If you define operators $a, a^\dagger$ which satisfy (e.g.) the relations $\{a,a\}=\{a^\dagger,a^\dagger\}=0$ and $\{a,a^\dagger\}=1$.

Will this uniquely define the operators such that $a |0\rangle \propto 0$, $a |1 \rangle \propto |0\rangle$, $a^\dagger |0\rangle \propto |1\rangle$, $a^\dagger |1\rangle \propto 0$?

  • 1
    $\begingroup$ It depends on what you mean by "uniquely define." Given the anticommutation relations that you wrote, we can construct an irreducible Hilbert-space representation that respects those commutation relations, and all such Hilbert-space representations are unitarily equivalent to each other. (This is obvious, because the Hilbert space is only two-dimensional.) But we don't normally consider just one $a,a^\dagger$ pair in isolation. It's usually in the context of a richer algebra, for which a two-dimensional Hilbert space is no longer sufficient. $\endgroup$ Nov 13 '21 at 16:01
  • $\begingroup$ Will it be uniquely defined if you provide all the respective anticommutators for a richer algebra? $\endgroup$
    – Alex Gower
    Nov 13 '21 at 16:33
  • $\begingroup$ In general, no: quantum field theories typically admit unitarily inequivalent Hilbert-space representations of the same algebra of field operators. A quick on-line search for the keywords "QFT unitarily inequivalent" turned up this paper and this paper and this paper and this Wikipedia page. (Disclaimer: I haven't quality-checked any of these.) $\endgroup$ Nov 13 '21 at 17:16
  • 1
    $\begingroup$ Ah, I think I might see the issue. Just because operators happen to satisfy the (anti)commutation relations of creation/annihilation operators doesn't mean they have anything at all to do with particles. Determining which operators create/annihilate particles is a (usually) difficult dynamics problem, involving the Hamiltonian among other things. The names "creation/annihilation" operators are borrowed from the fact that their commutation relations are the same as those of operators that do create/annihilate particles in trivial models, but in general it's just a name. $\endgroup$ Nov 15 '21 at 14:06
  • 1
    $\begingroup$ I don't know if any of my comments here would qualify as an answer to the intent of your original question, but I'd be happy to move the comments into an answer instead of you like, or I can just plain delete them if they're no longer needed. $\endgroup$ Nov 15 '21 at 14:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.