An example of a pair of anti-commutative Hermitian operators in a finite Hilbert space is $\sigma_x$ with $\sigma_z.$ Indeed $\sigma_z\sigma_x=i\sigma_y$, whereas $\sigma_x\sigma_z=-i\sigma_y$. My question is, do there exist pairs of anti-commutative Hermitian operators in an infinite dimensional Hilbert space? That is, Hermitian operators, a and b, satisfying, $[a,b]=2ab.$ If so, please enlighten me with an example! If not (which is what I suspect), why not?

  • $\begingroup$ Can it be any infinite-dimensional Hilbert space? Or did you have a specific one in mind? $\endgroup$ – probably_someone Jun 18 '17 at 6:15
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    $\begingroup$ An infinite direct sum of spaces $\mathbb{C}^2$ immediately produces an infinite dimensional couples of operators satisfying your constraint... $\endgroup$ – Valter Moretti Jun 18 '17 at 7:35
  • $\begingroup$ @probably_someone I'm thinking of operators such as momentum and position. $\endgroup$ – Henry Jun 18 '17 at 18:42
  • $\begingroup$ @ValterMoretti Surely a commutative space such as $\mathbb{C}^2$ would have $[a,b]=0$. I should have specified that I want nonzero operators too. $\endgroup$ – Henry Jun 18 '17 at 18:51
  • $\begingroup$ @Henry you are wrong. See my answer below. $\endgroup$ – Valter Moretti Jun 18 '17 at 19:18

Consider the infinite-dimensional Hilbert space $H$ over the algebraic direct sum $\oplus_{n=0}^{+\infty} H_n$ where $H_n = \mathbb{C}^2$, equipped with the scalar product $$\langle \{x_n\}_{n \in \mathbb{n}}| \{y_n\}_{n \in \mathbb{n}}\rangle = \sum_{n=0}^{+\infty} \overline{x_n}^t y_n\:.$$ and the elements of $H$ are the sequences $\{x_n\}_{n \in \mathbb{N}}$ such that $$\sum_{n=0}^\infty ||x_n||^2 < +\infty\:.$$ Next consider the densely-defined Hermitian operators $$A = \oplus_{n=0}^{+\infty} \sigma_z$$ $$B = \oplus_{n=0}^{+\infty} \sigma_x$$ with domain given by the space of definitively vanishing sequences. It evidently holds on that domain $$AB = i \oplus_{n=0}^{+\infty} \sigma_y$$ and $$BA = -i \oplus_{n=0}^{+\infty} \sigma_y\:.$$ Thus $$AB= -BA$$ as required.

  • $\begingroup$ Thank you. Do you have an example that doesn't involve Pauli matrices? $\endgroup$ – Henry Jun 19 '17 at 18:57
  • $\begingroup$ It depends on what you mean by "doesn't involve Pauli matrices". If $H_1$ is every separable Hilbert space like $L^2(\mathbb{R})$, since $H$ is separable, there is a surjective isometric map $U: H \to H_1$. Thus define $A' := UAU^{-1}$ and $B' := UBU^{-1}$. $A'$ and $B'$ are denseli defined and Hermitian and $A'B'= -B'A'$. $\endgroup$ – Valter Moretti Jun 19 '17 at 19:17
  • $\begingroup$ A natural question is whether or not this is the only example, modulo the choice of $U$. Under some further hypotheses it is...I think, but I do not have much time to think about. $\endgroup$ – Valter Moretti Jun 19 '17 at 19:18
  • $\begingroup$ a related question is here math.stackexchange.com/questions/1680239/… $\endgroup$ – Valter Moretti Jun 19 '17 at 19:23

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