# Classical canonical transformations and unitary transformations in quantum mechanics

1. Sep 14, 2010

### tom.stoer

Hello,

suppose one has a classical canonical transformation between two sets of canonical variables such that the new (primed) positions and momenta can be written as functions of the old (unprimed) ones.

$${\cal K}: x_i \to x_i^\prime(x); \quad p_i \to p_i^\prime(p)$$

Using these relations one can check immediately that the Poisson brackets remain unchanged.

This is translated to quantum mechanical language as follows: the positions and momenta are replaced by operators, the Poisson brackets are translated into canonical commutation relations, which remain unchanged as well:

$$[x_i,p_k] = [x_i^\prime,p_k^\prime] = i\delta_{ik}$$

The canonical transformation itself is replaced by an operator acting on the positions and momenta

$${\cal K}: x_i \to x_i^\prime = U_{\cal K}\, x_i\, U^\dagger_{\cal K}; \quad p_i \to p_i^\prime = U_{\cal K}\, p_i\, U^\dagger_{\cal K}$$

Questions:
• are there additional conditions for the canonical transformations $${\cal K}$$ in order to translate them into quantum mechanical operators $$U_{\cal K}$$?
• given a specific canonical transformations $${\cal K}$$, that means the functional relationship between new and old position and momenta, how does one construct the unitary operator $$U_{\cal K}$$ generating the quantum mechanical transformation?

Examples are rotations, translations, transformation to center-of-mass frame, ...

Last edited: Sep 14, 2010
2. Sep 14, 2010

### xepma

Re: Classical canonical transformations and unitary transformations in quantum mechan

Well, canonical transformations are usually defined as those transformations which preserve the symplectic structure of the (classical) phase space. As a correllary, one can always identify a generator G of a canonical transformation, which is some function of the phase space variables -- position and momentum. The Hamiltonian generates time translations, and so fort.

Note that the change of any other function due to a canonical transformation is given by the Poisson bracket of this function with the corresponding generator. So if F is a function and G is the generator of some canonical transformation, then the change in F is given by {F,G}.

The rules of quantization now state that an infinitesimal unitary transformation on the Hilbert space will change any operator F acting on it. The change of this operator F is given by -i[F,G], where G is the generator of the infinitesimal unitary transformation. The classical generator is some function of the phase space variables (position and momentum). Therefore, the quantum mechanical counterpart of the generator G is obtained by turning these phase space variables into operators.

I can't help but mention that if you choose G = H the Hamiltonian, then the change of an operator F is given by:

$$\delta F \equiv \frac{d F}{dt} = -i[F,H]$$

which is ofcourse the Heisenberg equation of motion.

By the way, the unitary transformation you are after is given by $$U = e^{iG}$$.

3. Sep 14, 2010

### strangerep

Re: Classical canonical transformations and unitary transformations in quantum mechan

One must also choose a representation, i.e., choose a vacuum state such that the
entire state space is generated by action of arbitrary polynomials in the operators
on it.

In general, it's possible to find canonical transformations which, when translated into
operators, throw you out of the Hilbert space into another one, so one must either
restrict the transformations, or handle this inconvenience in some other way.
(Are you familiar with Bogoliubov transformations?)

Xepma already mentioned $U = exp(iG)$, but I'm guessing you want
to know how to construct the generator G explicitly? If so, take a look at
Eugene Stefanovich's physics/0504062 -- the section about non-interacting
representation (of the Poincare algebra) in Fock space. He does the cases of
rotations and boosts explicitly, iirc.

4. Sep 15, 2010

### tom.stoer

Re: Classical canonical transformations and unitary transformations in quantum mechan

Thanks al lot; most of what you are saying is clear.

OK, this is the central equation

$$U_{\cal K} = e^{iG_{\cal K}}$$

to which my question is related: knowing the form of the canonical transformation (e.g. the generating function), how can I construct $$G_{\cal K}$$?

I will check the paper on arxiv.

5. Sep 15, 2010

### xepma

Re: Classical canonical transformations and unitary transformations in quantum mechan

Like I said: G is some function on the phase space variables -- it's some function of the coordinates and momentum, $$G(x,p)$$. Take that expression and turn the phase space variables into operators, $$G(\hat{x},\hat{p})$$ meaning the variables now satisfy the canonical commutation relations. This will give you the operator version of the generator G

Ofcourse, there's probably some ambiguity with the ordering of the operators, but this always arises when you go from classical to quantum mechanical.

6. Sep 15, 2010

### tom.stoer

Re: Classical canonical transformations and unitary transformations in quantum mechan

I tried to do this for the transformation to the c.o.m frame (1-dim) , but I run into trouble.

First I define

$$M = m_1 + m_2; \quad m = m_1 m_2 / M$$

$$X = (m_1 x_1 + m_2 x_2)/M; \quad x = x_1 - x_2$$

$$P = p_1 + p_2; \quad p = (m_2 p_1 - m_1 p_2) / M$$

Of course this transformation is canonical as one can check immediately from the Poisson brackets or the commutators.

Now I construct the canonical transformation using

$$F_2 = F_2(q, P)$$

$$Q_i = \frac{\partial F_2}{\partial P_i}; \quad p_i = \frac{\partial F_2}{\partial q_i}$$

with

$$q_{1,2} = x_{1,2}; \quad Q_{1,2} = X, x$$

$$p_{1,2}; \quad P_{1,2} = P, p$$

In the above mentioned case the transformation is linear and can be written as

$$Q_i = Q_i(q) = a_{ik} q_k$$

with an appropriate matrix. In this case the generating function is

$$F_2(q, P) = P_i a_{ik} q_k = \frac{1}{M}P(m_1 x_1 + m_2 x_2) + p(x_1 - x_2)$$

One can check that this function generates the required transformation via explicit calculation.

Now I want to construct an operator $$U_F = U_F(x, p)$$ defined entirely in terms of the old operators (!) which generates the required unitary transformation. That means I am looking for a generator $$G_F = G_F(x, p)$$ with

$$U_F = e^{iG_F}$$

such that the canonical transformation is implemented on the Hilbert space as

$$Q_i = U_F\, x_i\, U_F^\dagger$$

$$P_i = U_F\, p_i\, U_F^\dagger$$

i.e.

$$X = U_F\, x_1\, U_F^\dagger$$

$$x = U_F\, x_2\, U_F^\dagger$$

$$P = U_F\, p_1\, U_F^\dagger$$

$$p = U_F\, p_2\, U_F^\dagger$$

Of course the generator of a unitary transformation in quantum mechanics does not mix new and old variables (as the generating function in classical mechanics does).

If I understood you correctly you simply propose to write the function $$F(q, P)$$ entirely in terms of the old variables, translate this expression into an hermitean operator and use this operator as $$G_F(x, p)$$.

Doing that something goes wrong! Inserting the definition of the new variables and doing the calculation one finds

$$F(q, P) \to F(x,p) = x_1 p_1 + x_2 p_2 \to (x_1 p_1 + x_2 p_2) / 2 + c.c.$$

But this is not the generator of the corresponding unitary transformation. The unitary transformation implemented by

$$U = e^{iG} = e^{i (x_1 p_1 + x_2 p_2) / 2 +i c.c.}$$

is a simple rescaling of coordinates and momenta.

So the step from the generating function $$F_2(q, P)$$ to the corresponding q.m. generator $$G_F(x, p)$$ does not work as described. Either my calculation is wrong (I checked it several times and I am sure the canonical transformation itself is correct) or the “quantization” of the generating function does not work as proposed.

Last edited: Sep 15, 2010
7. Sep 15, 2010

### tom.stoer

Re: Classical canonical transformations and unitary transformations in quantum mechan

I found this reference http://www.unige.ch/fapse/dimage/SiteFR/Documents/Article%20Lacki.pdf [Broken] where the construction of the generator is explained. I still have to check the details but it looks as if the "exponentialordering" creates extra terms

Last edited by a moderator: May 4, 2017
8. Sep 15, 2010

### strangerep

Re: Classical canonical transformations and unitary transformations in quantum mechan

I don't understand how you got the last formula. I get:

$$\frac{1}{M}(p_1 + p_2)(m_1 x_1 + m_2 x_2) ~+~ \frac{(m_2 p_1 - m_1 p_2)}{M} (x_1 - x_2)$$

(or did I miss something?)

9. Sep 16, 2010

### tom.stoer

Re: Classical canonical transformations and unitary transformations in quantum mechan

Yes, you missed to continue with the calculation :-)

The terms with mixed indices 1,2 in x and p cancel each other. The terms with one index 1,1 or 2,2 in x and p come with two different mass terms 1,2 which add up to M.

I found the unitary transformation for two equal masses by "educated guessing". I hope I can do that for different masses as well. My trick was to use two unitary transformations.

$$U = e^{-ix_1 p_2 / 2} e^{ix_2 p_1}$$

Here the operator ordering does not cause any problems.

10. Sep 16, 2010

### tom.stoer

Re: Classical canonical transformations and unitary transformations in quantum mechan

One can combine the two exponentials into one using

$$e^X e^Y = e^{X+Y + [X,Y]/2 + \ldots}$$

which is related to the Baker-Campbell-Hausdorff formula:

$$e^{-ix_1p_2/2} e^{ix_2p_1} = e^{i(-x_1p2/2+x_2p_1) + i(-x_1p_1 + x_2p_2)/4 + \ldots}$$

I didn't check if the series terminates, but one can see immediately that the first few terms differ from the naive translation of the generating function of the canonical transformation.

--------------------------------------

In our special case for the co.m. frame it's clear that the exponent must be bilinear in x and p which allowes for the following ansatz

$$U = e^{ix_i A_{ik}p_k}$$

For the transformed positions and momenta one finds

$$Ux_iU^\dagger = \left(e^{A}\right)_{ik} x_k$$

$$Up_kU^\dagger = p_i \left(e^{-A}\right)_{ik}$$

Here I stopped with the calculation. Of course one could determine the matrix A using the well-known relations between the old and the new variables, but the ansatz cannot be generalized to more complicated transformations.

11. Sep 16, 2010

### strangerep

Re: Classical canonical transformations and unitary transformations in quantum mechan

Ugh, yes, I was about run out the door...

What type of "more complicated transformations" did you have in mind?
For more general homogeneous linear transformations, a more general quadratic
polynomial ansatz for the generator works (iirc). -- These are just standard Bogoliubov
transformations. For inhomogeneous transformations, add some linear terms to
the quadratic ansatz. These are usually called "field displacement" transformations.

Or did you want to handle something more pathological?

12. Sep 16, 2010

### xepma

Re: Classical canonical transformations and unitary transformations in quantum mechan

OK, I haven't read everything, but let me just make my point, and hopefully I'll explain what you are after.

I first focus on just a single type of transformation: a scale transformation of the coordinate $$x$$. In the following I'm not going to write a hat on every single operator, but it should be clear when I'm talking about a number or an operator.
======================
The transformation we're after is:

$$x \rightarrow x' = (1+a) x$$.

with a some positive number.

Now, as I stated before, if we have some transformation and we want to relate this to a generator, we need to solve:

$$x = x + \delta x$$

where $$\delta x$$ is the change in coordinate and it is related to some generator $$G$$ as:

$$\delta x = i\omega [G,x]$$

Here, $$\omega$$ is some number-valued parameter. In our case of a scale transformation we have:

$$x' = (1+a)x = x + a x$$

and so we want to find a G such that

$$i\omega [G,x] = a x$$

Therefore $$\omega = a$$. Now, you should check for yourself that the generator for a scale transformation is the so-called dilation operator D_x:

$$D_x = -i x \partial_x ( = xp_x)$$.

It's commutator with $$x$$ is

$$i[D_x,x] = x$$

Meaning we have now solved:

$$\delta x = ia [D_x,x]$$, such that $$x \rightarrow x' = x + \delta x = (1+a)x$$

======================

If we want to write this in exponentiated form, i.e. as a unitary transformation acting on the coordinates we can use $$U = e^{i\omega G}$$. To see that this works we look at:

$$UxU^\dag = e^{ia D_x} x e^{-ia D_x}$$

We only need to expand this to zeroth + first order. Higher order terms vanish. It's a good excersice to check that it should be:

$$e^{ia D_x} x e^{-ia D_x} = x + ia D_x x - iax D_x = x + ia[D_x,x]$$

which is precisely of the type $$x + \delta x$$. This shouldnt come as a suprise, since the method I started with is based on the assumption that the transformation looks like $$e^{ia D_x} x e^{-ia D_x}$$ in the first place.

Now that we have our unitary transformation in hand we should check what happens to the other operators. It should be clear that this transformation does not change other coordinates ($$y,p_y$$). The dilation operator commutes with these coordinates. It does change the momentum, however. The reason is simple: the momentum operator does not commute with the dilation operator. Specifically:

$$i[D_x, p_x] = -p$$.

Since the change in momentum $$\delta p$$ is again given by $$ia[D_x,p_x]$$ we can immediately write:

$$p' = p+\delta p = p - ap = (1-a)p$$.

Unfortunately, this is only the infinitesimal change of p. The full change of p is still given by $$UpU^\dag$$. But you should not be suprised that when we scale the coordinates as:

$$x\rightarrow (1+a)x$$

Then the momentum scales as

$$p \rightarrow \frac{1}{1+a}p$$.

I havent checked if you can obtain this from the relation $$UpU^\dag$$ though.

======================

Anyways, this is the recipse for performing a single type of transformation if you know the generators.

The transformation you are after can be done in two ways:

(1) Find the very specific generator that matches your transformation.

(2) Break the transformation down into a sequence of transformations of which you know the generator.

In the second case you only need two types of transformations: the scale transformation and the nameless transformation that linearly mixes two coordinates. Specifically:

$$D_x = -i x\partial_x = xp_x$$
$$G_{12} = -i x_1 \partial_{x_2} = x_1 p_{x_2}$$
$$G_{21} = -i x_2 \partial_{x_1} = x_2 p_{x_1}$$

As you can see the two new generators have the following commutation relations:
$$i[G_{12},x_2] = x_1$$
$$i[G_{12},p_1] = p_2$$
$$i[G_{12},x_1] = i[G_{12},p_2] = 0$$

So with the generator $$G_{12}$$ we generate transformations such as:

$$\lbrace x_1, x_2\rbrace \rightarrow \lbrace x_1 , x_2 + \omega x_1\rbrace$$

for some parameter $$\omega$$.

By fiddiling arount with these transformations in some order you can generate you transformation. For instance:

(1) Acting with $$G_{21}$$

$$\lbrace x_1, x_2\rbrace \rightarrow \lbrace x_1 + a x_2,x_2\rbrace$$

(2) Acting with $$G_{12}$$

$$\lbrace x_1 + a x_2,x_2\rbrace \rightarrow \lbrace (1+ab)x_1 + a x_2,x_2 + b x_1\rbrace$$

(3) Acting with $$D_1$$ and $$D_2$$

$$\lbrace (1+ab)x_1 + a x_2,x_2 + b x_1\rbrace \rightarrow \lbrace c(1+ab)x_1 + da x_2,dx_2 +c b x_1\rbrace$$

With each step you have some parameter a,b,c,d. You can now solve for these parameters and match them with the transformation you are after. In the end you have the unitary transformation that looks like:

$$U = e^{idD_2}e^{icD_1}e^{ibG_{12}}e^{iaG_{21}}$$

with the proper parameters for a,b,c,d. By acting with this operator on the momenta you can check that it generates the proper momentum relations as well. Why? Because the transformation is canonical! So it guarentees that the momentum transform in the proper way. But you should check this ofcourse.

13. Sep 16, 2010

### xepma

Re: Classical canonical transformations and unitary transformations in quantum mechan

Ok, now for the other method, which is faster.

We are after the transformation:

$$x_1 \rightarrow x_1 - x_2$$
$$x_2 \rightarrow \frac{m_2}{M} x_2 + \frac{m_1}{M} x_1 = x_2 + (\frac{m_2}{M}-1)x_2 + \frac{m_1}{M} x_1$$

We identify:
$$\delta x_1 = -x_2$$
$$\delta x_2 = (\frac{m_2}{M} - 1)x_2 +\frac{m_1}{M} x_1$$

So we need to find a G which solves:

$$\delta x_1 = -x_2 = i[G,x_1]$$
$$\delta x_2 = (\frac{m_2}{M} - 1)x_2 +\frac{m_1}{M} x_1 = i[G,x_2]$$

Here I have absorbed the free parameters into the definition of G. We can use:

$$i[x_1 p_1,x_1] = x_1$$
$$i[x_2 p_1,x_1] = x_2$$

Since the commutator is linear we have

$$G = -x_2 p_1 + (\frac{m_2}{M}-1)x_2 p_2 + \frac{m_1}{M}x_1p_2$$

And so the unitary transformation is $$\exp(iG)$$

I'm gonna leave it up to you to see how the momentum changes. It's best to look at the infinitesimal case first to see if it matches up, i.e. first determine

$$\delta p_1 = i[G,p_1]$$

Keep in mind though, that because you are dealing with scale transformations there is some tricky stuff going on if you go from the infinitesimal to the exponentiated form.

I hope I havent made too many serious errors here.

Last edited: Sep 17, 2010
14. Sep 16, 2010

### strangerep

Re: Classical canonical transformations and unitary transformations in quantum mechan

Fewer than I tend to make. :-(

Should the last term be $i[G,x_2]$ ?

15. Sep 17, 2010

### xepma

Re: Classical canonical transformations and unitary transformations in quantum mechan

I went over it a few times ;)

fixed!

16. Sep 17, 2010

### DrDu

Re: Classical canonical transformations and unitary transformations in quantum mechan

Maybe you'll find the discussion on quantization and symmetry transformations in the QM book by Balentine interesting.
Just a remark: The problems with inequivalent Hilbert spaces which strangerep mentioned in #3 do not occur in ordinary QM but are a specific problem of quantum field theory.

17. Sep 17, 2010

### tom.stoer

Re: Classical canonical transformations and unitary transformations in quantum mechan

Thanks for the interesting discussion.

My conclusion is that the path from classical to quantum mechanical canonical tranformations (unitary transformations) is thorny and that the explicit construction is by no means trivial - even if the generating function of the classical transformation is know.

The two different operators for the transformation to the c.o.m. frame are interesting as the operator ordering and the BCH formula can be studied explicitly.

18. Jan 25, 2011

### A. Neumaier

Re: Classical canonical transformations and unitary transformations in quantum mechan

This ambiguity is explained by remembering that classical mechanics is the limit hbar to zero of quantum mechanics. Thus every operator expression G(x,p,hbar) becomes classically G(x,p,0), i.e., one loses one degree of freedom. That one can determine the quantum theory from the classical limit is therefor as impossible as reconstructing a function of a single varible from its value at t=0.

On the other hand, different quantum mechanical orderings of the same classical expression differ only by a term O(hbar) that can be found by performing the reordering with the help of the CCR.

19. Jan 25, 2011

### tom.stoer

Re: Classical canonical transformations and unitary transformations in quantum mechan

I fully agree.

Afaik there are additional requirements for canonical transformations to be promoted to unitary operators which unfortunately I cannot remember.

20. Jan 26, 2011

### A. Neumaier

Re: Classical canonical transformations and unitary transformations in quantum mechan

well, questions of selfadjointness of the generators (needed to make the exponential well-defined) aren't handled by this formal substitution approach....