# Quantizing fields

1. Mar 16, 2009

### captain

I have seen that in order to quantize a field there is a need to establish commutation relation (or anti-commutation relations). I am wonder how does that in a sense quantize something. I feel as though I dont have a good grasp of how it works. The only thing that seems apparent is that this is going from numbers to operators (such as momentum becoming an operator when you go from classical mechanics to quantum mechanics). Thanks to anyone in advance who can elaborate on this.

2. Mar 16, 2009

### tiny-tim

Hi captain!

Commutation (or anti-commutation) relations are needed in any quantum field theory … they convert operators into numbers.

A field is made of operators.

But we don't want operators in a kinematic equation, we only want numbers!

3. Mar 16, 2009

### Ben Niehoff

Classically, everything is an ordinary number, so we must have (between any two conjugate variables x and p)

[x,p] = 0

Quantum mechanically, however, we suppose that [x,p] is not precisely zero, but a very small number instead. Then

$$[x,p] = i\hbar$$

The factor of i is chosen because it allows both x and p to be Hermitian:

$$[x,p]^{\dagger} = (i\hbar)^*$$

$$[p^{\dagger}, x^{\dagger}] = -i\hbar$$

$$[p^{\dagger}, x^{\dagger}] = -[x,p] = [p,x]$$

Simply giving [x,p] a nonzero value is enough to derive all the usual quantization rules via abstract linear algebra over a Hilbert space on which x and p act.

For fermions, however, things are trickier, because one is actually quantizing Grassman-valued fields; i.e., a "classical" fermion field has

{x,p} = 0

To quantize, we simply replace zero with a very small number, and write

$$\{x,p\} = \hbar$$

In this case, we do not need the factor of i, because

$$\{x,p\} = xp + px = \{p,x\}$$

4. Mar 18, 2009

### strangerep

"Quantization" essentially means passing from classical observables,
i.e., functions f(p,q) on phase space, to operators on a Hilbert space.
Typically, for any particular classical case, there's a
a dynamical algebra among the phase space functions (think Hamiltonian
dynamics and all that) expressed via a Poisson bracket on those
functions. Quantization (in simple cases) is then just a rule for
taking the classical Poisson bracket and changing it into a commutator
among operators on Hilbert space. So the quantum commutator is not
really postulated out of thin air, but is taken from a known classical
situation. The Planck constant that enters in this rule is just a
constant to keep the dimensions straight (since xp is not
dimensionless, but has units of action), and must be determined from
experiment.

There are variations on this general approach. E.g., for the
non-relativistic case, one can start from the classical Galilean
algebra (or Galilean group) and attempt to represent it as operators on
a Hilbert space. (This is called "constructing unitary irreducible
representation(s)" of the algebra.) Ballentine's QM book explains this
quite well. Well recommended for anyone who suspects they're "missing

Similarly, for the relativistic case, one can start from the Poincare
group and construct a unitary irreducible representations of that
algebra instead (this is sometimes called the "Wignerian" approach).
Weinberg's QFT book (vol-1) goes through this.

Further, when other groups are involved (such as the internal symmetry
groups of the Standard Model), one constructs unitary irreducible
representations of the direct product of those group(s) with the
Poincare group.

So "quantization" is really all about starting with an algebra of
observables and constructing unitary irreducible representation(s) of
the algebra.

BTW, the following can be a little misleading:

Commutation (or anti-commutation) relations only express the intrinsic
structure of the algebra under consideration. -- The result must be a
member of the algebra. For the Heisenberg algebra, although the rhs
looks like an ordinary number, that's a bit misleading because it's
really a "central" element of the algebra. ("Central" means "commutes
with everything in the algebra".) But more generally, a commutator
yields another member of the algebra. To "convert" from an operator to a
number more generally, one uses a "linear functional", e.g.,
$\langle\psi|A|\psi\rangle$ should be thought of as the numerical result
of acting with the linear functional $\psi$ upon the operator $A$.

5. Mar 19, 2009

### aristurtle

in quantum mechanics , we represent observables as abstract object ,the operator.we can tell the probobility of an observable to be a certain value.
in quantum field, we quantize the amplitute of a certain mode(not always Fourier mode,e.g. in curved spacetime) of our field.If you try to quantiza the mode amplitute in schrodinger picture,it will be more easy to see the essence.Of course heisenberg picture is practically more convenient.

6. Mar 19, 2009

### FredericGos

Can I ask a little side question here? I have seen this notation before but cannot remember. What does [x,p] = 0 mean? I mean what does the square brackets and the comma mean? Can you phrase it? It looks like a closed interval but surely it isnt.

Just curious. thx

/Frederic

7. Mar 19, 2009

### Staff: Mentor

[x,p] is shorthand for xp - px. We call it the "commutator of x and p."

8. Mar 19, 2009

### FredericGos

ah ok thx :) It's from Group Theory right?

9. Mar 19, 2009

### tiny-tim

Yes, the commutator does have to be a member of the algebra … but the commutator is a scalar, and a scalar is a member of the algebra … and the scalars of this algebra are ordinary numbers.

10. Mar 19, 2009

### Fra

"Quantization" can mean a few things, like already mentioned.

I got the impression reading that maybe you ask about the logic as to how does noncommutativity generally lead to discretization? Such as quantization of energy states? What is the general connection.

If you solve the schrödinger equation it's clear, so I assume you want a nontechnical answer.

The idea that two observables, doesn't commute, generally means that the two observables really aren't "independent" in the sense that information about one observables, changes the information about the other one. This implies a kind of "constraint" when you have other relations where these two observables are in the picture. The commutator is a kind of logical connection between the two observables. Why the commutator is what it is, is to me a separate question though.

For example in an atom, there is both kinetic energy and potential energy.

When you look for the stationary states of an atom, then a simple way of seeing it the combination of the energy conservation beeing a function effectively of the two mentioned observables, the constraint that some observables like x and p are related (ignored the details of HOW they are related) then the total energy can only take certain discrete values.

So maybe a simplified single phrase explanation could be that the nonzero-commutator implies an additional "constraint" on the stationary energy values. And this constraint together with the "normal constraints" such as energy conservation etc, implies discretization.

/Fredrik