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I Commutator of Parity operator and angular momentum

  1. Nov 28, 2016 #1

    dyn

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    Hi
    I have seen an example of commutator of the Parity operator of the x-coordinate , Px and angular momentum in the z-direction Lz calculated as [ Px , Lz ] ψ(x , y) = -2Lz ψ (-x , y)
    I have tried to calculate the commutator without operating on a wavefunction and just by expanding commutator brackets and I get [ Px , Lz ] = 2PxLz
    Are these 2 answers equivalent and if so how do I get from my answer to the given answer ?
    Thanks
     
  2. jcsd
  3. Nov 28, 2016 #2

    BvU

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    With $$
    P_x L_z = P_x (xp_y) - P_x (yp_x) = -x p_y P_x + yp_x P_x = - L_z P_x \Rightarrow P_x L_z + L_z P_x = 0
    $$you get $$
    [P_x, L_z] + 2 L_z P_x = 0 \Rightarrow [P_x, L_z] = - 2 L_z P_x
    $$So indeed your ##-2L_zP_x = 2 P_xL_z##
     
  4. Nov 28, 2016 #3

    dyn

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    Thanks for that. The example then goes on to state that [ H , Px ] = 0 , [ H , Lz ] = 0 and [ Px , Lz ]≠ 0 implies that energy eigenstates must be degenerate. Any ideas why ?
    The parity operator always confuses me. As a Hermitian operator I presume it is related to an observable. I understand (hopefully ! ) that when dealing with eg. the momentum operator and I take a measurement I get an eigenvalue giving me the momentum. But what does the parity operator give me ? can parity be measured ? Is it an observable ?
    Thanks
     
  5. Nov 29, 2016 #4

    BvU

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    If ##\ [H,P_x]=0\ ##, there exists a common set of eigenstates for these operators. Idem ##\ [H,L_z]=0\ ##. If energy eigenstates are nondegenerate, the common set for e.g. ##\ [H,P_x]=0\ ## with energy eigenvalue En would be one single state ##\ \left | n \right > \ ##. Since there is only one eigenstate of H, the common set of eigenstates for ##\ [H,L_z]=0\ ## with energy En would be that same state ##\ \left | n \right > \ ##.

    So ##\ \left | n \right > \ ## would be a common eigenstate for ##\ P_x\ ## and ##\ L_z \ ## and therefore you would have ##\ [P_x,L_z]=0\ ## , a contradiction.
     
  6. Nov 29, 2016 #5

    dyn

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    thanks for that clear explanation.
    Can anyone help with the last bit ? I see the point of measuring observables to get eg. position , momentum , energy etc. I presume if you measure parity you get the eigenvalue +1 or -1. Is there any point to measuring the parity observable ?
     
  7. Nov 30, 2016 #6

    BvU

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    Don't have the easy answer: you can't just look at parity on its own. @mfb ?
     
  8. Nov 30, 2016 #7

    vanhees71

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    It's not so clear, what you mean by "measuring parity". At least, I don't know, how to measure parity for a given system. It's a discrete quantum number of states for a discrete quantity that's conserved under the electromagnetic and the strong interaction (but not by the weak interaction). You can only analyse, e.g., reactions concerning the conservation or non-conservation of the corresponding quantum number of the system, but there's no measurement device, just measuring parity.
     
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