- #1
SoggyBottoms
- 59
- 0
At t = 0 a particle is in the (normalized) state:
[tex]\Psi(x, 0) = B \sin(\frac{\pi}{2a}x)\cos(\frac{7\pi}{2a}x)[/tex]
With [itex]B = \sqrt{\frac{2}{a}}[/itex]. Show that this can be rewritten in the form [itex] \Psi(x, 0) = c \psi_3(x) + d \psi_4(x)[/itex]
We can rewrite this to:
[tex]\Psi(x, 0) = \frac{B}{2}\left[ c \sin(\frac{4 \pi}{a}x) - d\sin(\frac{3\pi}{a}x)\right][/tex]
The answer sheet gives [itex]c = -d = \frac{1}{\sqrt{2}}[/itex]. I assume you can find this by calculating [itex]A^2 \int \left[ c \sin(\frac{4 \pi}{a}x) - d\sin(\frac{3\pi}{a}x)\right]^2 dx[/itex]. I attempted to do it this way, but it becomes a really long calculation and halfway through I just lose track of everything. Is there an easier way to find c and d?
[tex]\Psi(x, 0) = B \sin(\frac{\pi}{2a}x)\cos(\frac{7\pi}{2a}x)[/tex]
With [itex]B = \sqrt{\frac{2}{a}}[/itex]. Show that this can be rewritten in the form [itex] \Psi(x, 0) = c \psi_3(x) + d \psi_4(x)[/itex]
We can rewrite this to:
[tex]\Psi(x, 0) = \frac{B}{2}\left[ c \sin(\frac{4 \pi}{a}x) - d\sin(\frac{3\pi}{a}x)\right][/tex]
The answer sheet gives [itex]c = -d = \frac{1}{\sqrt{2}}[/itex]. I assume you can find this by calculating [itex]A^2 \int \left[ c \sin(\frac{4 \pi}{a}x) - d\sin(\frac{3\pi}{a}x)\right]^2 dx[/itex]. I attempted to do it this way, but it becomes a really long calculation and halfway through I just lose track of everything. Is there an easier way to find c and d?