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Hmmm, it would seem 0=1.

This contains some stuff that's a little advanced, but can be looked at by both theoretical physicists (2/3rd ish year) and by mathematicians (2/3rd ish year). I saw it on another forum and thought it was quite subtle. I'll pose it in the language of physics (kets, bras, thongs etc), but I hope the notation is obvious.

Let A be a self adjoint operator, with normalised eigenstates

|a \rangle

, with corresponding eigenvalue a (real since A is self adjoint). And so:

A |a \rangle = a |a \rangle

Introduce a the operator B, canonically conjugate to A, so that:

[A,B] = i I

Where I is the identitiy on whichever Fock (tweaked Hilbert) space we're working on. Then it follows that:

\langle a | [A,B] |a \rangle = a \langle a| B |a \rangle - a \langle a | B | a \rangle = 0

But since

[A,B]=I

, it follows that the LHS is equal to

\langle a |iI| a \rangle = i

. So are we to conclude that 1=0? What the Jones has gone wrong? Spoiler/hint in white: do there exist finite dimensional linear operators, A and B, such that [A,B]=I?

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Reply 1

Kolya

Wrangler

I'll pose it in the language of physics (kets, bras, thongs etc)

So that's what you get lectured on at Cambridge :rolleyes:

Reply 2

generalebriety

I thought this was going to be the age-old a=b rigmarole.

How wrong I was.

Reply 3

AlphaNumeric

Give Wrangler a little more credit than that :wink:

He'd not make it so easy.

I'll admit though, I'd no idea the answer when I read it on the forums I'm guessing Wrangler got it from until I read the rest of the thread where much more knowledgable people than I explained it.

Reply 4

yusufu

Ah, so simple...

Spoiler

Reply 5

FWoodhouse

This is pretty simple, actually. Clearly if [A,B] = I, then there must exist a self-homolytic adjoint space X in any suitable vector subsystem containing a stabilising Cauchy space S such that

\langle |S,X|[X] \rangle = \lfloor A,S \rfloor \times B

. The contradiction is obvious.

Reply 6

Wrangler

FWoodhouse

This is pretty simple, actually. Clearly if [A,B] = I, then there must exist a self-homolytic adjoint space X in any suitable vector subsystem containing a stabilising Cauchy space S such that

\langle |S,X|[X] \rangle = \lfloor A,S \rfloor \times B

. The contradiction is obvious.

Lol, nice try! :wink:

The starting point is accesible to A-Level students: can you find matrices A,B such that AB - BA = I?

Spoiler

And bingo was his name-o! So we can't be working on some finite dimensional space - so the operators are not necessarily bounded. Note that:

A^n B = A^{n-1}AB = A^{n-1}(BA + iI)

A little induction then gives:

in A^{n-1} = A^n B - B A^n \quad \Rightarrow \quad n|| A^{n-1} || \, \leq \, 2 || A^n||\,|| B ||

So one of A,B must be unbounded - i.e cannot have the whole of the space in it's domain. Our error was assuming that

|a\rangle \, \in \, \mathcal{D}( [A,B] )

. Just goes to show that we need to be a little careful when doing field theories (or just QM), because the maths can come round and bite you in the bum!

:smile:

Whimper.

wanderer

Whimper.

Seconded.

Reply 11

wanderer

Profesh

Seconded.

Thank you. But what are you doing in the maths forum?

Reply 12

AlphaNumeric

wanderer

Whimper.

Just think, should you make your offer and get into Peterhouse you'll probably have Wrangler supervise you at some point :p:

Reply 13

Profesh

wanderer

Thank you. But what are you doing in the maths forum?

It's the plural of 'math', don't you know.

*runs away*

Reply 14

wanderer

Profesh

It's the plural of 'math', don't you know.

*runs away*

Its an abbreviation of 'mathematics', actually. You're the last person I would expect to see bombarding us with Americanisms. :p:

Reply 15

wanderer

AlphaNumeric

Just think, should you make your offer and get into Peterhouse you'll probably have Wrangler supervise you at some point :p:

I'm not looking that far ahead. If I focus on passing time right now, then results day will be here before I know it. Or at least thats what I keep telling myself.

Reply 16

Profesh

wanderer

Its an abbreviation of 'mathematics', actually. You're the last person I would expect to see bombarding us with Americanisms. :p:

Clearly, I was obfuscating.

Reply 17

Cexy

Wrangler

And bingo was his name-o! So we can't be working on some finite dimensional space - so the operators are not necessarily bounded. Note that:

A^n B = A^{n-1}AB = A^{n-1}(BA + iI)

A little induction then gives:

in A^{n-1} = A^n B - B A^n \quad \Rightarrow \quad n|| A^{n-1} || \, \leq \, 2 || A^n||\,|| B ||

So one of A,B must be unbounded - i.e cannot have the whole of the space in it's domain. Our error was assuming that

|a\rangle \, \in \, \mathcal{D}( [A,B] )

. Just goes to show that we need to be a little careful when doing field theories (or just QM), because the maths can come round and bite you in the bum!

:smile:

That's quite cool. I have a possibly stupid question - why does it follow that because one of

A,B

is unbounded, they can't have the whole space as their domain? I can think of plenty of finite-dimensional operators that are unbounded yet have an entire Hilbert space as their domain - why does it fail in the infinite dimensional case?