# Canonical Quantization

1. Feb 2, 2010

### mtak0114

Hi,

I have recently been reading Dirac's book on Canonical Quantization of gauge theories, and I have a few questions:

So in the quantization procedure we need to identify all the constraints in the theory. Once this has been done (if we are dealing with a gauge theory) we need to check that all constraints are first class, i.e. that all constraints commute with each other, correct?

Now given that all the constraints commute we deduce that the constraints are satisfied we can be sure that the evolution in the hamiltonian system is equivalent to the evolution in the lagrangian theory i.e. that there equations of motion agree, correct?

Now when studying the Lagrangian theory we can see what are the gauge transformations take for example EM the gauge transformation is just:
$$A_\mu(x) \rightarrow A_\mu' = A_\mu(x) +\partial_\mu \theta(x)$$

To understand this in the canonical picture is where I have trouble, its something like the Lie algebra of the constraints generate gauge transformation but how can I see this?

M

2. Feb 2, 2010

Which book?

3. Feb 3, 2010

### mtak0114

sorry Lectures on Quantum Mechanics

4. Feb 4, 2010

### strangerep

Actually, we use the constraints initially to write down what's called a "total
Hamiltonian", which is different from the ordinary Hamiltonian, but the two
are "weakly equal" (meaning equal only if the equations of motion are satisfied).
At this stage, we're still working with classical quantities. We need to check
commutativity of the constraints regardless of whether we're working with a
gauge theory, since that tells us whether there's any second-class constraints
(in which case we must construct something called a Dirac-Bergman
bracket instead of the ordinary Poisson bracket as the classical starting point
for deciding what quantum commutators are appropriate).

Er, no... wait a minute. That's not really how it works. Setting the constraints
to zero just defines a phase space (hyper)surface, but we can also
consider them as functions over the whole phase space, and they are not
automatically zero everywhere. We use the constraint functions when working
with derivatives, etc, that potentially take us off the constraint surface.

But I get the feeling you need some other references to get a clearer picture
of what's going on...

Have you looked at this Wiki page:
http://en.wikipedia.org/wiki/Dirac_bracket ?
It gives a more condensed summary of what this constraint business

Do you have access to Henneaux & Teitelboim's, "Quantization of Gauge Systems"?
In sect 1.2, there's a general answer about how constraints generate gauge
transformations.

Maybe the best approach is to look at these other references, and then come
back with more specific questions if anything's still unclear.

Last edited: Feb 4, 2010
5. Feb 7, 2010

### Simon_Tyler

Another good introduction to the subject is the lectures given by Hans-Jurgen Matschull, quant-ph/9606031.

6. Feb 7, 2010

### mtak0114

thanks I'm getting a clearer idea now
but I still have one question:
what is the definition of a gauge transformation?

1) A symmetry of the Lagrangian which leaves the equations of motion invariant

or

2) a gauge transformation is whatever a 1st class constraint generates.

I would have thought 1) to be the commonly accepted definition in that case if you look at
quant-ph/9606031v1 equation 3.11 is not in general a gauge transformation how can we justify setting $$u_1 = \dot{u_2}$$?

thanks

M

7. Feb 7, 2010

### strangerep

A gauge freedom is an unphysical degree of freedom.
Changing to a different gauge has no effect on any of
the physically-observable quantities in the theory.

E.g., a Lorentz transformation between inertial frames
changes the Hamiltonian (i.e., energy), but in gauge theories
the (total) Hamiltonian is invariant under gauge transformations.

(This is mentioned briefly in that Wiki page I referenced earlier.
If any of the $u_k$ functions used in that treatment
remain undetermined, they indicate a gauge freedom.)

8. Feb 8, 2010

### mtak0114

Okay

how then can the two different notions of gauge freedom in the Lagrangian and Hamiltonian pictures be understood are they just different or is there some mechanism by which the arbitrary functions must satisfy the relation
$$u_1 = \dot{u_2}$$
so that they agree?

9. Feb 11, 2010

### strangerep

(mtak0114, you need to put a bit more detail into your questions. It took me quite
a while to guess what you're really asking. Even now, I'm not entirely sure I'm

I presume your
refers to stuff in Matschull's lecture notes (quant-ph/9606031) in his eqn(3.11)
and his paragraph thereafter (on p24)? If so, I think the answer is self-explanatory:
In one formulation of EM gauge invariance we use a single function w(x) and
the gauge transformation is of the form:

$$\delta A_\mu ~=~ \partial_\mu w(x) ~~,$$

whereas (following the Dirac programme) we end up with two
functions $$u_1(x) , u_2(x)$$ and the gauge transformation
takes the form:

$$\delta A_0 ~=~ u_1(x) ~~;~~~~ \delta A_i ~=~ \partial_i u_2(x) ~~.$$

So we conclude that the two forms of gauge invariance are related by

$$u_2(x) = w(x) ~~;~~~~ u_1(x) = \dot{u_2}(x)$$

We simply have two ways of expressing the same underlying
unphysical gauge freedom.