# Commutator of Parity operator and angular momentum

• I
• dyn
In summary: And there is no point, "measuring" parity alone, for parity is only conserved as a quantum number of states for a total system, not for the individual states of a system.In summary, the conversation discusses the calculation of the commutator of the Parity operator of the x-coordinate, Px, and angular momentum in the z-direction, Lz. It is shown that [Px, Lz] = -2LzPx and this leads to the conclusion that energy eigenstates must be degenerate. The concept of measuring parity is also discussed, with the understanding that it is a discrete quantum number of states for a system and cannot be measured on its own.
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

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##

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

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.

dyn, PeterDonis and vanhees71
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 ?

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

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.

dyn

## 1. What is the commutator of the Parity operator and angular momentum?

The commutator of the Parity operator and angular momentum is a mathematical operation that calculates the difference between the product of the two operators and the product of the operators in reverse order. It is denoted by [P,L].

## 2. What does the commutator of the Parity operator and angular momentum tell us?

The commutator tells us whether the two operators commute or do not commute. If the commutator is equal to zero, the operators commute, which means they can be applied in any order. If the commutator is not equal to zero, the operators do not commute, which means their order matters.

## 3. How is the commutator of the Parity operator and angular momentum calculated?

The commutator is calculated using the following formula: [P,L] = P*L - L*P, where P is the Parity operator and L is the angular momentum operator.

## 4. What is the physical significance of the commutator of the Parity operator and angular momentum?

The commutator of the Parity operator and angular momentum is important in quantum mechanics as it determines the uncertainty in the measurement of these operators. If the commutator is large, it means that the operators do not commute and their values cannot be measured simultaneously with high precision.

## 5. Can the commutator of the Parity operator and angular momentum be used to determine the spin of a particle?

No, the commutator of the Parity operator and angular momentum does not directly determine the spin of a particle. However, it can be used to derive the spin operator and its associated eigenvalues, which can then be used to determine the spin of a particle.

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