# Does the expectation value of an operator with respect to Quantum Vacuum State has experimental meaning?

I'm learning about the coherent state in Quantum Optics. And I'm confused with the statement in the book, which states that the coherent state is the minimum uncertainty state. And since the vacuum state is a special kind of coherent state, it is also a minimum uncertainty state.

My question is: I can understand the calculation of why the coherent state is the minimum uncertainty state, but what does the mean of an operator with respect to vacuum state mean in the language of measurement, since there is no photon in the vacuum state, and let alone a measurement with vacuum state?

• The crux of the question might be "since there is no... measurement with vacuum state". You're right that measurements cannot occur in the vacuum state -- nor can they occur in a single-particle state, nor in any model of a free quantum field (which is normally used to introduced coherent states because the lack of interactions simplifies the math). That's because measurement requires interaction with a sufficiently complex system. Is this your point? If so, then the question really isn't specific to coherent states... or did I misunderstand your point? Nov 3, 2021 at 13:45
• @ChiralAnomaly Yeah, my question is mainly about that I don't know what does measurements mean when it's a vacuum state. But I don't know that ''measurements cannot occur in a single-particle state'', so it makes measurements in a vacuum state strange to me. Doesn't the detector in the photonic experiment play the role of measurement, and so a single photon(if we can make it) can be measured? Nov 3, 2021 at 14:32
• "Doesn't the detector in the photonic experiment play the role of measurement" -- Yes, and that's exactly what I'm saying. There are no detectors in a world with only one photon, just like there are no detectors in a world with nothing at all (the vacuum state). But explicitly including detectors in the quantum model is too difficult, so we use a shortcut: we just declare that a given observable has magically been measured, without actually including the measuring device as part of the state. Nov 3, 2021 at 16:03
• That shortcut works just as well in the vacuum state as it does in a single-particle state. The vacuum state is not an eigenstate of most observables, so measuring most observables will give nontrivial results in the vacuum state. But again, that's a shortcut. In the real world, if a measuring device is present, then the state isn't the vacuum. It's a state with a physical device. Nov 3, 2021 at 16:03

To help answering your question perhaps it's in order to remember ourselves why people use vacuum expectation values in the first place.

Mathematically, the vacuum state is just another pure state. You could think of it as a basis element of $$\mathcal{H}$$ for a suitable basis, like the computational basis. It also has the property that when acting with its associated ladder operators (which are necessary to define the vacuum state in the first place) you get a $$0$$ if you try to lower it further or eventually exhaust the spectrum of possible states by rising it and combinations thereof.

Thus it has a clear physical/computational advantage: a priori, from the vacuum state you can get any state of physical relevance. So you only care about vacuum expectation values, as you can compute every expectation value from the vacuum state, at least in principle.

For completeness, your coherent state is no exception, quoting Wikipedia $$|\alpha\rangle=e^{-{|\alpha|^2\over2}}\sum_{n=0}^{\infty}{\alpha^n\over\sqrt{n!}}|n\rangle = e^{-{|\alpha|^2\over2}}e^{\alpha\hat a^\dagger}e^{-{\alpha^* \hat a}}|0\rangle.$$

Edit:

In a nutshell, my argument is for any $$|\psi\rangle = \mathcal{O}(a, a^\dagger )|0\rangle$$, where $$\mathcal{O}$$ is an operator constructed from $$a$$ and $$a^\dagger$$’s. An example is the coherent state above with $$\mathcal{O} \sim \exp(-\alpha a^\dagger)$$. So all I’m saying is that if you know what all operators expectation value on $$|0\rangle$$ are, then you know all expectation values on any state. I believe this argument is in Weinberg’s vol. I of his QFT books, in the cluster decomposition principle chapter.

• So you only care about vacuum expectation values, as you can compute everything from the vacuum state, at least in principle. - but can you calculate everything from the vacuum expectation value?! Nov 3, 2021 at 10:04
• @Jakob I would say every physical expectation value yes, although I'm in for a treat (I clarified that out). Things like entanglement etc obviously not. Nov 3, 2021 at 10:20
• But how does the knowledge of $\langle 0|A|0\rangle$ implies knowledge of $\langle\psi|B|\psi\rangle$ (even if $A=B$)? I don't see this right now. Nov 3, 2021 at 10:24
• This doesn't answer the question. The question is about expectation values of operators on the vacuum state, not the state itself. Nov 3, 2021 at 10:48
• My argument is for any $|\psi\rangle = \mathcal{O}(a, a^\dagger )|0\rangle$, where $\mathcal{O}$ is an operator constructed from $a$ and $a^\dagger$’s. An example is the coherent state above with $\mathcal{O} \sim \exp(-\alpha a^\dagger)$. So all I’m saying is that if you know what all operators expectation value on $|0\rangle$ are, then you know all expectation values on any state. I believe this argument is in Weinberg’s vol. I of his QFT books, in the cluster decomposition principle chapter. Nov 3, 2021 at 12:59