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.

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    $\begingroup$ You can derive it from Schur’s lemma. The key assumption, is that your Hilbert space is an irreducible representation of the canonical commutation relations (Weyl form). You were having a hard time because you weren’t exploiting irreducibility. So any braiding automorphism is a multiple of identity. $\endgroup$
    – LPZ
    Commented Dec 7, 2022 at 11:12
  • $\begingroup$ You are perfectly right, but I was hoping for something more direct and physical. I would like to avoid Schur's lemma and the concept of irreducible representation. I think I succeeded in this task, in a moment I will post my proof. $\endgroup$
    – Leonardo
    Commented Dec 7, 2022 at 11:46
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    $\begingroup$ If you believe that your attempt solves the problem, you should post it as a (self) answer instead of adding it to the question. $\endgroup$ Commented Dec 7, 2022 at 12:55
  • $\begingroup$ Oh, okay, sorry. I'm going to do it right now. $\endgroup$
    – Leonardo
    Commented Dec 7, 2022 at 13:00
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    $\begingroup$ @Leonardo What I am saying is: Whatever your proof does, a consequence of it will be that the Hilbert space is irreducible (otherwise, there would be other $U$s). So proving it must be at least as hard as proving irreducibility. $\endgroup$ Commented Dec 7, 2022 at 15:16

1 Answer 1


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


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


where $|x_1,\dots,x_n\rangle$ are simultaneous eigenvector of $q_1,\dots,q_n$, i.e.


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:


$=\int_{\mathbb{R}^n} \langle x_1,\dots,x_n \mid k_1',\dots,k_n'\rangle K(x_1,\dots,x_n)|x_1,\dots,x_n\rangle \text{d}x_1\dots\text{d}x_n$.

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?

  • $\begingroup$ Let us suppose $n=1$. Your proof assumes that there is a basis of the Hilbert space $|n>$ made of the eigenvectors of $p^2+q^2$ where $n$ is the eigenvalue. It is OK on $L^2(R)$. But now consider $H=L^2\otimes C^2$ and define $q':= q\otimes I$, $p':=p\otimes I$. Even if the new operators satisfy the CCR your proof does not work any longer. $\endgroup$ Commented Dec 7, 2022 at 17:43
  • $\begingroup$ That is because it is false that every eigenvalue of $p'^2+q'^2$ has a unique normalized eigenvector up to phases. $\endgroup$ Commented Dec 7, 2022 at 17:44
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    $\begingroup$ As a matter of fact CCRs are not enough. You must assume another further hypothesis, completeness of the set of observables or irreducibility of the represntation. $\endgroup$ Commented Dec 7, 2022 at 18:00
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    $\begingroup$ The theorem is the celebrated Stone von Neumann one... $\endgroup$ Commented Dec 7, 2022 at 18:02
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    $\begingroup$ It is "correct" in the non-rigorous approach by physicists. Strictly speaking it contains a number of issues. Regarding domains, selfadjointness problems etc. But, yes, all those issues can be fixed. $\endgroup$ Commented Dec 7, 2022 at 18:05

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