# What is physical meaning of anticommuting, not anticommuting operators

1. Jan 21, 2007

### Roman

hello everyone,

while studying QM you learn the physical meaning of commutating operators, namely they have simultaneous eigenstates. For observables it means, that they can be simultaneusly exactly mesured.

What is the physical meaning of anticommuting and not anticommuting operators? $$[A,B]_+=0, [A,B]_+\not=0$$
Is there any physical meaning or is it just a mathematical tool?

(sorry for my bad english)

Last edited: Jan 21, 2007
2. Jan 22, 2007

### dextercioby

Where did you get that ? That's incorrect.

The issues with anticommutators pertain to quantum field theory and they are the convenient mathematical tool which is necessary for the theory of fermionic fields to be valid.

Daniel.

3. Jan 22, 2007

### Roman

why isnt that correct? for example $$[L_i,L_j]_-\not=0$$ means that you cant exactly measure two components of angular momentum at the same time, the same for $$[x_i,p_i]_-\not=0$$.
can you find a counter-example?

so it is just a mathematical tool?

4. Jan 23, 2007

### dextercioby

The question and whole problematic of measurement is a thorny subject in quantum mechanics. This is subject to the different "interpretations". Basically i follow the "statistical interpretation" which guided Leslie Ballentine to write his excellent book on QM.
That's why i claim that what you wote above is incorrect.

Yes, anticommutators, just like commutators are nothing but a mathematical tool.

Daniel.

5. Feb 4, 2007

### ahrkron

Staff Emeritus
IIRC, a somewhat general expression for the uncertainty principle shows that the uncertainty in the measurement of two quantities is proportional to their commutator, which means that, regardless of interpretation, two observables can indeed be measured simultaneously if they have a vanishing commutator (with the possible exception in the case of time-dependent operators).

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