Questions about Commutators

  • Thread starter Drew Carey
  • Start date
  • #1
10
0
Hi all,
My motivation is understanding some derivations in Quantum Mechanics, but I think my questions are purely algebraic. I have a general question and then a specific one:

General Question - when writing the commutator of commuting vector and a scalar operators (for instance angular momentum and some Hamiltonian) - [itex][\vec A,H]=0[/itex] - what is meant by this *exactly*? I see two possible answers:

1. [itex][A_i,H]=0[/itex] for [itex]i=1,2,3[/itex]
2. [itex][A_1+A_2+A_3,H]=0[/itex] in which case we could have [itex][A_i,H]\ne0[/itex] for some [itex]i[/itex] .

It seems to me that in the QM context almost always what is meant is the first option but I'm not certain...

Specific Question - if [itex]\vec A[/itex] and [itex]\vec B[/itex] commute with [itex]H[/itex], does [itex]\vec A \cdot \vec B[/itex] also necessarily commute? If the answer to the question above is #1, then obviously it does. If the answer is #2 then I guess not?

Would greatly appreciate the clarifications. Thanks!
 
Last edited:

Answers and Replies

  • #2
maajdl
Gold Member
390
28
General question:
Definitively option 1.
Never in physics you will see such a thing as A1+A2+A3.
Such a quantity has no general meaning.
It is not independent of the frame of reference and physics is independent of the choice of the frame of reference.

Specific question:
The answer is no, not necessarily.
Do you think that A and B being vectors would play a role in the answer to your question?
At least, for scalar operators, the answer is clearly no.
 
  • #3
MathematicalPhysicist
Gold Member
4,514
285
H is a scalar operator, i.e:

[tex][\vec{A}, H ] = \vec{A}H - H\vec{A} = (A_1H,A_2 H , A_3 H) - (HA_1,HA_2,HA_3)[/tex]

So indeed it's the first case.
 
  • #4
10
0
General question:
Definitively option 1.
Never in physics you will see such a thing as A1+A2+A3.
Such a quantity has no general meaning.
It is not independent of the frame of reference and physics is independent of the choice of the frame of reference.

Specific question:
The answer is no, not necessarily.
Do you think that A and B being vectors would play a role in the answer to your question?
At least, for scalar operators, the answer is clearly no.

Can you please elaborate on the second part. What do you mean by " for scalar operators, the answer is clearly no."?
If [itex]A[/itex] and [itex]B[/itex] commute with [itex]H[/itex] then: [tex][AB,H]=A[B,H]+[A,H]B=0[/tex]
If the answer to my first question was #1, then for vector operators [itex]\vec A[/itex] and [itex]\vec B[/itex] that commute with [itex]H[/itex] we would have: [tex][\vec A \cdot \vec B,H]=[A_1B_1,H]+[A_2B_2,H]+[A_3B_3,H][/tex] and equals zero by the previous case.
 
  • #5
tom.stoer
Science Advisor
5,778
165
if [itex]\vec A[/itex] and [itex]\vec B[/itex] commute with [itex]H[/itex], does [itex]\vec A \cdot \vec B[/itex] also necessarily commute?
yes, because
[tex]\left[\sum_iA_iB_i,H\right] = \sum_i\left(A_i[B_i,H]+[A_i,H]B_i\right) = 0[/tex]
as you said.

(except for quantization anomalies in quantum field theories)

At least, for scalar operators, the answer is clearly no.
???

Where's the problem? Can you please provide a counterexample?
 
Last edited:
  • #6
maajdl
Gold Member
390
28
Bad reading of the initial post!
 
  • #7
tom.stoer
Science Advisor
5,778
165
Bad reading of the initial post!
proof style!
 

Related Threads on Questions about Commutators

Replies
2
Views
808
Replies
2
Views
521
Replies
4
Views
1K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
9
Views
2K
  • Last Post
Replies
9
Views
3K
  • Last Post
Replies
11
Views
846
  • Last Post
Replies
0
Views
2K
  • Last Post
Replies
6
Views
841
  • Last Post
Replies
4
Views
1K
Top