A paradox in commutation relation

[01030312]

The following paradox was put forward by "Fredrik" in a discussion on "time-uncertainity relation"-

Now suppose that two hermitian operators have a commutator that's proportional to the identity operator: [A,B]=cI, and eigenvectors satisfying A|a>=a|a> and B|b>=b|b>.

$1=\frac c c\langle a|a\rangle=\frac 1 c\langle a|cI|a\rangle=\frac 1 c\langle a|[A,B]|a\rangle=\frac 1 c\langle a|(aB-Ba)|a\rangle=0$


(Thanks to George Jones for posting this in some other thread).

The same thing happens if we use |b> instead of |a>. What this means is that two hermitan operators that satisfy such a commutation relation (like x and p) can't have eigenvectors. Bounded hermitian operators always do, so this is one way to see that x and p must be unbounded.

Lets look at this closely, using position momentum operators and a general quantum state-
$\langle U|xp - px|U \rangle$

This can be rewritten as-
$\sum_{x'}\sum_{x"}\sum_{p'} (\langle U|x|x' \rangle \langle x'|p|p' \rangle \langle p'|x" \rangle \langle x"|U\rangle - \langle U|x" \rangle \langle x"|p|p' \rangle \langle p'|x|x' \rangle \langle x'|U \rangle)$

At this point, putting $\langle U|x' \rangle = \delta(x-x')$ for some x, and $\delta$ being kroencker delta for discrete case and dirac-delta for continuous case, we get above expression as zero.

But if we proceed further, changing sum to integral, and taking $\langle x'|p' \rangle= e^{ip'x'}$, we get, after some trivial variable changes-
$\int p'(x'-x")e^{ip'(x'-x")}U(x')U^{*}(x")dx'dx"dp'$

Now we perform the integral over p' by writing $\int p'e^{ip'(x'-x")}dp' = \int (\partial (-ie^{ip(x'-x")})/\partial x') dp'$. This is $-i\partial \delta(x'-x")/\partial x'$. Now we perform an integration by parts and obtain $i$ as a result, even if $U(x')=\delta(x-x')$. But without the integration by parts, we would have obtained zero for this wave-function. Does this suggets something about the nature of operators for which a canonical conjugate can be defined. That is, they have to be continuous operators? For as we know, Angular momentum operators do not face the paradox put forward Fredrik, and their spectrum is discrete.

Moreover, something can be said about the canonical conjugate for Hamiltonian operator. Following paper deals extensively with this issue-
http://organizations.utep.edu/Portals/1475/Hilgevoord-Time%20in%20Quantum%20Mechanics.pdf

 

[eljose79]

Dear "Fredrik,"

Thank you for bringing up this interesting paradox. I am always intrigued by paradoxes and their potential to challenge our understanding of the laws of nature. In this case, the paradox you have presented highlights the limitations of our current understanding of the time-uncertainty relation in quantum mechanics.

It is true that the commutation relation [A,B]=cI leads to the paradox that you have described. This means that the two operators, A and B, cannot have simultaneous eigenstates. This is in contrast to the usual understanding of quantum mechanics, where it is assumed that all observables have well-defined eigenstates. However, as you have pointed out, this paradox can be resolved by considering the operators to be unbounded.

In quantum mechanics, operators are considered to be unbounded if they do not have a finite upper bound on their eigenvalues. This means that their eigenstates are not normalizable and cannot be described by a wave function. In the case of the position and momentum operators, this is due to the fact that they are related by a Fourier transform, and the Fourier transform of a wave function is not a wave function itself.

Your analysis of the paradox using the position and momentum operators is correct. The integral over p' can be evaluated using integration by parts, and this leads to the result of i, which is consistent with the uncertainty relation. This also suggests that the operators for which a canonical conjugate can be defined must be continuous, as discontinuous operators would not lead to a well-defined integral.

Regarding the canonical conjugate for the Hamiltonian operator, I agree with your suggestion to refer to the paper you have mentioned. The Hamiltonian operator is a self-adjoint operator, which means that it has a well-defined adjoint operator. However, in the case of the time-uncertainty relation, we are dealing with the Hamiltonian operator as an observable, and not as a generator of time translations. This means that the usual understanding of the canonical conjugate may not apply in this case.

In conclusion, the paradox you have presented highlights the limitations of our current understanding of the time-uncertainty relation in quantum mechanics. It suggests that the operators involved must be unbounded, and that the canonical conjugate for the Hamiltonian operator may not follow the usual understanding. Further research and analysis in this area may shed more light on this paradox and help us deepen our understanding of the fundamental principles of quantum mechanics.

Thank