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I’m trying to get a clearer physical and intuitive understanding of the number–phase uncertainty relation in quantum mechanics, especially in quantum optics.

$$\Delta \phi \cdot \Delta N \gtrsim 1$$

I know that phase and number are canonically conjugate (at least formally), and that there's an uncertainty relation like $\Delta \phi \cdot \Delta N \gtrsim 1$. I'm also aware that defining a proper phase operator is tricky due to the number operator’s discrete, bounded-from-below spectrum. I was also wondering if that can be solved by acting like a $N=-1$ corresponds to an antiparticle, this would remove the lowerbound.

What I’m really trying to get at is this:

  • What does it physically mean to say the phase is uncertain? I get what it means for number to fluctuate (photon counting), but what exactly is fluctuating in the case of phase?
  • If I measure something, what changes when the phase uncertainty is large? What does this look like in an experiment or in a real system?
  • Is there a clean way to picture this, like how we understand position–momentum uncertainty as a kind of Fourier duality?

I'm not looking for a full formal treatment, more of a solid conceptual or operational explanation. I've seen that interference patterns vanish in Fock states and appear in coherent states, but I'm still trying to connect that to the idea of phase being "uncertain" in a more concrete way.

Any good physical analogies, examples, or simple thought experiments would be appreciated.

Qmechanic
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bananenheld
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1 Answers1

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What does it physically mean to say the phase is uncertain? I get what it means for number to fluctuate (photon counting), but what exactly is fluctuating in the case of phase?

If we consider a coherent state: $$ a|\alpha\rangle=\alpha|\alpha\rangle\Rightarrow |\alpha\rangle = e^{-\frac{|\alpha|^2}{2}}\sum_{n=0}^\infty \frac{\alpha^n}{n!}|n\rangle, $$ where $\alpha =|\alpha|^{i\phi}$, then the expectation value of a typical EM field operator in respect to this state is $$ \langle \alpha|E_0(a^\dagger + a)|\alpha\rangle=E_0(\alpha^*+\alpha)=2E_0|\alpha|\cos\phi, $$ which is but a classical field with a definite phase (while a coherent state accounts for the quantum fluctuations about this average.)

On the other hand, the expectation value of the same operator in respect to a state with a definite number of photons, $|n\rangle$ is zero.

Is there a clean way to picture this, like how we understand position–momentum uncertainty as a kind of Fourier duality?

IMHO, it is possible to do this without defining a phase operator (that the suggested duplicate deals with.) We could define probability density of amplitude and phase of a field in state $|\Psi\rangle$ via $$ P(\alpha)=\frac{1}{\pi}|\langle \alpha|\Psi\rangle|^2, $$ which can be shown to be normalized using the completeness relation for coherent states $$ \int d^2\alpha P(|\alpha|,\phi)= \frac{1}{\pi}\int d^2\alpha|\langle \alpha|\Psi\rangle|^2= \langle\Psi|\left(\frac{1}{\pi}\int d^2\alpha |\alpha\rangle\langle \alpha|\right)|\Psi\rangle= \langle\Psi|\Psi\rangle. $$ Transforming to amplitude and phase and integrating out the amplitude, we can get the probability distribution of phases, and then compute its variance. We can then also compute the variance of the distribution of the photon number, which has probability distribution $$ P_n=|\langle n|\Psi\rangle|^2. $$

The uncertainty relation then should be within the reach (if I leave it as a homework, it is not because of arrogance towards students, but because the calculations are somewhat tedious and I am lazy.)

Roger V.
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