- #1
T-7
- 64
- 0
Hi,
I'm looking at this wave function:
[tex]\psi(x,t) = \frac{4}{5}{\psi}_{1} + \frac{3}{5}{\psi}_{2}[/tex]
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
[tex]A = 4/5.\sqrt{2/a}[/tex]
[tex]B = 3/5.\sqrt{2/a}[/tex]
[tex]K = \pi/a[/tex]
I might have rewritten this as
[tex]\psi(x,t) = Asin(Kx).exp(-\frac{iE_1.t}{\hbar}) + Bsin(2Kx).exp(-\frac{iE_2.t}{\hbar})[/tex]
However, the text restates this as
[tex]\psi(x,t) = Asin(Kx) + Bsin(2Kx).exp(-\frac{i\Delta.t}{\hbar})[/tex]
where
[tex]\Delta = E_{2} - E_{1}[/tex]
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!
I'm looking at this wave function:
[tex]\psi(x,t) = \frac{4}{5}{\psi}_{1} + \frac{3}{5}{\psi}_{2}[/tex]
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
[tex]A = 4/5.\sqrt{2/a}[/tex]
[tex]B = 3/5.\sqrt{2/a}[/tex]
[tex]K = \pi/a[/tex]
I might have rewritten this as
[tex]\psi(x,t) = Asin(Kx).exp(-\frac{iE_1.t}{\hbar}) + Bsin(2Kx).exp(-\frac{iE_2.t}{\hbar})[/tex]
However, the text restates this as
[tex]\psi(x,t) = Asin(Kx) + Bsin(2Kx).exp(-\frac{i\Delta.t}{\hbar})[/tex]
where
[tex]\Delta = E_{2} - E_{1}[/tex]
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!