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Every state in quantum mechanics is defined up to a global phase. In other words, quantum states $|\psi'\rangle = e^{i\varphi}|\psi\rangle$ which differ just by a phase factor are indistinguishable. Well then, what about Fock states and annihilation/creation operators?

Imagine that you have a system of $n+m$ indistinguishable bosonic particles that can located on two orthogonal modes $a$ and $b$. We know from quantum mechanics how annihilation operators $\hat{a}, \hat{b}$ act on the Fock state $|n,m\rangle$: $$\hat{a}|n,m\rangle = \sqrt{n}|n-1, m\rangle$$ $$\hat{b}|n,m\rangle = \sqrt{m}|n, m-1\rangle$$ And the overlap is equal to: $$\langle n-1, m| \hat{a} |n,m\rangle = \sqrt{n}$$ Is the above statement true or can the overlap actually be defined up to some phase factor: $$\langle n-1, m| \hat{a} |n,m\rangle = \sqrt{n}e^{i\theta}?$$ What would happen if I define an equivalent Fock state: $$|n,m\rangle_{!} = e^{i\varphi_{n,m}}|n,m\rangle$$ How does $\hat{a}$ act on $|n,m\rangle_{!}$, i.e. $$\hat{a}|n,m\rangle_{!} = \sqrt{n}|n-1,m\rangle_{!}?$$ $$\hat{a}|n,m\rangle_{!} = \sqrt{n}e^{i\varphi_{n,m}}|n-1, m\rangle?$$

I am curious about that, because there are a lot of spin squeezing experiments and theory where the value of the spin squeezing parameter $\xi_{R}$ depends on the overlap between Fock states with different particle numbers. For example, if you start with a state of the form: $$|\psi_{0}\rangle = \sum\limits_{k=0}^{N}C_{k}|k,N-k\rangle$$ and your Hamiltonian is such that the Fock states $|k,N-k\rangle$ are also eigenstates of the system, then time evolution takes the form: $$|\psi(t)\rangle = \sum\limits_{k=0}^{N}C_{k}e^{-itE_{k}}|k,N-k\rangle$$ Now, squeezing is based on the expectation values of spin operators. When you calculate $\langle \hat{S}_x \rangle$, where $$\hat{S}_x = \frac{1}{2}(\hat{a}^{\dagger}\hat{b} + \hat{b}^{\dagger}\hat{a})$$ you have a non-zero overlap of the form $$\langle k+1, N-k-1| \hat{S}_x | k, N-k\rangle$$ Following standard rules, one finds that the above quantity should be $1/2\sqrt{(k+1)(N-k)}$, but if it is defined up to a phase, then I will be able to cancel the energy phase factor and at the end get no squeezing (which is not true, as experiments have shown).

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Phase does play a role when it depends on some variable such as the modal index or time and one considers a superposition of such states. Although one can always remove a global phase factor, it is not possible to remove it for all states in such a superposition. In this sense, phase is always a relative concept; it can always be seen as a phase difference between the phase of the current state and some reference phase that is defined globally. This also ties in with how we observe phase; always as a relative phase with respect to some reference.

So anywhere in your analysis where you can remove a constant global phase you can do so because you are free to define any global reference phase. This would then give you the results that you would obtain experimentally.

If you would for instance use your spin operator to operate on the state, consisting of a superposition of states, each having a particular phase, you may find that the result would contain phases that cannot be removed. So regardless of your definition of a global phase, some phase factors will always remain.

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  • $\begingroup$ This is not true, a time-dependent global phase does not affect expectation values and can always be ignored. What you cannot ignore is a phase that is state-dependent (i.e. not a global phase). $\endgroup$ – Mark Mitchison Oct 4 '16 at 13:44
  • $\begingroup$ Depends how you measure it. A time-dependent phase can give rise to interference. A global phase cannot. $\endgroup$ – flippiefanus Oct 5 '16 at 4:01
  • $\begingroup$ Whether the phase is time-dependent or not is irrelevant. A global phase can always be removed by a gauge transformation, even if it depends on time. $\endgroup$ – Mark Mitchison Oct 5 '16 at 10:18
  • $\begingroup$ OK, under the assumption that one has such a gauge invariance, I would agree with you. I'll edit my answer accordingly. $\endgroup$ – flippiefanus Oct 5 '16 at 10:25
  • $\begingroup$ One does not need to make any assumptions about gauge invariance. Just consider any unitary transformation $\lvert \psi\rangle\to e^{i\theta(t)}\lvert \psi\rangle$, where $\theta$ is an arbitrary c-number function, it is easy to prove that all expectation values remain unchanged by such a phase factor. $\endgroup$ – Mark Mitchison Oct 5 '16 at 10:28

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