# Do Commuting Operators Always Form a Basis in QM and QFT?

• wangyi
In summary, the conversation discusses the relationship between operators in quantum mechanics and quantum field theory. The question is raised whether, if two operators commute with a third, will the two operators also commute with each other. The answer is no, as demonstrated with counterexamples in both QM and QFT. The conversation also touches on the concept of degeneracy in eigenvalues and its effect on the commutativity of operators.
wangyi
Hi, I have a question,
As it is said in QM, if two operators commute, they have so many common eigenstates that they form a basis. And the inverse is right.
Now there is the question,
if A,B,C are operators, [A,B]=0, [A,C]=0,
then is "[B,C]=0" also right?

If we simply say A and B, A and C both have common eigenstates, so B and C have common eigenstates, so [B,C]=0, it seems to be right.

But in QFT, if x,y spacelike, then [\phi(x),\phi(y)]=0,
if the above is right, then we can find a point z which is spacelike according to two non-spacelike point x,y to make any non-spacelike [\phi(x),\phi(y)]=0. It looks like a paradox.

thank you!

> If A,B,C are operators, [A,B]=0, [A,C]=0,
> then is "[B,C]=0" also right?

No, it's not right. For a counterexample in the usual QM variables, let $$A=x, B=y, C=p_y$$.

For a counterexample in the Dirac gamma matrices,
let $$A=\gamma^0, B=\gamma^1\gamma^2, C=\gamma^1\gamma^3$$.

For a counterexample in QFT, replace the gamma matrices with your favorite four anticommuting field variables.

In each of these counterexamples, A commutes with B and A commutes with C, but B and C do not commute.

Carl

wangyi said:
Now there is the question,
if A,B,C are operators, [A,B]=0, [A,C]=0,
then is "[B,C]=0" also right?

No. The angular momentum operators give a counterexample: A = L^2, B = L_x, and C = L_y. Then [A,B] = [A,C] = 0. But [B,C]= [L_x, L_y] = ih L_z.

It is true, however, that [B, C] commutes with A. This can be seen from the jacobi identity

[A, [B, C]] + [B, [C, A]] + [C, [A, B]] = 0.

> If we simply say A and B, A and C both have
> common eigenstates, so B and C have common
> eigenstates, so [B,C]=0, it seems to be right.

If A has no degeneracy in its eigenvalues, then your logic works. In the presence of degeneracy, A can arrange to share a different set of eigenstates with B than it shares with C.

Carl

[A,B]=0 means you can find a set of eigenstates common to A and B.
[A,C]=0 means you can find a set of eigenstates common to A and C.

That doesn't imply these two sets are the same, so it will in general not give a set of eigenstates common to B and C.

Thank you all, i see :)

## What is operator commutation?

Operator commutation is a mathematical concept that describes the behavior of operators when they are applied to a function or system. It refers to the order in which operators are applied and whether or not the result is affected by that order.

## Why is operator commutation important?

Operator commutation is important because it allows us to simplify and manipulate mathematical expressions involving operators. It also has practical applications in quantum mechanics and other fields of physics.

## What is the commutator of two operators?

The commutator of two operators A and B is defined as [A,B] = AB - BA. It measures the degree to which the operators do not commute, or do not produce the same result when applied in different orders.

## What is the difference between a commutator and an anticommutator?

The commutator and anticommutator are two different ways of measuring the commutativity of two operators. The commutator measures how much the operators do not commute, while the anticommutator measures how much they do commute.

## What is the Heisenberg uncertainty principle and how is it related to operator commutation?

The Heisenberg uncertainty principle states that the more precisely we know the position of a particle, the less precisely we can know its momentum, and vice versa. This is related to operator commutation because the operators representing position and momentum do not commute, and therefore cannot be measured simultaneously with arbitrary precision.

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