Rewriting a wave function superposition

[T-7]

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!

 

[jostpuur]

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

$$\psi(t,x)\mapsto e^{iE_1 t/\hbar}\psi(t,x)$$

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

$$\tilde{\psi}(t,x) = e^{iE_1 t/\hbar}\psi(t,x)$$

 

[T-7]

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!