Dirac Fields (1)

Dirac Basis

The wave equation for integer spins are already known as Klein-Gordon equation: $${\partial }^2\varphi +m^2\varphi =0$$We may develop half-ranked field to be $$\begin{array}{}\text{(1)}& (i{\gamma }^{\mu }{\partial }_{\mu }-m)\psi =0\end{array}$$when $$({\partial }^2+m^2)\psi =-(i{\gamma }^{\mu }{\partial }_{\mu }+m)(i{\gamma }^{\mu }{\partial }_{\mu }-m)\psi$$To satisfy the last equation, the Clifford algebra condition must be satisfied.$$\{{\gamma }^{\mu },{\gamma }^{\nu }\}=2{\eta }^{\mu \nu }$$Also the Lagrangian which derives the Dirac equation $(1)$ can be written as $$\mathfrak{L}=\overline{\psi }(i\text{⧸}\partial -m)\psi$$where $\text{⧸}\partial ={\gamma }^{\mu }{\partial }_{\mu }$. Then, we can see that the U(1) charge is $$\int \overline{\psi }{\gamma }^0\psi d^{3}x$$and usually we demand it to be $\int {\psi }^{†}\psi d^{3}x$ to guarantee finiteness so $$\overline{\psi }:={\psi }^{†}{\gamma }^0$$

     Any basis of gamma matrices and their following theories are all equivalent, so first the Dirac basis is taken which is handy in nonrelativistic case.$${\gamma }^0=\left[\begin{array}{cc}1& \\ & -1\end{array}\right];{\gamma }^k=\left[\begin{array}{cc}& {\sigma }^k\\ -{\sigma }^k& \end{array}\right]$$There is also Weyl basis which is written $${\gamma }^0=\left[\begin{array}{cc}& 1\\ 1& \end{array}\right];{\gamma }^k=\left[\begin{array}{cc}& {\sigma }^k\\ -{\sigma }^k& \end{array}\right]$$

Eigenspinors

As gamma matrices are constants, the plane wave solution $k_{\mu }=i{\partial }_{\mu }\Rightarrow \psi \sim e^{-i{k}_{\mu }{x}^{\mu }}$ can also be a solution for the Dirac equation. By applying the solution to the equation, the equation may transform into momentum space. Then, by producting ${\gamma }^0$ to the whole equation,
$$k^0\psi ={\gamma }^0(\overrightarrow{\gamma }\cdot \overrightarrow{k}+m)\psi$$where $k^{\mu }=(k^0,\overrightarrow{k})$. Here, the matrix on the right hand side is
$${\gamma }^0(\overrightarrow{\gamma }\cdot \overrightarrow{k}+m)=\left[\begin{array}{cc}m& \overrightarrow{\sigma }\cdot \overrightarrow{k}\\ \overrightarrow{\sigma }\cdot \overrightarrow{k}& -m\end{array}\right]$$The eigenvalue of $k^0$ can be obtained as $\pm \omega =\pm \sqrt{{\overrightarrow{k}}^2+m^2}$.

     For $k^0=\omega$, $\psi \sim \left[\begin{array}{c}m+\omega \\ \overrightarrow{\sigma }\cdot \overrightarrow{k}\end{array}\right]$. For $k^0=-\omega$, $\psi \sim \left[\begin{array}{c}-\overrightarrow{\sigma }\cdot \overrightarrow{k}\\ m+\omega \end{array}\right]$. Attached plane wave part, $$\begin{array}{rl}\psi & \sim \left[\begin{array}{c}m+\omega \\ \overrightarrow{\sigma }\cdot \overrightarrow{k}\end{array}\right]e^{-i\omega t+i\overrightarrow{k}\cdot \overrightarrow{x}},\left[\begin{array}{c}-\overrightarrow{\sigma }\cdot \overrightarrow{k}\\ m+\omega \end{array}\right]e^{i\omega t+i\overrightarrow{k}\cdot \overrightarrow{x}}\\ & =\left[\begin{array}{c}m+\omega \\ \overrightarrow{\sigma }\cdot \overrightarrow{k}\end{array}\right]e^{-i\omega t+i\overrightarrow{k}\cdot \overrightarrow{x}},\left[\begin{array}{c}\overrightarrow{\sigma }\cdot \overrightarrow{k}\\ m+\omega \end{array}\right]e^{i\omega t-i\overrightarrow{k}\cdot \overrightarrow{x}}\\ & =\left[\begin{array}{c}m+\omega \\ \overrightarrow{\sigma }\cdot \overrightarrow{k}\end{array}\right]e^{-i{k}_{\mu }{x}^{\mu }},\left[\begin{array}{c}\overrightarrow{\sigma }\cdot \overrightarrow{k}\\ m+\omega \end{array}\right]e^{i{k}_{\mu }{x}^{\mu }}\end{array}$$These are the eigenfunction of the Dirac equation. As the first two and the second two have positive frequency and negative frequency respectively, the first eigenfunctions are usually interpreted as particle while the second are interpreted as antiparticle.

     Now, to normalise the functions in the same sense in the quantum mechanics, Noether charge may be normalised to 1. U(1) Noether charge of Dirac field is $$\int -\overline{\psi }i{\gamma }^0\delta \psi d^{3}x=\int \overline{\psi }{\gamma }^0\psi d^{3}x=\int {\psi }^{†}\psi d^{3}x$$For both case, $${\psi }^{†}\psi =(m+\omega {)}^2+(\overrightarrow{\sigma }\cdot \overrightarrow{k}{)}^2=m^2+2m\omega +{\omega }^2+{\overrightarrow{k}}^2=2\omega (m+\omega )$$as $m^2+{\overrightarrow{k}}^2={\omega }^2$.

     Finally, the Dirac eigenfunctions are $$\sqrt{\frac{m+\omega }{2\omega V}}\left[\begin{array}{c}{\alpha }_i\\ \frac{\overrightarrow{\sigma }\cdot \overrightarrow{k}}{m+\omega }{\alpha }_i\end{array}\right]e^{-i{k}_{\mu }{x}^{\mu }},\sqrt{\frac{m+\omega }{2\omega V}}\left[\begin{array}{c}\frac{\overrightarrow{\sigma }\cdot \overrightarrow{k}}{m+\omega }{\beta }_j\\ {\beta }_j\end{array}\right]e^{i{k}_{\mu }{x}^{\mu }}$$where ${\alpha }_i$'s and ${\beta }_j$'s are the 2-component spinors for upper or lower space to handle $2\times 2$ Pauli matrices and show the additional degrees of freedom which will correspond to classical spin.

     Or, the eigenspinors can be expressed with the scalar waves and spinors.$$u^{(i)}(\overrightarrow{k})\frac{e^{-i{k}_{\mu }{x}^{\mu }}}{\sqrt{2\omega V}},v^{(j)}(\overrightarrow{k})\frac{e^{i{k}_{\mu }{x}^{\mu }}}{\sqrt{2\omega V}}$$so defined is$$u^{(i)}(\overrightarrow{k})=\sqrt{m+\omega }\left[\begin{array}{c}{\alpha }_i\\ \frac{\overrightarrow{\sigma }\cdot \overrightarrow{k}}{m+\omega }{\alpha }_i\end{array}\right];v^{(j)}=\sqrt{m+\omega }\left[\begin{array}{c}\frac{\overrightarrow{\sigma }\cdot \overrightarrow{k}}{m+\omega }{\beta }_j\\ {\beta }_j\end{array}\right]$$and$$u^{(i+2)}(\overrightarrow{k})=v^{(i)}(-\overrightarrow{k});v^{(j+2)}(\overrightarrow{k})=u^{(j)}(-\overrightarrow{k})$$

Spin

Spinor indices come from space-time symmetry, so the space-time rotation may have corresponding transformation in spinor space. By considering the rotational transform $x^{\mu }\to x^{\mu }+{{\omega }^{\mu }}_{\nu }{x}^{\nu }$ on the gamma matrices, which is a converter between spinor indices and space-time indices, we may find the spinor representation of rotation.

     First, we may consider general fields $f(x^{\mu })$, and then for the infinitesimal angle $\omega$, the scalar field may transforms to $f(x^{\mu })\to f(x^{\mu }+{{\omega }^{\mu }}_{\nu })$. Its generator may be $$L^{\mu \nu }=-i(x^{\mu }{\partial }^{\nu }-x^{\nu }{\partial }^{\mu })$$so that $f(x)\to e^{-\frac{i}{2}{L}^{\mu \nu }{\omega }_{\mu \nu }}f(x)$, and additional space-time indices may give other infinitesimal transforms.

     As the gamma matrices has 1 vector index and is (1,1)-rank tensor (linear operator) in spinor space, the matrices may transform, under rotation, $${\gamma }^{\mu }\to U({\gamma }^{\mu }+{{\omega }^{\mu }}_{\nu }{\gamma }^{\nu })U^{-1}$$Be ware of that $U$ is not guaranteed to be unitary. We may express with generators: $U=e^{-\frac{i}{2}{S}^{\rho \sigma }{\omega }_{\rho \sigma }}$ (one half factor by analogy to $L^{\mu \nu }$). We expect that the gamma matrices is numeric tensor so invariant under such total transform. $$e^{-\frac{i}{2}{S}^{\rho \sigma }{\omega }_{\rho \sigma }}({\gamma }^{\mu }+{{\omega }^{\mu }}_{\nu }{\gamma }^{\nu })e^{\frac{i}{2}{S}^{\rho \sigma }{\omega }_{\rho \sigma }}={\gamma }^{\mu }$$

     Applying Baker-Campbell formula $$e^{X}Y{e}^{-X}=Y+[X,Y]+O(X^2)$$gives the equation of commutators. $$\begin{array}{rl}& {{\omega }^{\mu }}_{\nu }{\gamma }^{\nu }-[\frac{i}{2}{S}^{\rho \sigma }{\omega }_{\rho \sigma },{\gamma }^{\mu }+{{\omega }^{\mu }}_{\nu }{\gamma }^{\nu }]+O({\omega }^2)\\ & ={{\omega }^{\mu }}_{\nu }{\gamma }^{\nu }-\frac{i}{2}[S^{\rho \sigma },{\gamma }^{\mu }]{\omega }_{\rho \sigma }+O({\omega }^2)=0\end{array}$$As $S^{\rho \sigma }$ has 2 space-time indices and has commutation relation with gamma matrices, the ansatz can be taken $S^{\rho \sigma }=A{\gamma }^{\rho }{\gamma }^{\sigma }$. Then, as $$[{\gamma }^{\rho }{\gamma }^{\sigma },{\gamma }^{\mu }]={\gamma }^{\rho }\{{\gamma }^{\sigma },{\gamma }^{\mu }\}-\{{\gamma }^{\mu },{\gamma }^{\rho }\}{\gamma }^{\sigma }=2{\eta }^{\mu \sigma }{\gamma }^{\rho }-2{\eta }^{\mu \rho }{\gamma }^{\sigma }$$the equation becomes $${{\omega }^{\mu }}_{\nu }{\gamma }^{\nu }-iA({{\omega }_{\rho }}^{\mu }{\gamma }^{\rho }-{{\omega }^{\mu }}_{\sigma }{\gamma }^{\sigma })={{\omega }^{\mu }}_{\nu }{\gamma }^{\nu }+2iA{{\omega }^{\mu }}_{\sigma }{\gamma }^{\sigma }=0$$As $A=\frac{i}{2}$ satifies the equation, the solution may be $S^{\rho \sigma }=\frac{i}{2}{\gamma }^{\rho }{\gamma }^{\sigma }$; however we demand the rotation generators to be antisymmetric generally, the solution may be antisymmetrised. $$S^{\rho \sigma }=\frac{i}{4}[{\gamma }^{\rho },{\gamma }^{\sigma }]$$

     Therefore, Dirac fields transform by generator $L+S$. Now we may see the eigenvalues of the total generator in the nonrelativistic limit to match the quantum mechanics. Before that, we may explicitly obtain the generator to calculate; here, in Dirac basis. $$[{\gamma }^i,{\gamma }^j]=-[{\sigma }^i,{\sigma }^j]=-2i{ϵ}^{ijk}{\sigma }^k=-4i{ϵ}^{ijk}{S}_k^{(Euclidean)}$$so$$S^{ij}={ϵ}^{ijk}{S}_k^{(Euclidean)}$$This is exactly the same with the familiar 3-d quantum mechanics. Thus, each upper and lower 2 component spinor in Dirac basis indicates the Pauli spinor for particle and antiparticle.

In Weyl basis, $$[{\gamma }^{\mu },{\gamma }^{\nu }]=\left[\begin{array}{cc}{\sigma }^{[\mu }{\overline{\sigma }}^{\nu ]}& \\ & {\overline{\sigma }}^{[\mu }{\sigma }^{\nu ]}\end{array}\right]$$
$$[{\gamma }^0,{\gamma }^i]=\left[\begin{array}{cc}-2{\sigma }^i& \\ & 2{\sigma }^i\end{array}\right]$$
$$[{\gamma }^i,{\gamma }^j]=-[{\sigma }^i,{\sigma }^j]$$