Rewriting a wave function superposition

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Hi,

I'm looking at this wave function:

\psi(x,t) = \frac{4}{5}{\psi}_{1} + \frac{3}{5}{\psi}_{2}

The functions involved here are the typical eigenfunctions for the ground state and first excited level in an infinitely-deep 1-D square well.

Defining
A = 4/5.\sqrt{2/a}
B = 3/5.\sqrt{2/a}
K = \pi/a

I might have rewritten this as

\psi(x,t) = Asin(Kx).exp(-\frac{iE_1.t}{\hbar}) + Bsin(2Kx).exp(-\frac{iE_2.t}{\hbar})

However, the text restates this as

\psi(x,t) = Asin(Kx) + Bsin(2Kx).exp(-\frac{i\Delta.t}{\hbar})

where
\Delta = E_{2} - E_{1}

Can someone tell me how the time element has been attached to only one of the eigenfunctions like that? I expect it's obvious, but I'm just not seeing it at the moment! (It's evidently been done this way to make the expectation value calculation that follows simpler).

Many thanks!
 
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You can always multiply a wave function with an arbitrary phase factor without changing the state physically. So there the author changes the wave function with

<br /> \psi(t,x)\mapsto e^{iE_1 t/\hbar}\psi(t,x)<br />

and it remains physically same. It could have been clearer to use a different symbol for the wave function with an additional phase factor. For example

<br /> \tilde{\psi}(t,x) = e^{iE_1 t/\hbar}\psi(t,x)<br />
 
jostpuur said:
You can always multiply a wave function with an arbitrary phase factor without changing the state physically. So there the author changes the wave function with...

Good. Thank you. Mathematically, that was the only way I could think of doing it, but the author didn't bother to use a different symbol for the shifted function, or to explain what he was doing physically.

Cheers!
 
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