What is the Frequency of a State Function Describing a Proton?

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Homework Help Overview

The discussion revolves around a state function for a proton, expressed in terms of exponential functions involving energy and time. The original poster is tasked with finding the frequency of this state function.

Discussion Character

  • Exploratory, Conceptual clarification

Approaches and Questions Raised

  • The original poster attempts to combine terms of the state function into a single exponential form to identify the angular frequency. They seek confirmation on whether this approach is appropriate.

Discussion Status

Participants are exploring the definition of the period of the function as a means to determine the frequency. One participant has provided guidance on finding the period, which may assist the original poster in their approach.

Contextual Notes

The original poster has indicated that the coefficients A and B are specific trigonometric functions, which may influence the periodicity of the overall state function.

Jelfish
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I have this state function that I've gotten to the form:

[tex]\Psi = A*\exp[iE_1 t/h] + B*\exp[4iE_1 t/h][/tex]

where A and B are functions of x. I know the energy. The h's are h-bars.

The state function is suppose to describe a proton and I'm asked to find the frequency.

My first thought is to somehow combine the terms such that I would have something like

[tex]\Psi = C*\exp[i\omega t][/tex]

Where [tex]\omega[/tex] would be the angular frequency.

I'm having some trouble trying to get it to this form.

What I want to know is if this is the correct approach.

Let me know if I need to give any more info. Thanks in advance.
 
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If it matters, A and B are trig functions, cos(Pi x / L) and sin(2 Pi x / L) respectively, both times sqrt(2/L).
 
Well, it helps to know the definition. :smile:

What you need to do is to find a period of your function -- that is, the smallest positive constant H such that Ψ(t) = Ψ(t + H). (For all t) Then, the period is just the reciprocal of that.
 
Thanks! I'll try that.
 

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