Domenico Fiorenza Dirac equation

The Dirac equation

Derivation

In this section, we will produce a covariant equation for a particle with spin non zero. This was first achieved by Dirac in 1928 for the electron, a particle with mass $m= 0,5 MeV$ and spin $1/2$. He moved his construction from two facts:

1. An equation for a wave function of a particle must be of order one in time coordinates, according to the fundamental principles of quantum mechanics
2. The square of the operator must be the K-G operator.

Forcing these two conditions, Dirac obtained an equation for a vectorial field $\Psi$ :

where $\slash{\partial} = \partial^{\mu} \gamma_{\mu}$. The matrices $\gamma_0,\gamma_1,\gamma_2,\gamma_3$ satisfy:

This is called the Feynman form for the Dirac equation. It results by an easy calculation that the lowest allowed dimension for the vectorial field $\Psi$ is four. This is the most surprising fact: we expected to find that a $1/2$-spin particle, such the electron is, could be represented by a two dimensional field, instead the dimension two is not even allowed. This was explained by introducting for each particle, the respective antiparticle.

Dictionary.

Antiparticle. We introduce three fundamental transformation over the Hilbert space $\mathcal{H}$ of the $L^2$ functions over the Minkowski space:

These three transformations are, respectevely, the charge coniugation, the parity and the time reversal. The antiparticle for the particle $a+$, represented by a wave function $\Psi$, is a particle $a-$ with wave function $CPT \Psi$. In concrete the antiparticle has the same mass, the same spin, but the opposite charge respect to the associated particle.

Once the antiparticles come into play, we can state that a solution for the Dirac equation is a field, describing the evolution for both the particle and the antiparticle, i.e. the electron and the positron.

The iter followed by Dirac, while it was the only suitable way to write such an equation in an early stage of the quatum field theory, it is quite brutal. Moreover we should verify a posteriori the covariance of the equation.

What we want to do now is to derive the same expression through the representation theory. We know that a particle, say $a$, with spin $s$ and mass $m \gt 0$ is an irreducible representation inducted by an irrep of the little group $SU(2)$ of dimension $2s+1$. Then a couple particle-antiparticle representation must be found in an irrep for $SU(2)$ of dimension $(2s+1)(2s+1)$. To fix the ideas, we study the case $s=1/2$. It’s crucial that we don’t want two indipendent particles. The physics suggests the following correlation in a rest frame:

where $\Psi_{a+}$ is a 2 component field for the particle, and $\Psi_{a-}$ is the analogous for the antiparticle. Giving such a condition “in a rest frame”, it’s the same than solving the equation over a single point $p=(m,0,0,0)$ over the mass hyperboloid. The same condition could also be written in a more fitting way:

where $I$ is the identity matrix $2\times2$

Now to produce the Dirac equation it is sufficient to act with a transformation of the rest frame over $\Psi$. Which is this transformation? Recall that by calculation (eq:act2) the action of the Poincaré group, is essentially the action of the little group: any other element contributes to move the mass eigenvalue over the hyperboloid $p^2=m^2$. We have to compute the action of $SU(2)$ over a four dimensional wave function. Remark that the infinitesimal generators for $SU(2)$ are the Pauli’s matrices $\sigma_i$, then the action of $SU(2)$ is written through the exponential $e^{ \phi_{i}\sigma^{i}}$, the sum is over the three Pauli’s matrices indeces. The action is the following

The dependancy from $\Lambda \in L$ is omissis.

Essentially we want to act over $\Psi_{a \pm}$ with $SU(2)$ simultaneously, and give an equation of $U \Psi$. Since $U^{-1}U \Psi = \Psi$ the following equations hold:

finally multiplying for $U$ we have:

We proceed to expand equation (3) and show that is exactly the Dirac equation. Remark that by construction the projection to the rest frame is equation (1).

First it results:

Recall the Hadamard lemma, valid for two square matrix $X,Y$:

For $X = U$ and $Y=\gamma_0$ it results:

and

Next we have to compute the exponential of a Pauli’s matrix. By utilizing the commutation rules we have:

this is achieved by utilizing the following relations between $\phi$ and $p$:

Finally $U$ results:

Call $U \Psi = \Psi(p)$ then we have obtained the explicit form of the equation (3):

the saturation of the indeces with the $\gamma$-matrices is also written as $\slash{p}$.

Identifying the representation

At this point we want to come back to the Dirac’s construction. Once the equation was given, we had no information on the representation the solutions should belong to. It is worth to come up with a strategy to identify the representation which satisfies a certain covariant equation. Again we proceed with the Dirac equation for $s=1/2$. The first step is finding a good space where the solutions can be written. The expression in Fourier coordinates for the Dirac operator is:

Remark that $(p^2 - m^2)Id = (\slash{p} - m)(\slash{p} + m)$. Then the equation

has non trivial solutions when $p^2 = m^2$. We can conclude that a state $\Psi$, solution for Dirac, solves K-G equation as well. Then a natural space where the solutions live is the set of the distributions $D_m$, taking values in $\mathbb{C}^4$, supported over the mass hyperboloid $p^2 = m^2$. This identifies the first index for an irrep of the Poincaré group: the mass $m$. Before proceeding furthermore, we introduce some more representation theory tool. What we want to do is understanding how factoring the K-G operator in Fourier can allow us to study representation associated to the factors. First we want to specify what we intend for “factoring” the K-G operator. Given two representation of a group $G$, $V$ and $W$, one can introduce the symmetric power of $V$, $Sym^n(V)$ and then define

Remark that such an homomorphism between representation is multilinear in the entries of $Sym^n(V)$, moreover the symmetric power has a concrete realization as the polynomials in the vectors $v \in V$. It is interesting to give the factorization notion for such an application.

Definition. Given a $\phi : Sym^n(V) \to W$ homomorphism between representation for a group $G$, we say that $\phi$ is factorizable if and only if exist $W_1,W_2,$ representations and two maps $p$ and $q$

such that $p$ and $q$ are a complete system of projectors over $W$, $W$ is isomorphic to $W_1 \otimes W_2$, $h+k=n$ and the following diagram commutes:

Then $p \otimes q = \phi.$ is a factorization for $\phi$.

We are going to explain why equation (4) is a factorization in this new sense. Consider $W=\mathbb{C}^4$, $V=\mathbb{C}^2$ and the application:

Recall that a Symmetric power of a vector space can be indetified with polynomial in the vectors. Consider $p=(\slash{p}+m)$ and $q=(\slash{p}-m)$: by an easy calculation it results that they are a complete system of projectors over $\mathbb{C}^4$, each projecting over $\mathbb{C}^2$. Then they can be identified as applications between $V$ and $\mathbb{C}^2$. Hence we can define:

Next, let $\phi$ be $\phi = pq$ and $\pi$ be the natural projection from the tensor product to the symmetric power. We have the following diagram:

We want to verify that is commutative, i.e. $f (p \otimes q) = \phi \pi$. Remark that since $p$ and $q$ are a complete set of projectors and $\phi$ is a multiple of the identity, it results $Dim(Ran(p \otimes q) ) = Dim(Ran(\phi))$. Then it is possibile to give an isomorphism $f$ such that $f (p \otimes q) = \phi \pi$, by defining $f(v \otimes w) = \phi \pi (v \otimes w)$. Such an $f$ is bi-linear in the entries, since so it is $\pi$, and due to the fact that $Ran(f)= W$, it results $Ker(f)= \{0\}$, then $f$ is an isomorphism between vector spaces. We left behind the calculation of the dimension of the image of the projectors $(\slash{p} \pm m )$. For convenience it can be calculated in a rest frame $p=(m,0,0,0)$. We have:

The operator is a $4 \times 4$ matrix with two identical rows, then $Dim(Ker(\gamma_0 - Id)) = 2$. Next we want an explicit action of $SU(2)$ over the solutions: this is done by forcing the covariance condition. Let $\Psi$ be a Dirac solution in a frame $O$, the solution in a new frame $O'$ is $\Psi^{'} = U(\Lambda)\Psi$, where $\Lambda$ is the lorentz transformation that connects $O$ and $O'$. For $\Psi^{'}$ we write the following equation:

Where $p^{'}$ is the momentum in $O'$. Multiplying for $U(\Lambda)$ we have:

This is the Dirac equation if the following condition is satisfied:

or the equivalent:

The expression for $U$ given in (2) solves the relation above. This is easily checked:

todo: typical Lorentz transformation

todo irreducibility