Angular momentum operators and eigenfunctions

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machofan
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Homework Statement



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Homework Equations


The Attempt at a Solution



I have tried inserting the first wavefunction into Lz which gets me 0 for the eigenvalue for the first wavefunction. Is this correct?

For the second wavefunction, I inserted it into Lz and this gets me -i*hbar*xAe^-r/a which is not equal to an eigenvalue*Ay*e-r/a, although I am not sure if this is the correct justification.

Any help is much appreciated, thank you.
 
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I'm also stuck on how to convert the wavefunctions into spherical polar coordinates, having tried to use the formula for spherical polar coordinates, I notice that the wavefunctions do not depend on the variables phi and theta, so I am unsure on how to approach the first problem.
 
Hello fan, and welcome to PF.

You get 0 for ##\Psi_0## because it is symmetric under rotations around the z axis: there is no angular momentum. So I think you are right. It is an eigenfunction.

For ##\Psi_1## you basically want to show that ##{\bf L}_z \left( \Psi_1 \right ) = L_z\, \Psi_1 ## with ##{\bf L}_z## the operator and ##L_z## an eigenvalue (i.e. a number, possibly complex), has no solutions.

You may assume |A| is not equal to zero (because of the normalization) so the eigenvalue zero is already excluded. So if you multiply the right hand side from the left by ##\Psi_1^*## and integrate, that yields ##L_z##, which is nonzero.

Do the same thing with the lefthand side and show that it does give zero. That way you prove that the assumption that ##\Psi_1## is an eigenfunction leads to a false equation.
 
machofan said:
I'm also stuck on how to convert the wavefunctions into spherical polar coordinates, having tried to use the formula for spherical polar coordinates, I notice that the wavefunctions do not depend on the variables phi and theta, so I am unsure on how to approach the first problem.
With ##z = r\cos\theta## you sure have a theta dependence!
 
But the exercise wants you to write out the z-component of the operator cross product ##{\bf r} \times {\bf p}##, not the wave functions!
[edit] sorry, it wants you to do both. In the first part the Lz, later on the ##\Psi##
 
BvU said:
Hello fan, and welcome to PF.

You get 0 for ##\Psi_0## because it is symmetric under rotations around the z axis: there is no angular momentum. So I think you are right. It is an eigenfunction.

For ##\Psi_1## you basically want to show that ##{\bf L}_z \left( \Psi_1 \right ) = L_z\, \Psi_1 ## with ##{\bf L}_z## the operator and ##L_z## an eigenvalue (i.e. a number, possibly complex), has no solutions.

You may assume |A| is not equal to zero (because of the normalization) so the eigenvalue zero is already excluded. So if you multiply the right hand side from the left by ##\Psi_1^*## and integrate, that yields ##L_z##, which is nonzero.

Do the same thing with the lefthand side and show that it does give zero. That way you prove that the assumption that ##\Psi_1## is an eigenfunction leads to a false equation.

That has really cleared things up, thank you!
 
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