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Bashyboy
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This is Problem comes from Griffiths Quantum Mechanics textbook; specifically, it is problem 2.5 (b).
A particle in an infinite square well has its initial wave function an even mixture of the first two stationary states:
[itex]\displaystyle \Psi(x,0) = A[\psi_1(x) + \psi_2(x)][/itex]
Here is the part of the problem that I am having a little trouble with:
(b) Find [itex]\displaystyle \Psi(x,t)[/itex] and [itex]\displaystyle |\Psi(x,t)|^2[/itex]|. Express the latter as a sinusoidal function of time, as in Example 2.1. To simplify the result, let [itex]\omega \equiv \frac{\pi^2 \hbar}{2ma^2}[/itex]
According to the answer key, even after t=0, the wave function continues to be a mixture of the first two stationary states. Why is that? I am having a little difficulty understanding this. Why can't it be a new 'mixture?'
As far as I understand, to calculate the wavefunction for all future times, we use the equations
[itex]\displaystyle \Psi (x,t) = \sum_{n=1}^{\infty} c_n \sqrt{\frac{2}{a}} \sin \left( \frac{n \pi}{a} x \right) e^{-i(n^2 \pi^2 \hbar/2ma^2)t}[/itex]
and
[itex]\displaystyle c_n = \sqrt{\frac{2}{a}} \int_0^a \sin \left( \frac{n \pi}{a} x \right) \Psi (x,0) dx [/itex]
A particle in an infinite square well has its initial wave function an even mixture of the first two stationary states:
[itex]\displaystyle \Psi(x,0) = A[\psi_1(x) + \psi_2(x)][/itex]
Here is the part of the problem that I am having a little trouble with:
(b) Find [itex]\displaystyle \Psi(x,t)[/itex] and [itex]\displaystyle |\Psi(x,t)|^2[/itex]|. Express the latter as a sinusoidal function of time, as in Example 2.1. To simplify the result, let [itex]\omega \equiv \frac{\pi^2 \hbar}{2ma^2}[/itex]
According to the answer key, even after t=0, the wave function continues to be a mixture of the first two stationary states. Why is that? I am having a little difficulty understanding this. Why can't it be a new 'mixture?'
As far as I understand, to calculate the wavefunction for all future times, we use the equations
[itex]\displaystyle \Psi (x,t) = \sum_{n=1}^{\infty} c_n \sqrt{\frac{2}{a}} \sin \left( \frac{n \pi}{a} x \right) e^{-i(n^2 \pi^2 \hbar/2ma^2)t}[/itex]
and
[itex]\displaystyle c_n = \sqrt{\frac{2}{a}} \int_0^a \sin \left( \frac{n \pi}{a} x \right) \Psi (x,0) dx [/itex]
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