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Suppose we have a particle with mass $m$ and energy $E$ in a gravitational field $V(z)=-mgz$. How can I find the wave function $\psi(z)$?

It should have an integral form on $dp$. Any help would be appreciated.

What I've tried

One way to solve the problem is use of change of variable
$$ x~:=~\left(\frac{\hbar^2}{2m^2g}\right)^{2/3}\frac{2m}{\hbar^2}(mgz-E) $$

we can reduce Schroedinger equation to

$$ \frac{d^2\phi}{dx^2}-x\phi(x)~=~0 $$

This is a standard equation, its solution is given by $$\phi(x)~=~B~\text{Ai}(x)$$ where $\text{Ai}$ is the Airy function. But my solution should be (not exactly) like this:

$$ \psi(z)= N\int_{-\infty}^\infty dp \exp\left[\left(\frac{E}{mg}+z\right)p-\frac{p^3}{6m^2g} \right] $$

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    $\begingroup$ Seems like a straight-forward application of Schroedinger's equation. What have you tried? $\endgroup$
    – Kyle Kanos
    Commented Dec 16, 2013 at 14:38
  • $\begingroup$ I have solved it by WKB approximation but my answer do not coincide with what it should be in test. $\endgroup$
    – Abolfazl
    Commented Dec 16, 2013 at 14:42
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    $\begingroup$ Perhaps, then, you should show what you did do, so we can point out what you had done wrong. $\endgroup$
    – Kyle Kanos
    Commented Dec 16, 2013 at 14:43
  • $\begingroup$ Wavefunction of a particle in linearly changing potential is given by Airy functions. $\endgroup$
    – Ruslan
    Commented Dec 16, 2013 at 15:07
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    $\begingroup$ This Wiki link should be relevant. $\endgroup$
    – Kyle Kanos
    Commented Dec 16, 2013 at 15:45

2 Answers 2

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The basic idea for this is to use the momentum space version of the Schroedinger equation: $$ \hat{p}\to p,\quad\hat{x}\to i\hbar\frac{\partial}{\partial p} $$ and then solve the system1, $$ \left[\frac{p^2}{2m}+img\hbar\frac{d}{dp}\right]\phi=E\phi $$

which should be solvable (e.g., complex exponentials). You can then Fourier transform to physical space to get $$ \psi(x)=\frac{1}{\sqrt{2\pi\hbar}}\int dp\,e^{ipx/\hbar}\phi(p) $$ This should match to what your professor expects.



More formally/generalized, you should get$$ \frac{p^2}{2m}\phi\left(p\right)+\int dp'\,U\left(p-p'\right)\phi\left(p'\right)=E\phi\left(p\right)$$ where $$U(p)=\frac{1}{\sqrt{2\pi\hbar}}\int dx\,e^{-ipx/\hbar}U(x)$$ but this will reduce to the above anyway

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  • $\begingroup$ I agree, I tried this method as well but I thought it was a bit odd.It should be solvable with more effort. Let me think more. Thank you. $\endgroup$
    – Abolfazl
    Commented Dec 16, 2013 at 17:13
  • $\begingroup$ @Abolfazl: two things to note to combine as a single hint: (1) $\frac{d}{dx}\exp[f(x)]=\exp[f(x)]\cdot\frac{df}{dx}$ and (2) your teachers solution has a $p^3$ term in it. $\endgroup$
    – Kyle Kanos
    Commented Dec 17, 2013 at 2:20
  • $\begingroup$ I think we can use this form of Airy function $$ Ai(t)=\frac{1}{{2\pi i}}\int dp\,exp(\frac{p^3}{3}-tp) $$ ,wiki, where $t=mgz+E$ $\endgroup$
    – Abolfazl
    Commented Dec 17, 2013 at 3:48
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    $\begingroup$ @Abolfazl: I would argue that you don't need to know the Airy function at all. So just completely forget that it was mentioned. Start by using a trial wave-function (in momentum space) of the form $\tilde{\phi}(p)=A\exp\left[f(p)\right]$ and see what conditions are required on $f(p)$ to satisfy this. $\endgroup$
    – Kyle Kanos
    Commented Dec 17, 2013 at 4:06
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$$ \left[\frac{p^2}{2m}+V(i\hbar\frac{d}{dp})\right]\phi(p)=E\phi(p) $$ $$ \left[\frac{p^2}{2m}+(-mg)(i\hbar\frac{d}{dp})\right]\phi(p)=E\phi(p) $$ $$ \frac{1}{i\hbar mg}(\frac{p^2}{2m}-E)\phi(p)=\frac{\phi(p)}{dp} $$ When integrate we have: $$ \frac{i}{\hbar mg}(Ep-\frac{p^3}{6m})=Ln\frac{\phi(p)}{\phi(p_{o})} $$ $$ \phi(p)=\phi(p_{0})e^{\frac{E}{mg}p-\frac{p^3}{6m^2g}} $$ $$ \psi(z)=\int dp e^{ipz/\hbar} \phi(p) $$ $$ \psi(z)=\phi(p_{0})\int_{-\infty}^\infty dp e^{i/\hbar \left[ (\frac{E}{mg}+z)p-\frac{p^3}{6m^2g}\right]} $$

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  • $\begingroup$ why -1? it seems to be true! $\endgroup$
    – Abolfazl
    Commented Jan 27, 2016 at 22:01

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