2 Spin

Let us try and solve the free Dirac equation in terms of plane waves. Since we expect both positive and negative energy solutions, we can make an ansatz

This leads to two different momentum-space version of the Dirac equation

In the restframe of the particle where $p=(m,0,0,0)$, we have

We can find explicit answers for the spinors using the explicit representation of (16)

Note that we have not one but two solution for each direction of $p$. One can show that these correspond to the two spin directions. At this point the normalisation factor $\mathcal{N}$ is a free parameter. It turns out that $\mathcal{N}=\sqrt{2m}$ but it is possible to modify the following equations to accommodate another normalisation.

To turn these into solutions for $u(p)$, we could perform a Lorentz boost. Alternatively, we can note that

to write

This is a solution of the Dirac equation because we get zero if we left-multiply with $\gamma\cdot p-m$ since $p^{2}=m^{2}$. A suitable normalisation would be

since it leads to be the “boost operator” to be normalised, i.e. for $\vec{p}\to 0$, the fraction has components of one.

Each of the two Dirac spinors has two linearly independent solutions which we claimed are the two possible spin states of a fermion. To verify this, we will try and define the spin operator that measures the spin of a fermion. In the particle’s rest frame, we had

These have eigenvalues $\pm\frac{1}{2}$ under the matrix

Repeating the same for anti-particles, we have the spin operator

You can easily verify that $\vec{S}^{2}=3/4$ as expected. We can therefore conclude that, at least in the rest frame, the top two components of $\psi$ describe a spin up ($S_{z}=+1/2$) and spin down ($S_{z}=-1/2$) state, respectively.

To generalise this to general $\vec{p}$, we can project the spin-operator along the direction of motion. This is called the helicity operator

This satisfies $h(\vec{p})^{2}=1$ and therefore has eigenvalues $\pm 1$.

To generate an equation for anti-particles, we first take the Hermitian conjugate of the Dirac equation and find

Here we have used the fact that $\gamma^{0{\dagger}}=\gamma^{0}$ and $\gamma^{i{\dagger}}=-\gamma^{i}$. The arrow over the derivative means that it acts to the left rather than to the right. We can rearrange this slightly by multiplying from the right with $\gamma^{0}$ and defining $\bar{\psi}=\psi^{\dagger}\gamma^{0}$

At $*$, we have used that $\{\gamma^{0},\gamma^{i}\}=0$. This can be understood by interpreting $\bar{\psi}$ as the wave function of an anti-particle.

One can show by explicit calculation that the choice of normalisation in (29) is orthonormal, i.e. that

Further, we can show that (completeness relation)

We still need to define how $\psi$ changes under Lorentz transformation, similarly to what we did for $\phi$ in (9) We define

with a suitable matrix $S(\Lambda)$. Similarly for $\bar{\psi}$

$S$ is determined by requiring that the Dirac equation is Lorentz invariant. It is easy to see that if $\{\gamma^{\mu},\gamma^{\nu}\}=g^{\mu\nu}$ then also the matrices $(\gamma^{\prime})^{\mu}=\Lambda^{\mu}_{\phantom{\mu}\nu}\gamma^{\nu}$ will. This means that they, too, are a representation of the Clifford algebra. One can now show that

This is called Pauli’s fundamental theorem. Therefore,

(40) is enough to fully specify the $S(\Lambda)$ matrix and we will not need an explicit form. It follows as well that

and therefore the transformation of $\bar{\psi}$ is also fine. The fact that $S^{-1}(\Lambda)\neq S^{\dagger}(\Lambda)$ is not surprising, and is due to the fact that the Lorentz group is non-compact, and therefore it does not admit unitary finite-dimensional representations.

It is very common to construct bi-linear products $\bar{\psi}\Gamma\psi$ with some $4\times 4$ matrix $\Gamma$. Since $\Gamma$ has $16=4^{2}$ degrees of freedom, we need 16 linearly independent elements to form a basis out of which we can construct any $\Gamma$. It turns out that it is possible to classify these 16 matrices very neatly by their behaviour under Lorentz transformations. We have

one scalar

The case of $\Gamma=1$ transforms as

$\displaystyle\bar{\psi}\psi\to\bar{\psi}S^{-1}(\Lambda)S(\Lambda)\psi=\bar{\psi}\psi\,.$

(45)

four vectors

The cases of $\Gamma=\gamma^{\mu}$ correspond to four different basis entries that all transform the same way

$\displaystyle\bar{\psi}\gamma^{\mu}\psi\to\bar{\psi}S^{-1}(\Lambda)\gamma^{\mu}S(\Lambda)\psi=\Lambda^{\mu}_{\phantom{\mu}\nu}(\bar{\psi}\gamma^{\nu}\psi)\,.$

(46)

six tensors

By combining two $\gamma$ matrices, we can construct more elements. Rather then setting $\Gamma=\gamma^{\mu}\gamma^{\nu}$ which is not linearly independent due to (18), we instead choose

$\displaystyle\Sigma^{\mu\nu}=\frac{i}{4}[\gamma^{\mu},\gamma^{\nu}]$

(47)

as our six basis elements. These transform as follows

$\displaystyle\bar{\psi}\Sigma^{\mu\nu}\psi\to\bar{\psi}S^{-1}(\Lambda)\frac{i}{4}[\gamma^{\mu},\gamma^{\nu}]S(\Lambda)\psi=\Lambda^{\mu}_{\phantom{\mu}\rho}\Lambda^{\nu}_{\phantom{\nu}\sigma}(\bar{\psi}\Sigma^{\rho\sigma}\psi)\,.$

(48)

To find the remaining five elements, we need to define one more matrix, in addition to our four normal $\gamma^{\mu}$. Traditionally called $\gamma^{5}$,

The last equality here is again assuming the Dirac representation. Independent of the explicit representation, one can show that

Using it, we can define

one pseudo-scalar

The case of $\Gamma=\gamma^{5}$ transforms as

	$\displaystyle\bar{\psi}\gamma^{5}\psi$	$\displaystyle\to\bar{\psi}S^{-1}(\Lambda)\frac{i}{4!}\varepsilon_{\mu\nu\rho\sigma}\gamma^{\mu}\gamma^{\nu}\gamma^{\rho}\gamma^{\sigma}S(\Lambda)\psi$		(51)
		$\displaystyle=\bar{\psi}\frac{i}{4!}\varepsilon_{\mu\nu\rho\sigma}\Lambda^{\mu}_{\phantom{\mu}\alpha}\Lambda^{\nu}_{\phantom{\nu}\beta}\Lambda^{\rho}_{\phantom{\rho}\gamma}\Lambda^{\sigma}_{\phantom{\sigma}\delta}\gamma^{\alpha}\gamma^{\beta}\gamma^{\gamma}\gamma^{\delta}\psi$		(52)
		$\displaystyle\stackrel{{\scriptstyle*}}{{=}}(\det\Lambda)\ \bar{\psi}i\Lambda^{0}_{\phantom{0}\alpha}\Lambda^{1}_{\phantom{1}\beta}\Lambda^{2}_{\phantom{2}\gamma}\Lambda^{3}_{\phantom{3}\delta}\gamma^{\alpha}\gamma^{\beta}\gamma^{\gamma}\gamma^{\delta}\psi=\det\Lambda\ \bar{\psi}\gamma^{5}\psi\,,$		(53)

where we have used at $*$ the fact that the determinant of a matrix can be written using the $\varepsilon$ tensor.

four pseudo-vectors

The cases of $\Gamma=\gamma^{5}\gamma^{\mu}$ correspond to four different basis entries that all transform the same way

$\displaystyle\bar{\psi}\gamma^{5}\gamma^{\mu}\psi\to(\det\Lambda)\Lambda^{\mu}_{\phantom{\mu}\nu}(\bar{\psi}\gamma^{5}\gamma^{\nu}\psi)\,.$

(54)

The presence of $\det\Lambda$ in these last two classes gives these objects the pseudo prefix that means they swap sign for improper transformations.

The most common use of $\gamma^{5}$ is in the projectors $P_{L}=(1-\gamma^{5})/2$ and $P_{R}=(1+\gamma^{5})/2$. You can check explicitly that these behave like projectors (ie. $P^{2}=P$ and $P_{L}P_{R}=0$). When these act upon a Dirac spinor they project out either the component with “left-handed” chirality or with “right-handed” chirality. These projectors therefore appear when considering weak interactions, for example, as $W$ bosons only couple to left-handed particles. One has to take care when defining the handedness of antiparticles because

A left-handed anti-particle appears with a right-handed projection operator next to it and vice-versa.