How to Find the Potential in Schrödinger's Equation?

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Homework Help Overview

The discussion revolves around finding the potential in Schrödinger's equation, focusing on the manipulation of wave functions and their derivatives within the context of quantum mechanics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the isolation of constants in wave functions and the implications of the Schrödinger equation. Questions arise regarding the criteria that wave functions must fulfill and the completeness of the original poster's attempts.

Discussion Status

Some participants provide guidance on checking derivatives and understanding the significance of the wave function in terms of probability. There is an ongoing exploration of the mathematical steps involved, with no explicit consensus reached on the approach to isolating the potential.

Contextual Notes

Participants note potential issues with the original poster's derivatives and the need for additional criteria for the wave function. The discussion reflects uncertainty regarding the correctness of the mathematical manipulations presented.

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Homework Statement


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Homework Equations

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
 
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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?
 
Overall: Check your derivatives.

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?
 
Also, the derivatives are not done correctly. Recheck what you have in the argument of the exponential.
 

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