Phase/Intensity uncertainty relation

Question

In quantum optic, we have the relationship:
$$Delta NDelta theta>1/2tag 1$$
But to my knowledge, $thetain[0,2pi[$ so $Deltatheta$ is finite, then for example if we know perfectly the number of photons: $Delta N=0$ and since $Delta theta$ is finite:
$$Delta NDelta theta=0$$
So my guess is that, $theta$ can take every value we want... But isn't $theta+2pi$ strictly equivalent physically to $theta$?
Graphically, isn't a fock state represented by a circle (infinitely thin) in the optical phase space? Justifying even more the idea of a finite $Delta theta$?
Where is the problem?

user34722 · Accepted Answer

The problem is that $Delta N Delta theta>1/2$ does not hold if you use the standard definition of variance, $Deltatheta = langle(theta-langlethetarangle)^2rangle$. However there is a nice generalization of variance to periodic variables, called the Holevo variance
$Deltatheta^2 equiv |langle e^{itheta}rangle|^{-2} - 1$
One then has $Delta N^2 Delta theta^2 = 1/4$. To your question, $Deltatheta^2$ diverges for a uniform distribution over $theta$, so it works for Fock states. This generalization of variance has the nice property that it reduces to the standard definition of variance for $|theta|llpi$.
The reason the standard uncertainty relation doesn't hold here is that phase is an unusual quantity compared to other quantum mechanical observables. In particular, there is no Hermitian observable whos eigenstates correspond to all the states of definite phase, which itself is an overcomplete basis. Therefore the usual derivation of the uncertainty principle doesn't work.

Phase/Intensity uncertainty relation

One Answer

Add your own answers!

Ask a Question