on ${\mathbf{r}}$. In the case where ${V}$ depends only on a power of ${r}$, we have

 $\displaystyle V$ $\displaystyle =$ $\displaystyle ar^{k}\ \ \ \ \ (4)$ $\displaystyle \mathbf{F}$ $\displaystyle =$ $\displaystyle -\frac{dV}{dr}=-kar^{k-1}\hat{\mathbf{r}} \ \ \ \ \ (5)$

In that case, we have

 $\displaystyle \frac{dG}{dt}$ $\displaystyle =$ $\displaystyle mv^{2}-\mathbf{r}\cdot kar^{k-1}\hat{\mathbf{r}}\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2T-kar^{k}\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2T-kV \ \ \ \ \ (8)$

where ${T}$ is the kinetic energy ${T=\frac{1}{2}mv^{2}}$. If the particle is moving in a circular orbit then its average position and average momentum (averaged over one orbit) do not change with time, so ${\frac{dG}{dt}=0}$ and we get

 $\displaystyle 2\left\langle T\right\rangle -k\left\langle V\right\rangle$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (9)$ $\displaystyle \left\langle T\right\rangle$ $\displaystyle =$ $\displaystyle \frac{k}{2}\left\langle V\right\rangle \ \ \ \ \ (10)$

Another way of seeing this is that, in a circular orbit at constant orbital speed, the only force acting is the centripetal force holding the particle in its orbit, which is

$\displaystyle F_{cen}=-\frac{mv^{2}}{r} \ \ \ \ \ (11)$

where the minus sign indicates that the force acts in the opposite direction to the outward pointing radius vector.

This force is provided by the gradient of the potential, so we have

$\displaystyle F_{cen}=-\frac{dV}{dr}=-kar^{k-1} \ \ \ \ \ (12)$

We therefore have

 $\displaystyle \frac{mv^{2}}{r}$ $\displaystyle =$ $\displaystyle kar^{k-1}\ \ \ \ \ (13)$ $\displaystyle mv^{2}=2T$ $\displaystyle =$ $\displaystyle kar^{k}=kV\ \ \ \ \ (14)$ $\displaystyle \left\langle T\right\rangle$ $\displaystyle =$ $\displaystyle \frac{k}{2}\left\langle V\right\rangle \ \ \ \ \ (15)$

For the case of a harmonic oscillator, ${V=\frac{1}{2}m\omega^{2}x^{2}}$ so the exponent is ${k=2}$ and we have ${T=V}$. We can verify this by calculating the mean kinetic and potential energies explicitly, using earlier results. In the oscillator state ${\left|n\right\rangle }$ we have

 $\displaystyle \left\langle x^{2}\right\rangle$ $\displaystyle =$ $\displaystyle \frac{\hbar}{m\omega}\left(n+\frac{1}{2}\right)\ \ \ \ \ (16)$ $\displaystyle \left\langle p^{2}\right\rangle$ $\displaystyle =$ $\displaystyle \hbar m\omega\left(n+\frac{1}{2}\right) \ \ \ \ \ (17)$

The energies are

 $\displaystyle \left\langle T\right\rangle$ $\displaystyle =$ $\displaystyle \frac{\left\langle p^{2}\right\rangle }{2m}=\frac{\hbar\omega}{2}\left(n+\frac{1}{2}\right)\ \ \ \ \ (18)$ $\displaystyle \left\langle V\right\rangle$ $\displaystyle =$ $\displaystyle \frac{1}{2}m\omega^{2}\left\langle x^{2}\right\rangle =\frac{\hbar\omega}{2}\left(n+\frac{1}{2}\right) \ \ \ \ \ (19)$

Therefore ${\left\langle T\right\rangle =\left\langle V\right\rangle }$ as required.

# Harmonic oscillator – eigenfunctions in momentum space

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 7.3, Exercise 7.3.7.

We’ve seen how to solve the Schrödinger equation for the harmonic oscillator in the position basis, where the independent variable is ${x}$. It’s actually fairly easy to adapt this solution to find the wave functions in momentum space. (We’ve also found these functions by using the Fourier transform of the position functions, but the present post shows an easier way.)

The Schrödinger equation for the stationary states of the harmonic oscillator is, in operator form:

$\displaystyle \frac{P^{2}}{2m}\psi+\frac{1}{2}m\omega^{2}X^{2}\psi=E\psi \ \ \ \ \ (1)$

To work in momentum space, we use the results

 $\displaystyle P$ $\displaystyle =$ $\displaystyle p\ \ \ \ \ (2)$ $\displaystyle X$ $\displaystyle =$ $\displaystyle i\hbar\frac{\partial}{\partial p} \ \ \ \ \ (3)$

This gives

$\displaystyle \frac{p^{2}}{2m}\psi-\frac{1}{2}\hbar^{2}m\omega^{2}\frac{d^{2}\psi}{dp^{2}}=E\psi \ \ \ \ \ (4)$

Dividing through by ${\left(m\omega\right)^{2}}$ we get

$\displaystyle -\frac{\hbar^{2}}{2m}\psi^{\prime\prime}+\frac{p^{2}}{2m^{3}\omega^{2}}=\frac{E}{\left(m\omega\right)^{2}}\psi \ \ \ \ \ (5)$

where a prime on ${\psi}$ indicates a derivative with respect to ${p}$.

This is similar to the Schrödinger equation in position space:

$\displaystyle -\frac{\hbar^{2}}{2m}\psi^{\prime\prime}+\frac{1}{2}m\omega^{2}x^{2}\psi=E\psi \ \ \ \ \ (6)$

(where a prime here indicates a derivative with respect to ${x}$). When we solved the position space equation, we introduced a dimensionless variable

$\displaystyle y\equiv\sqrt{\frac{m\omega}{\hbar}}x \ \ \ \ \ (7)$

Using this technique to solve 5, we try a definition for ${y}$ of

$\displaystyle y\equiv\frac{p}{\sqrt{\hbar m\omega}} \ \ \ \ \ (8)$

(You can check the units of ${\sqrt{\hbar m\omega}}$ to see they are the units of momentum, so ${y}$ is indeed dimensionless here.) Making this substitution, we get

 $\displaystyle \frac{d^{2}\psi}{dp^{2}}$ $\displaystyle =$ $\displaystyle \frac{1}{\hbar m\omega}\frac{d^{2}\psi}{dy^{2}}\ \ \ \ \ (9)$ $\displaystyle \frac{p^{2}}{2m^{3}\omega^{2}}$ $\displaystyle =$ $\displaystyle \frac{\hbar y^{2}}{2m^{2}\omega} \ \ \ \ \ (10)$

Thus 5 becomes

 $\displaystyle -\frac{\hbar^{2}}{2m}\frac{1}{\hbar m\omega}\frac{d^{2}\psi}{dy^{2}}+\frac{\hbar y^{2}}{2m^{2}\omega}\psi$ $\displaystyle =$ $\displaystyle \frac{E}{\left(m\omega\right)^{2}}\psi\ \ \ \ \ (11)$ $\displaystyle \frac{\hbar^{2}}{2m}\left[-\frac{d^{2}\psi}{dy^{2}}+y^{2}\psi\right]$ $\displaystyle =$ $\displaystyle \frac{\hbar E}{\omega}\psi \ \ \ \ \ (12)$

We can now use the same dimensionless parameter we used in the earlier derivation:

$\displaystyle \varepsilon\equiv\frac{E}{\hbar\omega} \ \ \ \ \ (13)$

This results in the differential equation

$\displaystyle \psi^{\prime\prime}+\left(2\varepsilon-y^{2}\right)\psi=0 \ \ \ \ \ (14)$

where a prime now indicates a derivative with respect to ${y}$. This is exactly the same differential equation that we got for the position basis, except that the independent variable ${y}$ is now defined in terms of ${p}$ by 8 instead of ${x}$. We can solve it in the same way, which results in the same quantization condition on the allowable energies of ${E_{n}=\left(n+\frac{1}{2}\right)\hbar\omega}$. The eigenfunctions look the same when expressed in terms of ${y}$:

$\displaystyle \psi_{n}(y)=A\frac{1}{\sqrt{2^{n}n!}}H_{n}(y)e^{-y^{2}/2} \ \ \ \ \ (15)$

where ${A}$ is a normalization constant with the value in the position basis of

$\displaystyle A=\left(\frac{m\omega}{\pi\hbar}\right)^{1/4} \ \ \ \ \ (16)$

and ${H_{n}}$ is a Hermite polynomial. We can get the eigenfunctions in momentum space by replacing ${y}$ by 8. We can see that this amounts to replacing ${x\rightarrow p}$ and ${m\omega\rightarrow\frac{1}{m\omega}}$, so we get

$\displaystyle \psi_{n}(p)=\frac{1}{\left(\pi\hbar m\omega\right)^{1/4}}\frac{1}{\sqrt{2^{n}n!}}H_{n}\left(\frac{p}{\sqrt{\hbar m\omega}}\right)e^{-p^{2}/2\hbar m\omega} \ \ \ \ \ (17)$

In particular, the ground state is

$\displaystyle \psi_{0}\left(p\right)=\frac{1}{\left(\pi\hbar m\omega\right)^{1/4}}e^{-p^{2}/2\hbar m\omega} \ \ \ \ \ (18)$

# Harmonic oscillator – zero-point energy from uncertainty principle

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 7.3.

There is a nice result derived in Shankar’s section 7.3 in which he shows that we can actually derive the ground state energy and wave function for the harmonic oscillator from the uncertainty principle. Classically, the energy of a harmonic oscillator is

$\displaystyle H=\frac{p^{2}}{2m}+\frac{1}{2}m\omega^{2}x^{2} \ \ \ \ \ (1)$

where both ${p}$ and ${x}$ are continuous variables that can, in principle, take on any values. Thus classically it is possible for an oscillator to have ${x=p=0}$ giving a ground state with zero energy. In quantum mechanics, because ${X}$ and ${P}$ don’t commute, the position and momentum cannot both have precise values, which means that the ground state must have an energy greater than zero. This so-called zero-point energy is (as found by Solving Schrödinger’s equation)

$\displaystyle E_{0}=\frac{\hbar\omega}{2} \ \ \ \ \ (2)$

To derive this without needing to solve Schrödinger’s equation, we first recall that a state in which the position-momentum uncertainty is a minimum must be a gaussian of form

$\displaystyle \Psi\left(x\right)=Ae^{-a(x-\langle x\rangle)^{2}/2\hbar}e^{i\langle p\rangle x/\hbar} \ \ \ \ \ (3)$

where ${a}$ is a positive real constant, ${A}$ is the normalization constant, ${\left\langle x\right\rangle }$ is the mean position and ${\left\langle p\right\rangle }$ is the mean momentum. For a harmonic oscillator centred at ${x=0}$, we have that both ${\left\langle x\right\rangle =\left\langle p\right\rangle =0}$, so we know that the ground state wave function has the form

$\displaystyle \psi\left(x\right)=Ae^{-ax^{2}/2\hbar} \ \ \ \ \ (4)$

To normalize this we require (assuming ${A}$ is real)

$\displaystyle \int_{-\infty}^{\infty}\psi^{2}\left(x\right)dx=1 \ \ \ \ \ (5)$

Using the standard result for a gaussian integral (see Appendix 2 in Shankar or use Google)

 $\displaystyle \int_{-\infty}^{\infty}\psi^{2}\left(x\right)dx$ $\displaystyle =$ $\displaystyle A^{2}\int_{-\infty}^{\infty}e^{-ax^{2}/\hbar}dx\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle A^{2}\sqrt{\frac{\pi\hbar}{a}} \ \ \ \ \ (7)$

Therefore

$\displaystyle A=\left(\frac{a}{\pi\hbar}\right)^{1/4} \ \ \ \ \ (8)$

We need to find ${a}$ such that ${\Delta X\Delta P}$ is minimized. The harmonic oscillator hamiltonian is

$\displaystyle H=\frac{P^{2}}{2m}+\frac{1}{2}m\omega^{2}X^{2} \ \ \ \ \ (9)$

Since ${\left\langle X\right\rangle =\left\langle P\right\rangle =0}$, the uncertainties become

 $\displaystyle \left(\Delta X\right)^{2}$ $\displaystyle =$ $\displaystyle \left\langle X^{2}\right\rangle -\left\langle X\right\rangle ^{2}=\left\langle X^{2}\right\rangle \ \ \ \ \ (10)$ $\displaystyle \left(\Delta P\right)^{2}$ $\displaystyle =$ $\displaystyle \left\langle P^{2}\right\rangle -\left\langle P\right\rangle ^{2}=\left\langle P^{2}\right\rangle \ \ \ \ \ (11)$

Averaging 9 we get

 $\displaystyle \left\langle H\right\rangle$ $\displaystyle =$ $\displaystyle \frac{\left\langle P^{2}\right\rangle }{2m}+\frac{1}{2}m\omega^{2}\left\langle X^{2}\right\rangle \ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\left(\Delta P\right)^{2}}{2m}+\frac{1}{2}m\omega^{2}\left(\Delta X\right)^{2} \ \ \ \ \ (13)$

At minimum uncertainty

$\displaystyle \Delta X\Delta P=\frac{\hbar}{2} \ \ \ \ \ (14)$

so we have

 $\displaystyle \Delta P$ $\displaystyle =$ $\displaystyle \frac{\hbar}{2\Delta X}\ \ \ \ \ (15)$ $\displaystyle \left\langle H\right\rangle$ $\displaystyle =$ $\displaystyle \frac{\hbar^{2}}{8m\left(\Delta X\right)^{2}}+\frac{1}{2}m\omega^{2}\left(\Delta X\right)^{2} \ \ \ \ \ (16)$

The minimum energy can now be found by finding the value of ${\left(\Delta X\right)^{2}}$ that minimizes this function. Treating ${\left(\Delta X\right)^{2}}$ (not just ${\Delta X}$) as the independent variable, we have

 $\displaystyle \frac{\partial\left\langle H\right\rangle }{\partial\left(\Delta X\right)^{2}}$ $\displaystyle =$ $\displaystyle -\frac{\hbar^{2}}{8m\left[\left(\Delta X\right)^{2}\right]^{2}}+\frac{1}{2}m\omega^{2}\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{\hbar^{2}}{8m\left(\Delta X\right)^{4}}+\frac{1}{2}m\omega^{2}=0\ \ \ \ \ (18)$ $\displaystyle \left(\Delta X\right)^{2}$ $\displaystyle =$ $\displaystyle \frac{\hbar}{2m\omega} \ \ \ \ \ (19)$

This gives a minimum value for the mean energy of

$\displaystyle \left\langle H\right\rangle _{min}=\frac{\hbar\omega}{2} \ \ \ \ \ (20)$

To complete the derivation, we need to find the gaussian 4 that gives the correct value 19 for ${\left(\Delta X\right)^{2}}$. That is, we need to find ${a}$ such that

$\displaystyle \left(\Delta X\right)^{2}=\left\langle X^{2}\right\rangle =\frac{\hbar}{2m\omega} \ \ \ \ \ (21)$

This requires doing another gaussian integral:

 $\displaystyle \left\langle X^{2}\right\rangle$ $\displaystyle =$ $\displaystyle \int_{-\infty}^{\infty}x^{2}\psi^{2}\left(x\right)dx\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{a}{\pi\hbar}}\int_{-\infty}^{\infty}x^{2}e^{-ax^{2}/\hbar}dx\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{a}{\pi\hbar}}\sqrt{\frac{\pi\hbar}{a}}\frac{h}{2a}\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\hbar}{2a} \ \ \ \ \ (25)$

We therefore get

 $\displaystyle \frac{\hbar}{2a}$ $\displaystyle =$ $\displaystyle \frac{\hbar}{2m\omega}\ \ \ \ \ (26)$ $\displaystyle a$ $\displaystyle =$ $\displaystyle m\omega \ \ \ \ \ (27)$

which gives a normalized minimum energy wave function

$\displaystyle \psi_{min}\left(x\right)=\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}e^{-m\omega x^{2}/2\hbar} \ \ \ \ \ (28)$

This is the lowest possible value for the energy, but is it actually the ground state energy? What we have shown so far is that

$\displaystyle \left\langle \psi_{min}\left|H\right|\psi_{min}\right\rangle \le\left\langle \psi_{0}\left|H\right|\psi_{0}\right\rangle =E_{0} \ \ \ \ \ (29)$

where ${\left|\psi_{0}\right\rangle }$ is the ground state energy. However, we can invoke the variational principle which states that if ${\psi}$ is any normalized function, then the ground state energy ${E_{0}}$ of any hamiltonian ${H}$ satisfies

$\displaystyle E_{0}\le\left\langle \psi\left|H\right|\psi\right\rangle \ \ \ \ \ (30)$

Using ${\psi=\psi_{min}}$ we therefore have

$\displaystyle E_{0}\le\left\langle \psi_{min}\left|H\right|\psi_{min}\right\rangle \ \ \ \ \ (31)$

Combining 29 and 31 we have

$\displaystyle \left\langle \psi_{min}\left|H\right|\psi_{min}\right\rangle \le E_{0}\le\left\langle \psi_{min}\left|H\right|\psi_{min}\right\rangle \ \ \ \ \ (32)$

which means that

$\displaystyle E_{0}=\left\langle \psi_{min}\left|H\right|\psi_{min}\right\rangle \ \ \ \ \ (33)$

and therefore that ${\left|\psi_{0}\right\rangle =\left|\psi_{min}\right\rangle }$, that is, 28 is actually the ground state wave function.

Although this clever little derivation gives us the ground state energy and wave function, it doesn’t say anything about the higher energy states, or tell us that they are all equally spaced with a spacing of ${\hbar\omega}$. Nevertheless, it’s a pleasant exercise.

# Harmonic oscillator – mean position and momentum

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 7.3, Exercise 7.3.5.

The energy eigenfunctions of the harmonic oscillator are

$\displaystyle \psi_{n}(y)=\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\frac{1}{\sqrt{2^{n}n!}}H_{n}(y)e^{-y^{2}/2} \ \ \ \ \ (1)$

where

$\displaystyle y\equiv\sqrt{\frac{m\omega}{\hbar}}x \ \ \ \ \ (2)$

and ${H_{n}}$ is the Hermite polynomial of order ${n}$. Since an even (odd) Hermite polynomial is an even (odd) function, ${\psi_{n}}$ is even (odd) if ${n}$ is even (odd), so we can use this fact to show that

$\displaystyle \left\langle n\left|X\right|n\right\rangle =\int_{-\infty}^{\infty}x\psi_{n}^{2}\left(x\right)\;dx=0 \ \ \ \ \ (3)$

This follows because the square of either an even or odd function gives an even function, and ${x}$ itself is odd, so the integrand is the product of an odd and even function, which is odd. The integral over any interval symmetric about ${x=0}$ of an odd function is odd. Thus the mean position ${\left\langle X\right\rangle }$ of a particle in any of the harmonic oscillator’s energy eigenstates is zero.

For the momentum ${P}$, we have

$\displaystyle \left\langle n\left|P\right|n\right\rangle =-i\hbar\int_{-\infty}^{\infty}\psi_{n}\frac{d\psi_{n}}{dx}\;dx \ \ \ \ \ (4)$

As we showed earlier

$\displaystyle -i\hbar\frac{d\psi_{n}}{dx}=i\sqrt{\hbar m\omega}\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\frac{1}{\sqrt{2^{n}n!}}e^{-y^{2}/2}\left[nH_{n-1}-\frac{1}{2}H_{n+1}\right] \ \ \ \ \ (5)$

If ${n}$ is even (odd), then ${nH_{n-1}-\frac{1}{2}H_{n+1}}$ is odd (even), so the product ${\psi_{n}\frac{d\psi_{n}}{dx}}$ is always the product of one odd and one even function, making it odd. Thus

$\displaystyle \left\langle n\left|P\right|n\right\rangle =0 \ \ \ \ \ (6)$

Thus the mean momentum ${\left\langle P\right\rangle =0}$ in all energy eigenstates.

This means that the uncertainties in position and momentum are determined entirely by the mean square values:

 $\displaystyle \left(\Delta X\right)^{2}$ $\displaystyle =$ $\displaystyle \left\langle X^{2}\right\rangle -\left\langle X\right\rangle ^{2}=\left\langle X^{2}\right\rangle \ \ \ \ \ (7)$ $\displaystyle \left(\Delta P\right)^{2}$ $\displaystyle =$ $\displaystyle \left\langle P^{2}\right\rangle \ \ \ \ \ (8)$

We can work out these values for a couple of specific states. For ${n=1}$ we have

 $\displaystyle \left\langle 1\left|X^{2}\right|1\right\rangle$ $\displaystyle =$ $\displaystyle \int_{-\infty}^{\infty}x^{2}\psi_{1}^{2}\left(x\right)\;dx\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\sqrt{\frac{m\omega}{\pi\hbar}}\int_{-\infty}^{\infty}x^{2}H_{1}^{2}\left(y\right)e^{-y^{2}}dx\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\frac{\hbar}{\sqrt{\pi}m\omega}\int_{-\infty}^{\infty}y^{2}H_{1}^{2}\left(y\right)e^{-y^{2}}dy\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2\hbar}{\sqrt{\pi}m\omega}\int_{-\infty}^{\infty}y^{4}e^{-y^{2}}dy\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2\hbar}{\sqrt{\pi}m\omega}\frac{3\sqrt{\pi}}{4}\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{3\hbar}{2m\omega} \ \ \ \ \ (14)$

We’ve used

$\displaystyle H_{1}\left(y\right)=2y \ \ \ \ \ (15)$

and formula just before A.2.3 from the appendix in Shankar, which gives

 $\displaystyle I_{4}\left(\alpha\right)$ $\displaystyle =$ $\displaystyle \int_{-\infty}^{\infty}x^{4}e^{-\alpha x^{2}}dx\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\partial^{2}}{\partial\alpha^{2}}\int_{-\infty}^{\infty}e^{-\alpha x^{2}}dx\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\partial^{2}}{\partial\alpha^{2}}I_{0}\left(\alpha\right) \ \ \ \ \ (18)$

From formula A.2.2

 $\displaystyle I_{4}\left(\alpha\right)$ $\displaystyle =$ $\displaystyle \frac{\partial^{2}}{\partial\alpha^{2}}\sqrt{\frac{\pi}{\alpha}}\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{3\sqrt{\pi}}{4\alpha^{5/2}} \ \ \ \ \ (20)$

Setting ${\alpha=1}$ gives

$\displaystyle \int_{-\infty}^{\infty}y^{4}e^{-y^{2}}dy=\frac{3\sqrt{\pi}}{4} \ \ \ \ \ (21)$

For ${P}$, we have

$\displaystyle \left\langle 1\left|P^{2}\right|1\right\rangle =-\hbar^{2}\int_{-\infty}^{\infty}\psi_{1}\frac{d^{2}}{dx^{2}}\psi_{1}\;dx \ \ \ \ \ (22)$

The derivative is

 $\displaystyle \frac{d^{2}}{dx^{2}}\psi_{1}$ $\displaystyle =$ $\displaystyle \frac{d^{2}\psi_{1}}{dy^{2}}\left(\frac{dy}{dx}\right)^{2}\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2}}\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\frac{m\omega}{\hbar}\frac{d^{2}}{dy^{2}}\left[H_{1}(y)e^{-y^{2}/2}\right]\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2}}\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\frac{m\omega}{\hbar}\frac{d^{2}}{dy^{2}}\left[2ye^{-y^{2}/2}\right]\ \ \ \ \ (25)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(\frac{m\omega}{4\pi\hbar}\right)^{1/4}\frac{m\omega}{\hbar}e^{-y^{2}/2}\left[2y^{3}-6y\right] \ \ \ \ \ (26)$

We can now evaluate 22:

 $\displaystyle -\hbar^{2}\int_{-\infty}^{\infty}\psi_{1}\frac{d^{2}}{dx^{2}}\psi_{1}\;dx$ $\displaystyle =$ $\displaystyle -\hbar^{2}\sqrt{\frac{m\omega}{4\pi\hbar}}\frac{m\omega}{\hbar}\sqrt{\frac{\hbar}{m\omega}}\int_{-\infty}^{\infty}2ye^{-y^{2}}2\left[y^{3}-3y\right]dy\ \ \ \ \ (27)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{2m\omega\hbar}{\sqrt{\pi}}\int_{-\infty}^{\infty}e^{-y^{2}}\left[y^{4}-3y^{2}\right]dy\ \ \ \ \ (28)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{2m\omega\hbar}{\sqrt{\pi}}\left[\frac{3\sqrt{\pi}}{4}-\frac{3\sqrt{\pi}}{2}\right]\ \ \ \ \ (29)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{3}{2}m\omega\hbar \ \ \ \ \ (30)$

Thus for the ${n=1}$ state

 $\displaystyle \Delta X$ $\displaystyle =$ $\displaystyle \sqrt{\frac{3\hbar}{2m\omega}}\ \ \ \ \ (31)$ $\displaystyle \Delta P$ $\displaystyle =$ $\displaystyle \sqrt{\frac{3}{2}m\omega\hbar}\ \ \ \ \ (32)$ $\displaystyle \Delta X\Delta P$ $\displaystyle =$ $\displaystyle \frac{3}{2}\hbar>\frac{\hbar}{2} \ \ \ \ \ (33)$

For the ${n=0}$ (ground) state, we can use ${H_{0}=1}$ to get

 $\displaystyle \left\langle X^{2}\right\rangle$ $\displaystyle =$ $\displaystyle \int_{-\infty}^{\infty}x^{2}\psi_{0}^{2}\left(x\right)\;dx\ \ \ \ \ (34)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{m\omega}{\pi\hbar}}\int_{-\infty}^{\infty}x^{2}H_{0}^{2}\left(y\right)e^{-y^{2}}dx\ \ \ \ \ (35)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\hbar}{\sqrt{\pi}m\omega}\int_{-\infty}^{\infty}y^{2}e^{-y^{2}}dy\ \ \ \ \ (36)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\hbar}{2m\omega} \ \ \ \ \ (37)$

For ${P}$:

 $\displaystyle \frac{d^{2}}{dx^{2}}\psi_{0}$ $\displaystyle =$ $\displaystyle \frac{d^{2}\psi_{0}}{dy^{2}}\left(\frac{dy}{dx}\right)^{2}\ \ \ \ \ (38)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\frac{m\omega}{\hbar}e^{-y^{2}/2}\left(y^{2}-1\right)\ \ \ \ \ (39)$ $\displaystyle \left\langle P^{2}\right\rangle =-\hbar^{2}\int_{-\infty}^{\infty}\psi_{0}\frac{d^{2}}{dx^{2}}\psi_{0}\;dx$ $\displaystyle =$ $\displaystyle -\hbar^{2}\sqrt{\frac{m\omega}{\pi\hbar}}\frac{m\omega}{\hbar}\sqrt{\frac{\hbar}{m\omega}}\int_{-\infty}^{\infty}e^{-y^{2}}\left(y^{2}-1\right)dy\ \ \ \ \ (40)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{m\omega\hbar}{\sqrt{\pi}}\left[\frac{\sqrt{\pi}}{2}-\sqrt{\pi}\right]\ \ \ \ \ (41)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}m\omega\hbar \ \ \ \ \ (42)$

The uncertainty principle in this case gives

$\displaystyle \Delta X\Delta P=\frac{\hbar}{2} \ \ \ \ \ (43)$

so it saturates the condition ${\Delta X\Delta P\ge\frac{\hbar}{2}}$.