Time Dependence of Wave Function

AI Thread Summary
The discussion focuses on solving equations related to the time dependence of wave functions in quantum mechanics, specifically using the relationship $$e^{i\omega_a t}=e^{i\omega_b t}$$. It emphasizes the importance of understanding the underlying trigonometric nature of these equations, as expressed through Euler's identity. The conversation suggests that finding the smallest time value involves manipulating the equation to derive $$|(\omega_a-\omega_b)| t=2\pi$$. Additionally, it encourages plotting the real part of the wave function for better comprehension. Overall, a solid grasp of these concepts is deemed essential for tackling related problems in quantum mechanics.
sarahjohn
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Homework Statement
An electron is trapped in a well (neither square nor harmonic). It is in a superposition of three energy states, ψa, ψb, and ψc, with energies:
Ea = 2.2 eV,
Eb = 2.6 eV.
Ec = 3.2 eV.

1) At t = 0, all phases (the e iωt factors) are equal. After how much time (in seconds) will the phases of ψa, and ψb become equal again? Choose the smallest positive time.
Δtab =

2) After how much time (in seconds) will the phases of ψa, and ψc become equal again? Choose the smallest positive time.
Δtac =

3) After how much time (in seconds) will all three phases again be equal? Choose the smallest positive time.
Δtabc =
Relevant Equations
E = h*w/(2*pi)
I started out by finding the w (omega) value for all of the three states but I'm not sure where to go from there.
 
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I am not so good in quantum mechanics but I think all you got to do is to solve equations like $$e^{i\omega_a t}=e^{i\omega_b t}$$ for t and since these are actually trigonometric equations (i believe you know euler's identity $$e^{i\omega t}=\cos \omega t+i\sin \omega t$$) there going to be many values of t, you just have to choose the smallest t.
 
Part (a) should be pretty easy. If you haven't seen it before, it is worth spending some effort to solidly understand. Plot out the real (sin) part. Or play with complex exponentials $$e^{i\omega_a t}=e^{i\omega_b t}$$ $$e^{i(\omega_a -\omega_b) t}=1$$so $$|(\omega_a-\omega_b)| t=2\pi $$ for the shortest time
Part (c) is similar.
 
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