Do Functions of Commuting Operators Always Commute?

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    Commutator Property
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Discussion Overview

The discussion centers on whether functions of commuting operators always commute with each other, specifically in the context of quantum mechanics. Participants explore the implications of commuting operators and provide examples and counterexamples related to this concept.

Discussion Character

  • Debate/contested
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • Some participants propose that if two operators A and B commute, then any function of A will commute with any function of B.
  • One participant provides a specific example involving the operators L_z and r^2, claiming that since [L_z, r^2] = 0, it follows that [L_z, r] = 0, and seeks proof for this assertion.
  • Another participant discusses the expansion of functions of operators as power series and suggests that if A and B commute, then A^n commutes with B^m for any integers n and m.
  • A participant questions how to prove the commutation for decreasing powers of operators, indicating a need for clarification on the reverse argument.
  • One participant presents a mathematical manipulation involving L and r, attempting to derive that [L, r] = 0, but expresses uncertainty about the validity of their reasoning.
  • A counterexample is provided involving the Pauli matrices, where it is noted that while [σ_x, σ_y^2] = 0, [σ_x, σ_y] ≠ 0, suggesting that the initial claim does not hold in general.
  • Another participant acknowledges the counterexample and expresses understanding that proving [L, r] = 0 may not have shortcuts.

Areas of Agreement / Disagreement

Participants express disagreement regarding the generality of the claim that functions of commuting operators always commute. Some provide examples supporting the claim, while others present counterexamples that challenge it, indicating that the discussion remains unresolved.

Contextual Notes

The discussion includes various assumptions about the properties of operators and their functions, with some participants relying on specific examples that may not universally apply. There are also unresolved mathematical steps in the arguments presented.

McLaren Rulez
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Hi,

If we have two commuting operators A and B, is true that any function of A will commute with any function of B? I have a result which takes [itex][L_{z},r^{2}]=0[/itex] and claims that [itex][L_{z},r]=0[/itex]. How can this be proved? Thank you
 
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McLaren Rulez said:
Hi,

If we have two commuting operators A and B, is true that any function of A will commute with any function of B? I have a result which takes [itex][L_{z},r^{2}]=0[/itex] and claims that [itex][L_{z},r]=0[/itex]. How can this be proved? Thank you

Note that A[itex]^{2}[/itex]B[itex]^{2}[/itex] = AABB = A(AB)B = A(BA)B since A and B commute.
= (AB)AB = (BA)(BA) = B(AB)A = B(BA)A = B[itex]^{2}[/itex]A[itex]^{2}[/itex]

i.e. A[itex]^{2}[/itex] commutes with B[itex]^{2}[/itex].

In like manner one can prove that A[itex]^{n}[/itex] commutes with B[itex]^{m}[/itex].

Note also that many functions of A can be expanded as power series in A. Similarly for functions of B.
 
Thank you for replying grzz.

I understand how to go from a lower power to a higher power. You showed that if A and B commute, then A^2 and B^2 also commute. But the reverse is what is being done in my question so how can I prove it for decreasing powers?

Thank you.
 
I did not read your qustion carefully!
 
Can anyone else help? I have this idea but I am not sure if its correct.

We have [itex]Lr^{2}-r^{2}L=0[/itex]

So, [itex]Lr^{2}-rLr+rLr-r^{2}L=0[/itex] where I just added and subtracted the middle terms.

Then [itex][L,r]r+r[L,r]=0[/itex].

Would it be correct to say that since this is true for arbitrary r, [L,r] must be zero? I am not sure of this argument, especially the last step. Thank you

EDIT: No, that's not right. So any help?
 
It's not true in general.
Counterexample: take the pauli matrices [itex][\sigma_x,\sigma_y^2]=[\sigma_x,1]=0[/itex], but [itex][\sigma_x,\sigma_y]=2i\sigma_z \neq0[/itex]
 
Thank you aesir. I get it now.

Looks like there aren't any shortcuts to prove [L,r]=0.
 

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