Propagator for quantum linear potential

[Peeter]

Homework Statement

Attempting a problem related to a quantum particle in free fall due to constant force (old exam question)

$$\begin{align*}H = \frac{1}{{2m}} P^2 + m g X\end{align*}$$

The last part of the question asks to verify that the position space propagator has the following form

$$\begin{align*}\phi(x,t) &= \int_{-\infty}^\infty \phi(x', 0) G(x,x';t) dx' \\ G(x,x';t) &=\sqrt{\frac{m}{2 \pi \hbar i t}} \exp\left(\frac{im}{2 h t}\left( x - x' + gt^2/2\right)^2\right)\exp\left(-\frac{img}{h}\left( x t + gt^3/6\right)\right)\end{align*}$$

(working in the momentum representation to get that far).

Homework Equations

One is able to show

$$\begin{align*}i \hbar \frac{\partial {}}{\partial {t}}\tilde{\phi}(p,t)= (mg) i \hbar \frac{\partial {}}{\partial {p}}\tilde{\phi}(p,t)+ \frac{p^2}{2m}\frac{\partial {}}{\partial {p}}\tilde{\phi}(p,t)\end{align*}$$

I find as a solution, using separation of variables

$$\begin{align*}\tilde{\phi}(p,t) = C \exp \left( \frac{1}{{m g i \hbar}} \left( E p - \frac{p^3}{6m}\right) - \frac{i E t}{\hbar} \right)\end{align*}$$

and verify that this has the form
$$\begin{align*}\tilde{\phi}(p,t) &= f( p + mg t ) \exp \left( - i \frac{ p^3} {6 m^2 \hbar g}\right) \\ f( p ) &=\tilde{\phi}(p,0) \exp \left( - i \frac{ p^3} {6 m^2 \hbar g}\right) \end{align*}$$

The Attempt at a Solution

Relating the position and momentum representation with Fourier transforms

$$\begin{align*}\tilde{\phi}(p,t) &= \tilde{\phi}(p + m g t,0) \exp \left( - i \frac{ p^3 + (p + m g t)^3} {6 m^2 \hbar g}\right) \\ \phi(x,t) &= \frac{1}{{2 \pi \hbar}} \int dp\tilde{\phi}(p,t) e^{i p x/\hbar} \\ \tilde{\phi}(p + m g t,0) &=\frac{1}{{2 \pi \hbar}} \int dx'\phi(x', 0) e^{-i x'(p + mg t)/\hbar }\end{align*}$$

Putting these all together and making some changes of variables I can reduce this to the propagator form:
$$\begin{align*}G(x,x'; t) = \frac{1}{{2 \pi}} e^{-i x' mg t/\hbar} \int dk \exp\left( i k (x-x') -\frac{i \hbar^2 }{6 m^2 g}\left( (k+ mgt/\hbar)^3 + k^3 \right)\right)\end{align*}$$

But appear out of luck for any easy way to integrate this, so I suspect that I've messed this up in some fundamental way, and am looking to be pointed onto the right path.

 

[Peeter]

With the help of the Airy function
$$\begin{align*}Ai(x) = \frac{1}{{2\pi}} \int_{-\infty}^\infty e^{i(u^3/3 + ux)} du\end{align*}$$

and it's Fourier transform
$$\begin{align*}F(Ai(x)) = \frac{1}{{2\pi}} e^{i k^3/3}\end{align*}$$

I was able to compute

$$\begin{align*}G(x,x'; t) = \frac{1}{{2 \pi}} \frac{6 m^2 g}{\hbar^2} e^{-i x' mg t/\hbar} \int_{-\infty}^\infty du \exp\left( i \left( u^3/3 + (x-x') u \right)\right) du\end{align*}$$

(although I think I may have lost some factors of 3 along the way). This has some similarities to the expected form, but is not a match. We also haven't covered Airy functions in class, so I think there must be some simpler approach.

I'm wondering now if I've improperly assumed that $\tilde{\phi}(p,t)$ is related to $\phi(x,t)$ via Fourier transformation. Perhaps that only makes sense for the free particle case where $\left\langle{{x}} \vert {{p}}\right\rangle \propto e^{i p x/\hbar}$?

 

[gabbagabbahey]

I haven't gone over your calculations, so I can't tell you where your error is, but I can suggest an easier way to do things... In order to verify a solution, all you really need to do is show that the solution satisfies the Time-dependent Schrödinger's Equation... so why not just show that

$$i\hbar\frac{\partial}{\partial t} \phi(x,t) = H\phi(x,t)$$

by calculating the appropriate derivatives?

 

[Peeter]

I was loose in my description of the problem, which asked to show that this is the value for G, and not just verify that it works. I suppose the verification you suggest would do that indirectly, but I think they really did want a derivation.