Unitary operators that implement the same canonical transformation In quantum mechanics a transformation of the spatial coordinate operators and conjugate momentums of the type: $$(q_1,\dots,q_n,p_1,\dots,p_n) \to (Q_1,\dots,Q_n,P_1,\dots,P_n),$$ is called canonical transformation if the commutators are conserved, i.e. if $$[q_i,p_j]=iħ\delta_{ij}=[Q_i,P_j]$$ for each $i,j=1,\dots,n$. It can be demonstrated (von Neumann's theorem, see for example the Picasso lectures in quantum mechanics, chapter 6, page 108,) that a unitary operator $U$ can be associated to each canonical transformation of this type such that $$Q_i=Uq_iU^{\dagger}$$ and $$P_i =Up_iU^{\dagger}$$ for each $i=1,...,n$ (here $U^{\dagger}$ represents the adjoint of $U$). My question is: how is it possible to prove that if $U$ and $V$ are two unitary operators associated with the same canonical transformation, then they differ only by a phase factor? i.e. $U=e^{i\phi}V$, where $\phi$ is a real number.
 A: My own attempt (I noticed that I made a mistake in my previous attempt, so I fixed my proof)
Let $U_1$ and $U_2$ be the unitary operators implementing the same canonical transformation
$(q_1,\dots,q_n,p_1,\dots,p_n) \to (Q_1,\dots,Q_n,P_1,\dots,P_n)$
We define the operator $V=U_2^{\dagger}U_1$. Clearly $V$ is unitary. Furthermore $Vq_iV^{\dagger}=q_i$ and $Vp_iV^{\dagger}=p_i$ for each $i=1,\dots,n$. So in particular for every polynomial function $f(q,p)$ we have that $Vf(q,p)V^{\dagger}=f(q,p)$, that is $[V,f(q,p)]=0$. Now we define $H_i=\frac{p_i^2}{2m}+\frac{1}{2}m\omega^2q_i^2$, i.e. the Hamiltonian of the one-dimensional harmonic oscillator associated to the pair $(q_i,p_i)$. As known, this non-degenerate self-adjoint operator gives a Hilbert basis of eigenvectors $|k_i\rangle$ with $k_i \in \mathbb{N}$ and $H_i|k_i\rangle=E_i|k_i\rangle$ where $E_i=\hbar\omega(k_i+\frac{1}{2})$. However, we are working in a Hilbert space with n degrees of freedom, so its basis will rather be given by $|k_1,\dots,k_n\rangle$, where $H_i|k_1,\dots,k_n\rangle=E_i|k_1,\dots,k_n\rangle$ for each $i=1,\dots,n$, i.e. $|k_1,\dots,k_n\rangle$ are simultaneous eigenvectors of $H_1,\dots,H_n$ (here $k_1,\dots,k_n \in \mathbb{N}^n$).
Clearly $V$ commutes with all $H_i$ and therefore $V|k_1,\dots,k_n\rangle$ is an eigenvector of each $H_i$, i.e.
$H_iV|k_1,\dots,k_n\rangle=VH_i|k_1,\dots,k_n\rangle=E_iV|k_1,\dots,k_n\rangle$ for each $i=1,\dots,n$
and, being $H_1,\dots,H_n$ a complete system of compatible observables (i.e. each $|k_1,\dots,k_n\rangle$ is univocally identified by a $n$-tuple $(k_1,\dots,k_n)$), we can say that
$V|k_1,\dots,k_n\rangle=c(k_1,\dots,k_n)|k_1,\dots,k_n\rangle$,
with $c(k_1,\dots,k_n)$ a complex number which, however, generally depends on the $n$-tuple $(k_1,\dots,k_n) \in \mathbb{N}^n$. Moreover clearly $V$ also commutes with all the $q_i$, therefore in particular, since these all together form a complete system of compatible observables too, we have that
$V|x_1,\dots,x_n\rangle=K(x_1,\dots,x_n)|x_1,\dots,x_n\rangle$,
where $|x_1,\dots,x_n\rangle$ are simultaneous eigenvector of $q_1,\dots,q_n$, i.e.
$q_i|x_1,\dots,x_n\rangle=x_i|x_1,\dots,x_n\rangle$
for each $i=1,\dots,n$. Here $K(x_1,\dots,x_n)$ is a complex number depending in general on the $n$-tuple $x_1,\dots,x_n$. Now it suffices to observe that, since $|x_1,\dots,x_n\rangle$ is a (generalized) Hilbert basis as $(x_1,\dots,x_n)$ varies in $\mathbb{R}^n$, then:
$|k_1',\dots,k_n'\rangle=\int_{\mathbb{R}^n} \langle X_1,\dots,x_n \mid k_1',\dots,k_n'\rangle |x_1,\dots,x_n\rangle \text{d}x_1\dots\text{d}x_n$,
where $|k_1,\dots,k_n\rangle$ is fixed. Applying $V$ to both sides we obtain:
$c(k_1',\dots,k_n')|k_1',\dots,k_n'\rangle=$
$=\int_{\mathbb{R}^n} \langle x_1,\dots,x_n \mid k_1',\dots,k_n'\rangle K(x_1,\dots,x_n)|x_1,\dots,x_n\rangle \text{d}x_1\dots\text{d}x_n$.
Multiplying on the left by an arbitrary fixed $|x_1',\dots,x_n'\rangle$ we finally obtain:
$c(k_1',\dots,k_n') \langle x_1',\dots,x_n' \mid k_1',\dots,k_n'\rangle=K(x_1',\dots,x_n') \langle x_1',\dots,x_n' \mid k_1,\dots,k_n\rangle$,
from which $c(k_1',\dots,k_n')=K(x_1',\dots,x_n')$, i.e. $K$ does not depend on the $n$-tuple $(x_1,\dots,x_n)$ chosen. So $V$ is a multiple of the identity, given that, taking the generic vector:
$|A\rangle=\int_{\mathbb{R}^n} \langle x_1,\dots,x_n \mid A \rangle |x_1,\dots,x_n \rangle \text{d}x_1\dots\text{d}x_n$,
we have that:
$V|A\rangle=\int_{\mathbb{R}^n} \langle x_1,\dots,x_n \mid A \rangle V|x_1,\dots,x_n\rangle \text{d}x_1\dots\text{d}x_n=$
$=\int_{\mathbb{R}^n} \langle x_1,\dots,x_n \mid A \rangle K|x_1,\dots,x_n\rangle \text{d}x_1\dots\text{d}x_n=K|A\rangle$.
Since $V=KI$ (I is the identity) is unitary, we must have that $K=e^{i\phi}$. So $U_1=e^{i\phi}U_2$. What do you think? Is there any simpler approach?
