Finding potential to satisfy SE

Gold Member

The Attempt at a Solution

(a) Well, I just isolate A, so
$$A = \Psi (x,t) e^{a[(mx^{2}/ \hbar) + it]}$$

I am not sure if this is what is meant, seems too obvious.

(b) So I know the Schrödinger equation can be written
$$i \hbar \frac {\partial \Psi}{ \partial t} = \Big [ - \frac {\hbar}{2m} \frac {\partial^{2}}{\partial x^{2}} + V \Big ] \Psi$$

So I take the given wave function,
$$\Psi = Ae^{-a[\frac {mx^{2}}{\hbar} + it]}$$

And find the derivatives with respect to x and t,

$$\frac {\partial \Psi}{\partial t} = -A[a(\frac {mx^{2}}{\hbar}) + i]e^{-a[\frac {mx^{2}}{\hbar} + it]}$$
$$\frac {\partial \Psi}{\partial x} = -A[a(\frac {2mx}{\hbar}) + it]e^{-a[\frac {mx^{2}}{\hbar} + it]}$$
$$\frac {\partial^{2} \Psi}{\partial x^{2}} = A[a^{2}(\frac {4m^{2}x^{2}}{\hbar^{2}}) + i^{2}t^{2}]e^{-a[\frac {mx^{2}}{\hbar} + it]}$$

And I substitute back into the SE,
$$i \hbar(-A[a(\frac {mx^{2}}{\hbar}) + i]e^{-a[\frac {mx^{2}}{\hbar} + it]}) = - \frac {\hbar^{2}}{2m} \Big( A[a^{2}(\frac {4m^{2}x^{2}}{\hbar^{2}}) + i^{2}t^{2}]e^{-a[\frac {mx^{2}}{\hbar} + it]} \Big ) + V( Ae^{-a[\frac {mx^{2}}{\hbar} + it]})$$

From here I can isolate V

Last edited:

Orodruin
Staff Emeritus
Homework Helper
Gold Member
(a) No, this is not what is intended. You do not know the wave function up to the constant A so you obviously cannot have an expression for A involving the wave function. You need to apply some criteria that the wave function should fulfil. What such criteria do you know?

(b) Your post seems incomplete as it ends "So I". What did you intend to say?

DEvens