What happened to the phase of creation and annihilation operators?

Question

It's the whole day that I try to figure out at what point exactly the phase factor appearing in both creation and annihilation operators, went away maybe absorbed inside occupation number states. Here the procedure I followed: sorry in advanced for the "heavy" notation, I'm dumb and I need to have all informations in front of my eyes all the time.

So everything starts from these commutation relations \begin{gather*} \hat{n}_\mu(t,\boldsymbol{k}) \doteq \hat{a}_\mu^\dagger(t,\boldsymbol{k}) \hat{a}_\mu(t,\boldsymbol{k}) \\ \begin{cases} \left[ \hat{a}_\mu(t,\boldsymbol{k}), \hat{n}_\nu\left(\tilde{t},\tilde{\boldsymbol{k}}\right) \right] = \delta_{\mu\nu} \delta\left(t,\tilde{t}\right) \delta\left(\boldsymbol{k},\tilde{\boldsymbol{k}}\right) \hat{a}_\nu\left(\tilde{t},\tilde{\boldsymbol{k}}\right) \\ \left[ \hat{a}_\mu^\dagger(t,\boldsymbol{k}), \hat{n}_\nu\left(\tilde{t},\tilde{\boldsymbol{k}}\right) \right] = - \delta_{\nu\mu} \delta\left(t,\tilde{t}\right) \delta\left(\boldsymbol{k},\tilde{\boldsymbol{k}}\right) \hat{a}_\nu^\dagger\left(\tilde{t},\tilde{\boldsymbol{k}}\right) \end{cases} \end{gather*} Because of that, after some calculations \begin{gather*} \begin{cases} \hat{a}_\mu(t,\boldsymbol{k}) |n_\mu(t,\boldsymbol{k})\rangle \equiv c^-_\mu(t,\boldsymbol{k}) |(n-1)_\mu(t,\boldsymbol{k})\rangle \\ \hat{a}_\mu^\dagger(t,\boldsymbol{k}) |n_\mu(t,\boldsymbol{k})\rangle \equiv c^+_\mu(t,\boldsymbol{k}) |(n+1)_\mu(t,\boldsymbol{k})\rangle \end{cases} \\ \begin{cases} |c^-_\mu(t,\boldsymbol{k})|^2 = n_\mu(t,\boldsymbol{k}) \\ |c^+_\mu(t,\boldsymbol{k})|^2 = n_\mu(t,\boldsymbol{k})+1 \end{cases} \\ \begin{cases} \hat{a}_\mu(t,\boldsymbol{k}) |n_\mu(t,\boldsymbol{k})\rangle = e^{i \alpha^-_\mu\left(n_\mu(t,\boldsymbol{k}),t,\boldsymbol{k}\right)} \sqrt{n_\mu(t,\boldsymbol{k})} |(n-1)_\mu(t,\boldsymbol{k})\rangle \\ \hat{a}_\mu^\dagger(t,\boldsymbol{k}) |n_\mu(t,\boldsymbol{k})\rangle = e^{i \alpha^+_\mu\left(n_\mu(t,\boldsymbol{k}),t,\boldsymbol{k}\right)} \sqrt{n_\mu(t,\boldsymbol{k})+1} |(n+1)_\mu(t,\boldsymbol{k})\rangle \end{cases} \end{gather*} where $\alpha^-,\alpha^+$ are dinstinct functions with $(n_\mu(t,\boldsymbol{k}),t,\boldsymbol{k})$ arguments (right?). Using the above commutation relations I'm only able to prove that, unless a multiple of $2\pi$, the following equality holds \begin{equation*} \alpha^+\left(n_\mu(t,\boldsymbol{k}),t,\boldsymbol{k}\right) = -\alpha^-\left((n+1)_\mu(t,\boldsymbol{k}),t,\boldsymbol{k}\right) \end{equation*} but what I desire to prove is also that \begin{equation*} \alpha^+\left(n_\mu(t,\boldsymbol{k}),t,\boldsymbol{k}\right) = \alpha^-\left(n_\mu(t,\boldsymbol{k}),t,\boldsymbol{k}\right) \end{equation*} Is there a way to prove it or is not possible? In this other case how can I make the phase factor go away?

Answer 1

The logic for a single oscillator is as follows, the extension to a Hamiltonian $H=\sum_k \omega_k a_k^\dagger a_k$ is immediate I think.

Let's say we've constructed operators $a,a^\dagger$ such that $[a,a^\dagger]=1$. Then defining the number operator $n=a^\dagger a$ we also have $[n,a^\dagger] = [a,a^\dagger]a^\dagger = a^\dagger$. Thus we can calculate:

$$n\left(a^\dagger |n\rangle\right ) = \left([n,a^\dagger]+a^\dagger n\right)|n\rangle$$ $$n\left(a^\dagger |n\rangle \right) = (1+n)\left(a^\dagger |n\rangle\right)$$

So we conclude that acting with $a^\dagger$ takes us from eigenstates of $n$ to eigenstates of $n$ and increases the eigenvalue by $1$. At this point we simply define:

$$|n+1\rangle = \mathcal{N}a^\dagger |n\rangle$$

where $\mathcal{N}$ is a normalisation factor whose phase is not defined by the above considerations. We simply choose $\mathcal{N}$ to be real because that is simpler.

Answer 2

I maybe arrived to a result, but it is to be discussed so I post it as an answer to my own question. The problem with this result is that it come from a comparison and not an equality...

I thought that the main problem with the equality between $\alpha^-,\alpha^+$ I wanted to arrive at, was that is not possible to create an equality with singularly creation or annihilation operators, an equality in the sense of the occupation number state: I think this statement is evident from the nature of the creation and annihilation operators themselves, even because the occupation number operator is hermitian and so has orthogonal eigenstates: in other words $|n_\mu(t,\boldsymbol{k})\rangle$ has absolutely anything to do with $|(n-1)_\mu(t,\boldsymbol{k})\rangle$ or $|(n+1)_\mu(t,\boldsymbol{k})\rangle$.

Due to this I cannot say $\hat{a}=e^{i\alpha^-(n_\mu)}\sqrt{n_\mu}$ because the states at the two sides of the equality are not the same; if it was possible to do this, after this equality I would do $\hat{a}^\dagger=e^{-i\alpha^-(n_\mu)}\sqrt{n_\mu}$ (that as you can see is wrong) and I would say: "fantastic! $-\alpha^-(n_\mu)=\alpha^+(n_\mu)$ problem solved."

But that's not possible

So in the despair I invented something a bit odd and maybe wrong: I did a redefinition of the occupation number states. Consider that, first of all the following holds \begin{gather*} \hat{a}_\mu |n_\mu\rangle = e^{i\alpha^-(n_\mu)} \sqrt{n_\mu} |(n-1)_\mu\rangle \\ \hat{a}_\mu^\dagger |n_\mu> = e^{i\alpha^+(n_\mu)} \sqrt{n_\mu+1} |(n+1)_\mu\rangle \\ e^{i\alpha^+(n_\mu-1)} = e^{-i\alpha^-(n_\mu)} \end{gather*} But what if I redefine the occupation number states with the following substitution $e^{-i\alpha^+(n_\mu)}|n_\mu\rangle\to|n_\mu\rangle$. Thanks to the preceding equalities I would have \begin{gather*} \hat{a}_\mu e^{-i\alpha^+(n_\mu)} |n_\mu\rangle = e^{-i\alpha^+(n_\mu)} e^{-i\alpha^+((n-1)_\mu)} \sqrt{n_\mu} |(n-1)_\mu\rangle \\ e^{-i\alpha^+((n+1)_\mu)} \hat{a}_\mu^\dagger e^{-i\alpha^+(n_\mu)} |n_\mu\rangle = e^{-i\alpha^+((n+1)_\mu)} \sqrt{n_\mu+1} |(n+1)_\mu\rangle \end{gather*} So that everything can be newly written \begin{gather*} \hat{a}_\mu |n_\mu\rangle = e^{-i\alpha^+(n_\mu)} \sqrt{n_\mu} |(n-1)_\mu\rangle \\ \hat{a}_\mu^\dagger |n_\mu\rangle = e^{i\alpha^+((n+1)_\mu)} \sqrt{n_\mu+1} |(n+1)_\mu\rangle \end{gather*} Or equivalently \begin{gather*} \hat{a}_\mu |n_\mu\rangle = e^{i\alpha^-((n+1)_\mu)} \sqrt{n_\mu} |(n-1)_\mu\rangle \\ \hat{a}_\mu^\dagger |n_\mu\rangle = e^{i\alpha^+((n+1)_\mu)} \sqrt{n_\mu+1} |(n+1)_\mu\rangle \end{gather*} So it's like $e^{i\alpha^-((n+1)_\mu)}\equiv e^{i\alpha^-(n_\mu)}\equiv e^{i\alpha^-}$ and the same for $e^{i\alpha^+}$! They just remain function of $(t,\boldsymbol{k})$. Due also to the preceding equality that was used several times here $e^{i\alpha^+(n_\mu-1)}= e^{-i\alpha^-(n_\mu)}$, we would also have $e^{i\alpha^+}=e^{-i\alpha^-}$.

Once determined that the phase factors are not dependent on occupation number states (being equivalent to exponential operators themselves), they can be "absorbed" inside creation and annihilation operators without problems, without changing any of the commutation relations that define them.

Is this true? Is this wrong? I completely don't know, I would love to have somebody's opinion. As always, thanks a lot!