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My background in quantum mechanics is minimal, and I had seen the canonical commutation relation $$[\hat x, \hat p_x] = i\hbar\, \mathbb{I}$$ in a course, I took about two years ago. I'm doing pure mathematics, and in Operator Theory, I learned that the commutator of two bounded linear operators on a Hilbert space cannot be a non-zero scalar multiple of the identity. For references, see Halmos' exposition here. The result was first proved by Wintner, though.

Question: What's going wrong in the case of the canonical commutation relation? Is at least one of the two operators not bounded or linear? Also, what's the Hilbert space on which the operators $\hat x$ and $\hat p_x$ act?

Qmechanic
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stoic-santiago
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Just to develop what people already said in the comment: in fact, you put your finger on the proof that position and momentum operators can't have bounded spectra.

Since those operators have an infinite number of eigenvalues, their trace is undefined (the sum doesn't converge), so the commutation relation can hold without violating any algebric property.

Such operators actually act inside a space that is larger than a simple Hilbert space, a structure called a rigged Hilbert space.

Fun fact: a pair of operators governed by a relation like $[A,B]=i$, then they also verify a Heisenberg-type property, $\Delta a\,\Delta b\geqslant\frac{\lvert[A,B]\rvert}{2}$.

Miyase
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  • Is there a name for the "rigged" Hilbert space on which these operators act? – stoic-santiago Jun 03 '22 at 19:32
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    In English, I believe that the proper name is rigged Hilbert space. In my language (French), it can be translated as a "Gelfand triplet". – Miyase Jun 03 '22 at 19:34
  • @connected-subgroup It should really be '"rigged Hilbert" space,' not '"rigged" Hilbert space.' The rigged space is not a Hilbert space, since it contains non-normalizable states: https://physics.stackexchange.com/a/614466/70245 – Buzz Jun 03 '22 at 21:29
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    Minor remark, there are plenty of bounded operators with an infinite number of eigenvalues (even continuum infinite if you pass me the jargon, since in this case they are not eigenvalues). – lcv Jun 03 '22 at 23:16
  • @lcv This is of course correct. I could have made the distinction between discrete and continuous/residual spectra instead, but I didn't want to get into too much details (and the limits of my knowledge on this topic lies no that far in this direction...) – Miyase Jun 03 '22 at 23:21
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    No problem. But to be precise there are also plenty of bounded operators with purely continuous spectrum. – lcv Jun 03 '22 at 23:27