-1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.