Is the commutator of two operators always a scalar?

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Discussion Overview

The discussion revolves around the nature of the commutator of two operators in quantum mechanics, specifically whether it is always a scalar or can be a matrix. Participants explore the implications of the commutator in the context of quantum mechanics and its representation in mathematical terms.

Discussion Character

  • Technical explanation, Conceptual clarification, Debate/contested

Main Points Raised

  • One participant states that the commutator [A,B] is generally a matrix, but notes that the specific case of the position and momentum operators results in a scalar, i*hbar.
  • Another participant clarifies that in this context, i*hbar is shorthand for i*hbar times the identity operator.
  • A later reply expresses surprise that this point was not made clear in Griffiths' text, suggesting a gap in the explanation of the mathematics involved.
  • Another participant points out that the commutator of two scalars is a scalar, while the commutator of two vectors results in a matrix, providing an example involving the Kronecker delta.
  • This participant also mentions that the trace of the commutator in four dimensions would yield a specific numerical result.

Areas of Agreement / Disagreement

Participants express differing views on the nature of the commutator, with some asserting it is a scalar in specific cases while others argue it is generally a matrix. The discussion remains unresolved regarding the broader implications of these interpretations.

Contextual Notes

There is a lack of clarity regarding the definitions and assumptions surrounding the commutator, particularly in relation to the identity operator and the dimensionality of the operators involved.

Aziza
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[A,B] = AB-BA, so the commutator should be a matrix in general, but yet
[x,p]=i*hbar...which is just a scalar. Unless by this commutator, we mean i*hbar*(identity matrix) ?

I am asking because I see in a paper the following:

tr[A,B]

Which I interpret to mean the trace of the commutator [A,B]. But if [A,B] is just a scalar, then trace of a scalar should always just be the scalar..
 
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Yes, in this case ##i \hbar## is shorthand for "##i \hbar## times the identity operator."
 
Ohhh ok thanks! I can't believe I went through all of Griffiths and this point was never made clear.
 
Since Griffiths doesn't get into the mathematics of the CCRs, nor specifically treats the old matrix mechanics, it's quite understandable that he leaves out the unit operator/matrix.
 
The commutator of two scalars is a scalar.
The commutator of two vectors is a matrix. e.g.

[xi,pj]=iħδij

in this case it is the unity matrix. The trace of this in 4 dimensions would be 4.
 

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