# I Commutator of Parity operator and angular momentum

1. Nov 28, 2016

### dyn

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. Nov 28, 2016

### BvU

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 $-2L_zP_x = 2 P_xL_z$

3. Nov 28, 2016

### dyn

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

4. Nov 29, 2016

### BvU

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.

5. Nov 29, 2016

### dyn

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 ?

6. Nov 30, 2016

### BvU

Don't have the easy answer: you can't just look at parity on its own. @mfb ?

7. Nov 30, 2016

### vanhees71

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.