The angular momentum operator acting on a wave function

[Jerrynap]

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?

 

[kith]

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>*.

 

[Jerrynap]

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]

 

[kith]

Well, A* = A (hermitian)

Hermitian refers to the adjoint operator A⁺ and not to the complex conjugate A*. If you look at L_z in spherical coordinates, you see that it isn't invariant under complex conjugation because it contains an "i".

 

[Jerrynap]

Oh... I see where the negative sign came about. L_z* = -L_z. This can be seen in Cartesian coordinates as well since p* = -p. Thanks kith