The angular momentum operator acting on a wave function

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Jerrynap
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Hi guys, I need help on interpreting this solution.

Let me have two wave functions:
[itex]\phi_1 = N_1(r) (x+iy)[/itex]
[itex]\phi_2 = N_2(r) (x-iy)[/itex]

If the angular momentum acts on both of them, the result will be:

[itex]L_z \phi_1 = \hbar \phi_1[/itex]
[itex]L_z \phi_2 = -\hbar \phi_2[/itex]

My concern is, [itex]\phi_1[/itex] and [itex]\phi_2[/itex] look really like the complex conjugate of each other, so why do they have different eigenvalue?
 
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Jerrynap said:
My concern is, [itex]\phi_1[/itex] and [itex]\phi_2[/itex] look really like the complex conjugate of each other, so why do they have different eigenvalue?
Why should they have the same eigenvalue? Have a look at the complex conjugated eigenvalue equation (A|λ>)* = (λ|λ>)* <=> A*|λ>*=λ|λ>*.
 
kith said:
Why should they have the same eigenvalue? Have a look at the complex conjugated eigenvalue equation (A|λ>)* = (λ|λ>)* <=> A*|λ>*=λ|λ>*.

Well, A* = A (hermitian) and λ is real. So wouldn't it be

[itex] \hat{A}^*\left| λ\right\rangle^* = \hat{A}\left| λ\right\rangle^* = λ\left| λ\right\rangle^* ?[/itex]
 
Jerrynap said:
Well, A* = A (hermitian)
Hermitian refers to the adjoint operator A+ and not to the complex conjugate A*. If you look at Lz in spherical coordinates, you see that it isn't invariant under complex conjugation because it contains an "i".
 
Oh... I see where the negative sign came about. Lz* = -Lz. This can be seen in Cartesian coordinates as well since p* = -p. Thanks kith