# Physics:Dirac spinor

In quantum field theory, the Dirac spinor is the bispinor in the plane-wave solution

$\psi = \omega_\vec{p}\;e^{-ipx} \;$

of the free Dirac equation,

$\left(i\gamma^\mu \partial_\mu - m\right)\psi = 0 \;,$

where (in the units $c \,=\, \hbar \,=\, 1$)

$\psi$ is a relativistic spin-1/2 field,
$\omega_\vec{p}$ is the Dirac spinor related to a plane-wave with wave-vector $\vec{p}$,
$px \;\equiv\; p_\mu x^\mu \;\equiv\; E t - \vec{p} \cdot \vec{x}$,
$p^\mu \;=\; \left\{\pm\sqrt{m^2 + \vec{p}^2},\, \vec{p}\right\}$ is the four-wave-vector of the plane wave, where $\vec{p}$ is arbitrary,
$x^\mu$ are the four-coordinates in a given inertial frame of reference.

The Dirac spinor for the positive-frequency solution can be written as

$\omega_\vec{p} = \begin{bmatrix} \phi \\ \frac{\vec{\sigma} \cdot \vec{p}}{E_\vec{p} + m} \phi \end{bmatrix} \;,$

where

$\phi$ is an arbitrary two-spinor,
$\vec{\sigma}$ are the Pauli matrices,
$E_\vec{p}$ is the positive square root $E_\vec{p} \;=\; + \sqrt{m^2 + \vec{p}^2}$

## Derivation from Dirac equation

The Dirac equation has the form

$\left(-i \vec{\alpha} \cdot \vec{\nabla} + \beta m \right) \psi = i \frac{\partial \psi}{\partial t} \,$

In order to derive the form of the four-spinor $\scriptstyle\omega$ we have to first note the value of the matrices α and β:

$\vec\alpha = \begin{bmatrix} \mathbf{0} & \vec{\sigma} \\ \vec{\sigma} & \mathbf{0} \end{bmatrix} \quad \quad \beta = \begin{bmatrix} \mathbf{I} & \mathbf{0} \\ \mathbf{0} & -\mathbf{I} \end{bmatrix}$

These two 4×4 matrices are related to the Dirac gamma matrices. Note that 0 and I are 2×2 matrices here.

The next step is to look for solutions of the form

$\psi = \omega e^{-i p \cdot x}$,

while at the same time splitting ω into two two-spinors:

$\omega = \begin{bmatrix} \phi \\ \chi \end{bmatrix} \,$.

### Results

Using all of the above information to plug into the Dirac equation results in

$E \begin{bmatrix} \phi \\ \chi \end{bmatrix} = \begin{bmatrix} m \mathbf{I} & \vec{\sigma}\cdot\vec{p} \\ \vec{\sigma}\cdot\vec{p} & -m \mathbf{I} \end{bmatrix}\begin{bmatrix} \phi \\ \chi \end{bmatrix}$.

This matrix equation is really two coupled equations:

\begin{align} \left(E - m \right) \phi &= \left(\vec{\sigma} \cdot \vec{p} \right) \chi \\ \left(E + m \right) \chi &= \left(\vec{\sigma} \cdot \vec{p} \right) \phi \end{align}

Solve the 2nd equation for $\chi \,$ and one obtains

$\omega = \begin{bmatrix} \phi \\ \frac{\vec{\sigma} \cdot \vec{p}}{E + m} \phi \end{bmatrix} \,$.

Note that this solution needs to have $E = +\sqrt{p^2 + m^2}$ in order for the solution to be valid in a frame where the particle has $\vec p = \vec 0$.

Alternatively, solve the 1st equation for $\phi \,$ and one finds

$\omega = \begin{bmatrix} -\frac{\vec{\sigma} \cdot \vec{p}}{-E + m} \chi \\ \chi \end{bmatrix} \,$.

In this case one needs to enforce that $E = -\sqrt{p^2 + m^2}$ for this solution to be valid in a frame where the particle has $\vec p = \vec 0$. This can be shown analogously to the previous case.

This solution is useful for showing the relation between anti-particle and particle.

## Details

### Two-spinors

The most convenient definitions for the two-spinors are:

$\phi^1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \quad \quad \phi^2 = \begin{bmatrix} 0 \\ 1 \end{bmatrix} \,$

and

$\chi^1 = \begin{bmatrix} 0 \\ 1 \end{bmatrix} \quad \quad \chi^2 = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \,$

### Pauli matrices

The Pauli matrices are

$\sigma_1 = \begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix} \quad \quad \sigma_2 = \begin{bmatrix} 0 & -i \\ i & 0 \end{bmatrix} \quad \quad \sigma_3 = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}$

Using these, one can calculate:

$\vec{\sigma}\cdot\vec{p} = \sigma_1 p_1 + \sigma_2 p_2 + \sigma_3 p_3 = \begin{bmatrix} p_3 & p_1 - i p_2 \\ p_1 + i p_2 & - p_3 \end{bmatrix}$

## Four-spinors

### For particles

Particles are defined as having positive energy. The normalization for the four-spinor ω is chosen so that the total probability is invariant under Lorentz transformation. The total probability is:

$P = \int_{V} \omega^\dagger \omega dV$

where $V$ is the volume of integration. Under Lorentz transformation, the volume scales as the inverse of Lorentz factor: $(E/m)^{-1}$. This implies that the probability density must be normalized proportional to $E$ so the total probability is Lorentz invariant. The usual convention is to choose $\omega^\dagger \omega \;=\; 2 E \,$. Hence the spinors, denoted as u are:

$u\left(\vec{p}, s\right) = \sqrt{E + m} \begin{bmatrix} \phi^{(s)}\\ \frac{\vec{\sigma} \cdot \vec{p}}{E + m} \phi^{(s)} \end{bmatrix} \,$

where s = 1 or 2 (spin "up" or "down")

Explicitly,

$u\left(\vec{p}, 1\right) = \sqrt{E+m} \begin{bmatrix} 1 \\ 0 \\ \frac{p_3}{E + m} \\ \frac{p_1 + i p_2}{E + m} \end{bmatrix} \quad \mathrm{and} \quad u\left(\vec{p}, 2\right) = \sqrt{E + m} \begin{bmatrix} 0 \\ 1 \\ \frac{p_1 - i p_2}{E + m} \\ \frac{-p_3}{E + m} \end{bmatrix}$

### For anti-particles

Anti-particles having positive energy $E$ are defined as particles having negative energy and propagating backward in time. Hence changing the sign of $E$ and $\vec{p}$ in the four-spinor for particles will give the four-spinor for anti-particles:

$v(\vec{p},s) = \sqrt{E+m} \begin{bmatrix} \frac{\vec{\sigma} \cdot \vec{p}}{E + m} \chi^{(s)} \\ \chi^{(s)} \end{bmatrix}$

Here we choose the $\scriptstyle\chi$ solutions. Explicitly,

$v\left(\vec{p}, 1\right) = \sqrt{E + m} \begin{bmatrix} \frac{p_1 - i p_2}{E + m} \\ \frac{-p_3}{E + m} \\ 0 \\ 1 \end{bmatrix} \quad \mathrm{and} \quad v(\vec{p}, 2) = \sqrt{E+m} \begin{bmatrix} \frac{p_3}{E+m} \\ \frac{p_1 + i p_2}{E+m} \\ 1 \\ 0 \end{bmatrix}$

Note that these solutions are readily obtained by substituting the ansatz $\psi = v e^{+ i p x}$ into the Dirac equation.

## Completeness relations

The completeness relations for the four-spinors u and v are

\begin{align} \sum_{s=1,2}{u^{(s)}_p \bar{u}^{(s)}_p} &= {p\!\!\!/} + m \\ \sum_{s=1,2}{v^{(s)}_p \bar{v}^{(s)}_p} &= {p\!\!\!/} - m \end{align}

where

${p\!\!\!/} = \gamma^\mu p_\mu \,$      (see Feynman slash notation)
$\bar{u} = u^\dagger \gamma^0 \,$

## Dirac spinors and the Dirac algebra

The Dirac matrices are a set of four 4×4 matrices that are used as spin and charge operators.

### Conventions

There are several choices of signature and representation that are in common use in the physics literature. The Dirac matrices are typically written as $\gamma^\mu$ where $\mu$ runs from 0 to 3. In this notation, 0 corresponds to time, and 1 through 3 correspond to x, y, and z.

The + − − − signature is sometimes called the west coast metric, while the − + + + is the east coast metric. At this time the + − − − signature is in more common use, and our example will use this signature. To switch from one example to the other, multiply all $\scriptstyle\gamma^\mu$ by $i$.

After choosing the signature, there are many ways of constructing a representation in the 4×4 matrices, and many are in common use. In order to make this example as general as possible we will not specify a representation until the final step. At that time we will substitute in the "chiral" or "Weyl" representation.

### Construction of Dirac spinor with a given spin direction and charge

First we choose a spin direction for our electron or positron. As with the example of the Pauli algebra discussed above, the spin direction is defined by a unit vector in 3 dimensions, (a, b, c). Following the convention of Peskin & Schroeder, the spin operator for spin in the (a, b, c) direction is defined as the dot product of (a, b, c) with the vector

\begin{align} (i\gamma^2\gamma^3,\;\;i\gamma^3\gamma^1,\;\;i\gamma^1\gamma^2) &= -(\gamma^1,\;\gamma^2,\;\gamma^3)i\gamma^1\gamma^2\gamma^3 \\ \sigma_{(a,b,c)} &= ia\gamma^2\gamma^3 + ib\gamma^3\gamma^1 + ic\gamma^1\gamma^2 \end{align}

Note that the above is a root of unity, that is, it squares to 1. Consequently, we can make a projection operator from it that projects out the sub-algebra of the Dirac algebra that has spin oriented in the (a, b, c) direction:

$P_{(a,b,c)} = \tfrac{1}{2}\left(1 + \sigma_{(a,b,c)}\right)$

Now we must choose a charge, +1 (positron) or −1 (electron). Following the conventions of Peskin & Schroeder, the operator for charge is $Q \,=\, -\gamma^0$, that is, electron states will take an eigenvalue of −1 with respect to this operator while positron states will take an eigenvalue of +1.

Note that $Q$ is also a square root of unity. Furthermore, $Q$ commutes with $\scriptstyle\sigma_{(a, b, c)}$. They form a complete set of commuting operators for the Dirac algebra. Continuing with our example, we look for a representation of an electron with spin in the (a, b, c) direction. Turning $Q$ into a projection operator for charge = −1, we have

$P_{-Q} = \frac{1}{2}\left(1 - Q\right) = \frac{1}{2}\left(1 + \gamma^0\right)$

The projection operator for the spinor we seek is therefore the product of the two projection operators we've found:

$P_{(a, b, c)}\;P_{-Q}$

The above projection operator, when applied to any spinor, will give that part of the spinor that corresponds to the electron state we seek. So we can apply it to a spinor with the value 1 in one of its components, and 0 in the others, which gives a column of the matrix. Continuing the example, we put (a, b, c) = (0, 0, 1) and have

$P_{(0, 0, 1)} = \frac{1}{2}\left(1 + i\gamma_1\gamma_2\right)$

and so our desired projection operator is

$P = \frac{1}{2}\left(1+ i\gamma^1\gamma^2\right) \cdot \frac{1}{2}\left(1 + \gamma^0\right) = \frac{1}{4}\left(1 + \gamma^0 + i\gamma^1\gamma^2 + i\gamma^0\gamma^1\gamma^2\right)$

The 4×4 gamma matrices used in the Weyl representation are

\begin{align} \gamma_0 &= \begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix} \\ \gamma_k &= \begin{bmatrix}0 & \sigma^k \\ -\sigma^k & 0\end{bmatrix} \end{align}

for k = 1, 2, 3 and where $\sigma^i$ are the usual 2×2 Pauli matrices. Substituting these in for P gives

$P = \frac{1}{4}\begin{bmatrix}1 + \sigma^3 & 1 + \sigma^3 \\ 1 + \sigma^3 & 1 + \sigma^3 \end{bmatrix} = \frac{1}{2}\begin{bmatrix}1 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0\end{bmatrix}$

Our answer is any non-zero column of the above matrix. The division by two is just a normalization. The first and third columns give the same result:

$\left|e^-,\, +\frac{1}{2}\right\rangle = \begin{bmatrix} 1 \\ 0 \\ 1 \\ 0 \end{bmatrix}$

More generally, for electrons and positrons with spin oriented in the (a, b, c) direction, the projection operator is

$\frac{1}{4}\begin{bmatrix} 1 + c & a - ib & \pm(1 + c) & \pm(a - ib) \\ a + ib & 1 - c & \pm(a + ib) & \pm(1 - c) \\ \pm(1 + c) & \pm(a - ib) & 1 + c & a - ib \\ \pm(a + ib) & \pm(1 - c) & a + ib & 1 - c \end{bmatrix}$

where the upper signs are for the electron and the lower signs are for the positron. The corresponding spinor can be taken as any non zero column. Since $a^2 + b^2 + c^2 \,=\, 1$ the different columns are multiples of the same spinor. The representation of the resulting spinor in the Dirac basis can be obtained using the rule given in the bispinor article.