Arbitrariness of phase in the definition of the Fock state leads to lack of squeezing

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).

• 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. – Mark Mitchison Oct 5 '16 at 10:28