- #1
noamriemer
- 50
- 0
Hello again!
Say I have a potential well, between 0 and a. I also know how the wave function looks like for (t=0):
[itex] \psi(x,0)= \frac {2bx} {a} [/itex] for [itex] 0<x<\frac {a} {2} [/itex]
and
[itex] \psi(x,0)= 2b(1- \frac {x} {a} ) [/itex] for [itex] \frac {a} {2} <x<a [/itex]
Now, I wish to find the wave function of a general time, t.
I know I need to find symmetry. Therefore I move the x axis:
[itex] x\rightarrow x', x'\equiv x-\frac {a} {2} [/itex]
So now I have the same well, only symmetrical.
That means:
[itex] \varphi_n =\sqrt {\frac {2} {a} } sin(\frac {\pi lx'} {a}) \rightarrow l=2n
\varphi_n =\sqrt {\frac {2} {a} } cos(\frac {\pi lx'} {a}) \rightarrow l=2n+1 [/itex]
For the question before the axis transition, I know I have to look for a sin- solution.
But now, when I have this symmetry, I should look for a solution which vanishes at x=-a/2 instead of x=0
Meaning, the cosine must stay. But in this problem's solution, I see that the sine stays...
why is that so?
Thank you!
Say I have a potential well, between 0 and a. I also know how the wave function looks like for (t=0):
[itex] \psi(x,0)= \frac {2bx} {a} [/itex] for [itex] 0<x<\frac {a} {2} [/itex]
and
[itex] \psi(x,0)= 2b(1- \frac {x} {a} ) [/itex] for [itex] \frac {a} {2} <x<a [/itex]
Now, I wish to find the wave function of a general time, t.
I know I need to find symmetry. Therefore I move the x axis:
[itex] x\rightarrow x', x'\equiv x-\frac {a} {2} [/itex]
So now I have the same well, only symmetrical.
That means:
[itex] \varphi_n =\sqrt {\frac {2} {a} } sin(\frac {\pi lx'} {a}) \rightarrow l=2n
\varphi_n =\sqrt {\frac {2} {a} } cos(\frac {\pi lx'} {a}) \rightarrow l=2n+1 [/itex]
For the question before the axis transition, I know I have to look for a sin- solution.
But now, when I have this symmetry, I should look for a solution which vanishes at x=-a/2 instead of x=0
Meaning, the cosine must stay. But in this problem's solution, I see that the sine stays...
why is that so?
Thank you!