The angular momentum operator acting on a wave function

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SUMMARY

The discussion centers on the action of the angular momentum operator \( L_z \) on two wave functions, \( \phi_1 = N_1(r) (x+iy) \) and \( \phi_2 = N_2(r) (x-iy) \). The results show that \( L_z \phi_1 = \hbar \phi_1 \) and \( L_z \phi_2 = -\hbar \phi_2 \), indicating different eigenvalues despite the wave functions being complex conjugates. The key insight is that the operator \( L_z \) is not invariant under complex conjugation due to the presence of the imaginary unit \( i \), leading to the observed negative sign in the eigenvalue for \( \phi_2 \).

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

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

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

L_z \phi_1 = \hbar \phi_1
L_z \phi_2 = -\hbar \phi_2

My concern is, \phi_1 and \phi_2 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, \phi_1 and \phi_2 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

<br /> \hat{A}^*\left| λ\right\rangle^* = \hat{A}\left| λ\right\rangle^* = λ\left| λ\right\rangle^* ?<br />
 
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
 

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