- #1

jfy4

- 649

- 3

## Homework Statement

Consider a quantum system that acts on an N-dimensional space. We showed that any operator could be expressed as a polynomial of the form

[tex]

O=\sum_{m,n=1}^{\infty}o_{mn}U^m V^n

[/tex]

where [itex]U[/itex] and [itex]V[/itex] are complementary unitary operators satisfying [itex](U^N = V^N =1)[/itex] Show that if [itex]O[/itex] commutes with [itex]U[/itex] this polynomial can only be a function of [itex]U[/itex]; if [itex]O[/itex] commutes with [itex]V[/itex] this polynomial can only be a function of [itex]V[/itex]; and if [itex]O[/itex] commutes with [itex]U[/itex] and [itex]V[/itex] this polynomial must be a constant times the identity

## Homework Equations

[tex]UV=VU e^{i2\pi /N}[/tex]

[tex]V^n U=UV^n e^{i2\pi n/N}[/tex]

## The Attempt at a Solution

Me and some buddies thought we had a solution to this, but then another clever buddy pointed out something that we are all unsure of... can a fellow PF lurker lend a hand here

I solved this by assuming the commutation and writing it out explicitly and using the above relavent equations

[tex]

[O,U]=\sum_{m,n=1}^{N}o_{mn}(U^m V^n U-UU^m V^n)=\sum_{m,n=1}^{N}o_{mn}U^{m+1}V^n (e^{i2\pi n/N}-1)=0

[/tex]

I then confidently took

[tex]

e^{i2\pi n/N}=1\implies n=N

[/tex]

which, as stated in the beginning, makes [itex]V^n=V^N=1[/itex] which gives me my polynomial in terms of [itex]U[/itex] only.

Now here is the confusion, there is a summation there in my solution, which seems to make my solution incorrect, since what I wrote doesn't need to be true, just that the sum of the terms needs to add up to zero. Are we missing something obvious, or is that solution simply not valid?

Thanks in advance,