Simultaneous observables for hydrogen

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  • #1
bobred
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Homework Statement


Is there a state that has definite non-zero values of [itex]E, L^2[/itex] and [itex]L_x[/itex]


Homework Equations



[itex]L^2[/itex] and [itex]L_z[/itex] commute with the Hamiltonian so we can find eigenfunctions for these


The Attempt at a Solution


I would say that there is a state with simultaneous eigenfunctions of [itex]L_x,L_y,L_z[/itex] and [itex]L^2[/itex], but with eigenvalues equal to zero. This being the state with [itex]l=0[/itex] and [itex]m=0[/itex], so there are no definite non-zero values of [itex]E, L^2[/itex] and [itex]L_x[/itex]. For other states [itex]L_x,L_y,L_z[/itex] and [itex]L^2[/itex] do not commute.
 

Answers and Replies

  • #2
TSny
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Hello.

Something to think about. Should the z axis be special in the hydrogen atom? That is, if there exist states with definite non-zero eigenvalues of ##E, L^2,## and ##L_z##, why shouldn't there exist states with definite non-zero eigenvalues of ##E, L^2,## and ##L_x##?

Suppose you had a wavefunction ##\psi(r, \theta, \phi)## that represents an eigenstate of ##E, L^2,## and ##L_z##. Can you think of how you could transform ##\psi(r, \theta, \phi)## into another function ##\psi'(r, \theta, \phi)##that would be an eigenstate of ##E, L^2,## and ##L_x## with the same eigenvalues for ##E## and ## L^2## and with an eigenvalue of ##L_x## equal to the eigenvalue that ##\psi## had for ##L_z##?

[Edit: It might be easier to think in terms of Cartesian coordinates ##\psi(x, y, z)]##
 
Last edited:
  • #3
bobred
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Thanks, z is an arbitrary choice.
 

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