- #1
Peeter
- 305
- 3
Homework Statement
Attempting a problem related to a quantum particle in free fall due to constant force (old exam question)
[tex]\begin{align*}H = \frac{1}{{2m}} P^2 + m g X\end{align*} [/tex]
The last part of the question asks to verify that the position space propagator has the following form
[tex]\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*} [/tex]
(working in the momentum representation to get that far).
Homework Equations
One is able to show
[tex]\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*} [/tex]
I find as a solution, using separation of variables
[tex]\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*} [/tex]
and verify that this has the form
[tex]\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*} [/tex]
The Attempt at a Solution
Relating the position and momentum representation with Fourier transforms
[tex]\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*} [/tex]
Putting these all together and making some changes of variables I can reduce this to the propagator form:
[tex]\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*} [/tex]
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.