How to dimensionally reduce the 3+1D Dirac equation into the 1+1D Dirac equation?

Question

In 3+1D the Dirac equation looks like $$i\partial_\mu \gamma^\mu \Psi -m\Psi=0.$$

If we only consider $x$-direction, then it should reduce to $$i\partial_t\gamma^0\Psi =(-i\partial_x\gamma^1+m)\Psi.$$

Here $\Psi$ is still four component object. Although, here, they have given the one-dimensional Dirac equation as $$i\partial_t\Psi =(-i\sigma_1\partial_x+m\sigma_0)\Psi $$ where the wave function $\Psi$ is a complex two-component vector.

How are the two pictures equivalent?

---

Edit. In particular, I want to know how you reduce the $3+1$D Dirac equation, which has four components, to the $1+1$D Dirac equation, which has only two.

I tried putting the $\gamma$ matrices values $$\gamma^0 =\begin{bmatrix}I_2 & 0 \\ 0 & -I_2 \end{bmatrix} \ \ \ \ \ \gamma^i = \begin{bmatrix}0 & \sigma^i \\ -\sigma^i & 0 \end{bmatrix}$$ Putting this, we get

$$i\partial_t \begin{bmatrix} \phi \\ -\chi \end{bmatrix}=-i\sigma^1\begin{bmatrix} \chi \\ -\phi \end{bmatrix}+m \begin{bmatrix} \phi \\ \chi \end{bmatrix}$$

where $$\Psi = \begin{bmatrix} \phi \\ \chi \end{bmatrix}$$ As you see, the equation comes out to be coupled.

Answer 1

The method suggested by @Qmechanic. It can be done in three lines.

The 4-component Dirac equation can be re-written as $$ i\partial_t \psi = \left [ \gamma^0 \gamma^{a} (-i\nabla_a) + \gamma^0 m \right ] \psi $$

Assuming $(+---)$ signature, I will use the basis: $$ \gamma^{0} = \sigma_z\otimes\mathbb{I}_{\tau}, \; \gamma^{0}\gamma^{a} = \sigma_x \otimes \tau_a $$ where $\tau$ and $\sigma$ are Pauli matrices .

If you restrict to only $x$-direction, there will only be $\tau_x$. The eigenspaces of $\tau_x$ decouple.

But I want to add that, in principle, if you are handed a problem in (1+1)-D, you don't have to (actually, you shouldn't) dimensionally reduce from (3+1)-D. Just write down the Clifford algebra in (1+1)-D and be done with it.

Answer 2

I have actually done some mistakes in the following answer, since the mass term has to mix them, and you have to get a different relative in between the derivatives, I'm pretty sure the mistakes are in the redefinition of the field containing the two different components that behave equally (maybe the redefinition has to be a combination of diferent components, I don't recall now).

But while someone writes a better answer, I hope this helps a bit understanding the path to follow. Try to check it yourself, left to the reader as an exercise :)

---

Let's start from the original Dirac equation:

$$ (i(\partial_t\gamma^0+\partial_x\gamma^1)-m)\Psi=0$$

which using:

$$\gamma^0 =\begin{bmatrix}I_2 & 0 \\ 0 & -I_2 \end{bmatrix}= \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{bmatrix} \ \ \ \text{ and } \ \ \ \ \ \gamma^i = \begin{bmatrix}0 & \sigma^i \\ -\sigma^i & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & -1 & 0 & 0 \\ -1 & 0 & 0 & 0 \end{bmatrix} $$ and defining your fields with two fields of 2 components each as:

$$\Psi = \begin{bmatrix} \phi \\ \chi \end{bmatrix}$$

ends up like:

$$ (\pm i\partial_t-m) \ \phi/\chi= \pm i \sigma_1\partial_x \chi/\phi $$

which choosing these two equations out of the four, we see how it affects this two components, we see that we get these two equations:

$$ (i\partial_t-m) \ \phi_0=-\sigma_1\partial_x \chi_1 \\ (i\partial_t-m) \ \chi_1=-\sigma_1\partial_x \phi_0 $$

which if you can put together into:

$$ (i\partial_t-m)^2 \ \phi_0=-\sigma_1\partial_x ((i\partial_t-m)\phi_1))= \sigma_1^2 \partial_x^2 \phi_0 $$

(if you did the same but substituting them in the other way you would get the same with $\phi_1$), ending in:

$$ (i\partial_t-m)^2 \ \phi_{0}=\partial_x^2 \phi_{0}$$

where we see that the evolution of both components is completely indendent and exactly equal, meaning you can compact them into one components for a linear differential equation (where you can just add up the previously different components into one, and behaves correctly).

Because of this in the end you only care about two components in the Dirac equation with $\phi_0$ and $\chi_0$, and the only relevant equation is then:

$$ (\pm i\partial_t-m) \ \phi_0= \pm i \sigma_1\partial_x \chi_1 \equiv \pm i \sigma_1\partial_x \phi_0$$

which don't mix in the derivatives at least! (the mass term actually should mix them if you do it with more patience).

So the idea to take from this all is that if you take two components of the four Dirac components, you can naturally reduce them to two because you will actually only be left with two different behaving degrees of freedom. And if you do all the process correctly you should arrive to something like:

$$ i(\partial_t + \partial_x) \ \phi_0= m \chi_1$$ $$ i(\partial_t - \partial_x) \ \chi_1= m \phi_0$$

Where now you realize that if you take $\Psi$ to only have these two components, can be written in the form of the 1+1d Dirac equation you wrote.

---

Edit: Maybe all of this would have been way easier doing as Qmechanic said, and using the 4 Dirac gamma matrices as Kronecker tensor products of 2 Pauli sigma matrices, where you would arrive at the same. without needing to write so many components :)