# Quantizating a symmetric Dirac Lagrangian

## Main Question or Discussion Point

As is well known, a Dirac Lagrangian can be written in a symmetric form:

L = i/2 (\bar\psi \gamma \partial (\psi) - \partial (\bar\psi) \gamma \psi ) - m \bar\psi \psi

Let \psi and \psi^\dagger be independent fields. The corresponding canonical momenta are

p = i/2 \psi^\dagger;

p^\dagger = - i/2 \psi.

The anticommutation relations would be

{(\psi)_k, i/2 (\psi^dagger)_m}= i delta_km,

{(\psi^\dagger)_k, - i/2 (\psi)_m}= i delta_km,

which are in contradiction with one another. How can this happen?

## Answers and Replies

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dextercioby
Homework Helper
Actually the quantization procedure forces you to use only one type of fields, either the coordinate fields, or their momenta. So one has only one set of anticommutation relations between the fields (operator-valued distributions).

OK, I thought so. But, with this Lagrangian, I get an extra factor 2 in my anticommutation relations compared to the standard anticomm. relations.

dextercioby
Homework Helper
Then you must have made an error somewhere. The fundamental Dirac parentheses are

$$\left[\Psi^{a}(x),\bar{\Psi}_b (y)\right]_{+, x^0 = y^0} ^{*} = - \delta^{a}_{b} \delta \left(\vec{x}-\vec{y}\right)$$

and using the canonical anticommutation rule, one finds the needed anticommutation relations, no other numerical factor needed.

The mistake is quite subtle. You forgot to pick up an additional minus sign from passing $\frac{\partial}{\partial\dot\psi}$ through $\bar\psi$ when defining your canonical momentum $p$.

You should have gotten
$$p = -\frac{i}{2} \psi^\dagger.$$

How can I have anticommuting variables before I define my canonical momentum which defines canonical commutation relations?

It turns out $\psi$ is a Grassman-valued function of spacetime -- even at the classical level. When you move to quantum mechanics, all this means is that its matrix elements will be Grassman-valued too.

This was never taught at my relativistic quantum mechanics class :( Even Greiner (a classic textbook) doesn't mention it. So, what is the basis for such an assumption?

dextercioby
Homework Helper
Well, the so-called <classical level> of spinors or spinor fields doesn't exist. The Dirac field is already a quantum field. So it must obey the spin-statistics theorem, therefore it describes fermions. Quantizing a fermion field through commuting field operators doesn't produce a Hamiltonian bounded from below. This is the original argument of 1940(Wolfgang Pauli in the Physical Review).

Is there a classical analogy of Poisson brackets which anticommute?

dextercioby
Homework Helper
Poisson brackets are part of the classical mechanics. Classical mechanics doesn't use anticommuting momenta and coordinates.

This was never taught at my relativistic quantum mechanics class :( Even Greiner (a classic textbook) doesn't mention it. So, what is the basis for such an assumption?
The assumption comes from quantum mechanics (yes, I know it's backwards, but that's the way it goes). You don't get a local quantum field theory with commuting spinors; you get it by using anticommuting spinors.

Well, the so-called <classical level> of spinors or spinor fields doesn't exist. The Dirac field is already a quantum field. So it must obey the spin-statistics theorem, therefore it describes fermions. Quantizing a fermion field through commuting field operators doesn't produce a Hamiltonian bounded from below. This is the original argument of 1940(Wolfgang Pauli in the Physical Review).
Classical level of spinors do exist. They are just Grassman-valued, which makes it pretty hard to observe.

Is there a classical analogy of Poisson brackets which anticommute?
Yes, there is; see below.

Poisson brackets are part of the classical mechanics. Classical mechanics doesn't use anticommuting momenta and coordinates.
Not true. If $\theta$ is a Grassman coordinate and $\pi$ is its conjugate momentum. Then the Poisson bracket between two Grassman-valued dynamical quantities, $A(\theta,\pi)$ and $B(\theta,\pi)$ is defined to be
$$\{A,B\}=-\left(\frac{\partial A}{\partial \theta}\frac{\partial B}{\partial \pi}+\frac{\partial A}{\partial \pi}\frac{\partial B}{\partial \theta}\right),$$
for the convention of left derivatives.

dextercioby