Theory Behind Dirac Lagrangian: Reasons Nature Didn't Choose Mine

[TriTertButoxy]

We all know that the free Lagrangian for a spin-1/2 Dirac field is

$$\mathcal{L}=\bar\psi(i\gamma_\mu\partial^\mu-m)\psi.$$​

But, if I were to invent a Lagrangian, I would have tried

$$\mathcal{L}=\partial_\mu\bar\psi\partial^\mu\psi-m^2\bar\psi\psi.$$​

What's wrong with this second Lagrangian? Why didn't nature choose my Lagrangian? (I'm looking for theoretical reasons. I don't want 'because it doesn't match up with observations')

 

[tom.stoer]

Your Lagrangian is perfectly valid but one finds that it doesn't describe spin 1/2 particles.

You need to introduce spinors, Dirac: 4-component bi-spinors in the (1/2, 1/2) representation. But I think this is explained in nearly every derivation of the Dirac Lagrangian (in books or scripts regarding relativistic quantum mechanics)

 

[dextercioby]

That's the $\left(\frac{1}{2},0\right)\oplus\left(0,\frac{1}{2}\right)$ representation of SL(2,C).

 

[samalkhaiat]

You need to introduce spinors, Dirac: 4-component bi-spinors in the (1/2, 1/2) representation.

(1/2,1/2) is the vector representation. Dirac 4-spinor is the dirct sum of the two fundamental, two-component, spinors (0,1/2) and (1/2,0) of SL(2,C).

 

[tom.stoer]

Thanks, sorry for that

 

[TriTertButoxy]

Hmm... none of your responses helped; but it's partly my fault. I should have stated from the beginning: Let $\psi(x)$ be a Dirac spinor transforming under the $\textstyle(\frac{1}{2},0)\oplus(0,\frac{1}{2})$ representation of the Lorentz group.

What now is the logical flaw in writing down the following Lagrangian for this field?
$$\mathcal{L}=\partial_\mu\bar\psi\partial^\mu\psi-m^2\bar\psi\psi$$

It is definitely Lorentz invariant, and it is certainly describing spin-1/2 particles by virtue of the Lorentz transformation of the field.

 

[The_Duck]

In section 36 of Srednicki's text it is claimed that if we try to write down a Lagrangian for a single left-handed spinor field chi with the kinetic term
$$\partial_\mu \chi \partial^\mu \chi$$
then we get a Hamiltonian that is unbounded below. Your Lagrangian is not exactly the same but perhaps it has the same problem?

 

[samalkhaiat]

It is definitely Lorentz invariant, and it is certainly describing spin-1/2 particles by virtue of the Lorentz transformation of the field.

It also describes 4 spin-0 particles!
On the level of field equations, there is no problem. One can show that
$$\{i\gamma^{\mu}_{ij}\partial_{\mu}\psi_{j}=0\}\ \ \Leftrightarrow \ \ \{\partial_{\mu}\partial^{\mu}\psi_{i}=0\}$$
This “equivalence” though does not extend to the corresponding Lagrangians:
1) all Noether currents derived from your Lagrangian don’t agree with those derived from the Dirac Lagrangian. For example, the U(1) Dirac current is
$$J^{\mu}=\bar{\psi}_{i}\gamma^{\mu}_{ij}\psi_{j}$$
while your Lagrangian, leads to
$$j^{\mu}=\bar{\psi}_{i}\partial^{\mu}\psi_{i}$$
This current, when coupled to potential $A_{\mu}$, describes SCALAR QED.
2) the dimension of the Dirac field is [3/2] in mass unit. The field in your Lagrangian is [1] mass unit which is (again) the dimension of a boson field.
On the representation level, we proceed as follow
1) We have the two fundamental rep.
$$\phi^{a}\sim (1/2,0), \ \ \psi_{\dot{c}}\sim (0,1/2)$$
2) we have the operator $\partial_{\mu}$ and the matrices $\sigma^{\mu}$
3) for massive particles, we also have the mass of the representation $m$
4) we form the vector rep.
$$(1/2,0)\otimes (0,1/2) = (1/2,1/2) \sim \partial^{a\dot{c}}= (\sigma_{\mu}\partial^{\mu})^{a\dot{c}}$$
Clearly, this can be translated into a pair of coupled differential equations
$$\partial_{a\dot{c}}\phi^{a} = m \psi_{\dot{c}},$$
$$\partial^{a\dot{c}}\psi_{\dot{c}}=-m\phi^{a}$$
5) in order to form the bispinor of Dirac, we first extend SL(2,C) with parity operation then we put the above two equations together. This leads to the Dirac equation.
Notice again that each component satisfies the K-G equation
$$(\partial_{\mu}\partial^{\mu}+m^{2}) \left( \begin{array}{c} \phi^{a} \\ \psi_{\dot{c}} \\ \end{array}\right)=0$$

 

[TriTertButoxy]

Thanks! that is very informative, but I'd like to pose a few more questions:

It also describes 4 spin-0 particles!

I definitely see how the interpretation comes about; but there's something special going on here: these fields get mixed up by Lorentz transformations, and not some internal O(4) symmetry.

On the level of field equations, there is no problem. One can show that
$$\{i\gamma^{\mu}_{ij}\partial_{\mu}\psi_{j}=0\}\ \ \Leftrightarrow \ \ \{\partial_{\mu}\partial^{\mu}\psi_{i}=0\}$$

Shouldn't the arrow be going one way here (to the right only)? This is just a minor comment.

On the representation level, we proceed as follow
1) We have the two fundamental rep.
$$\phi^{a}\sim (1/2,0), \ \ \psi_{\dot{c}}\sim (0,1/2)$$
2) we have the operator $\partial_{\mu}$ and the matrices $\sigma^{\mu}$
3) for massive particles, we also have the mass of the representation $m$
4) we form the vector rep.
$$(1/2,0)\otimes (0,1/2) = (1/2,1/2) \sim \partial^{a\dot{c}}= (\sigma_{\mu}\partial^{\mu})^{a\dot{c}}$$
Clearly, this can be translated into a pair of coupled differential equations
$$\partial_{a\dot{c}}\phi^{a} = m \psi_{\dot{c}},$$
$$\partial^{a\dot{c}}\psi_{\dot{c}}=-m\phi^{a}$$
5) in order to form the bispinor of Dirac, we first extend SL(2,C) with parity operation then we put the above two equations together. This leads to the Dirac equation.
Notice again that each component satisfies the K-G equation
$$(\partial_{\mu}\partial^{\mu}+m^{2}) \left( \begin{array}{c} \phi^{a} \\ \psi_{\dot{c}} \\ \end{array}\right)=0$$

So how do I generalize this procedure to derive the correct equation of motion for spin-3/2 fields, or spin-5/2 fields, or ...?

 

[Avodyne]

Weinberg (vol I) explains how starting with spin-1/2 particles, and writing down the simplest free field that you can build out of their creation and annihilation operators, leads to a field that obeys the Dirac equation. He also discusses the procedure for higher spin fields.

 

[samalkhaiat]

Shouldn't the arrow be going one way here (to the right only)? This is just a minor comment

.

No, you can always find an equivalent set of 1st order DE's for a given set of 2nd order DE's.

So how do I generalize this procedure to derive the correct equation of motion for spin-3/2 fields, or spin-5/2 fields, or ...?

Once, you have constructed the object you are after and identified all physical constrains, writing the equations of motion becomes an easy task. See posts 24,25 and 29 in
www.physicsforums.com/showthread.php?t=192572

 

[Matterwave]

Wouldn't your Lagrangian give rise to second order time derivatives? Wouldn't that mean then that it would be much more difficult to interpret that wave function as a probability distribution because its evolution would no longer be unitary?