Commutators of vector operators

In summary, [AB,C] = A[B,C] + [A,C]B is a useful identity for expressing the commutator of vector operators \left[\mathbf{\hat A}\cdot\mathbf{\hat B}, \mathbf{\hat C}\right] in terms of simpler commutators. However, the validity of this identity depends on how one defines vector multiplication and whether they are being rigorous with their notation.
  • #1
thecommexokid
70
2
I've been trying to work out some expressions involving commutators of vector operators, and I am hoping some of y'all might know some identities that might make my job a little easier.

Specifically, what is [itex]\left[\mathbf{\hat A}\cdot\mathbf{\hat B}, \mathbf{\hat C}\right][/itex]? Are there any useful identities to express this in terms of simpler commutators?

Any help is appreciated.
 
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  • #2
Are you sure you mean a "vector operator", typically we talk about matrix operators when discussing the commutator relationships (or group elements in a more general setting).
Vector operator: http://en.wikipedia.org/wiki/Vector_operator

However, your question is straightforward,
[S,T] = ST - TS (by definition)

Start with [AB,C] = ABC - CAB (+ ACB - ACB )
= ABC - ACB + ACB - CAB
= A(BC - CB) + (AC - CA)B
= A[B,C] + [A,C]B

Therefore we conclude [AB,C] = A[B,C] + [A,C]B
to be an identity.
Does that answer your question... you could have looked anywhere on the internet to get this... so I'm guessing this isn't what you want.
 
  • #3
  • #4
Thank you for the [STRIKE]reply[/STRIKE]replies.

brydustin said:
[S,T] = ST - TS (by definition)

Start with [AB,C] = ABC - CAB (+ ACB - ACB )
= ABC - ACB + ACB - CAB
= A(BC - CB) + (AC - CA)B
= A[B,C] + [A,C]B

Therefore we conclude [AB,C] = A[B,C] + [A,C]B

It seems to me that you're being pretty cavalier about vector multiplication, what with the way you're just putting vectors in a row next to each other without any dots or parentheses. For instance, what do you mean when you write “ABC”, when A, B and C are vector operators?

I would think that you should define [itex][{\bf{\hat S}},{\bf{\hat T}}] = {\bf{\hat S}} \cdot {\bf{\hat T}} - {\bf{\hat T}} \cdot {\bf{\hat S}}[/itex], and therefore start your derivation with
[tex][{\bf{\hat A}} \cdot {\bf{\hat B}},{\bf{\hat C}}] = ({\bf{\hat A}} \cdot {\bf{\hat B}}){\bf{\hat C}} - {\bf{\hat C}}({\bf{\hat A}} \cdot {\bf{\hat B}}).[/tex]
But from there, I'm not sure how you can safely proceed, if you're being rigorous with your dots and parens. For instance — and correct me if I'm wrong on this — but I don't think [itex]({\bf{\hat A}} \cdot {\bf{\hat C}}){\bf{\hat B}}[/itex] is equal to [itex]{\bf{\hat A}}({\bf{\hat C}} \cdot {\bf{\hat B}})[/itex], so your next step seems iffy.
 
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  • #5
Yeah, sorry I don't know. I thought you were intending for matrix operators. Good luck.
 

1. What is a commutator of vector operators?

A commutator of vector operators is a mathematical operation that measures the degree of non-commutativity between two vector operators. It is defined as the difference between the product of the two operators and the product of the operators in reverse order.

2. Why are commutators of vector operators important?

Commutators of vector operators are important in quantum mechanics because they determine the uncertainty in the measurements of physical quantities. They also play a crucial role in understanding the underlying symmetries and properties of physical systems.

3. How are commutators of vector operators calculated?

Commutators of vector operators are calculated using the commutator algebra, which involves applying the operators to a test function and then taking the difference between the two resulting functions. The resulting function is then integrated over all space to obtain the commutator.

4. What is the physical significance of a zero commutator?

A zero commutator between two vector operators indicates that the operators commute, meaning that their order does not affect the outcome of a measurement. In other words, the two operators can be measured simultaneously with no uncertainty in their values.

5. Are there any rules or properties that govern commutators of vector operators?

Yes, there are several rules and properties that govern commutators of vector operators, such as the linearity property, the product rule, and the Jacobi identity. These rules can be used to simplify and manipulate commutator expressions in order to solve problems in quantum mechanics.

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