# 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
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(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
Gold Member

Re: Finding A. Remember what the wave function means. When you take psi* times psi, you get the probability of finding the particle in the range dx. So, what is the total probability of finding the particle somewhere? And so, how can you determine A?

For the rest: Can you see any factors to divide out of your last equation?

Orodruin
Staff Emeritus