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# WordPress help requested

As regular visitors will know, this blog occasionally goes off line due to problems connecting to the WordPress database that stores the posts. I have contacted my hosting provider and they say that this is due to an excessive number of database connections that are opened but then not closed again, but are unable to provide any help beyond that.

As I do not want to mess with any of the WordPress code (for two reasons: 1 – changing the code could introduce security problems and 2 – I don’t know enough about either WordPress or PHP to mess with their code) I was wondering if any readers have experience with running WordPress blogs and know of any ways to prevent excessive database access. I have just installed the “W3 Total Cache” plugin which may help, but if anyone has any other suggestions I’d be very grateful.

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# Welcome to Physics Pages

This blog consists of my notes and solutions to problems in various areas of mainstream physics. An index to the topics covered is contained in the links in the sidebar on the right, or in the menu at the top of the page.

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# 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 ${1/a_{0}=me^{2}/\hbar^{2}}$.

# Path integral to Schrödinger equation for a vector potential

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 8. Section 8.6, Exercise 8.6.4.

When we showed that the path integral approach is equivalent to the Schrödinger equation, we did so for a scalar potential ${V}$, so that the Lagrangian is the usual ${L=T-V}$, and we can use that to calculate the action over an infinitestimal time interval ${\varepsilon}$, during which time the particle moves from ${x^{\prime}}$ to ${x}$. In the calculation, we chose the value of ${V}$ at the midpoint of this interval, that is ${V\left(\frac{x+x^{\prime}}{2}\right)}$. In fact, in this derivation it didn’t matter where in the interval ${\left[x^{\prime},x\right]}$ we chose to evaluate ${V}$, since we took only terms up to first order in ${\varepsilon}$, and moving the point at which we evaluate ${V}$ introduced terms only of order ${\varepsilon^{2}}$ or higher.

Things get a bit more complicated if we consider a system such as the electromagnetic force, where the Lagrangian is no longer just ${T-V}$, but becomes

$\displaystyle L=\frac{1}{2}m\mathbf{v}\cdot\mathbf{v}-q\phi+\frac{q}{c}\mathbf{v}\cdot\mathbf{A} \ \ \ \ \ (1)$

To examine the effect this has on the demonstration that the path integral approach is equivalent to the Schrödinger equation, we’ll consider only one dimension, and leave out the electrostatic potential ${\phi}$ since it’s just a scalar potential and we already know that such potentials do indeed convert to the Schrödinger equation. Thus the Lagrangian we’ll consider is

$\displaystyle L=\frac{1}{2}mv^{2}+\frac{q}{c}vA \ \ \ \ \ (2)$

Over the infinitesimal time interval ${\varepsilon}$ we have

$\displaystyle v=\frac{x-x^{\prime}}{\varepsilon} \ \ \ \ \ (3)$

The propagator over this time interval is

 $\displaystyle U\left(x,\varepsilon;x^{\prime},0\right)$ $\displaystyle =$ $\displaystyle \sqrt{\frac{m}{2\pi\hbar\varepsilon i}}\exp\left[\frac{i}{\hbar}\left(\frac{1}{2}m\frac{\left(x-x^{\prime}\right)^{2}}{\varepsilon}+\varepsilon\frac{q}{c}\frac{x-x^{\prime}}{\varepsilon}A\left(x+\alpha\left(x-x^{\prime}\right)\right)\right)\right]\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{m}{2\pi\hbar\varepsilon i}}\exp\left[\frac{i}{\hbar}\left(\frac{1}{2}m\frac{\eta^{2}}{\varepsilon}-\frac{q}{c}\eta A\left(x+\alpha\eta\right)\right)\right]\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{m}{2\pi\hbar\varepsilon i}}\exp\left(\frac{im\eta^{2}}{2\hbar\varepsilon}\right)\exp\left[-\frac{iq}{\hbar c}\eta A\left(x+\alpha\eta\right)\right] \ \ \ \ \ (6)$

where ${\alpha}$ is a parameter that we can vary between 0 and 1 in order to vary the point along the path from ${x^{\prime}}$ to ${x}$ at which we evaluate the vector potential ${A}$. Also,

$\displaystyle \eta\equiv x^{\prime}-x \ \ \ \ \ (7)$

Using the same argument as before, we require

$\displaystyle \left|\eta\right|\lesssim\sqrt{\frac{2\hbar\varepsilon\pi}{m}} \ \ \ \ \ (8)$

so calculations to first order in ${\varepsilon}$ must include terms up to second order in ${\eta}$.

Once we have ${U\left(x,\varepsilon;x^{\prime},0\right)}$, we can find ${\psi\left(x,\varepsilon\right)}$ from

$\displaystyle \psi\left(x,\varepsilon\right)=\int_{-\infty}^{\infty}U\left(x,\varepsilon;x^{\prime},0\right)\psi\left(x^{\prime},0\right)dx^{\prime} \ \ \ \ \ (9)$

To find ${U}$ to first order in ${\varepsilon}$, we need to expand the second exponential in 6 out to terms in ${\eta^{2}}$, so we first look at the argument of the exponential:

$\displaystyle -\frac{iq}{\hbar c}\eta A\left(x+\alpha\eta\right)=-\frac{iq}{\hbar c}\left(\eta A\left(x\right)+\alpha\eta^{2}\frac{\partial A}{\partial x}+\ldots\right) \ \ \ \ \ (10)$

where the derivative is evaluated at the endpoint ${x}$ and is constant in the integral. The second exponential in 6 now becomes, to second order in ${\eta}$:

$\displaystyle \exp\left[-\frac{iq}{\hbar c}\eta A\left(x+\alpha\eta\right)\right]=1-\frac{iq}{\hbar c}\left(\eta A\left(x\right)+\alpha\eta^{2}\frac{\partial A}{\partial x}\right)-\left(\frac{q}{\hbar c}\right)^{2}\frac{\eta^{2}A^{2}\left(x\right)}{2} \ \ \ \ \ (11)$

We also need the expansion of the wave function in 9 up to second order in ${\eta}$:

$\displaystyle \psi\left(x+\eta,0\right)=\psi\left(x,0\right)+\eta\frac{\partial\psi}{\partial x}+\frac{\eta^{2}}{2}\frac{\partial^{2}\psi}{\partial x^{2}} \ \ \ \ \ (12)$

Again, both derivatives are evaluated at the endpoint ${x}$ and are constants in the integral.

We now need to do the integral 9, which consists of several standard Gaussian integrals. From 7, ${dx^{\prime}=d\eta}$, so

 $\displaystyle \int_{-\infty}^{\infty}U\left(x,\varepsilon;x^{\prime},0\right)\psi\left(x^{\prime},0\right)dx^{\prime}$ $\displaystyle =$ $\displaystyle \sqrt{\frac{m}{2\pi\hbar\varepsilon i}}\psi\left(x,0\right)\int_{-\infty}^{\infty}\exp\left(\frac{im\eta^{2}}{2\hbar\varepsilon}\right)d\eta+\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle$ $\displaystyle \sqrt{\frac{m}{2\pi\hbar\varepsilon i}}\left(\frac{\partial\psi}{\partial x}-\frac{iq}{\hbar c}A\left(x\right)\psi\left(x,0\right)\right)\int_{-\infty}^{\infty}\exp\left(\frac{im\eta^{2}}{2\hbar\varepsilon}\right)\eta\;d\eta+\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle$ $\displaystyle \sqrt{\frac{m}{2\pi\hbar\varepsilon i}}\left(\frac{1}{2}\frac{\partial^{2}\psi}{\partial x^{2}}-\frac{iq}{\hbar c}A\left(x\right)\frac{\partial\psi}{\partial x}+\psi\left(x,0\right)\left(-\frac{iq\alpha}{\hbar c}\frac{\partial A}{\partial x}-\frac{1}{2}\left(\frac{qA\left(x\right)}{\hbar c}\right)^{2}\right)\right)\times\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle$ $\displaystyle \int_{-\infty}^{\infty}\exp\left(\frac{im\eta^{2}}{2\hbar\varepsilon}\right)\eta^{2}d\eta \ \ \ \ \ (16)$

We can now do the integrals:

 $\displaystyle \int_{-\infty}^{\infty}\exp\left(\frac{im\eta^{2}}{2\hbar\varepsilon}\right)d\eta$ $\displaystyle =$ $\displaystyle \sqrt{\frac{2\pi\hbar\varepsilon i}{m}}\ \ \ \ \ (17)$ $\displaystyle \int_{-\infty}^{\infty}\exp\left(\frac{im\eta^{2}}{2\hbar\varepsilon}\right)\eta\;d\eta$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (18)$ $\displaystyle \int_{-\infty}^{\infty}\exp\left(\frac{im\eta^{2}}{2\hbar\varepsilon}\right)\eta^{2}d\eta$ $\displaystyle =$ $\displaystyle -\frac{\hbar\varepsilon}{im}\sqrt{\frac{2\pi\hbar\varepsilon i}{m}} \ \ \ \ \ (19)$

Plugging these in we get

 $\displaystyle \psi\left(x,\varepsilon\right)$ $\displaystyle =$ $\displaystyle \psi\left(x,0\right)-\frac{\hbar\varepsilon}{im}\left[\frac{1}{2}\frac{\partial^{2}\psi}{\partial x^{2}}-\frac{iq}{\hbar c}A\left(x\right)\frac{\partial\psi}{\partial x}+\psi\left(x,0\right)\left(-\frac{iq\alpha}{\hbar c}\frac{\partial A}{\partial x}-\frac{1}{2}\left(\frac{qA\left(x\right)}{\hbar c}\right)^{2}\right)\right]\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \psi\left(x,0\right)+\frac{\varepsilon}{i\hbar}\left[-\frac{\hbar^{2}}{2m}\frac{\partial^{2}\psi}{\partial x^{2}}+\frac{i\hbar q}{mc}A\left(x\right)\frac{\partial\psi}{\partial x}+\psi\left(x,0\right)\left(\frac{i\hbar q\alpha}{mc}\frac{\partial A}{\partial x}+\frac{1}{2m}\left(\frac{qA\left(x\right)}{c}\right)^{2}\right)\right] \ \ \ \ \ (21)$

We can compare this with the quantum version of the Hamiltonian for the vector potential part of the electromagnetic force. The classical Hamiltonian is

$\displaystyle H=\frac{\left|\mathbf{p}-q\mathbf{A}/c\right|^{2}}{2m} \ \ \ \ \ (22)$

Because ${\mathbf{A}}$ depends on ${x}$, it doesn’t commute with ${\mathbf{p}}$ so to get the quantum version we need to symmetrize when we expand the square. The one dimensional version is

$\displaystyle H=\frac{P^{2}}{2m}-\frac{q}{2mc}PA-\frac{q}{2mc}AP+\frac{q^{2}A^{2}}{2mc^{2}} \ \ \ \ \ (23)$

In the coordinate basis, we have, using ${P=-i\hbar\partial/\partial x}$

 $\displaystyle H\psi$ $\displaystyle =$ $\displaystyle -\frac{\hbar^{2}}{2m}\frac{\partial^{2}\psi}{\partial x^{2}}+\frac{i\hbar q}{2mc}\left(\frac{\partial\left(A\psi\right)}{\partial x}+A\frac{\partial\psi}{\partial x}\right)+\frac{q^{2}A^{2}}{2mc^{2}}\psi\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{\hbar^{2}}{2m}\frac{\partial^{2}\psi}{\partial x^{2}}+\frac{i\hbar q}{2mc}\left(2A\frac{\partial\psi}{\partial x}+\psi\frac{\partial A}{\partial x}\right)+\frac{q^{2}A^{2}}{2mc^{2}}\psi\ \ \ \ \ (25)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{\hbar^{2}}{2m}\frac{\partial^{2}\psi}{\partial x^{2}}+\frac{i\hbar q}{mc}\left(A\frac{\partial\psi}{\partial x}+\frac{1}{2}\psi\frac{\partial A}{\partial x}\right)+\frac{q^{2}A^{2}}{2mc^{2}}\psi \ \ \ \ \ (26)$

Returning to the result we got from the path integral, upon rearranging 21 we get

 $\displaystyle i\hbar\frac{\psi\left(x,\varepsilon\right)-\psi\left(x,0\right)}{\varepsilon}$ $\displaystyle =$ $\displaystyle -\frac{\hbar^{2}}{2m}\frac{\partial^{2}\psi}{\partial x^{2}}+\frac{i\hbar q}{mc}\left(A\left(x\right)\frac{\partial\psi}{\partial x}+\alpha\psi\left(x,0\right)\frac{\partial A}{\partial x}\right)+\frac{\psi\left(x,0\right)}{2m}\left(\frac{qA\left(x\right)}{c}\right)^{2}\ \ \ \ \ (27)$ $\displaystyle i\hbar\frac{\partial\psi}{\partial t}$ $\displaystyle =$ $\displaystyle -\frac{\hbar^{2}}{2m}\frac{\partial^{2}\psi}{\partial x^{2}}+\frac{i\hbar q}{mc}\left(A\left(x\right)\frac{\partial\psi}{\partial x}+\alpha\psi\frac{\partial A}{\partial x}\right)+\frac{\psi}{2m}\left(\frac{qA\left(x\right)}{c}\right)^{2} \ \ \ \ \ (28)$

where in the last line we took the limit as ${\varepsilon\rightarrow0}$ on the LHS to get Schrödinger’s equation in the form

$\displaystyle i\hbar\frac{\partial\psi}{\partial t}=H\psi \ \ \ \ \ (29)$

Comparing the RHS of 28 with 26, we see that they are equal provided we take ${\alpha=\frac{1}{2}}$. Thus in this case, we really do need to evaluate the vector potential ${A}$ at the midpoint of the path.

# 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.

# Path integrals for special potentials; use of classical action

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 8. Section 8.6, Exercises 8.6.1 – 8.6.2.

We’ve seen that if we use the path integral formulation for a free particle, we get the exact propagator by considering only one path (the classical path) between the starting point ${\left(x^{\prime},t^{\prime}\right)}$ and the end point ${\left(x,t\right)}$. In this case, the propagator has the form

$\displaystyle U\left(x,t;x^{\prime},t^{\prime}\right)=A\left(t\right)e^{iS_{cl}/\hbar} \ \ \ \ \ (1)$

where ${S_{cl}}$ is the classical action. It turns out that this form is true for a wider set of potentials, beyond just the free particle. The general form of the potential for which this is true is

$\displaystyle V=a+bx+cx^{2}+d\dot{x}+ex\dot{x} \ \ \ \ \ (2)$

where ${a,b,c,d}$ and ${e}$ are constants. The general expression for the propagator is (where we’re taking the starting time to be ${t^{\prime}=0}$):

$\displaystyle U\left(x,t;x^{\prime}\right)=\int_{x^{\prime}}^{x}e^{iS\left[x\left(t^{\prime\prime}\right)\right]/\hbar}\mathfrak{D}\left[x\left(t^{\prime\prime}\right)\right] \ \ \ \ \ (3)$

where the notation ${\mathfrak{D}\left[x\left(t^{\prime\prime}\right)\right]}$ means an integration over all possible paths from ${x^{\prime}}$ to ${x}$ in the given time interval.

For a given path, we can write the location of the particle ${x\left(t^{\prime\prime}\right)}$ as composed of its position on the classical path ${x_{cl}\left(t^{\prime\prime}\right)}$ plus the deviation ${y\left(t^{\prime\prime}\right)}$ from the classical path:

$\displaystyle x\left(t^{\prime\prime}\right)=x_{cl}\left(t^{\prime\prime}\right)+y\left(t^{\prime\prime}\right) \ \ \ \ \ (4)$

As the endpoints are fixed

$\displaystyle y\left(0\right)=y\left(t\right)=0 \ \ \ \ \ (5)$

Also, since for any given potential and choice of endpoints, ${x_{cl}\left(t^{\prime\prime}\right)}$ is fixed for all times, it is effectively a constant with regard to the path integration. Therefore

$\displaystyle dx=dy \ \ \ \ \ (6)$

Making these substitutions into 3, we get, using Shankar’s slightly misleading notation:

$\displaystyle U\left(x,t;x^{\prime}\right)=\int_{0}^{0}e^{iS\left[x_{cl}\left(t^{\prime\prime}\right)+y\left(t^{\prime\prime}\right)\right]/\hbar}\mathfrak{D}\left[y\left(t^{\prime\prime}\right)\right] \ \ \ \ \ (7)$

Usually, when the limits on an integral are the same, the integral evaluates to zero. However, in this case, the notation ${\int_{0}^{0}\mathfrak{D}\left[y\left(t^{\prime\prime}\right)\right]}$ means that ${y}$ starts and ends at zero, but covers all possible paths between these endpoints.

The action is the integral of the Lagrangian which, for the potential 2 is

 $\displaystyle L$ $\displaystyle =$ $\displaystyle T-V\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}m\dot{x}^{2}-a-bx-cx^{2}-d\dot{x}-ex\dot{x} \ \ \ \ \ (9)$

Because ${L}$ is quadratic in both ${x}$ and ${\dot{x}}$, we can expand it in a Taylor series up to second order without any approximation. That is

 $\displaystyle L\left(x_{cl}+y,\dot{x}_{cl}+\dot{y}\right)$ $\displaystyle =$ $\displaystyle L\left(x_{cl},\dot{x}_{cl}\right)+\left.\frac{\partial L}{\partial x}\right|_{x_{cl}}y+\left.\frac{\partial L}{\partial\dot{x}}\right|_{x_{cl}}\dot{y}+\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle$ $\displaystyle \frac{1}{2}\left(\left.\frac{\partial^{2}L}{\partial x^{2}}\right|_{x_{cl}}y^{2}+2\left.\frac{\partial^{2}L}{\partial x\partial\dot{x}}\right|_{x_{cl}}y\dot{y}+\left.\frac{\partial^{2}L}{\partial\dot{x}^{2}}\right|_{x_{cl}}\dot{y}^{2}\right) \ \ \ \ \ (11)$

Look first at the last two terms on the RHS of the first line. Using the equations of motion, we have

$\displaystyle \left.\frac{\partial L}{\partial x}\right|_{x_{cl}}=\frac{d}{dt}\left(\left.\frac{\partial L}{\partial\dot{x}}\right|_{x_{cl}}\right) \ \ \ \ \ (12)$

To get the action, we need to integrate the Lagrangian over the time interval of interest. Integrating these two terms gives

 $\displaystyle \int_{0}^{t}\left[\left.\frac{\partial L}{\partial x}\right|_{x_{cl}}y+\left.\frac{\partial L}{\partial\dot{x}}\right|_{x_{cl}}\dot{y}\right]dt^{\prime\prime}$ $\displaystyle =$ $\displaystyle \int_{0}^{t}\left[\frac{d}{dt}\left(\left.\frac{\partial L}{\partial\dot{x}}\right|_{x_{cl}}\right)y+\left.\frac{\partial L}{\partial\dot{x}}\right|_{x_{cl}}\dot{y}\right]dt^{\prime\prime}\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left.\left(\left.\frac{\partial L}{\partial\dot{x}}\right|_{x_{cl}}\right)y\right|_{0}^{t}-\int_{0}^{t}\left.\frac{\partial L}{\partial\dot{x}}\right|_{x_{cl}}\dot{y}dt^{\prime\prime}+\int_{0}^{t}\left.\frac{\partial L}{\partial\dot{x}}\right|_{x_{cl}}\dot{y}dt^{\prime\prime}\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (15)$

where we integrated the first term by parts. The integrated term in the second line is zero because ${y=0}$ at both endpoints, and the last two terms cancel each other.

Returning to 11, we can calculate the three second derivatives explicitly:

 $\displaystyle \frac{1}{2}\frac{\partial^{2}L}{\partial x^{2}}$ $\displaystyle =$ $\displaystyle -c\ \ \ \ \ (16)$ $\displaystyle \left.\frac{\partial^{2}L}{\partial x\partial\dot{x}}\right|_{x_{cl}}$ $\displaystyle =$ $\displaystyle -e\ \ \ \ \ (17)$ $\displaystyle \frac{1}{2}\frac{\partial^{2}L}{\partial\dot{x}^{2}}$ $\displaystyle =$ $\displaystyle m \ \ \ \ \ (18)$

The integral of the first term on the RHS of 10 is just the classical action, so we get for the propagator 7:

$\displaystyle U\left(x,t;x^{\prime}\right)=e^{iS_{cl}/\hbar}\int_{0}^{0}\exp\left[\frac{i}{\hbar}\int_{0}^{t}\left(\frac{m\dot{y}^{2}}{2}-cy^{2}-ey\dot{y}\right)dt^{\prime\prime}\right]\mathfrak{D}\left[y\left(t^{\prime\prime}\right)\right] \ \ \ \ \ (19)$

The remaining path integral can still be difficult to evaluate, but we can observe a few properties that it has. First, for any given path in the path integral, we must be able to express both ${y}$ and ${\dot{y}}$ as functions of time ${t^{\prime\prime}}$, so the complete path integral can depend only on the end time ${t}$ (and, of course, on the constants ${m}$, ${c}$ and ${e}$). That is, the propagator will always have the form 1:

$\displaystyle U\left(x,t;x^{\prime},t^{\prime}\right)=A\left(t\right)e^{iS_{cl}/\hbar} \ \ \ \ \ (20)$

We have already evaluated the integral for the free particle where ${c=e=0}$ and we found there that

$\displaystyle U\left(x,t;x^{\prime}\right)=\sqrt{\frac{m}{2\pi\hbar it}}e^{iS_{cl}/\hbar} \ \ \ \ \ (21)$

Since the constant ${b}$ doesn’t appear in 19, the propagator must have the same form for the more general case where ${V=a+bx}$. For more complex potentials, such as the harmonic oscillator, the function ${A\left(t\right)}$ will in general have a different form and will have to be calculated explicitly in these cases.

As an example, we’ll consider the case of a particle subject to a constant force in the ${x}$ direction, so that the potential is given by

$\displaystyle V\left(x\right)=-fx \ \ \ \ \ (22)$

This gives a constant force of

$\displaystyle F=-\frac{dV}{dx}=f \ \ \ \ \ (23)$

and thus a constant acceleration of ${f/m}$. For such a particle, its classical position is (from first year physics)

 $\displaystyle x_{cl}\left(t^{\prime\prime}\right)$ $\displaystyle =$ $\displaystyle x_{0}+v_{0}t^{\prime\prime}+\frac{1}{2}\frac{f}{m}t^{\prime\prime2}\ \ \ \ \ (24)$ $\displaystyle \dot{x}_{cl}\left(t^{\prime\prime}\right)$ $\displaystyle =$ $\displaystyle v_{0}+\frac{f}{m}t^{\prime\prime} \ \ \ \ \ (25)$

To find ${x_{0}}$ and ${v_{0}}$, we impose boundary conditions. At ${t^{\prime\prime}=0}$

$\displaystyle x_{cl}\left(0\right)=x_{0}=x^{\prime} \ \ \ \ \ (26)$

At ${t^{\prime\prime}=t}$, its position is

 $\displaystyle x_{cl}\left(t\right)$ $\displaystyle =$ $\displaystyle x=x^{\prime}+v_{0}t+\frac{f}{2m}t^{2} \ \ \ \ \ (27)$

This gives

$\displaystyle v_{0}=\frac{x-x^{\prime}}{t}-\frac{f}{2m}t \ \ \ \ \ (28)$

The classical Lagrangian is

 $\displaystyle L$ $\displaystyle =$ $\displaystyle T-V\ \ \ \ \ (29)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}m\dot{x}_{cl}^{2}+fx_{cl}\ \ \ \ \ (30)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}m\left(v_{0}+\frac{f}{m}t^{\prime\prime}\right)^{2}+f\left(x_{0}+v_{0}t^{\prime\prime}+\frac{1}{2}\frac{f}{m}t^{\prime\prime2}\right)\ \ \ \ \ (31)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}m\left(\frac{x-x^{\prime}}{t}-\frac{f}{2m}t+\frac{f}{m}t^{\prime\prime}\right)^{2}+f\left(x^{\prime}+\left(\frac{x-x^{\prime}}{t}-\frac{f}{2m}t\right)t^{\prime\prime}+\frac{1}{2}\frac{f}{m}t^{\prime\prime2}\right) \ \ \ \ \ (32)$

Note that ${t}$ is a constant, as it is the time of the endpoint of the motion. To find the classical action, we must integrate this from ${t^{\prime\prime}=0}$ to ${t}$. The integral is a straightforward integral of a quadratic in ${t^{\prime\prime}}$, although the algebra is tedious if done by hand, so is best done with Maple.

 $\displaystyle S_{cl}$ $\displaystyle =$ $\displaystyle \int_{0}^{t}L\;dt^{\prime\prime}\ \ \ \ \ (33)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{3}{\frac{{f}^{2}{t}^{3}}{m}}+\left({\frac{x-x^{\prime}}{t}}-\frac{1}{2}{\frac{ft}{m}}\right)f{t}^{2}+\frac{1}{2}m\left({\frac{x-x^{\prime}}{t}}-\frac{1}{2}{\frac{ft}{m}}\right)^{2}t+fxt\ \ \ \ \ (34)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{f^{2}t^{3}}{24m}+\frac{1}{2}ft\left(x+x^{\prime}\right)+\frac{m\left(x-x^{\prime}\right)^{2}}{2t} \ \ \ \ \ (35)$

From 21, this gives a propagator of

$\displaystyle U\left(x,t;x^{\prime}\right)=\sqrt{\frac{m}{2\pi\hbar it}}\exp\left[\frac{i}{\hbar}\left(-\frac{f^{2}t^{3}}{24m}+\frac{1}{2}ft\left(x+x^{\prime}\right)+\frac{m\left(x-x^{\prime}\right)^{2}}{2t}\right)\right] \ \ \ \ \ (36)$

This agrees with Shankar’s result in his equation 5.4.31.

As another example, consider the harmonic oscillator, where the potential is

$\displaystyle V=\frac{1}{2}m\omega^{2}x^{2} \ \ \ \ \ (37)$

This potential is also of the form 2, so the propagator must have the form 20. This time, however, since ${c\ne0}$, the function ${A\left(t\right)}$ will probably not have the form used in 21. The best we can say therefore is that

$\displaystyle U\left(x,t;x^{\prime},t^{\prime}\right)=A\left(t\right)e^{iS_{cl}/\hbar} \ \ \ \ \ (38)$

where ${A\left(t\right)}$ has the form (from 19):

$\displaystyle A\left(t\right)=\int_{0}^{0}\exp\left[\frac{i}{\hbar}\int_{0}^{t}\left(\frac{m\dot{y}^{2}}{2}-\frac{1}{2}m\omega^{2}y^{2}\right)dt^{\prime\prime}\right]\mathfrak{D}\left[y\left(t^{\prime\prime}\right)\right] \ \ \ \ \ (39)$

We worked out the classical action for the harmonic oscillator earlier and found

$\displaystyle S_{cl}=\frac{m\omega}{2\sin\omega t}\left[\left(x^{\prime2}+x^{2}\right)\cos\omega t-2x^{\prime}x\right] \ \ \ \ \ (40)$

where the particle is at ${x^{\prime}}$ at ${t^{\prime\prime}=0}$ and at ${x}$ at ${t^{\prime\prime}=t}$. The propagator is therefore

$\displaystyle U\left(x,t;x^{\prime}\right)=A\left(t\right)\exp\left[\frac{im\omega}{2\hbar\sin\omega t}\left(\left(x^{\prime2}+x^{2}\right)\cos\omega t-2x^{\prime}x\right)\right] \ \ \ \ \ (41)$

with ${A\left(t\right)}$ given by 39.

# The path integral is equivalent to the Schrödinger equation

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 8. Section 8.5.

We’ve seen that we can produce the propagator for the free particle by means of a complete path integral over all paths between some specified initial state at ${\left(x_{0},t_{0}\right)}$ and specified final state ${\left(x_{N},t_{N}\right)}$. Here we’ll show that the path integral approach is formally equivalent to the Schrödinger equation, even for an arbitrary potential ${V}$.

The Schrödinger equation is a differential equation that allows us to calculate the wave function as a function of position ${x}$ and time ${t}$, when solved in the position basis. To find the same thing from the path integral, we’ll consider an infinitesimal time interval ${\varepsilon}$ and try to find ${\psi\left(x,\varepsilon\right)}$ given the wave function at ${t=0}$, that is, given ${\psi\left(x^{\prime},0\right)}$ for some arbitrary ${x^{\prime}}$. To use a path integral in this way, we’re effectively asking for the contribution to the propagator from all possible paths between ${t=0}$ and ${t=\varepsilon}$. That is, we’re considering that the particle may have started at any position ${x^{\prime}}$ at ${t=0}$ and stilll ended up at position ${x}$ at ${t=\varepsilon}$. In terms of the propagator, this is

$\displaystyle \psi\left(x,\varepsilon\right)=\int_{-\infty}^{\infty}U\left(x,\varepsilon;x^{\prime},0\right)\psi\left(x^{\prime},0\right)dx^{\prime} \ \ \ \ \ (1)$

Looking at our previous derivation of the propagator, we saw that there we fixed the initial and final states and integrated over all possible paths between these two states. In this case, all we’re specifying is the final state so in principle, the particle could have been anywhere at ${t=0}$.

The general form for the propagator is

$\displaystyle U\left(t\right)=A\int_{all\;paths}e^{iS\left[x\left(t\right)\right]/\hbar} \ \ \ \ \ (2)$

where ${A}$ is a scale factor and ${S\left[x\left(t\right)\right]}$ is the action for travelling along path ${x\left(t\right)}$:

$\displaystyle S=\int_{0}^{\varepsilon}L\;dt \ \ \ \ \ (3)$

We can approximate the action by taking the Lagrangian to be

 $\displaystyle L$ $\displaystyle =$ $\displaystyle T-V\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}mv^{2}-V\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}m\frac{\left(x-x^{\prime}\right)^{2}}{\varepsilon^{2}}-V\left(\frac{x+x^{\prime}}{2},0\right) \ \ \ \ \ (6)$

Here we take the velocity over the interval ${\varepsilon}$ to be constant at ${v=\frac{x-x^{\prime}}{\varepsilon}}$, and we take the potential to be constant, with its value at the midpoint between ${x}$ and ${x^{\prime}}$ at time ${t=0}$. The reason we can approximate ${V}$ by its value at ${t=0}$ is that in calculating the action 3, we will multiply ${L}$ by ${\varepsilon}$, and we’re interested only in terms of first order in ${\varepsilon}$. The action to this order is then

$\displaystyle S=\varepsilon L=\frac{1}{2}m\frac{\left(x-x^{\prime}\right)^{2}}{\varepsilon}-\varepsilon V\left(\frac{x+x^{\prime}}{2},0\right) \ \ \ \ \ (7)$

which gives a propagator of

$\displaystyle U\left(x,\varepsilon;x^{\prime},0\right)=A\exp\left[\frac{i}{\hbar}\left(\frac{1}{2}m\frac{\left(x-x^{\prime}\right)^{2}}{\varepsilon}-\varepsilon V\left(\frac{x+x^{\prime}}{2},0\right)\right)\right] \ \ \ \ \ (8)$

We can try the same value for ${A}$ that we had for the free particle

$\displaystyle A=\left(\frac{m}{2\pi\hbar\varepsilon i}\right)^{N/2} \ \ \ \ \ (9)$

In this case, we have only one step so ${N=1}$ and

$\displaystyle U\left(x,\varepsilon;x^{\prime},0\right)=\sqrt{\frac{m}{2\pi\hbar\varepsilon i}}\exp\left[\frac{i}{\hbar}\left(\frac{1}{2}m\frac{\left(x-x^{\prime}\right)^{2}}{\varepsilon}-\varepsilon V\left(\frac{x+x^{\prime}}{2},0\right)\right)\right] \ \ \ \ \ (10)$

We now need to do some approximating. The kinetic energy term is

$\displaystyle \exp\left[\frac{i}{\hbar}\left(\frac{1}{2}m\frac{\left(x-x^{\prime}\right)^{2}}{\varepsilon}\right)\right] \ \ \ \ \ (11)$

The exponent is pure imaginary so for infinitesimal ${\varepsilon}$, it oscillates very rapidly away from the stationary point at ${x=x^{\prime}}$. When this term is placed in the integral 1, it multiplies ${\psi\left(x^{\prime},0\right)}$ which we’ll assume is a smooth function that doesn’t oscillate much, at least over the scale at which 11 oscillates. We define

$\displaystyle \eta\equiv x^{\prime}-x \ \ \ \ \ (12)$

to be the distance from the minimum phase. Once the phase approaches ${\pi}$, the oscillations will be rapid enough that the contributions to the integral effectively cancel out, so we’re looking at the region

$\displaystyle \frac{m\eta^{2}}{2\hbar\varepsilon}\lesssim\pi \ \ \ \ \ (13)$

or

$\displaystyle \left|\eta\right|\lesssim\sqrt{\frac{2\hbar\varepsilon\pi}{m}} \ \ \ \ \ (14)$

If we work to first order in ${\varepsilon}$ we therefore must retain terms up to second order in ${\eta}$. In terms of ${\eta}$, 1 now becomes

$\displaystyle \psi\left(x,\varepsilon\right)=\sqrt{\frac{m}{2\pi\hbar\varepsilon i}}\int_{-\infty}^{\infty}\exp\left(\frac{im\eta^{2}}{2\hbar\varepsilon}\right)\exp\left(-\frac{i\varepsilon}{\hbar}V\left(x+\frac{\eta}{2},0\right)\right)\psi\left(x+\eta,0\right)d\eta \ \ \ \ \ (15)$

We now expand the last two factors as a Taylor series in ${\eta}$ and ${\varepsilon}$ up to first order in ${\varepsilon}$ or second order in ${\eta}$:

 $\displaystyle \exp\left(-\frac{i\varepsilon}{\hbar}V\left(x+\frac{\eta}{2},0\right)\right)$ $\displaystyle =$ $\displaystyle 1-\frac{i\varepsilon}{\hbar}V\left(x+\frac{\eta}{2},0\right)+\ldots\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1-\frac{i\varepsilon}{\hbar}V\left(x,0\right)+\ldots \ \ \ \ \ (17)$

We can drop terms in the expansion of ${V\left(x+\frac{\eta}{2},0\right)}$ beyond ${V\left(x,0\right)}$ since they will be of order ${\mathcal{O}\left(\varepsilon\eta\right)=\mathcal{O}\left(\varepsilon^{3/2}\right)}$ or higher.

For the second term, we have

$\displaystyle \psi\left(x+\eta,0\right)=\psi\left(x,0\right)+\eta\frac{\partial\psi}{\partial x}+\frac{\eta^{2}}{2}\frac{\partial^{2}\psi}{\partial x^{2}}+\ldots \ \ \ \ \ (18)$

where the partial derivatives are evaluated at ${\eta=0}$.

Inserting these into the integral 15 we get

 $\displaystyle \psi\left(x,\varepsilon\right)$ $\displaystyle =$ $\displaystyle \sqrt{\frac{m}{2\pi\hbar\varepsilon i}}\int_{-\infty}^{\infty}\exp\left(\frac{im\eta^{2}}{2\hbar\varepsilon}\right)\times\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle$ $\displaystyle \left[\psi\left(x,0\right)+\eta\frac{\partial\psi}{\partial x}+\frac{\eta^{2}}{2}\frac{\partial^{2}\psi}{\partial x^{2}}\right]\times\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle$ $\displaystyle \left[1-\frac{i\varepsilon}{\hbar}V\left(x,0\right)\right]d\eta \ \ \ \ \ (21)$

Again, retaining only terms up to first order in ${\varepsilon}$ or second order in ${\eta}$:

 $\displaystyle \psi\left(x,\varepsilon\right)$ $\displaystyle =$ $\displaystyle \sqrt{\frac{m}{2\pi\hbar\varepsilon i}}\int_{-\infty}^{\infty}\exp\left(\frac{im\eta^{2}}{2\hbar\varepsilon}\right)\times\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle$ $\displaystyle \left[\psi\left(x,0\right)-\frac{i\varepsilon}{\hbar}V\left(x,0\right)\psi\left(x,0\right)+\eta\frac{\partial\psi}{\partial x}+\frac{\eta^{2}}{2}\frac{\partial^{2}\psi}{\partial x^{2}}\right]d\eta \ \ \ \ \ (23)$

Everything in the integrand is constant with respect to ${\eta}$ except for the first exponential and the factors of ${\eta}$ and ${\eta^{2}}$ in the last two terms. We are therefore faced with a couple of Gaussian integrals. We have

 $\displaystyle \int_{-\infty}^{\infty}\exp\left(\frac{im\eta^{2}}{2\hbar\varepsilon}\right)d\eta$ $\displaystyle =$ $\displaystyle \int_{-\infty}^{\infty}\exp\left(-\frac{m\eta^{2}}{2\hbar i\varepsilon}\right)d\eta\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{2\pi\hbar i\varepsilon}{m}}\ \ \ \ \ (25)$ $\displaystyle \int_{-\infty}^{\infty}\eta\exp\left(\frac{im\eta^{2}}{2\hbar\varepsilon}\right)d\eta$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (26)$ $\displaystyle \int_{-\infty}^{\infty}\eta^{2}\exp\left(\frac{im\eta^{2}}{2\hbar\varepsilon}\right)d\eta$ $\displaystyle =$ $\displaystyle \int_{-\infty}^{\infty}\eta^{2}\exp\left(-\frac{m\eta^{2}}{2\hbar i\varepsilon}\right)d\eta\ \ \ \ \ (27)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\hbar i\varepsilon}{m}\sqrt{\frac{2\pi\hbar i\varepsilon}{m}}\ \ \ \ \ (28)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{\hbar\varepsilon}{im}\sqrt{\frac{2\pi\hbar i\varepsilon}{m}} \ \ \ \ \ (29)$

Putting it all together, we have

 $\displaystyle \psi\left(x,\varepsilon\right)$ $\displaystyle =$ $\displaystyle \sqrt{\frac{m}{2\pi\hbar\varepsilon i}}\sqrt{\frac{2\pi\hbar i\varepsilon}{m}}\left[\left(1-\frac{i\varepsilon}{\hbar}V\left(x,0\right)\right)-\frac{\hbar\varepsilon}{2im}\frac{\partial^{2}}{\partial x^{2}}\right]\psi\left(x,0\right)\ \ \ \ \ (30)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \psi\left(x,0\right)-\frac{i\varepsilon}{\hbar}\left(-\frac{\hbar^{2}}{2m}\frac{\partial^{2}}{\partial x^{2}}+V\left(x,0\right)\right)\psi\left(x,0\right) \ \ \ \ \ (31)$

Rearranging, we get

$\displaystyle i\hbar\frac{\psi\left(x,\varepsilon\right)-\psi\left(x,0\right)}{\varepsilon}=\left(-\frac{\hbar^{2}}{2m}\frac{\partial^{2}}{\partial x^{2}}+V\left(x,0\right)\right)\psi\left(x,0\right) \ \ \ \ \ (32)$

In the limit ${\varepsilon\rightarrow0}$, the LHS becomes ${i\hbar\frac{\partial\psi}{\partial t}}$ and we get the Schrödinger equation:

$\displaystyle i\hbar\frac{\partial\psi}{\partial t}=\left(-\frac{\hbar^{2}}{2m}\frac{\partial^{2}}{\partial x^{2}}+V\left(x,0\right)\right)\psi\left(x,0\right) \ \ \ \ \ (33)$

# Free particle propagator from a complete path integral

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 8. Section 8.4.

We’ve seen that the free-particle propagator can be obtained in the path integral approach by using only the classical path in the sum over paths. It turns out that it’s not too hard to calculate the propagator for a free particle properly, by summing over all possible paths. The notation used by Shankar is as follows.

We want to evaluate the path integral

$\displaystyle \int_{x_{0}}^{x_{N}}e^{iS\left[x\left(t\right)\right]/\hbar}\mathfrak{D}\left[x\left(t\right)\right] \ \ \ \ \ (1)$

The notation ${\mathfrak{D}\left[x\left(t\right)\right]}$ means an integration over all possible paths from ${x_{0}}$ to ${x_{N}}$ in the given time interval. This includes paths where the particle might move to the right for a while, then jog back to the left, then back to the right again and so on. This might seem like a hopeless task, but we can make sense of this method by splitting the time interval between ${t_{0}}$ and ${t_{N}}$ into ${N}$ small intervals, each of length ${\varepsilon}$. Thus an intermediate time ${t_{n}=t_{0}+n\varepsilon}$, and the final time is ${t_{N}=t_{0}+N\varepsilon}$.

For a free particle, there is no potential energy so the Lagrangian is just the kinetic energy:

$\displaystyle L=\frac{1}{2}m\dot{x}^{2} \ \ \ \ \ (2)$

We can estimate the velocity in each time slice by

$\displaystyle \dot{x}_{i}=\frac{x_{i+1}-x_{i}}{\varepsilon} \ \ \ \ \ (3)$

Note that this assumes that the velocity within each time slice is constant, but as we make ${\varepsilon}$ smaller and smaller, this is increasingly accurate. Also note that it is possible for ${\dot{x}_{i}}$ to be both positive (if the particle moves to the right in the interval) or negative (if it moves to the left).

The action for a given path is given by the integral of the Lagrangian:

$\displaystyle S=\int_{t_{0}}^{t_{N}}L\left(t\right)dt \ \ \ \ \ (4)$

In our discretized approximation, we evaluate ${L}$ within each time slice, and ${dt}$ becomes the interval length ${\varepsilon}$, so the action becomes a sum:

 $\displaystyle S$ $\displaystyle =$ $\displaystyle \sum_{i=0}^{N-1}L\left(t_{i}\right)\varepsilon\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{m}{2}\sum_{i=0}^{N-1}\left(\frac{x_{i+1}-x_{i}}{\varepsilon}\right)^{2}\varepsilon\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{m}{2}\sum_{i=0}^{N-1}\frac{\left(x_{i+1}-x_{i}\right)^{2}}{\varepsilon} \ \ \ \ \ (7)$

The key point here is to notice that we can label any given path by choosing values for all the ${x_{i}}$s between the two times, and that each ${x_{i}}$ can vary independently of the others, over a range from ${-\infty}$ to ${+\infty}$. We can therefore implement the multiple integration required by ${\mathfrak{D}\left[x\left(t\right)\right]}$ by integrating over all the ${x_{i}}$ variables separately. That is,

$\displaystyle \int_{x_{0}}^{x_{N}}e^{iS\left[x\left(t\right)\right]/\hbar}\mathfrak{D}\left[x\left(t\right)\right]=A\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\ldots\int_{-\infty}^{\infty}\exp\left[\frac{im}{2\hbar}\sum_{i=0}^{N-1}\frac{\left(x_{i+1}-x_{i}\right)^{2}}{\varepsilon}\right]dx_{1}dx_{2}\ldots dx_{N-1} \ \ \ \ \ (8)$

where ${A}$ is some constant to make the scale come out right.

We don’t integrate over ${x_{0}}$ or ${x_{N}}$ since these are fixed as the end points of the path. To get the final version, we need to take the limit of this expression as ${N\rightarrow\infty}$ and ${\varepsilon\rightarrow0}$. This still looks pretty scary, but in fact it is doable. We define the variable

 $\displaystyle y_{i}$ $\displaystyle \equiv$ $\displaystyle \sqrt{\frac{m}{2\hbar\varepsilon}}x_{i}\ \ \ \ \ (9)$ $\displaystyle dx_{i}$ $\displaystyle =$ $\displaystyle \sqrt{\frac{2\hbar\varepsilon}{m}}dy_{i} \ \ \ \ \ (10)$

This gives us

$\displaystyle A\left(\frac{2\hbar\varepsilon}{m}\right)^{\left(N-1\right)/2}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\ldots\int_{-\infty}^{\infty}\exp\left[i\sum_{i=0}^{N-1}\left(y_{i+1}-y_{i}\right)^{2}\right]dy_{1}dy_{2}\ldots dy_{N-1} \ \ \ \ \ (11)$

We can do the integral in stages in order to spot a pattern. Consider first the integral over ${y_{1}}$, which involves only two of the factors in the integrand:

$\displaystyle \int_{-\infty}^{\infty}e^{i\left[\left(y_{1}-y_{0}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}\right]}dy_{1} \ \ \ \ \ (12)$

We first simplify the exponent

 $\displaystyle \left(y_{1}-y_{0}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}$ $\displaystyle =$ $\displaystyle y_{2}^{2}+y_{0}^{2}+2\left(y_{1}^{2}-y_{0}y_{1}-y_{1}y_{2}\right)\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle y_{2}^{2}+y_{0}^{2}+2y_{1}^{2}-2\left(y_{0}+y_{2}\right)y_{1} \ \ \ \ \ (14)$

We get

$\displaystyle \int_{-\infty}^{\infty}e^{i\left[\left(y_{1}-y_{0}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}\right]}dy_{1}=e^{i\left(y_{2}^{2}+y_{0}^{2}\right)}\int_{-\infty}^{\infty}e^{2i\left[y_{1}^{2}-\left(y_{0}+y_{2}\right)y_{1}\right]}dy_{1} \ \ \ \ \ (15)$

We can evaluate this using a standard Gaussian integral

$\displaystyle \int_{-\infty}^{\infty}e^{-ax^{2}+bx}dx=e^{b^{2}/4a}\sqrt{\frac{\pi}{a}} \ \ \ \ \ (16)$

This gives

 $\displaystyle \int_{-\infty}^{\infty}e^{i\left[\left(y_{1}-y_{0}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}\right]}dy_{1}$ $\displaystyle =$ $\displaystyle e^{i\left(y_{2}^{2}+y_{0}^{2}\right)}e^{4\left(y_{0}+y_{2}\right)^{2}/8i}\sqrt{-\frac{\pi}{2i}}\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle e^{i\left(y_{2}^{2}+y_{0}^{2}\right)}e^{\left(y_{0}+y_{2}\right)^{2}/2i}\sqrt{\frac{\pi i}{2}} \ \ \ \ \ (18)$

To simplify the exponents on the RHS:

 $\displaystyle i\left(y_{2}^{2}+y_{0}^{2}\right)+\frac{\left(y_{0}+y_{2}\right)^{2}}{2i}$ $\displaystyle =$ $\displaystyle \frac{1}{2i}\left[\left(y_{0}+y_{2}\right)^{2}-2y_{2}^{2}-2y_{0}^{2}\right]\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{1}{2i}\left(y_{0}-y_{2}\right)^{2} \ \ \ \ \ (20)$

Thus we have

$\displaystyle \int_{-\infty}^{\infty}e^{i\left[\left(y_{1}-y_{0}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}\right]}dy_{1}=\sqrt{\frac{\pi i}{2}}e^{-\left(y_{0}-y_{2}\right)^{2}/2i} \ \ \ \ \ (21)$

Having eliminated ${y_{1}}$ we can now do the integral over ${y_{2}}$:

$\displaystyle \sqrt{\frac{\pi i}{2}}\int_{-\infty}^{\infty}e^{-\left(y_{3}-y_{2}\right)^{2}/i-\left(y_{2}-y_{0}\right)^{2}/2i}dy_{2} \ \ \ \ \ (22)$

Again, we can simplify the exponent:

$\displaystyle -\frac{\left(y_{3}-y_{2}\right)^{2}}{i}-\frac{\left(y_{2}-y_{0}\right)^{2}}{2i}=\frac{1}{2i}\left[-\left(2y_{3}^{2}+y_{0}^{2}\right)-3y_{2}^{2}+y_{2}\left(4y_{3}+2y_{0}\right)\right] \ \ \ \ \ (23)$

The integral now becomes

 $\displaystyle \sqrt{\frac{\pi i}{2}}\int_{-\infty}^{\infty}e^{-\left(y_{3}-y_{2}\right)^{2}/i-\left(y_{2}-y_{0}\right)^{2}/2i}dy_{2}$ $\displaystyle =$ $\displaystyle \sqrt{\frac{\pi i}{2}}e^{-\left(2y_{3}^{2}+y_{0}^{2}\right)/2i}\int_{-\infty}^{\infty}e^{\left(-3y_{2}^{2}+y_{2}\left(4y_{3}+2y_{0}\right)\right)/2i}dy_{2} \ \ \ \ \ (24)$

Doing the Gaussian integral on the RHS using 16:

 $\displaystyle \int_{-\infty}^{\infty}e^{\left(-3y_{2}^{2}+y_{2}\left(4y_{3}+2y_{0}\right)\right)/2i}dy_{2}$ $\displaystyle =$ $\displaystyle e^{-\left(4y_{3}+2y_{0}\right)^{2}i/24}\sqrt{\frac{2\pi i}{3}}\ \ \ \ \ (25)$ $\displaystyle$ $\displaystyle =$ $\displaystyle e^{\left(2y_{3}+y_{0}\right)^{2}/6i}\sqrt{\frac{2\pi i}{3}} \ \ \ \ \ (26)$

Thus the combined integral over ${y_{1}}$ and ${y_{2}}$ is

 $\displaystyle \sqrt{\frac{\pi i}{2}}e^{-\left(2y_{3}^{2}+y_{0}^{2}\right)/2i}e^{\left(2y_{3}+y_{0}\right)^{2}/6i}\sqrt{\frac{2\pi i}{3}}$ $\displaystyle =$ $\displaystyle \sqrt{\frac{\left(\pi i\right)^{2}}{3}}e^{\left(-6y_{3}^{2}-3y_{0}^{2}+\left(2y_{3}+y_{0}\right)^{2}\right)/6i}\ \ \ \ \ (27)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{\left(\pi i\right)^{2}}{3}}e^{-\left(y_{3}-y_{0}\right)^{2}/3i} \ \ \ \ \ (28)$

The general pattern after ${N-1}$ integrations is (presumably this could be proved by induction, but we’ll accept the result):

$\displaystyle \frac{\left(\pi i\right)^{\left(N-1\right)/2}}{\sqrt{N}}e^{-\left(y_{N}-y_{0}\right)^{2}/Ni}=\frac{\left(\pi i\right)^{\left(N-1\right)/2}}{\sqrt{N}}e^{-m\left(x_{N}-x_{0}\right)^{2}/2\hbar\varepsilon Ni} \ \ \ \ \ (29)$

where we reverted back to ${x_{i}}$ using 9.

Going back to 11, we must multiply the result by ${A\left(\frac{2\hbar\varepsilon}{m}\right)^{\left(N-1\right)/2}}$ to get the final expression for the propagator:

 $\displaystyle U$ $\displaystyle =$ $\displaystyle A\left(\frac{2\hbar\varepsilon}{m}\right)^{\left(N-1\right)/2}\frac{\left(\pi i\right)^{\left(N-1\right)/2}}{\sqrt{N}}e^{-m\left(x_{N}-x_{0}\right)^{2}/2\hbar\varepsilon Ni}\ \ \ \ \ (30)$ $\displaystyle$ $\displaystyle =$ $\displaystyle A\left(\frac{2\pi\hbar\varepsilon i}{m}\right)^{N/2}\sqrt{\frac{m}{2\pi\hbar iN\varepsilon}}e^{im\left(x_{N}-x_{0}\right)^{2}/2\hbar\varepsilon N} \ \ \ \ \ (31)$

In the limit as ${N\rightarrow\infty}$ and ${\varepsilon\rightarrow0}$, ${N\varepsilon=t_{N}-t_{0}}$ so we have

$\displaystyle U=A\left(\frac{2\pi\hbar\varepsilon i}{m}\right)^{N/2}\sqrt{\frac{m}{2\pi\hbar i\left(t_{N}-t_{0}\right)}}e^{im\left(x_{N}-x_{0}\right)^{2}/2\hbar\left(t_{N}-t_{0}\right)} \ \ \ \ \ (32)$

The expression we got earlier using the Schrödinger method is

$\displaystyle U\left(x,t;x^{\prime},t^{\prime}\right)=\sqrt{\frac{m}{2\pi\hbar i\left(t-t^{\prime}\right)}}e^{im\left(x-x^{\prime}\right)^{2}/2\hbar\left(t-t^{\prime}\right)} \ \ \ \ \ (33)$

Thus the full path integral gives the same result, with ${t^{\prime}=t_{0}}$ and ${t=t_{N}}$ (similarly for ${x}$), provided that we can set

$\displaystyle A=\left(\frac{m}{2\pi\hbar\varepsilon i}\right)^{N/2}\equiv B^{-N} \ \ \ \ \ (34)$

Shankar then says that it is conventional to associate one factor of ${B^{-1}}$ with each integration over an ${x_{i}}$, and the remaining factor with the overall process. This seems to overlook a basic problem, in that as ${N\rightarrow\infty}$ and ${\varepsilon\rightarrow0}$, ${A\rightarrow\infty}$, so we seem to be cancelling two infinities when we multiply the path integral by ${A}$.

# Path integral formulation of quantum mechanics: free particle propagator

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 8.

Although all the non-relativistic quantum mechanics we’ve done so far has started with the Schrödinger equation, a different approach was devised by Richard Feynman in the 1940s. The Schrödinger method requires us to find the eigenvalues (allowed energies) and eigenstates of the hamiltonian ${H}$ and then use these to construct the unitary operator known as the propagator. For discrete energies, this propagator is

$\displaystyle U\left(t\right)=\sum e^{-iEt/\hbar}\left|E\right\rangle \left\langle E\right| \ \ \ \ \ (1)$

and for continuous energies, we have

$\displaystyle U\left(t\right)=\int e^{-iEt/\hbar}\left|E\right\rangle \left\langle E\right|dE \ \ \ \ \ (2)$

Given the state of the system at an initial time ${t=0}$, the general solution as a function of time is then

$\displaystyle \left|\psi\left(t\right)\right\rangle =U\left(t\right)\left|\psi\left(0\right)\right\rangle \ \ \ \ \ (3)$

Feynman’s method allows us to compute the propagator directly, without first solving the Schrödinger equation. It is known as the path integral forumulation.

The idea is based on the observation that the exponential ${e^{-iEt/\hbar}}$ that appears in the propagator contains the ratio of two quantities with the dimensions of action, that is, energy times time. In classical mechanics, the actual trajectory of a particle is found by minimizing the action ${S}$ over all possible paths available to the particle. The path integral formulation of quantum mechanics works in a similar way, although at first sight, it looks like a completely impractical method.

The formulation works like this, for a single particle:

1. Find all paths available for the particle to travel between its initial point ${\left(x^{\prime},t^{\prime}\right)}$ and its final point ${\left(x,t\right)}$. This is actually similar to what we do in classical mechanics, where ${S}$ is defined as ${S=\int L\;dt}$ where ${L}$ is the Lagrangian. We then use the functional derivative to minimize ${S}$ over all these paths and find the path that gives the minimum action.
2. For each path, calculate the action ${S}$. (This is where things sound terribly impractical, since there are an infinite number of paths of all possible shape, so how can we find the action for all these paths? It turns out that, in most cases, we don’t need to.)
3. Calculate the propagator as

$\displaystyle U\left(x,t;x^{\prime},t^{\prime}\right)=A\sum_{all\;paths}e^{iS\left[x\left(t\right)\right]/\hbar} \ \ \ \ \ (4)$

The notation ${S\left[x\left(t\right)\right]}$ indicates that ${S}$ is a functional of the path ${x\left(t\right)}$.

The key to the success of this method is that since the action is real, the exponential ${e^{iS\left[x\left(t\right)\right]/\hbar}}$ is an oscillatory function, so we can expect contributions from the actions for different paths to cancel each other to some extent. Although the quantum path of a particle can’t be defined precisely due to the uncertainty principle, we expect that the particle is much more likely to be found following a path that is close to the classical path, and the classical path occurs when ${S\left[x\left(t\right)\right]}$ is a minimum. Paths sufficiently far from this minimum will tend to cancel each other, so for practical purposes, we need calculate 4 only for paths near to the classical path.

The example given by Shankar is of a particle of mass 1 gram moving from ${\left(x,t\right)=\left(0,0\right)}$ to ${\left(1,1\right)}$ by two different paths. In the first path, the particle moves with constant speed so ${x=t}$. The action is

 $\displaystyle S$ $\displaystyle =$ $\displaystyle \int_{0}^{1}L\;dt\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int_{0}^{1}\left(T-V\right)dt\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int_{0}^{1}\frac{1}{2}mv^{2}dt\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{m}{2}\int_{0}^{1}\left(\frac{dx}{dt}\right)^{2}dt\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{m}{2}\int_{0}^{1}dt\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{m}{2} \ \ \ \ \ (10)$

In the second path, we have ${x=t^{2}}$, so the velocity is

$\displaystyle v=\frac{dx}{dt}=2t \ \ \ \ \ (11)$

with associated action

 $\displaystyle S$ $\displaystyle =$ $\displaystyle \int_{0}^{1}\frac{1}{2}mv^{2}dt\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2m\int_{0}^{1}t^{2}dt\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2m}{3} \ \ \ \ \ (14)$

The guideline for when the phases of the paths start to cancel each other is when ${S/\hbar}$ is about ${\pi}$ out of phase with ${S_{cl}/\hbar}$. In this example, the second path is ${\pi}$out of phase with the first when

$\displaystyle \left(\frac{2m}{3}-\frac{m}{2}\right)=\pi\hbar\approx3\times10^{-34}\mbox{ m}^{2}\mbox{kg s}^{-1} \ \ \ \ \ (15)$

Thus for any mass larger than about ${6\pi\hbar\approx1.8\times10^{-33}\mbox{ kg}}$ the second path will contribute essentially nothing to 4 and can be ignored. This mass is smaller than the mass of the electron.

For the free particle, we worked out the propagator earlier and found that (where we’ve generalized the earlier result for an arbitrary initial time ${t^{\prime}}$):

$\displaystyle U\left(t,t^{\prime}\right)=\int_{-\infty}^{\infty}e^{-ip^{2}\left(t-t^{\prime}\right)/2m\hbar}\left|p\right\rangle \left\langle p\right|dp \ \ \ \ \ (16)$

The matrix elements of ${U}$ in the ${x}$ basis are worked out by evaluating a Gaussian integral

 $\displaystyle U\left(x,t;x^{\prime},t^{\prime}\right)$ $\displaystyle =$ $\displaystyle \int_{-\infty}^{\infty}e^{-ip^{2}\left(t-t^{\prime}\right)/2m\hbar}\left\langle x\left|p\right.\right\rangle \left\langle p\left|x^{\prime}\right.\right\rangle dp\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2\pi\hbar}\int_{-\infty}^{\infty}e^{ip\left(x-x^{\prime}\right)/\hbar}e^{-ip^{2}\left(t-t^{\prime}\right)/2m\hbar}dp\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{m}{2\pi\hbar i\left(t-t^{\prime}\right)}}e^{im\left(x-x^{\prime}\right)^{2}/2\hbar\left(t-t^{\prime}\right)} \ \ \ \ \ (19)$

We can try to estimate ${U}$ using the path integral approach by assuming that only the classical path contributes to the propagator. For a free particle travelling between ${\left(x^{\prime},t^{\prime}\right)}$ to ${\left(x,t\right)}$, the constant velocity is

$\displaystyle v=\frac{x-x^{\prime}}{t-t^{\prime}} \ \ \ \ \ (20)$

The Lagrangian is a constant

$\displaystyle L=\frac{mv^{2}}{2}=\frac{m}{2}\left(\frac{x-x^{\prime}}{t-t^{\prime}}\right)^{2} \ \ \ \ \ (21)$

The classical action is thus

 $\displaystyle S_{cl}$ $\displaystyle =$ $\displaystyle \int_{t^{\prime}}^{t}L\;dt^{\prime\prime}\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{m}{2}\left(\frac{x-x^{\prime}}{t-t^{\prime}}\right)^{2}\int_{t^{\prime}}^{t}dt^{\prime\prime}\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{m}{2}\left(\frac{x-x^{\prime}}{t-t^{\prime}}\right)^{2}\left(t-t^{\prime}\right)\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{m}{2}\frac{\left(x-x^{\prime}\right)^{2}}{t-t^{\prime}} \ \ \ \ \ (25)$

The propagator in this approximation is

$\displaystyle U\left(x,t;x^{\prime},t^{\prime}\right)=A\exp\left[\frac{im}{2\hbar}\frac{\left(x-x^{\prime}\right)^{2}}{\left(t-t^{\prime}\right)}\right] \ \ \ \ \ (26)$

Comparing with 19 we see that the exponential factors match; all that is left is to determine the constant ${A}$. To do this, we require ${\lim_{t\rightarrow t^{\prime}}U\left(x,t;x^{\prime},t^{\prime}\right)=\delta\left(x-x^{\prime}\right)}$, since if the time interval ${t^{\prime}-t}$ goes to zero, the particle cannot move so must be in the same place. By comparing 26 with the form of a delta function as the limit of a gaussian integral, which is

$\displaystyle \lim_{\Delta^{2}\rightarrow0}\frac{1}{\left(\pi\Delta^{2}\right)^{1/2}}\int_{-\infty}^{\infty}e^{-\left(x-x^{\prime}\right)/\Delta^{2}}dx=\delta\left(x-x^{\prime}\right) \ \ \ \ \ (27)$

we see that

$\displaystyle \Delta^{2}=\frac{2\hbar i\left(t-t^{\prime}\right)}{m} \ \ \ \ \ (28)$

so the final propagator is the same as 19.

# Dirac delta function as limit of a gaussian integral

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 1.10.

Yet another form of the Dirac delta function is as the limit of a Gaussian integral. We start with

$\displaystyle g_{\Delta}\left(x-x^{\prime}\right)=\frac{1}{\left(\pi\Delta^{2}\right)^{1/2}}e^{-\left(x-x^{\prime}\right)^{2}/\Delta^{2}} \ \ \ \ \ (1)$

If ${\Delta^{2}}$ is real and positive, we have

$\displaystyle \frac{1}{\left(\pi\Delta^{2}\right)^{1/2}}\int_{-\infty}^{\infty}e^{-\left(x-x^{\prime}\right)^{2}/\Delta^{2}}dx=1 \ \ \ \ \ (2)$

Thus the area under the curve is always 1, for any real value of ${\Delta^{2}}$. Now as ${\Delta^{2}\rightarrow0}$ the exponential becomes zero except when ${x=x^{\prime}}$. The factor ${1/\left(\pi\Delta^{2}\right)^{1/2}}$ tends to infinity as ${\Delta^{2}\rightarrow0}$, but the exponential always tends to zero faster than any power of ${\Delta}$, so ${g_{\Delta}\left(x-x^{\prime}\right)}$ tends to zero everywhere except at ${x=x^{\prime}}$. Thus it satisfies the requirements of a delta function: it is zero everywhere except when ${x-x^{\prime}=0}$ and has an integral of 1. Thus

$\displaystyle \lim_{\Delta\rightarrow0}g_{\Delta}\left(x-x^{\prime}\right)=\delta\left(x-x^{\prime}\right) \ \ \ \ \ (3)$

However, if we plug the integral into Maple without any restrictions on ${\Delta^{2}}$, it informs us that the integral is still 1 even if ${\Delta^{2}}$ is pure imaginary, provided that the imaginary number is positive, that is, we can write ${\Delta^{2}=i\beta^{2}}$ for real ${\beta}$. Thus it would appear that ${g_{\Delta}}$ still gives a delta function in the limit ${\Delta^{2}\rightarrow0}$ even if ${\Delta^{2}}$ is a positive imaginary number.

Shankar provides a rationale for this in his footnote to equation 1.10.19. In terms of ${\beta}$ we can integrate some smooth function ${f\left(x^{\prime}\right)}$ multiplied by ${g_{\Delta}}$ over a region that includes ${x^{\prime}=x}$.

$\displaystyle \frac{1}{\left(\pi i\beta^{2}\right)^{1/2}}\int_{-\infty}^{\infty}e^{i\left(x-x^{\prime}\right)^{2}/\beta^{2}}f\left(x^{\prime}\right)dx \ \ \ \ \ (4)$

As ${\beta^{2}\rightarrow0}$, the exponent becomes a very large positive imaginary number everywhere except at ${x=x^{\prime}}$, so the exponential oscillates very rapidly. Provided that ${f\left(x^{\prime}\right)}$ doesn’t vary as rapidly, the integral will contain equal positive and negative contributions everywhere except at ${x=x^{\prime}}$ so in the limit of ${\beta^{2}=0}$, only the point ${x=x^{\prime}}$ contributes, which means we can pull ${f\left(x\right)}$ out of the integral and get

$\displaystyle \lim_{\beta^{2}\rightarrow0}\frac{1}{\left(\pi i\beta^{2}\right)^{1/2}}\int_{-\infty}^{\infty}e^{i\left(x-x^{\prime}\right)^{2}/\beta^{2}}f\left(x^{\prime}\right)dx=f\left(x\right) \ \ \ \ \ (5)$

Thus 3 is valid for all real ${\Delta}$ and for ${\Delta^{2}}$ positive imaginary.

# Thermodynamics of harmonic oscillators – classical and quantum

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 7.5, Exercise 7.5.4.

One application of harmonic oscillator theory is in the behaviour of crystals as a function of temperature. A reasonable model of a crystal is of a number of atoms that vibrate as harmonic oscillators. From statistical mechanics, the probability ${P\left(i\right)}$ of finding a system in a state ${i}$ is given by the Boltzmann formula

$\displaystyle P\left(i\right)=\frac{e^{-\beta E\left(i\right)}}{Z} \ \ \ \ \ (1)$

where ${\beta=1/kT}$, with ${k}$ being Boltzmann’s constant and ${T}$ the absolute temperature, and ${Z}$ is the partition function

$\displaystyle Z=\sum_{i}e^{-\beta E\left(i\right)} \ \ \ \ \ (2)$

The thermal average energy of the system is then

 $\displaystyle \bar{E}$ $\displaystyle =$ $\displaystyle \sum_{i}E\left(i\right)P\left(i\right)\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\sum_{i}E\left(i\right)e^{-\beta E\left(i\right)}}{Z}\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{\partial\left(\ln Z\right)}{\partial\beta} \ \ \ \ \ (5)$

For a classical harmonic oscillator, the energy is a continuous function of the position ${x}$ and momentum ${p}$:

$\displaystyle E_{cl}=\frac{p^{2}}{2m}+\frac{1}{2}m\omega^{2}x^{2} \ \ \ \ \ (6)$

The classical partition function is then

 $\displaystyle Z_{cl}$ $\displaystyle =$ $\displaystyle \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-\beta p^{2}/2m}e^{-\beta m\omega^{2}x^{2}/2}dp\;dx\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int_{-\infty}^{\infty}e^{-\beta p^{2}/2m}dp\int_{-\infty}^{\infty}e^{-\beta m\omega^{2}x^{2}/2}dx\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{2\pi m}{\beta}}\sqrt{\frac{2\pi}{\beta m\omega^{2}}}\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2\pi}{\omega\beta} \ \ \ \ \ (10)$

Where we used the standard formula for Gaussian integrals to get the third line. The average classical energy is, from 5

$\displaystyle \bar{E}_{cl}=-\frac{\partial\left(\ln Z_{cl}\right)}{\partial\beta}=\frac{1}{\beta}=kT \ \ \ \ \ (11)$

The average energy of a classical oscillator thus depends only on the temperature, and not on the frequency ${\omega}$.

For a quantum oscillator, the energies are quantized with values of

$\displaystyle E\left(n\right)=\hbar\omega\left(n+\frac{1}{2}\right) \ \ \ \ \ (12)$

The quantum partition function is therefore

$\displaystyle Z_{qu}=e^{-\beta\hbar\omega/2}\sum_{n=0}^{\infty}e^{-\beta\hbar\omega n} \ \ \ \ \ (13)$

The sum is a geometric series, so we can use the standard result for ${\left|x\right|<1}$:

$\displaystyle \sum_{n=0}^{\infty}x^{n}=\frac{1}{1-x} \ \ \ \ \ (14)$

This gives

$\displaystyle Z_{qu}=\frac{e^{-\beta\hbar\omega/2}}{1-e^{-\beta\hbar\omega}} \ \ \ \ \ (15)$

The mean quantum energy is again found from 5, although this time the derivative is a bit messier, so is most easily done using Maple. However, by hand, you’d get

 $\displaystyle \bar{E}_{qu}$ $\displaystyle =$ $\displaystyle -\frac{\partial\left(\ln Z_{qu}\right)}{\partial\beta}\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1-e^{-\beta\hbar\omega}}{e^{-\beta\hbar\omega/2}}\left[-\frac{1}{2}\frac{\hbar\omega e^{-\beta\hbar\omega/2}}{1-e^{-\beta\hbar\omega}}-\frac{\hbar\omega e^{-\beta\hbar\omega/2}e^{-\beta\hbar\omega}}{\left(1-e^{-\beta\hbar\omega}\right)^{2}}\right]\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\hbar\omega}{2}\left(\frac{1+e^{-\beta\hbar\omega}}{1-e^{-\beta\hbar\omega}}\right)\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\hbar\omega}{2}\left(\frac{1-e^{-\beta\hbar\omega}+2e^{-\beta\hbar\omega}}{1-e^{-\beta\hbar\omega}}\right)\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \hbar\omega\left(\frac{1}{2}+\frac{1}{e^{\beta\hbar\omega}-1}\right) \ \ \ \ \ (20)$

The average energy is the ground state energy ${\hbar\omega/2}$ plus a quantity that increases with increasing temperature (decreasing ${\beta}$). For small ${\beta}$ we have

 $\displaystyle \bar{E}_{qu}$ $\displaystyle \rightarrow$ $\displaystyle \hbar\omega\left(\frac{1}{2}+\frac{1}{1+\beta\hbar\omega-1}\right)\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\hbar\omega}{2}+\frac{1}{\beta}\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle \rightarrow$ $\displaystyle kT \ \ \ \ \ (23)$

since as ${\beta\rightarrow0}$, ${\frac{1}{\beta}\gg\frac{\hbar\omega}{2}}$. Thus the quantum energy reduces to the classical energy 11 for high temperatures. The ‘high temperature’ condition is that

 $\displaystyle \frac{1}{\beta}$ $\displaystyle \gg$ $\displaystyle \frac{\hbar\omega}{2}\ \ \ \ \ (24)$ $\displaystyle T$ $\displaystyle \gg$ $\displaystyle \frac{\hbar\omega}{2k} \ \ \ \ \ (25)$

So far, we’ve considered the average behaviour of only one oscillator. Suppose we now have a 3-d crystal with ${N_{0}}$ atoms. Assuming small oscillations we can approximate its behaviour by a system of ${3N_{0}}$ decoupled oscillators. In the classical case, the average energy is found from 11:

$\displaystyle \bar{\mathcal{E}}_{cl}=3N_{0}\bar{E}_{cl}=3N_{0}kT \ \ \ \ \ (26)$

The heat capacity per atom is the amount of heat (energy) ${\Delta E}$ required to raise the temperature by ${\Delta T}$, so

$\displaystyle C_{cl}=\frac{1}{N_{0}}\frac{\partial\bar{\mathcal{E}}_{cl}}{\partial T}=3k \ \ \ \ \ (27)$

For the quantum system, we have from 20

 $\displaystyle \bar{\mathcal{E}}_{qu}$ $\displaystyle =$ $\displaystyle 3N_{0}\bar{E}_{qu}\ \ \ \ \ (28)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 3N_{0}\hbar\omega\left(\frac{1}{2}+\frac{1}{e^{\beta\hbar\omega}-1}\right) \ \ \ \ \ (29)$

The quantum heat capacity is therefore

 $\displaystyle C_{qu}$ $\displaystyle =$ $\displaystyle \frac{1}{N_{0}}\frac{\partial\bar{\mathcal{E}}_{qu}}{\partial T}\ \ \ \ \ (30)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 3\hbar\omega\frac{\partial}{\partial\beta}\left(\frac{1}{e^{\beta\hbar\omega}-1}\right)\frac{d\beta}{dT}\ \ \ \ \ (31)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 3\frac{\hbar^{2}\omega^{2}}{kT^{2}}\frac{e^{\hbar\omega/kT}}{\left(e^{\beta\hbar\omega}-1\right)^{2}} \ \ \ \ \ (32)$

We can define the Einstein temperature as

$\displaystyle \theta_{E}\equiv\frac{\hbar\omega}{k} \ \ \ \ \ (33)$

which gives the heat capacity as

$\displaystyle C_{qu}=3k\frac{\theta_{E}^{2}}{T^{2}}\frac{e^{\theta_{E}/T}}{\left(e^{\theta_{E}/T}-1\right)^{2}} \ \ \ \ \ (34)$

For large temperatures, the exponent ${\theta_{E}/T}$ becomes small, so we have

 $\displaystyle C_{qu}$ $\displaystyle \underset{T\gg\theta_{E}}{\longrightarrow}$ $\displaystyle 3k\frac{\theta_{E}^{2}}{T^{2}}\frac{1+\theta_{E}/T}{\left(1+\theta_{E}/T-1\right)^{2}}\ \ \ \ \ (35)$ $\displaystyle$ $\displaystyle \rightarrow$ $\displaystyle 3k \ \ \ \ \ (36)$

For low temperatures ${e^{\theta_{E}/T}\gg1}$ so we have

 $\displaystyle C_{qu}$ $\displaystyle \underset{T\ll\theta_{E}}{\longrightarrow}$ $\displaystyle 3k\frac{\theta_{E}^{2}}{T^{2}}\frac{e^{\theta_{E}/T}}{e^{2\theta_{E}/T}}\ \ \ \ \ (37)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 3k\frac{\theta_{E}^{2}}{T^{2}}e^{-\theta_{E}/T} \ \ \ \ \ (38)$

The heat capacity again reduces to the classical value for high temperatures. The observed behaviour at low temperatures is that ${C_{qu}\rightarrow T^{3}}$, so this simple model fails for very low temperatures. However, as is shown by Shankar’s figure 7.3 Einstein’s quantum model is actually quite good for all but the lowest temperatures.