- #1
StarWombat
- 2
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OP warned about not using the homework template
Hi,
I'm trying to work my way through some problems and am stuck on one for a symmetric infinite square well, of width 2a, so -a<x<+a. Since this is the symmetric case, the wavefunction should be a linear combination of the terms
(a)-½ cos (nπx/2a) for odd n,
(a)-½ sin (nπx/2a) for even n.
The question gives a wavefunction Ψ(x,0) = ε(x) (3a)-½(1 - cos (2πx/a)) where ε(x) = +1 for x>0, -1 for x<0. I want to rearrange this so it's a linear combination of cos and sin terms, for various values of n. It's easy to get 1-cos(2y) to a single term proportional to sin2(y) using trig identities, but I don't want the square. I want a linear combination of cosine and sine terms where I can read off the value of n for each term (and hence find the probabilities of the various energy eigenvalues). I don't think I can just read n out of the existing cosine because that would imply that n=4, but the cos terms correspond to odd values of n.
And I'm probably sleepy and will kick myself once I see the solution, but at the moment I just can't think of what way to massage this into the linear combination I want. So if anyone has any ideas please shout them out.
Thanks in advance
I'm trying to work my way through some problems and am stuck on one for a symmetric infinite square well, of width 2a, so -a<x<+a. Since this is the symmetric case, the wavefunction should be a linear combination of the terms
(a)-½ cos (nπx/2a) for odd n,
(a)-½ sin (nπx/2a) for even n.
The question gives a wavefunction Ψ(x,0) = ε(x) (3a)-½(1 - cos (2πx/a)) where ε(x) = +1 for x>0, -1 for x<0. I want to rearrange this so it's a linear combination of cos and sin terms, for various values of n. It's easy to get 1-cos(2y) to a single term proportional to sin2(y) using trig identities, but I don't want the square. I want a linear combination of cosine and sine terms where I can read off the value of n for each term (and hence find the probabilities of the various energy eigenvalues). I don't think I can just read n out of the existing cosine because that would imply that n=4, but the cos terms correspond to odd values of n.
And I'm probably sleepy and will kick myself once I see the solution, but at the moment I just can't think of what way to massage this into the linear combination I want. So if anyone has any ideas please shout them out.
Thanks in advance