Time Dependant One Dimensional Square well

[JPurdie7]

Given the wave function: Ψtot(x,t) = ((√2)/2)ψ₃e^(-(iE₃t)/[STRIKE]h[/STRIKE]) + ((√2)/2)ψ₅e^(-(iE₅t)/[STRIKE]h[/STRIKE]) @ |x|≤a/2 and ψtot(x,t) = 0 @ |x|≥a/2 where a=100nm , E=((([STRIKE]h[/STRIKE])^2)((k_n)^2))/2m , k_n=pi*n/a , & T₁ = 2pi[STRIKE]h[/STRIKE]/E₁

How would I find the Period of the wave function in terms of T₁??

 

[collinsmark]

Hello JPurdie7,

Welcome to physics forums!

Given the wave function: Ψtot(x,t) = ((√2)/2)ψ₃e^(-(iE₃t)/[STRIKE]h[/STRIKE]) + ((√2)/2)ψ₅e^(-(iE₅t)/[STRIKE]h[/STRIKE]) @ |x|≤a/2 and ψtot(x,t) = 0 @ |x|≥a/2 where a=100nm , E=((([STRIKE]h[/STRIKE])^2)((k_n)^2))/2m , k_n=pi*n/a , & T₁ = 2pi[STRIKE]h[/STRIKE]/E₁

How would I find the Period of the wave function in terms of T₁??

I take it you mean the time period of the expectation value of the wavefunction with respect to T₁?

Hypothetically, you could calculate the expectation value of the wavefunction
$$\int_{x=- \infty} ^{\infty} \Psi^*(x, t) x \Psi(x, t) dx,$$
but you probably don't need to actually fully evaluate the integral, if you keep an eye on the math.

But go ahead and evaluate the integrand. Keep an eye on terms that are only a function of t that you can pull out from under the integral. [Edit: you'll need to do a bit of multiplication first, but no integration.]

And keep the exponential representations of sine and cosine in your back pocket. They might come in handy, if you recognize them.

Edit: Just in case, for your reference:

$$\sin(\omega t) = \frac{e^{i \omega t} - e^{-i \omega t}}{2i}$$

$$\cos(\omega t) = \frac{e^{i \omega t} + e^{-i \omega t}}{2}$$