
#1
Mar1212, 03:45 PM

P: 61

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? 



#2
Mar1212, 07:54 PM

Sci Advisor
HW Helper
PF Gold
P: 2,606

The difference between the two expressions is just a trig identity that you should be able to work out with no unknown coefficients. Therefore, if you can verify that B is the correct normalization factor for the first version, c and d are just determined by the trig identity.
However, if you are going to be studying QM, you should really learn how to do integrals like the one that gave you trouble. That is fairly simple compared to integrals you will come across later. QM is not a forgiving subject if you don't have a good calculus background. 


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