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A particle of mass $m$ is subjected to a force $F(r) = -\nabla V(r)$ such that the wave function $\varphi(p, t)$ satisfies the momentum-space Schrödinger equation

$$\left(\frac{p^2}{2m}-a\Delta^2_p\right) \varphi (p,t) = i\frac{\partial \varphi (p,t)}{ \partial t}$$

where $\hbar = 1$, $a$ is some real constant and

$$\Delta^2_p \equiv \frac{\partial^2 }{ \partial p^2_x} + \frac{\partial^2}{ \partial p^2_y } + \frac{\partial^2 }{\partial^2_z} \, .$$

How do we find force $F(r) \equiv -\nabla V(r)$?


We know that the coordinate and momentum representations of a wave function are related by

$$\\psi (r,t) = \left(\frac {1}{2\pi}\right)^{\frac {3}{2}} \int \varphi (k,t) e^{ik\cdot r} \mathrm dk \tag {1}$$

$$\varphi (k,t) = \left(\frac {1}{2\pi}\right)^{\frac {3}{2}} \int \psi (r,t) e^{-ik\cdot r} \mathrm dr \tag {2}$$

where $k \equiv p / \hbar$ with $Ii = 1$.

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  • $\begingroup$ Could someone take a look? Help is needed. $\endgroup$ – Maxwell Oct 17 '17 at 12:59
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    $\begingroup$ I assume you mean $F(r) = -\nabla V(r)$? $\endgroup$ – caverac Oct 17 '17 at 13:07
  • $\begingroup$ @caverac Yes, I mean that. $\endgroup$ – Maxwell Oct 17 '17 at 16:48
  • $\begingroup$ Maybe it would be more useful if you tell us exactly what you're looking for. @Giuseppe already gave you an answer that seems to solve your problem $\endgroup$ – caverac Oct 18 '17 at 13:41
  • $\begingroup$ @caverac I just want to find an answer which I can get easily. Couldn't work it out. $\endgroup$ – Maxwell Oct 18 '17 at 13:42
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HINT:

https://en.wikipedia.org/wiki/Hamiltonian_(quantum_mechanics)

As you can see, the Schrödinger Hamiltonian operator $H$ can be written as the sum of a kinetic term $T$ and of a potential term $V$. For a system of one particle, the kinetic term takes the form $T=\frac{p^2}{2m}$. Your task is to determine the potential term $V$; once this is done, the force will be given by the formula $F=-\nabla V$.

The first step should be the rewriting of the Schrödinger equation in the following form: $$ -i\partial_t \psi = H\psi, $$ where $\psi$ is the wavefunction in the position representation. From this you can obtain the explicit expression of $H$, which gives $V$ by the formula $V=H-T$.

The obstruction is the fact that the Schrödinger equation that has been given to you is expressed in the momentum representation. Convert it to the position representation first.

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  • $\begingroup$ I've tried it again but could not work it out. $\endgroup$ – Maxwell Oct 17 '17 at 17:48
  • $\begingroup$ Could you be more clear? $\endgroup$ – Maxwell Oct 17 '17 at 18:50
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    $\begingroup$ @Maxwell: I am sorry, I cannot do the computations now. (Moreover, I think that it would be better for you if you solved the exercise yourself). Transform your equation in the position representation. You will see that one of the terms is exactly the kinetic energy $\hat{T}=-\frac{p^2}{2m}$. The other one will be the potential. Taking the nabla of that term, you get the force. $\endgroup$ – Giuseppe Negro Oct 17 '17 at 18:54
  • $\begingroup$ can you add it into your answer? $\endgroup$ – Maxwell Oct 18 '17 at 18:10
  • $\begingroup$ @Maxwell: it should be clearer now. $\endgroup$ – Giuseppe Negro Oct 19 '17 at 8:51
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$\newcommand{\vect}[1]{{\bf #1}}$

Call $$ \vect{q} = \frac{\vect{p}}{\sqrt{2ma}} $$

such that your can write your equation as

$$ \left(-\frac{1}{2m}\nabla_{\vect{q}}^2 + a\vect{q}^2\right)\phi(\vect{q},t) = -i\partial_t\phi(\vect{q},t) $$

which is just Schrodinger's equation with a harmonic potential $V(\vect{q}) = a\vect{q}^2$. Solutions to this problems are well known in terms of Hermite polynomials.

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  • $\begingroup$ I did not get what you mean. Could you be more clear? $\endgroup$ – Maxwell Oct 17 '17 at 17:38
  • $\begingroup$ @Maxwell Could you tell me exactly where are you getting stuck? $\endgroup$ – caverac Oct 18 '17 at 13:44
  • $\begingroup$ How did you call $$ \vect{q} = \frac{\vect{p}}{\sqrt{2ma}} $$? Also If you be more clear about the equation, I'd be happy. It seems like I can do it now. $\endgroup$ – Maxwell Oct 18 '17 at 13:46
  • $\begingroup$ Because $$ \frac{1}{2m}\frac{\partial^2}{\partial{q_x^2}} = \frac{1}{2m}\frac{\partial^2}{\partial{p_x}^2/(2ma)} = a\frac{\partial^2}{\partial p_x^2} $$ $\endgroup$ – caverac Oct 18 '17 at 13:49
  • $\begingroup$ Shouldn't the potential be $$V(r) = ar^2$$ $\endgroup$ – Maxwell Oct 18 '17 at 13:53

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