5 Fermions and Photons

The $\gamma$ matrices are fully defined through (267), i.e. we may use any set of matrices that fulfil these requirements. This means that if we have a different set of matrices $(\gamma^{\prime})^{\mu}$ that also fulfil (267), they must be related to $\gamma^{\mu}$ through a constant invertible matrix $S$

This is called Pauli’s fundamental theorem. Its proof is not that important but it makes use of an important fact: we can write any product of $\gamma$ matrices using a basis of $16=4\times 4$ elements.

To pick these, it is customary to define a fifth $\gamma$ matrix

with the totally anti-symmetric tensor $\epsilon^{\mu\nu\rho\sigma}$, defined to be $\varepsilon^{0123}=1$.

Suggested Exercise

Show using the anti-commutation relations and the definition of $\gamma^{5}$

	$\displaystyle(\gamma^{5})^{\dagger}$	$\displaystyle=\gamma^{5}\,,$		(282a)
	$\displaystyle(\gamma^{5})^{2}$	$\displaystyle=1\,,$		(282b)
	$\displaystyle\{\gamma^{5},\gamma^{\mu}\}$	$\displaystyle=0\,.$		(282c)

For example, for the anti-commutator

$\displaystyle\{\gamma^{5},\gamma^{\lambda}\}=-\frac{i}{4!}\varepsilon_{\mu\nu\rho\sigma}\{\gamma^{\mu}\gamma^{\nu}\gamma^{\rho}\gamma^{\sigma},\gamma^{\lambda}\}\,,$

(283)

we write

$\displaystyle\gamma^{\mu}\gamma^{\nu}\gamma^{\rho}\gamma^{\sigma}\gamma^{\lambda}$

$\displaystyle=\gamma^{\mu}\gamma^{\nu}\gamma^{\rho}2\eta^{\lambda\sigma}-\gamma^{\mu}\gamma^{\nu}\gamma^{\sigma}2\eta^{\lambda\rho}+\gamma^{\mu}\gamma^{\rho}\gamma^{\sigma}2\eta^{\nu\lambda}-\gamma^{\nu}\gamma^{\rho}\gamma^{\sigma}2\eta^{\mu\lambda}+\gamma^{\lambda}\gamma^{\mu}\gamma^{\nu}\gamma^{\rho}\gamma^{\sigma}\,.$

(284)

Therefore

$\displaystyle\{\gamma^{5},\gamma^{\lambda}\}=-\frac{2i}{4!}\varepsilon_{\mu\nu\rho\sigma}\Big{(}\gamma^{\mu}\gamma^{\nu}\gamma^{\rho}\eta^{\lambda\sigma}-\gamma^{\mu}\gamma^{\nu}\gamma^{\sigma}\eta^{\lambda\rho}+\gamma^{\mu}\gamma^{\rho}\gamma^{\sigma}\eta^{\nu\lambda}-\gamma^{\nu}\gamma^{\rho}\gamma^{\sigma}\eta^{\mu\lambda}+\gamma^{\lambda}\gamma^{\mu}\gamma^{\nu}\gamma^{\rho}\gamma^{\sigma}\Big{)}$

(285)

We can write any product of $\gamma$ matrices using the following basis

The numbers indicate that the number of basis elements of this form. We refer to these as scalar, vector, tensor, pseudo-scalar and pseudo-vector respectively.

If we simultaneously transform the spinor $\psi$ as $\psi\to S^{-1}\psi$, the Dirac equation transforms to

This means physics will always be invariant under basis change.

Even if the exact form of the $\gamma$ matrices does not matter, it is sometimes helpful to have one

with the Pauli matrices $\sigma^{i}$

If we want to view the spinor $\psi$ as physically meaningful, we need to understand how it transforms in different frames. Consider therefore two frames described by $x$ and $x^{\prime}$ with $(x^{\prime})^{\mu}=\Lambda^{\mu}_{\phantom{\mu}\nu}x^{\nu}$. If $\psi(x)$ is a solution in the $x$ frame and $\psi^{\prime}(x^{\prime})$ is a solution in the $x^{\prime}$ frame, we must have

We can use (27) to transform the derivative

The new $\gamma$ matrices $(\gamma^{\prime})^{\mu}=\Lambda^{\mu}_{\phantom{\mu}\nu}\gamma^{\nu}$ still need to fulfil (267a)

since (23) guarantees $\eta_{\mu\nu}\Lambda^{\mu}_{\phantom{\mu}\rho}\Lambda^{\nu}_{\phantom{\nu}\sigma}=\eta_{\sigma\rho}$. Pauli’s fundamental theorem proves the existence of a spinor matrix $S(\Lambda)$ such that

Therefore,

If we left-multiply with $S(\Lambda)$ and identify $\psi^{\prime}(x)^{\prime}=S(\Lambda)\psi(\Lambda^{-1}x^{\prime})$, we arrive at

the transformed Dirac equation (290)

To study $S(\Lambda)$ consider an infinitesimal transformation $\Lambda$ which should also be infinitesimal in $S$, i.e. we have

Since $\omega_{\mu\nu}+\omega_{\nu\mu}=0$, we have the same anti-symmetry for $\Sigma$. Let us now calculate what happens to $\gamma$

This is satisfied by

which can be integrated to

Suggested Exercise

Show that

$\displaystyle[\gamma^{\mu},\gamma_{\lambda}\gamma_{\rho}]=2(\delta^{\mu}_{\phantom{\mu}\lambda}\gamma_{\rho}-\delta^{\mu}_{\phantom{\mu}\rho}\gamma_{\lambda})\,,$

(304)

by using the anticommutator. Use this to show that (302) is a solution of (301).

The above discussion is only valid for proper transformations that have $\det\Lambda=+1$. To also cover improper transformations, consider

This transformation flips the spatial components, i.e. it is a parity transformation. We need to fulfil

There are two possible choices for $S(P)$

correctly transforms $\gamma^{\mu}$. When acting on the wavefunction, we can have two solutions as well

The sign is called intrinsic parity and only starts to matter once we consider system with changing numbers of particles.

To eventually construct a field theory, we will need a basis of solutions to the wave equation (264). As always, we begin with a Fourier-transformation of $\psi(x)$

Here the $u$ and $v$ objects are vectors in spinor space. They fulfil the momentum-space Dirac equation

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

We can find explicit answers for the spinors using the explicit representation of (288)

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

Suggested Exercise

(313) relies on the fact that $(\gamma\cdot a)(\gamma\cdot a)=a^{2}$. Proof this.

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. This choice results in the

Further, we can show that (completeness relation)

To define the Lagrangian of the free Dirac field, it is helpful to first define the adjoint spinor

With this, the Lagrangian can be written as

since it results in the correct Euler-Lagrange equation for $\bar{\psi}$. To see this, we calculate

and write

Here we have used the notation $\overleftarrow{\partial_{\mu}}$ to indicate that the derivative is acting to the left rather than the right. We can Hermitian-conjugate this to arrive at (264).

The assosciated Hamiltonian is

Since $u$ and $v$ are vectors, we will extend our Fourier-decomposition of the fermion field slightly to split the creation and annihilation operators from the spinor vectors, i.e. we write

We have further flipped the propagation direction of the $v$ spinors to ensure positive energy. Recall how we used CCRs to quantise the KG field

Here we would have

However, this will lead to a contradiction. We can show that the Hamiltonian of this theory is

Since we can view $a_{r}^{\dagger}a_{r}$ and $b_{r}^{\dagger}b_{r}$ as particle numbers for particles of type $a$ and $b$, this would mean that creating more $b$-type particles decreases the energy of the system.

If we instead swap $b$ and $b^{\dagger}$

and chose anitcommutation relations

we would find

which means that both $a^{\dagger}$ and $b^{\dagger}$ create a particle of mass $m^{2}=p^{2}$.

Suggested Exercise

Show (325) and (329).

Suggested Exercise

You may find the following quantum mechanics problem instructive. We normally consider the bosonic harmonic oscillator defined as

$\displaystyle H_{B}=\frac{\omega}{2}(a^{\dagger}a+aa^{\dagger})\qquad\text{with}\qquad[a,a^{\dagger}]=1\,,\quad[a,a]=[a^{\dagger},a^{\dagger}]=0\,.$

(330)

Now define the fermionic oscillator with

$\displaystyle H_{F}=\frac{\omega}{2}(b^{\dagger}b-bb^{\dagger})\qquad\text{with}\qquad\{b,b^{\dagger}\}=1\,,\quad\{b,b\}=\{b^{\dagger},b^{\dagger}\}=0\,.$

(331)

Write $H_{F}$ and $H_{B}$ in terms of number operators $N_{F}=b^{\dagger}b$ and $N_{B}=a^{\dagger}a$. What are the allowed eigenvalues of $N_{F}$ and $N_{B}$?

You can also define a combined system $H=H_{B}+H_{F}$ with $|{n}\rangle=|{n_{B}}\rangle\otimes|{n_{F}}\rangle\equiv|{n_{B},n_{F}}\rangle$. This system treats bosons and fermions the same and is therefore supersymmetric. Show that the supercharge operator $Q=ab^{\dagger}$ fulfils

$\displaystyle\{Q,Q\}=\{Q^{\dagger},Q^{\dagger}\}=0\,,\quad\omega\{Q,Q^{\dagger}\}=H\,,\quad[Q,H]=[Q^{\dagger},H]=0\,.$

(332)

Therefore, $Q$ is a conserved quantity. Finally, apply $Q|{n_{B},n_{F}}\rangle$ and explain what the operator does to a fermion or boson.

We can now define a few states. As before, we define $|{0}\rangle$ as the state destroyed by $a$ and $b$

We can also define two different one-particles states

These states are properly normalised such that