Understanding Commutators in Quantum Mechanics: General and Specific Questions

[Drew Carey]

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) - $[\vec A,H]=0$ - what is meant by this *exactly*? I see two possible answers:

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

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 $\vec A$ and $\vec B$ commute with $H$, does $\vec A \cdot \vec B$ 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!

 

[maajdl]

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.

 

[MathematicalPhysicist]

H is a scalar operator, i.e:

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

So indeed it's the first case.

 

[Drew Carey]

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 $A$ and $B$ commute with $H$ then: $$[AB,H]=A[B,H]+[A,H]B=0$$
If the answer to my first question was #1, then for vector operators $\vec A$ and $\vec B$ that commute with $H$ we would have: $$[\vec A \cdot \vec B,H]=[A_1B_1,H]+[A_2B_2,H]+[A_3B_3,H]$$ and equals zero by the previous case.

 

[tom.stoer]

if $\vec A$ and $\vec B$ commute with $H$, does $\vec A \cdot \vec B$ also necessarily commute?

yes, because
$$\left[\sum_iA_iB_i,H\right] = \sum_i\left(A_i[B_i,H]+[A_i,H]B_i\right) = 0$$
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?

 

[maajdl]

Bad reading of the initial post!

 

[tom.stoer]

Bad reading of the initial post!

proof style!