Proof of Covariance

To prove Lorentz covariance two conditions must be satisfied:

1. If $(i\hbar\gamma^\mu\partial_\mu - mc)\psi(x) = 0$ then $(i\hbar\tilde{\gamma}^\mu\partial_\mu^\prime - mc)\psi^\prime(x^\prime) = 0$.
2. Given $\psi(x)$ of observer $O$, there must be a prescription for observer $O^\prime$ to compute $\psi^\prime(x^\prime)$, which describes to $O^\prime$ the same physical state.

It can be shown that all $4\times 4$ matrices (with $\gamma^0$ hermitian and $\gamma^i$ anti-hermitian) are equivalent up to a unitary transformation:

$$\tilde{\gamma}_\mu = U^\dagger \gamma_\mu U .$$ (5.113)

where $U^\dagger=U^{-1}$. We drop the distinction between $\tilde{\gamma}^\mu$ and $\gamma^\mu$ and write

$$(\not{\!\hat{p}}^\prime -mc)\psi^\prime(x^\prime) = 0 ,$$ (5.114)

where $\not{\!\hat{p}}^\prime \equiv i\hbar\gamma^\mu\partial_\mu^\prime$.

We require that the transformation between $\psi$ and $\psi^\prime$ be linear since the Dirac equation and Lorentz transformation are linear.

$$\psi^\prime(x^\prime) = \psi^\prime(ax) = S(a) \psi(x) = S(a) \psi(a^{-1}x^\prime) .$$ (5.115)

$S(a)$ is a $4\times 4$ matrix which depends only on the relative velocities of $O$ and $O^\prime$. $S(a)$ has an inverse if $O\rightarrow O^\prime$ and also $O^\prime\rightarrow O$. The inverse is

$$\psi(x) = S^{-1}(a)\psi^\prime(x^\prime) = S^{-1}(a)\psi^\prime(ax)$$ (5.116)

or we could write

$$\psi(x) = S(a^{-1})\psi^\prime(ax)$$ (5.117)

$$\Rightarrow S(a^{-1}) = S^{-1}(a) .$$ (5.118)

We can now write

$$(i\hbar\gamma^\mu\partial_\mu - mc) \psi(x) = 0$$

$$(i\hbar\gamma^\mu\partial_\mu - mc) S^{-1}(a)\psi^\prime(x^\prime) = 0$$

$$S(a)(i\hbar\gamma^\mu\partial_\mu - mc) S^{-1}(a)\psi^\prime(x^\prime) = 0$$

$$(i\hbar S(a)\gamma^\mu S^{-1}(a)\partial_\mu - mc) \psi^\prime(x^\prime) = 0$$ (5.119)

Using $\frac{\partial}{\partial x^\mu} = \frac{\partial {x^\prime}^\nu}{\partial x^\mu... ...artial {x^\prime}^\nu} = a^\nu_{\ \mu} \frac{\partial}{\partial {x^\prime}^\nu}$ we have

$$\left( i\hbar S(a)\gamma^\mu S^{-1}(a)a^\nu_{\ \mu} \frac{\p... ...ial {x^\prime}^\nu} - mc \right) \psi^\prime(x^\prime) = 0 .$$ (5.120)

Therefore we require

$$S(a) \gamma^\mu S^{-1}(a) a^\nu_{\ \mu} = \gamma^\nu ,$$ (5.121)

or

$$\fbox{$\displaystyle S^{-1}(a) \gamma^\nu S(a) = a^\nu_{\ \mu} \gamma^\mu $}\ .$$ (5.122)

This relationship defines $S(a)$ only up to an arbitrary factor. this factor is further restricted to a $\pm$ sign if we require that the $S(a)$ form a representation of the Lorentz group. We obtain thus the two-valued spinor representation, in agreement with our previous assumptions. A wave function transforming according to equation 5.115 and equation 5.116 by means of equation 5.122 is a four-component Lorentz spinor. Such a spinor is also frequently called a bi-spinor, since it consists of two 2-component spinors, known to us from the Pauli equation.

Consider an infinitesimal proper Lorentz transformation

$$a^\nu_{\ \mu} = g^\nu_{\ \mu} + \Delta\omega^\nu_{\ \mu},$$ (5.123)

where $\Delta\omega^{\nu\mu}$ is anti-symmetric for an invariant proper time interval. Each of the six independent non-vanishing $\Delta\omega^{\mu\nu}$ generates an infinitesimal Lorentz transformation.

$$\Delta\omega^{01} = \Delta\beta$$ (5.124)

for a transformation to a coordinate system moving with velocity $c\Delta\beta$ along the $x$-direction.

$$\Delta\omega^1_{\ 2} = - \Delta\omega^{12} = \Delta\phi$$ (5.125)

for a rotation through an angle $\Delta\phi$ about the $z$-direction.

We expand $S$ in powers of $\Delta\omega^{\mu\nu}$ to first order,

$$S = 1 -\frac{i}{4} \sigma_{\mu\nu} \Delta \omega^{\mu\nu} ,$$ (5.126)

$$S^{-1} = 1 +\frac{i}{4} \sigma_{\mu\nu} \Delta \omega^{\mu\nu} ,$$ (5.127)

with $\sigma_{\mu\nu} = -\sigma_{\nu\mu}$.

We now solve for $\sigma_{\mu\nu}$. Equation 5.122 becomes

$$(g^\nu_{\ \mu} + \Delta\omega^\nu_{\ \mu})\gamma^\mu = \left( 1 + \frac{i}{4}\sigma_{\alpha\beta}\Delta\omega^{\alpha\be... ... \left( 1 - \frac{i}{4}\sigma_{\alpha\beta}\Delta\omega^{\alpha\beta} \right) ,$$

$$\Delta\omega^\nu_{\ \mu}\gamma^\mu = \frac{i}{4}\sigma_{\alpha\beta}\Delta\omega^{\alpha\beta}\gamma^\nu -\gamma^\nu\frac{i}{4}\sigma_{\alpha\beta}\Delta\omega^{\alpha\beta} ,$$

$$= -\frac{i}{4} \Delta\omega^{\alpha\beta} (\gamma^\nu\sigma_{\alpha\beta} - \sigma_{\alpha\beta}\gamma^\nu) ,$$

$$= -\frac{i}{4} \Delta\omega^{\alpha\beta} [\gamma^\nu ,\sigma_{\alpha\beta}].$$ (5.128)

Also

$$\Delta\omega^\nu_{\ \mu} \gamma^\mu = 1/2 (\Delta\omega^\nu_{\ \mu} \gamma^\mu + \Delta\omega^\nu_{\ \mu} \gamma^\mu)$$

$$= 1/2 (\Delta\omega^\alpha_{\ \mu} g^\nu_{\ \alpha}\gamma^\mu + \Delta\omega^\beta_{\ \mu} g^\nu_{\ \beta}\gamma^\mu)$$

$$= 1/2 (\Delta\omega^\alpha_{\ \beta} g^\nu_{\ \alpha}\gamma^\beta + \Delta\omega^\beta_{\ \alpha} g^\nu_{\ \beta}\gamma^\alpha)$$

$$= 1/2 (\Delta\omega^{\alpha\beta} g^\nu_{\ \alpha}\gamma_\beta + \Delta\omega^{\beta\alpha} g^\nu_{\ \beta}\gamma_\alpha)$$

$$= 1/2 \Delta\omega^{\alpha\beta} (g^\nu_{\ \alpha}\gamma_\beta - g^\nu_{\ \beta}\gamma_\alpha)$$

$$= -i/4 \Delta\omega^{\alpha\beta} 2i (g^\nu_{\ \alpha}\gamma_\beta - g^\nu_{\ \beta}\gamma_\alpha).$$ (5.129)

Combining equation 5.128 and equation 5.129 gives,

$$2i[g^\nu_{\ \alpha}\gamma_\beta - g^\nu_{\ \beta }\gamma_\alpha] = [\gamma^\nu,\sigma_{\alpha\beta}].$$ (5.130)

We must find six matrices $\sigma_{\alpha\beta}$ which satisfy the above equation. We try the anti-symmetric product of two matrices

$$\sigma_{\mu\nu} = \frac{i}{2} [\gamma_\mu,\gamma_\nu] .$$ (5.131)

Substituting (5.131) into the right-hand side of (5.130) gives

$$[\gamma^\nu,\sigma_{\alpha\beta}] = \frac{i}{2} [\gamma^\nu, [\gamma_\alpha, \gamma_\beta]]$$

$$= \frac{i}{2} ([\gamma^\nu, \gamma_\alpha \gamma_\beta] - [\gamma^\nu, \gamma_\beta \gamma_\alpha])$$

$$= \frac{i}{2} ([\gamma^\nu, \gamma_\alpha\gamma_\beta] + [\gamma^\nu, \gamma_\alpha \gamma_\beta] -2 [\gamma^\nu,g_{\beta\alpha}])$$

$$= i[\gamma^\nu, \gamma_\alpha \gamma_\beta]$$

$$= i( \gamma^\nu \gamma_\alpha \gamma_\beta - \gamma_\alpha \gamma_\beta \gamma^\nu)$$

$$= i (\gamma^\nu \gamma_\alpha \gamma_\beta + \gamma_\alpha \gamma^\nu \gamma_\beta -2 \gamma_\alpha g_\beta^{\ \nu}$$

$$= i ( \gamma^\nu \gamma_\alpha \gamma_\beta - \gamma^\nu \gamma_\al... ...\gamma_\beta + 2g_\alpha^{\ \nu} \gamma_\beta -2 \gamma_\alpha g_\beta^{\ \nu})$$

$$= 2i [g_\alpha^{\ \nu} \gamma_\beta - g_\beta^{\ \nu} \gamma_\alpha] ,$$ (5.132)

which is the left-hand side of (5.130). Therefore

$$\sigma_{\alpha\beta} = \frac{i}{2} [\gamma_\alpha,\gamma_\beta]$$ (5.133)

is a solutions to equation 5.130.

Thus

$$S = 1 - \frac{i}{4} \sigma_{\mu\nu} \Delta \omega^{\mu\nu} = 1 + \frac{1}{8} [\gamma_\mu,\gamma_\nu] \Delta \omega^{\mu\nu} .$$ (5.134)

We now construct finite proper transformations. We define

$$\Delta \omega^\nu_{\ \mu} = \Delta \omega (I_n)^\nu_{\ \mu}$$ (5.135)

where $\Delta\omega$ is an infinitesimal parameter of the Lorentz group. $I_n$ is a $4\times 4$ matrix for a general unit space-time rotation around an axis in the direction labelled by $n$. For proper Lorentz transformations $I$ has the property $I_n^4=I_n$. $\nu$ labels the row and $\mu$ labels the column.

We can write the finite transformation using $\Delta\omega\rightarrow \omega/N$ as

$${x^\prime}^\nu = \lim_{N\rightarrow\infty} \left( g+\frac{\omega}{N}I \right)^\nu_... ...dots \left( g+\frac{\omega}{N}I\right)^{\alpha_{N-1}}_{\ \alpha_N} x^{\alpha_N}$$

$$= (e^{\omega I})^\nu_{\ \mu} x^\mu .$$ (5.136)

For Lorentz translations, $I_n^3=I_n$ and we can write

$${x^\prime}^\nu = (\cosh\omega I + \sinh\omega I)^\nu_{\ \mu} x^\mu$$

$$= \left[\left( 1 + \frac{(\omega I)^2}{2!} + \frac{(\omega I)^4}{4!... ...{\omega I}{1!} + \frac{(\omega I)^3}{3!} + ... \right)\right]^\nu_{\ \mu} x^\mu$$

$$= \left[\left( 1 + I^2\frac{\omega^2}{2!} + I^2\frac{\omega^4}{4!} ... ... \frac{\omega}{1!} + \frac{\omega^3}{3!} + ... \right)\right]^\nu_{\ \mu} x^\mu$$

$$= (1 - I^2 + I^2\cosh\omega + I\sinh\omega)^\nu_{\ \mu} x^\mu .$$ (5.137)

Similarly for space-rotations, $I_n^3=-I_n$ and we can write

$${x^\prime}^\nu = (\cosh\omega I + \sinh\omega I)^\nu_{\ \mu} x^\mu$$

$$= \left[\left( 1 + \frac{(\omega I)^2}{2!} + \frac{(\omega I)^4}{4!... ...{\omega I}{1!} + \frac{(\omega I)^3}{3!} + ... \right)\right]^\nu_{\ \mu} x^\mu$$

$$= \left[\left( 1 + I^2\frac{\omega^2}{2!} - I^2\frac{\omega^4}{4!} ... ... \frac{\omega}{1!} - \frac{\omega^3}{3!} + ... \right)\right]^\nu_{\ \mu} x^\mu$$

$$= (1 + I^2 - I^2\cos\omega + I\sin\omega)^\nu_{\ \mu} x^\mu .$$ (5.138)

Turning now to the construction of a finite spinor transformation $S$, we have

$$S = \lim_{N\rightarrow\infty} \left( 1 - \frac{i}{4} \frac{\... ...u\nu}\right)^N = e^{-(i/4)\omega\sigma_{\mu\nu}I_n^{\mu\nu}} .$$ (5.139)

The following sections consider finite transformations for a rotation in 3-space, a general Lorentz boost, and spatial inversion.