- #1

ShayanJ

Gold Member

- 2,794

- 596

I know this is raised several times in this forum but I still don't see what's the solution! So I want to discuss it again.

Consider two self-adjoint operators A and B which obey [itex] [A,B]=c I[/itex]. Now I take the normalized state [itex] |a\rangle [/itex] such that [itex] A|a\rangle=a |a\rangle [/itex]. Now I can write:

[itex]

1=\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 |a \rangle-\langle a | BA |a \rangle]=\\ \frac 1 c [(A|a\rangle)^\dagger B|a\rangle-a\langle a |B|a\rangle]=\frac 1 c [(a|a\rangle)^\dagger B|a\rangle-a\langle a |B|a\rangle]=\frac a c [\langle a | B |a \rangle-\langle a | B |a \rangle]=0

[/itex]

So there should be something wrong with the reasoning. There were several suggestions before but I don't think they work.

P.S.

1) Please

2) I know that the commutation relation can't be realized in a finite dimensional Hilbert space. So take an infinite-dimensional Hilbert space! Can this be used here?

3) I know that at least one of the operators should be unbounded. Can this be used here?

4) I know domains are important. But I can still find a vector in the intersection of the domains of A and B. Can you prove their domains are disjoint? Or can you prove there is no eigenvector of A in the intersection of the domains of A and B?

5) I know that here we should consider a rigged Hilbert space. But a rigged Hilbert space is actually a triplet like [itex] D \subset L^2(\mathbb R,dx) \subset D' [/itex]. So I can still choose vectors from D which are actually ordinary nice vectors. Can you prove there is no eigenvector of A in D? Or can you prove for such an eigenvector, B necessarily gives ill-defined results so that [itex] \langle a | B |a \rangle-\langle a | B |a \rangle [/itex] is indeterminate instead of 0?

6) An important point here is the symmetry between A and B. So even if we can use one of the above suggestions, we then can swap the role of A and B and retain the argument. Can you break this symmetry?

7) Hey @Greg Bernhardt , can you set a prize for solving this?:D

I hope this time I'll get a definite solution for this and get this case closed.

Thanks

Consider two self-adjoint operators A and B which obey [itex] [A,B]=c I[/itex]. Now I take the normalized state [itex] |a\rangle [/itex] such that [itex] A|a\rangle=a |a\rangle [/itex]. Now I can write:

[itex]

1=\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 |a \rangle-\langle a | BA |a \rangle]=\\ \frac 1 c [(A|a\rangle)^\dagger B|a\rangle-a\langle a |B|a\rangle]=\frac 1 c [(a|a\rangle)^\dagger B|a\rangle-a\langle a |B|a\rangle]=\frac a c [\langle a | B |a \rangle-\langle a | B |a \rangle]=0

[/itex]

So there should be something wrong with the reasoning. There were several suggestions before but I don't think they work.

P.S.

1) Please

**DO NOT**specialize to momentum and position!(Or any other pair, unless you can prove there is only a few pairs of operators that satisfy the requirements and you can say what is wrong about each pair.)2) I know that the commutation relation can't be realized in a finite dimensional Hilbert space. So take an infinite-dimensional Hilbert space! Can this be used here?

3) I know that at least one of the operators should be unbounded. Can this be used here?

4) I know domains are important. But I can still find a vector in the intersection of the domains of A and B. Can you prove their domains are disjoint? Or can you prove there is no eigenvector of A in the intersection of the domains of A and B?

5) I know that here we should consider a rigged Hilbert space. But a rigged Hilbert space is actually a triplet like [itex] D \subset L^2(\mathbb R,dx) \subset D' [/itex]. So I can still choose vectors from D which are actually ordinary nice vectors. Can you prove there is no eigenvector of A in D? Or can you prove for such an eigenvector, B necessarily gives ill-defined results so that [itex] \langle a | B |a \rangle-\langle a | B |a \rangle [/itex] is indeterminate instead of 0?

6) An important point here is the symmetry between A and B. So even if we can use one of the above suggestions, we then can swap the role of A and B and retain the argument. Can you break this symmetry?

7) Hey @Greg Bernhardt , can you set a prize for solving this?:D

I hope this time I'll get a definite solution for this and get this case closed.

Thanks

Last edited: