Linkage of a classical canonical transformation to a quantum unitary transformation

Question

Suppose I have an classical Hamiltonian $H(q, p)$ and the quantum equivalent $\hat{H}(\hat{q}, \hat{p})$.

We can apply linear canonical transformation defined by a type-II generating function $S_2(q,p^\prime, t)$ to the classical Hamiltonian yielding: \begin{equation} K(q^\prime, p^\prime, t) = H(q, p, t) + \frac{\partial S_2}{\partial t}\end{equation}

Similarly, we have can apply a unitary transformation $\hat{U}$ to quantum Hamiltonian: \begin{equation} \hat{K}=UHU^\dagger + i\hbar \frac{\partial \hat{U}}{\partial t} \hat{U}^\dagger \end{equation}

Given I have shown that in the quantum to classical limit $\hat{K}\to K$, how do I show that the unitary transfomation is the same as the canonical transformation?

Suppose I have a quantum unitary transformation, how do I find the classical mechanical equivalent canonical tranformation?

Answer 1

You might delve into the thicket of 23 references at the beginning of Ch 0.17 of this booklet, but your examples are sufficiently uncomplicated to bypass deformation quantization, the language you properly identified as most suitable for the classical limit. So...

Instead, since you provide an unambiguous quantum hamiltonian ("the", e.g., $\hat H=\hat p^2 + V(\hat q)$ ), I'll provide a trivial example of phase-space translations, to mark the path of ideas involved.

Choose a time-independent quantum canonical transformation, for simplicity, $$ \hat U= e^{i(\hat p a -\hat q b)/\hbar}, \qquad \leadsto \\ \hat q '=\hat U \hat q \hat U ^\dagger= \hat q +a \\ \hat p '= \hat U \hat p \hat U ^\dagger= \hat p +b , $$ by Campbell's adjoint action identity. Thus, $$ [\hat q ', \hat p ']= [\hat q , \hat p ]=i\hbar,\\ \hat K=\hat U \hat H \hat U^\dagger = \hat p ^{'~2}+V(\hat q’), $$ whose classical limit presents as $$ q'=q+a, \qquad p'= p+b , \\ \{ q', p' \}=\{ q, p\}, \qquad \hbox {in both primed and unprimed bases},\\ K=(p+b)^2 +V(q+a), ~~~H=p^2+ V(q). $$

In classical phase space, the adjoint action of the above canonical transformation is the customary $$ q'= e^{\{qb-ap,} q= q+ \{qb-ap ,q\},\\ p'= e^{\{qb-ap,} p= p+ \{qb-ap ,p\}, $$ since the series of the exponential trivially terminates here, after only one PB term!

If, for whatever reason, you must convert it to a type II transformation, you have $$ S_2=q p'+ ap'-bq, ~~\implies \\ p=\partial S_2/\partial q = p'-b, \qquad q'=\partial S_2/\partial p'= q+a. $$

Answer 2

I had some fun trying to answer your question, so I'll post my answer here. I'm not a specialist, just a pedestrian, so if you want to thrust in someone, thrust in @CosmasZachos

Let $(\hat q,\hat p)$ and $(\hat Q,\hat P)$ the pairs of conjugated variables related by the unitary transformation $U$. We have

$$ \hat Q = U\hat qU^\dagger $$

and

$$ P= U\hat pU^\dagger $$

If $\hat H = \hat H(\hat q,\hat p)$ is the Hamiltonian operator, we have(this can be verified through Taylor expansion of $\hat H $and unitarity property of $U$):

$$ U\hat H(\hat q,\hat p)U^\dagger = \hat H(\hat Q,\hat P) $$

Now let $U=e^{-iF/\hbar}$ for some Hermitian operator $F$. The expression of $K$ turns out to be

$$ \hat K = \hat H(\hat Q,\hat P) + {\partial F\over \partial t} $$

Now we need to find the Classical counterpart for this transformed Hamiltonian. I thought about two possibilities: Path integral formulation and the Principle of Correspondence. The first is a little bit tricky and, to be honest, I think we should not expect it will work everytime. But I'll let it here, because it was my best try to connect the generating function with the function $F$.

Path integral approach

We can use this relation to define the time Evolution operator. We have

$$ U(t)= e^{-i\hat K t/\hbar} = e^{-i(\hat H + \partial F/\partial t)/\hbar} $$

Now we can use the basis $|Q\rangle$ for a representation of the above operator and the idea is to build the path integral representation with trajectories in the $Q$ space, yo get the expression for the action related to this sustem. Since the classical action and the quantum action are the same, we can use it to find the classical action. I'll not work all the calculations explicitly, but since ${\partial F\over \partial t}$ could be thought as a function of $(Q,P)$, we can work with it the same way we deal with external potentials when building the path integral.

The kernel is

$$ K(Q,t;Q_0,0) = \langle Q|U(t)| Q_0\rangle $$

and in the path integral formalism it could be written as

$$ K(Q,t;Q_0,0) = \int \mathcal D Q \,\, e^{iS[Q]/\hbar} $$

where

$$ S[Q] = \int \left(P\dot Q - H(Q,\dot Q) - {\partial F\over \partial t}\right ) dt $$

Now I think we need to impose the Hamilton condition for generating functions to connect the above expression with the Hamiltonian $H(q,p)$ that is related to $\hat H(\hat q,\hat p)$ through the Principle of Correspondence. It means we should find a function $G$ such that

$$ P\dot Q - H(Q,P) - {\partial F\over \partial t} = p\dot q - H(q,p) + {dG\over dt} $$

Now, some considerations:

• There is no reason to think that $F$ could be the generating function since, in the way I worked, it will be given as a function of $(Q,P)$, and a generating function should be a function of old and new variables.
• It seems, by this way, that the function $F$ could be discarded from the expression of the classical action. I'm not sure if it makes sense, but if so, we could discard this term in the last equation above.

Principle of Correspondence approach

We just say that the classical transformed Hamiltonian is

$$ K(Q,P) = H(Q,P) + {\partial F\over \partial t} $$

and the Hamiltonian condition follows as before. If the unitary evolution does not depend explictly on time, we have

$$ K(Q,P) = H(Q,P) $$

and the Hamiltonian condition reads

$$ P\dot Q - K(Q,P) = p\dot q - H(q,p) + {dG\over dt} $$

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Some sanity check. For $F=0$, the unitary transformation is the identity and $P=p$, $Q=q$ and we can take $G=0$.

If the unitary transformation is a translation, $\hat F = \alpha \hat p$, so $P=p$ and $Q = q+\alpha$, so

$$ p\dot q - K(q+\alpha,p) = p \dot q - H(q,p) + {dG\over dt} $$

It's enought to take $G = \alpha p$.