Recognitions:
Gold Member
Homework Help
Staff Emeritus

## 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.
 Thanks guys! have been very helpful!

Blog Entries: 7
Recognitions:
Gold Member
Homework Help
 Quote by Plutoniummatt 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

$$ae^{i\phi_a}; \:be^{i\phi_b}$$

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.
 Blog Entries: 7 Recognitions: Gold Member Homework Help Finally, vela and I converged.
 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
 Recognitions: Gold Member Homework Help Science Advisor Staff Emeritus 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.