# Lagrangian, Hamiltonian and Legendre transform of Dirac field.

1. Jun 23, 2011

### kof9595995

In most of the physical systems, if we have a Lagrangian $L(q,\dot{q})$, we can define conjugate momentum $p=\frac{\partial L}{\partial{\dot{q}}}$, then we can obtain the Hamiltonian via Legendre transform $H(p,q)=p\dot{q}-L$. A important point is to write $\dot{q}$ as a function of $p$.
However, for the Dirace field with Lagrangian ${\cal L}=\bar{\psi}({i\gamma}^{\mu}{\partial}_{\mu}+m) \psi$, it's impossible to write $\dot{q}$ as a function of $p$, because the conjugate momentum is $i{\psi}^{\dagger}$, this means the Hamitonian has to contain $\dot{\psi}$? How do we make sense of it?

2. Jun 23, 2011

### dextercioby

Well, we obtain the so-called primary constraints. The procedure of dealing with constraints is presented in a number of books, the best one being Henneaux & Teitelboim's <Quantization of Gauge Systems>.

3. Jun 23, 2011

### strangerep

For a briefer overview of constrained quantization (i.e., before diving into H&T),
try this Wiki page:

http://en.wikipedia.org/wiki/Dirac_bracket

The bottom line is that (in general) we must use a "Dirac bracket" instead
of the usual Poisson bracket to obtain the quantization.

4. Jun 23, 2011

### Bill_K

All right, well I guess I'm the only one here who does not understand this, because I think it's a lot simpler than that. So please set me straight (without just pointing me to a book to read!) The Lagrangian is

L = ψ(iγμμ - m)ψ (overscore written as an underscore)

The conjugate momentum is

π ≡ ∂L/∂ψ· = i ψ+

and so the Hamiltonian is

H ≡ πψ· - L = i ψ+ψ· - ψ(iγμμ - m)ψ

The ψ· s cancel and we are left with

H = - i ψ+ α·∇ψ + βmψ+ψ

where γ0 = β and γi = β αi

Obviously the Hamiltonian does not contain ψ·. This is completely standard. So what's wrong with it, in your view?

5. Jun 23, 2011

### dextercioby

Alright then, what are then the Hamilton equations of motion ?

6. Jun 23, 2011

### kof9595995

I simply didn't realize the first 2 terms cancel, but according dextercioby and strangerep it's more subtle than this. So I have the same question with Bill_K. Especially I don't remember imposing any constraints on the Dirac field quantization, like gauge fixing or Gauss's law constraint for EM field, and we simply give a anticommutation realton to $\psi$ and ${\psi}^{\dagger}$

7. Jun 23, 2011

### strangerep

Take it easy. I think you're right: it is a "lot simpler than that".

(I thought the OP was asking a somewhat deeper question, but in the
light of a new day I realized he wasn't. Thanks for giving him a better

8. Jun 24, 2011

### Bill_K

There's still a question here, at least in my mind. The canonical variables are q = ψ and p = ih ψ+. Hamilton's equations are δH/δp = ∂q/∂t and δH/δq = - ∂p/∂t, which lead to Dirac's equation for ψ and ψ+ respectively. Yet there does appear to be a constraint, in that the 'coordinate' ψ and the 'momentum' ψ+ are algebraically related.

Maybe what this is telling you to do is start with L = φ(iγμμ - m)ψ + ψ(iγμμ - m)φ and identify φ and ψ at the end. Or maybe split ψ up into a pair of coupled 2-spinors?

9. Jun 24, 2011

### kof9595995

emm, I'm getting more confused, what's the precise definition of "constriant", for an ordinary Lagrangian, the canonical momentum is $p=\frac{\partial L}{\partial \dot{q}}$, and p is usually a function of q dot, then why isn't this a constraint, since p and q are related by a time derivative?

10. Jun 25, 2011

### strangerep

Here's \$0.02 worth...

The Hessian of the Lagrangian (matrix of 2nd derivatives wrt to velocities)
is not invertible. In our case, the Hessian is trivial:
$$\frac{ \partial^2 L}{ (\partial \dot \psi)^2} ~=~ 0$$
This means we have a "singular" Lagrangian: the velocities can not be
determined uniquely from the momenta and positions. Thus, the Hamiltonian
form of dynamics (in its naive form) does not determine time evolution
uniquely.

[kof9595995: this is why ordinary Lagrangians don't produce the kind of
constraints we're talking about here. In the ordinary case, you can get
the velocities from position and momenta because the Hessian above
is invertible. But in the singular case, the constraints contain the detailed
information which is preventing the Hessian being invertible.]

Since q = ψ and p = ih ψ+, there's a constraint:
$$\phi ~:=~ p - iq^+ ~=~ 0$$
This is called a "first class" constraint because its Poisson bracket with
all other constraints vanishes. (In our case, this is trivial because there's
only 1 constraint.)

By standard theory (cf. http://en.wikipedia.org/wiki/Dirac_bracket )
a first class constraint generates a gauge transformation.
Therein lies the non-uniqueness of the time evolution: we can apply
an arbitrary gauge transformation to ψ at each point in time.

(This gets more complicated if one wants to handle the anticommutativity,
since this involves classical Grassman variables and a suitably
generalized notion of Poisson bracket. Henneaux & Teitelboim give details.)

But for a basic understanding, one can simply note that the singular nature
of the Dirac Lagrangian is associated with its well-known gauge freedom,
i.e., multiplication by a (generally time-dependent) phase factor.

Last edited: Jun 25, 2011
11. Jun 25, 2011

### Bill_K

The Dirac Lagrangian has no such gauge freedom. And if you multiply ψ by a time-dependent phase factor it no longer satisfies the Dirac Equation. I think you're anticipating the minimal coupling with the electromagnetic field, which does have such a gauge invariance, but we're talking about the vacuum case here.

12. Jun 25, 2011

### dextercioby

Since the existence of the Dirac field is required by a quantum theory, it makes sense to perform the Hamiltonian analysis assuming the fields to be anticommutative in the Lagrangian version. One shows then that the constraints of the Hamiltonian theory are second class, not first class.

13. Jun 25, 2011

### strangerep

That's a symptom of the fact that the Lagrangian is singular: we can't go uniquely
from the conjugate pair (q,p) of the Hamiltonian formalism back to the (q,v) of the
Lagrangian formalism.

Sorry. My last paragraph was misleading. I'll try to clarify...

Are you familiar with the distinction between "weak" and "strong" equality
in constrained Hamiltonian dynamics? One says that two dynamical quantities
are "weakly" equal if they're equal on the constraint surface, and "strongly"
equal if they're equal on the whole phase space. The field equations
include a restriction to the constraint surface.

For consistency, one insists that the constraint functions are time-independent,
and this has the consequence that one can add constraint functions to the
Hamiltonian without affecting the physics.

[Hmm, this answer needs to be longer before it can be satisfactory but I don't have
enough time right now.]

Thanks for mentioning that. Do you have time to elaborate?

14. Jun 25, 2011

### kof9595995

I see, so the constraint is a constraint on phase space of (p,q), and that's why $p=f(\dot{q})$ is not called a constraint but $p=f(q)$ is , am I right?

15. Jun 26, 2011

### dextercioby

Well, the so-called <classical Lagrangian> is built up from objects which have no significance at a classical level, the Dirac fields, psi and psibar.

Why, though, formulate a Lagrangian with such objects ? Well, I can give 3 reasons.

1. If one formulates a valid Hamiltonian dynamics, one can obtain the quantum theory of the Dirac field via the so-called <canonical quantization> (for the sake of my argument and avoiding complications which I don't know how to treat, the canonical quantization assumes no Gribov ambiguities or other anomalies).

2. The <classical> Lagrangian is used to write down the Lagrangian path integral which is also key ingredient in the quantum theory.

3. The <classical> Lagrangian exhibits a global phase (U(1)) symmetry which governs its couplings to gauge fields (=> QED and QCD). [Gauge fields have significance at a classical level (electromagnetism, gravity and gauge multiplets of e-m potentials (classical gluon fields)), so one can discuss the possible couplings purely at a classical level.]

So it makes sense to write down a Lagrangian with Psi and Psibar. After writing it down with commutative fields (just like Schroedinger fields but with spinor indices) and going to point 1. above with the Hamiltonian and further going into the quantum theory you end up with a Hamiltonian unbounded from below. So to keep things flowing rightfully, one should have written down a Lagrangian with fermionic fields. Then the Hamiltonian analysis is different and leads through canonical quantization to acceptable quantum observables.

As for the Hamiltonian formalism for the <classical> anti-commutative Dirac fields being 2nd class, this is textbook material.

16. Jun 26, 2011

### strangerep

That's my understanding. [But the subject is large and subtle.]

17. Jun 26, 2011

### strangerep

Rats. That's the bit on which I was hoping you'd give a helpful overview.
I guess I'll just have to slog through the textbook properly without help. :-(

18. Jun 27, 2011

### dextercioby

Well, if you're familiar with the Dirac algorithm and the Dirac bracket, then you should have no problem if you start with the symmetrized Lagrangian:

$$\mathcal{S}_{L} =\int d^4 x {}\left[\frac{1}{2}\bar{\Psi}i\gamma^{\mu}\partial_{\mu}^{\leftrightarrow} \Psi - m\bar{\Psi}\Psi\right]$$

and do all the calculations by yourself.

19. Aug 23, 2011

### kof9595995

Recently I have a new related question: how to prove the equivalence between canonical and path integral quantization for Dirac field?
For bosons the lagrangian is quadratic in time derivative, so canonical variables are well-defined, so are the corresponding kets |q> and |p>. In addition <q|p>=exp(iqp).......(*),
and this in the end will give $p\dot{q}$ term in $p\dot{q}-H$. However for Dirac field canonical variables are just related by Hermitian transpose($\psi$ and $i\psi^{\dagger}$), so the relations of canonical kets should be very different from the Bosonic ones, I seriously doubt if there's still a relation like (*), if not, how can we finally arrive at $p\dot{q}-H$