Relativistic Covariance

It is important to show that the Dirac equation, with its constant $\gamma$ matrices, can be covariant. This will come down to finding the right transformation of the Dirac spinor $\psi$. Remember that spinors transform under rotations in a way quite different from normal vectors. The four components if the Dirac spinor do not represent x, y, z, and t. We have already solved a similar problem. We derived the rotation matrices for spin ${1\over 2}$ states, finding that they are quite different than rotation matrices for vectors. For a rotation about the $j$ axis, the result was.

$$\bgroup\color{black}R(\theta)= \cos{\theta\over 2}+i\sigma_j\sin{\theta\over 2} \egroup$$

We can think of rotations and boosts as the two basic symmetry transformations that we can make in 4 dimensions. We wish to find the transformation matrices for the equations.

$$\begin{eqnarray*} \psi'&=&S_{rot}(\theta)\psi \\ \psi'&=&S_{boost}(\beta)\psi \\ \end{eqnarray*}$$

We will work with the Dirac equation and its transformation. We know how the Lorentz vectors transform so we can derive a requirement on the spinor transformation. (Remember that $a_{\mu\nu}$ works in an entirely different space than do $\gamma_\mu$ and $S$.)

$$\begin{eqnarray*} \gamma_\mu{\partial\over\partial x_\mu}\psi(x)+{mc\over\hbar}\... ...ght){\partial\over\partial x_\nu}\psi+{mc\over\hbar}\psi&=&0 \\ \end{eqnarray*}$$

The transformed equation will be the same as the Dirac equation if $S^{-1}\gamma_\mu S a_{\mu\nu}=\gamma_\nu$. Multiply by the inverse Lorentz transformation.

$$\begin{eqnarray*} a_{\mu\nu}(a)^{-1}_{\nu\lambda}&=&\delta_{\mu\lambda} \\ a_{\... ...a\nu} \\ S^{-1}\gamma_\lambda S&=&\gamma_\nu a_{\lambda\nu} \\ \end{eqnarray*}$$

$$S^{-1}\gamma_\mu S=\gamma_\nu a_{\mu\nu}$$

This is the requirement on $S$ for covariance of the Dirac equation.

Rotations and boosts are symmetry transformations of the coordinates in 4 dimensions. Consider the cases of rotations about the z axis and boosts along the x direction, as examples.

$$\begin{eqnarray*} a^{(rot)}_{\mu\nu}&=&\pmatrix{\cos\theta & \sin\theta & 0 & 0 ... ... & 0\cr 0 & 0 & 1 & 0\cr -i\sinh\chi & 0 & 0 & \cosh\chi\cr} \\ \end{eqnarray*}$$

The boost is just another rotation in Minkowski space through and angle $i\chi=i\tanh^{-1}\beta$. For example a boost with velocity $\beta$ in the x direction is like a rotation in the 1-4 plane by an angle $i\chi$. Let us review the Lorentz transformation for boosts in terms of hyperbolic functions. We define $\tanh\chi=\beta$.

$$\begin{eqnarray*} \tanh\chi&=&{e^\chi-e^{-\chi}\over e^\chi+e^{-\chi}}=\beta ... ... 2}=\cosh\chi \\ \beta\gamma&=&\tanh\chi\cosh\chi=\sinh\chi \\ \end{eqnarray*}$$

$$\begin{eqnarray*} a^{(boost)}_{\mu\nu} &=&\pmatrix{\cosh\chi & 0 & 0 & i\sinh\ch... ...a x+\beta\gamma i(ict) \\ x_4'&=&x_4\cosh\chi-ix_1\sinh\chi \\ \end{eqnarray*}$$

We verify that a boost along the $i$ direction is like a rotation in the $i4$ plane through an angle $i\chi$.

We need to find the transformation matrices $S$ that satisfy the equation $S^{-1}\gamma_\mu S=\gamma_\nu a_{\mu\nu}$ for the Dirac equation to be covariant. Recalling that the 4 component equivalent of $\sigma_z$ is $\Sigma_z={[\gamma_1,\gamma_2]\over 2i}= {\gamma_1\gamma_2\over i}$, we will show that these matrices are (for a rotation in the xy plane and a boost in the x direction).

$$\begin{eqnarray*} S_{rot}&=&\cos{\theta\over 2}+\gamma_1\gamma_2\sin{\theta\over... ...st}&=&\cosh{\chi\over 2}+i\gamma_1\gamma_4\sinh{\chi\over 2} \\ \end{eqnarray*}$$

Note that this is essentially the transformation that we derived for rotations of spin one-half states extended to 4 components. For the case of the boost the angle is now $i\chi$.

Lets verify that this choice works for a boost.

$$\begin{eqnarray*} \left(\cosh{\chi\over 2}+i\gamma_1\gamma_4\sinh{\chi\over 2}\r... ...\gamma_1\gamma_4\sinh^2{\chi\over 2} &=&a_{\mu\nu}\gamma_\nu \\ \end{eqnarray*}$$

The equation we must satisfy can be checked for each $\gamma$ matrix. First check $\gamma_1$. The operations with the $\gamma$ matrices all come from the anticommutator, $\{\gamma_\mu,\gamma_\nu\}=2\delta_{\mu\nu}$, which tells us that the square of any gamma matrix is one and that commuting a pair of (unequal) matrices changes the sign.

$$\begin{eqnarray*} \gamma_1\cosh^2{\chi\over 2}+i\gamma_1\gamma_1\gamma_4\cosh{\c... ...i\over 2}+\gamma_1\sinh^2{\chi\over 2} &=&a_{1\nu}\gamma_\nu \\ \end{eqnarray*}$$

$$\begin{eqnarray*} \cosh^2{\chi\over 2}+\sinh^2{\chi\over 2}={1\over 4}((e^{\chi\... ...\gamma_4\sinh{\chi}=\gamma_1\cosh\chi+i\gamma_4\sinh\chi& & \\ \end{eqnarray*}$$

That checks for $\gamma_1$. Now, try $\gamma_4$.

$$\begin{eqnarray*} \gamma_4\cosh^2{\chi\over 2}+i\gamma_4\gamma_1\gamma_4\cosh{\c... ...i\gamma_1\sinh{\chi}&=&-i\sinh\chi\gamma_1+\cosh\chi\gamma_4 \\ \end{eqnarray*}$$

That one also checks. As a last test, try $\gamma_2$.

$$\begin{eqnarray*} \gamma_2\cosh^2{\chi\over 2}+i\gamma_2\gamma_1\gamma_4\cosh{\c... ...2{\chi\over 2} &=&a_{2\nu}\gamma_\nu \\ \gamma_2&=&\gamma_2 \\ \end{eqnarray*}$$

The Dirac equation is therefore shown to be invariant under boosts along the $x_i$ direction if we transform the Dirac spinor according to $\psi'=S_{boost}\psi$ with the matrix

$$S_{boost}=\cosh{\chi\over 2}+i\gamma_i\gamma_4\sinh{\chi\over 2}$$

and $\tanh\chi=\beta$.

The pure rotation about the z axis should also be verified.

$$\begin{eqnarray*} \left(\cos{\theta\over 2}+\gamma_1\gamma_2\sin{\theta\over 2}\... ...gamma_1\gamma_2\sin^2{\theta\over 2} &=&a_{\mu\nu}\gamma_\nu \\ \end{eqnarray*}$$

For $\mu=3$ or 4, $a_{\mu\nu}=\delta_{\mu\nu}$ and the requirement is fairly obviously satisfied. Checking the requirement for $\mu=1$, we get.

$$\begin{eqnarray*} \gamma_1\cos^2{\theta\over 2} +\gamma_1\gamma_1\gamma_2\cos{\t... ...gamma_2\sin{\theta} &=&\cos\theta\gamma_1+\sin\theta\gamma_2 \\ \end{eqnarray*}$$

This also proves to be the right transformation of $\psi$ so that the Dirac equation is invariant under rotations about the $k$ axis if we transform the Dirac spinor according to $\psi'=S_{rot}\psi$ with the matrix

$$S_{rot}=\cos{\theta\over 2}+\gamma_i\gamma_j\sin{\theta\over 2}$$

and $ijk$ is a cyclic permutation.

Despite the fact that we are using a vector of constant matrices, $\gamma_\mu$, the Dirac equation is covariant if we choose the right transformation of the spinors. This allows us to move from one coordinate system to another.

As an example, we might try our solution for a free electron with spin up along the z axis at rest.

$$\bgroup\color{black} \psi^{(1)}=\psi_{E=+mc^2,+\hbar/2}={1\ov... ...}}\pmatrix{1\cr 0\cr 0\cr 0\cr}e^{ip_\rho x_\rho/\hbar} \egroup$$

The solution we found for a free particle with momentum $\vec{p}$ was.

$$\bgroup\color{black} \psi^{(1)}_{\vec{p}}=\sqrt{E+mc^2\over 2... ...r {(p_x+ip_y)c\over E+mc^2}\cr}e^{ip_\rho x_\rho/\hbar} \egroup$$

Imagine we boost the coordinate system along the $-x$ direction with ${v\over c}=\beta$. We can transform the momentum of the electron to the new frame.

$$\bgroup\color{black} p_\nu'=a^{(boost)}_{\mu\nu}p_\nu =\pmatr... ...cr}=\pmatrix{\beta\gamma mc\cr 0\cr 0\cr i\gamma mc\cr} \egroup$$

The momentum along the x direction is $p_x=\beta\gamma mc=mc\sinh\chi$. We now have two ways to get the free particle state with momentum in the x direction. We can use our free particle state

$$\begin{eqnarray*} \psi^{(1)} &=&\sqrt{E+mc^2\over 2EV'}\pmatrix{1\cr 0\cr 0\cr {... ...i \over \cosh\chi +1}\cr}e^{i(\vec{p}\cdot\vec{x}-Et)/\hbar} \\ \end{eqnarray*}$$

$$\begin{eqnarray*} &=&\sqrt{E+mc^2\over 2EV'}\pmatrix{1\cr 0\cr 0\cr {2\sinh{\chi... ...cr \tanh{\chi\over 2}\cr}e^{i(\vec{p}\cdot\vec{x}-Et)/\hbar} \\ \end{eqnarray*}$$

where the normalization factor is now set to be ${1\over \sqrt{V'}}$, defining this as the primed system.

We can also find the same state by boosting the at rest solution. Recall that we are boosting in the x direction with $-\beta$, implying $\chi\rightarrow -\chi$.

$$\begin{eqnarray*} S_{boost}&=&\cosh{\chi\over 2}-i\gamma_1\gamma_4\sinh{\chi\ove... ...} & 0\cr \sinh{\chi\over 2} & 0 & 0 & \cosh{\chi\over 2}\cr} \\ \end{eqnarray*}$$

$$\begin{eqnarray*} \psi^{(1)}_{\vec{p}}&=&\pmatrix{\cosh{\chi\over 2} & 0 & 0 & \... ...\cr 0\cr 0\cr \tanh{\chi\over 2}\cr}e^{ip_\rho x_\rho/\hbar} \\ \end{eqnarray*}$$

$$\begin{eqnarray*} {1+\cosh\chi\over 2}&=&{1+{e^\chi+e^{-\chi}\over 2}\over 2}={e... ...\cr 0\cr 0\cr \tanh{\chi\over 2}\cr}e^{ip_\rho x_\rho/\hbar} \\ \end{eqnarray*}$$

In the last step the simple Lorentz contraction was used to set $V'={V\over\gamma}$. This boosted state matches the plane wave solution including the normalization.