What is physical meaning of anticommuting, not anticommuting operators

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Discussion Overview

The discussion revolves around the physical meaning of anticommuting and non-anticommuting operators in quantum mechanics (QM), particularly in relation to their implications for measurement and observables. Participants explore whether these concepts serve merely as mathematical tools or if they hold deeper physical significance, with references to quantum field theory and interpretations of QM.

Discussion Character

  • Debate/contested
  • Conceptual clarification
  • Technical explanation

Main Points Raised

  • One participant asserts that anticommuting operators have a physical meaning related to measurement, while others challenge this view, suggesting that the concepts are primarily mathematical tools.
  • Another participant references the statistical interpretation of QM, indicating that the measurement problem is complex and subject to various interpretations.
  • Concerns are raised about the implications of non-anticommuting operators, with examples provided to illustrate the inability to measure certain observables simultaneously.
  • A later reply mentions a general expression for the uncertainty principle, suggesting that the commutator's behavior influences the simultaneous measurability of observables.

Areas of Agreement / Disagreement

Participants express disagreement regarding the interpretation of anticommuting and non-anticommuting operators. While some view them as merely mathematical constructs, others argue for their physical relevance, leading to an unresolved discussion.

Contextual Notes

Participants reference different interpretations of quantum mechanics, indicating that the discussion is influenced by subjective views on measurement and observables. The complexity of the measurement problem and the role of commutators and anticommutators in this context remain unresolved.

Roman
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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)
 
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Roman said:
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.

Where did you get that ? That's incorrect.


Roman said:
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?

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.
 
why isn't that correct? for example [L_i,L_j]_-\not=0 means that you can't 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?
 
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.
 
dextercioby said:
...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.

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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