- #1

01030312

- 30

- 0

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

Lets look at this closely, using position momentum operators and a general quantum state-

[itex]\langle U|xp - px|U \rangle [/itex]

This can be rewritten as-

[itex]\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)[/itex]

At this point, putting [itex]\langle U|x' \rangle = \delta(x-x')[/itex] for some x, and [itex]\delta[/itex] 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 [itex]\langle x'|p' \rangle= e^{ip'x'}[/itex], we get, after some trivial variable changes-

[itex]\int p'(x'-x")e^{ip'(x'-x")}U(x')U^{*}(x")dx'dx"dp'[/itex]

Now we perform the integral over p' by writing [itex]\int p'e^{ip'(x'-x")}dp' = \int (\partial (-ie^{ip(x'-x")})/\partial x') dp'[/itex]. This is [itex]-i\partial \delta(x'-x")/\partial x'[/itex]. Now we perform an integration by parts and obtain [itex]i[/itex] as a result, even if [itex]U(x')=\delta(x-x')[/itex]. 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

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>.

[itex]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[/itex]

(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-

[itex]\langle U|xp - px|U \rangle [/itex]

This can be rewritten as-

[itex]\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)[/itex]

At this point, putting [itex]\langle U|x' \rangle = \delta(x-x')[/itex] for some x, and [itex]\delta[/itex] 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 [itex]\langle x'|p' \rangle= e^{ip'x'}[/itex], we get, after some trivial variable changes-

[itex]\int p'(x'-x")e^{ip'(x'-x")}U(x')U^{*}(x")dx'dx"dp'[/itex]

Now we perform the integral over p' by writing [itex]\int p'e^{ip'(x'-x")}dp' = \int (\partial (-ie^{ip(x'-x")})/\partial x') dp'[/itex]. This is [itex]-i\partial \delta(x'-x")/\partial x'[/itex]. Now we perform an integration by parts and obtain [itex]i[/itex] as a result, even if [itex]U(x')=\delta(x-x')[/itex]. 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

Last edited: