# 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.

• 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.
– LPZ
Commented Dec 7, 2022 at 11:12
• 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. Commented Dec 7, 2022 at 11:46
• If you believe that your attempt solves the problem, you should post it as a (self) answer instead of adding it to the question. Commented Dec 7, 2022 at 12:55
• Oh, okay, sorry. I'm going to do it right now. Commented Dec 7, 2022 at 13:00
• @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. Commented Dec 7, 2022 at 15:16

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?

• 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. Commented Dec 7, 2022 at 17:43
• That is because it is false that every eigenvalue of $p'^2+q'^2$ has a unique normalized eigenvector up to phases. Commented Dec 7, 2022 at 17:44
• 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. Commented Dec 7, 2022 at 18:00
• The theorem is the celebrated Stone von Neumann one... Commented Dec 7, 2022 at 18:02
• 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. Commented Dec 7, 2022 at 18:05