The essential fact about the Klein Gordon equation is that it is a non-linear equation. But, equations are consistent with Lorentz transformation when they are linear in space and time. Dirac came forward with a new way of resolving the problem of this non-linearity by factorizing the relativistic energy momentum relation into two linear equations. The equation is given by the format,
$$ i\frac{\partial\psi}{\partial{t}} = \left(-i \vec{\alpha}\cdot\vec{p}+\beta{m}\right)\psi $$
But, still it should satisfy the energy momentum relation $$ E^2 \psi = \left(m^2+|\vec{p}|^2\right)\psi $$
that is,
$$ E (E\psi) = i \frac{\partial}{\partial{t}}\left(\frac{i\partial\psi}{\partial{t}}\right) \\ = \left(-i \vec{\alpha}\cdot\vec{p}+\beta{m}\right) \left(-i \vec{\alpha}\cdot\vec{p}+\beta{m}\right) \psi $$ [the differentials commute]
In summation convention,
$$\frac{\partial^2\psi}{\partial{t^2}} = \left[\alpha_i\alpha_j\partial_i\partial_j + i(\alpha_i\beta+\beta\alpha_i)m\partial_i - \beta^2m^2\right]\psi $$
[It is assumed $\beta$ commutes with differentials]
The above can be evaluated separately from Klein Gordon equation as, $$\frac{\partial^2\psi}{\partial{t^2}} = \left[\vec{\nabla}^2 - m^2\right]\psi $$
Comparing both of them, we can see that $\alpha_i,\beta$ shouldn't commute and so they cannot be numbers. Dirac proposed the idea of matrices instead of numbers. This leads to the following properties that should be satisfied by $\alpha_i$ and $\beta$ as,
$$\{\alpha_i,\alpha_j\} = 2 \delta_{ij} \\ \{\alpha_i,\beta\} = 0 \\ \alpha_i^2=\beta^2 =1$$
$$Tr[\alpha_i^2{X}\, (or)\, \beta^2{X}=\lambda^2{X}=IX]$$ gives us the eigenvalues to be $\pm 1$
$$Tr[\alpha_i] \\ = Tr[\alpha_i{\beta^2}] = Tr[\beta\alpha_i{\beta}] (using \,the \,property\, Tr(AB)=Tr(BA)) \\ = Tr[-\beta\alpha_i\beta] (using \,anti-commutation \,relation)$$
Doesn't matter whether we start with $\alpha_i$ or $\beta$, this proves the trace of the matrices $\alpha_i,\beta$ should be zero.
From the above two properties, dimension of the matrix should be even.
Starting with the dimension N=2, there are no sufficient number of independent matrices to describe our four matrices (since pauli matrices span the space completely).
Thus, we are left with the next possibility of N=4 by constructing our matrices as,
$$ \alpha_i = \left(\begin{matrix}0 & {\sigma_i}\\ {\sigma_i}&0\end{matrix}\right) \\ \beta = \left(\begin{matrix}\mathbb{I} & 0 \\ 0 & \mathbb{-I}\end{matrix}\right) $$
To make it simple, we introduce the symbol $\gamma$ and defined everything in terms of new Dirac $\gamma$ matrices. And, the Dirac equation is finally given in the simplest form as,
$$\gamma^0 =\beta = \left(\begin{matrix}\mathbb{I} & 0 \\ 0& \mathbb{-I}\end{matrix}\right) \\ \gamma^i = \beta\alpha_i = \left(\begin{matrix}0 &{\sigma_i}\\ {-\sigma_i}&0\end{matrix}\right)$$
with the properties,
$$\gamma^0 = (\gamma^0)^\dagger \\ (\gamma^0)^2 = \mathbb{I} \\ \gamma^i = -(\gamma^i)^\dagger \\ (\gamma^i)^2=-\mathbb{I} \\ \{\gamma^\mu,\gamma^\nu\} = 2g_{\mu\nu} $$
Using this new representation, the Dirac equation is written as, $$\left(i\gamma^\mu\partial_\mu - m \right)\psi = 0 $$
In slashed notation, any Lorentz vector is $\slashed{x} = \gamma^\mu{x}_\mu$ and so, Dirac equation is, $$ \left(i\slashed{\partial}-m\right)\psi = 0$$
$$ i\frac{\partial\psi}{\partial{t}} = \left(-i \vec{\alpha}\cdot\vec{p}+\beta{m}\right)\psi $$
But, still it should satisfy the energy momentum relation $$ E^2 \psi = \left(m^2+|\vec{p}|^2\right)\psi $$
that is,
$$ E (E\psi) = i \frac{\partial}{\partial{t}}\left(\frac{i\partial\psi}{\partial{t}}\right) \\ = \left(-i \vec{\alpha}\cdot\vec{p}+\beta{m}\right) \left(-i \vec{\alpha}\cdot\vec{p}+\beta{m}\right) \psi $$ [the differentials commute]
In summation convention,
$$\frac{\partial^2\psi}{\partial{t^2}} = \left[\alpha_i\alpha_j\partial_i\partial_j + i(\alpha_i\beta+\beta\alpha_i)m\partial_i - \beta^2m^2\right]\psi $$
[It is assumed $\beta$ commutes with differentials]
The above can be evaluated separately from Klein Gordon equation as, $$\frac{\partial^2\psi}{\partial{t^2}} = \left[\vec{\nabla}^2 - m^2\right]\psi $$
Comparing both of them, we can see that $\alpha_i,\beta$ shouldn't commute and so they cannot be numbers. Dirac proposed the idea of matrices instead of numbers. This leads to the following properties that should be satisfied by $\alpha_i$ and $\beta$ as,
$$\{\alpha_i,\alpha_j\} = 2 \delta_{ij} \\ \{\alpha_i,\beta\} = 0 \\ \alpha_i^2=\beta^2 =1$$
$$Tr[\alpha_i^2{X}\, (or)\, \beta^2{X}=\lambda^2{X}=IX]$$ gives us the eigenvalues to be $\pm 1$
$$Tr[\alpha_i] \\ = Tr[\alpha_i{\beta^2}] = Tr[\beta\alpha_i{\beta}] (using \,the \,property\, Tr(AB)=Tr(BA)) \\ = Tr[-\beta\alpha_i\beta] (using \,anti-commutation \,relation)$$
Doesn't matter whether we start with $\alpha_i$ or $\beta$, this proves the trace of the matrices $\alpha_i,\beta$ should be zero.
From the above two properties, dimension of the matrix should be even.
Starting with the dimension N=2, there are no sufficient number of independent matrices to describe our four matrices (since pauli matrices span the space completely).
Thus, we are left with the next possibility of N=4 by constructing our matrices as,
$$ \alpha_i = \left(\begin{matrix}0 & {\sigma_i}\\ {\sigma_i}&0\end{matrix}\right) \\ \beta = \left(\begin{matrix}\mathbb{I} & 0 \\ 0 & \mathbb{-I}\end{matrix}\right) $$
To make it simple, we introduce the symbol $\gamma$ and defined everything in terms of new Dirac $\gamma$ matrices. And, the Dirac equation is finally given in the simplest form as,
$$\gamma^0 =\beta = \left(\begin{matrix}\mathbb{I} & 0 \\ 0& \mathbb{-I}\end{matrix}\right) \\ \gamma^i = \beta\alpha_i = \left(\begin{matrix}0 &{\sigma_i}\\ {-\sigma_i}&0\end{matrix}\right)$$
with the properties,
$$\gamma^0 = (\gamma^0)^\dagger \\ (\gamma^0)^2 = \mathbb{I} \\ \gamma^i = -(\gamma^i)^\dagger \\ (\gamma^i)^2=-\mathbb{I} \\ \{\gamma^\mu,\gamma^\nu\} = 2g_{\mu\nu} $$
Using this new representation, the Dirac equation is written as, $$\left(i\gamma^\mu\partial_\mu - m \right)\psi = 0 $$
In slashed notation, any Lorentz vector is $\slashed{x} = \gamma^\mu{x}_\mu$ and so, Dirac equation is, $$ \left(i\slashed{\partial}-m\right)\psi = 0$$
No comments:
Post a Comment
Let everyone know what you think about this