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Quantum Operators (or just operators in general)

 
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Feb15-10, 05:36 PM   #18
 
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Quantum Operators (or just operators in general)


The normalization condition always turns out to be real because the squared terms always involve the multiplication of a complex number by its conjugate and the cross terms are complex conjugates of each other so when they combine, the imaginary part cancels out.

In this problem, because c is assumed to be real, a and b can be real simultaneously. You could always multiply the state by an arbitrary phase and get complex coefficients if you want, but the relative phase of a and b will remain unchanged.
Feb15-10, 05:38 PM   #19
 
Thanks guys! have been very helpful!
Feb15-10, 05:40 PM   #20
 
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Quote by Plutoniummatt View Post
Yeah, I realize now that I cannot do that, is it possible to do it without resorting to the Gram-Schmit process? Kuruman was suggesting using simultaneous equations, but are the coefficients a and b necessarily real?
They don't have to be. You can use coefficients

[tex]ae^{i\phi_a}; \:be^{i\phi_b}[/tex]

where a and b are real. Then the normalization condition will be in terms of a, b and the (arbitrary) phase difference φab. Still two equations and two unknowns.
Feb15-10, 05:41 PM   #21
 
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Finally, vela and I converged.
Feb15-10, 06:10 PM   #22
 
ok guys, so:

a + bc = 0


aa* + ab*c + ba*c + bb* = 1
2aa* + bb* - ccbb* = 1

are my equations, where do I go from here? If a and b are in general complex, how do I solve for them? does the phase not matter like whatsoever? so I can just write aa* as a^2
Feb15-10, 06:38 PM   #23
 
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Use the first equation to eliminate a from the second equation. You should write bb* = |b|2. The best you can do is get the relative phase between a and b, so you can assume one is real.
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