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In quantum mechanics, we convert Poisson brackets to commutators for the observables to account for the uncertainty principle. However, I do not understand why do we do this.

  • What motivates us to choose commutators?
  • Why not anti-commutators or some other mathematical structure?
  • How does uncertainty in simultaneous measurements of certain observables lead to commutation relations?
Saurabh Shringarpure
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    It is a consequence of the Dirac postulate. In a quantum measurement, the post-measurement state is an eigenstate of the observable(A) you are measuring. Now if you want to measure an incompatible observable for this post-measurement state, it better not be an eigenstate of the new observable(B). This condition is satisfied only if [A,B] doesn't equal to 0 – Arnab Apr 19 '21 at 23:25
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    Anti-commutators are important in quantum theories describing fermions like QFT. – Charlie Apr 19 '21 at 23:27
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    Agree with @Arnab. The Dirac Quantization Condition is a fundamental postulate of QM (like the Principle of Extremized Action, often referred to as the Principle of Least Action, is a fundamental postulate of classical mechanics). Heisenberg Uncertainty is a consequence of the DQC, not the other way around. If memory serves, https://arxiv.org/abs/quant-ph/9907069v2 is a good resource for reading more about how uncertainty manifests. Also agree with @Charlie; see also anyons in 2+1D QFTs. – astronautgravity Apr 20 '21 at 00:17
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    Related: https://physics.stackexchange.com/q/573908 – Nihar Karve Apr 20 '21 at 03:07

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