Dirac equation

References: Mark Srednicki, Quantum Field Theory, (Cambridge University Press, 2007) – Chapter 1, Problem 1.1.

The Klein-Gordon equation is an early attempt at a relativistic quantum theory, but it contains a second-order time derivative which leads to probability not being conserved over time. Dirac proposed another equation that attempts to solve this problem for particles of spin 1/2. The Dirac equation is essentially a modification of the Schrödinger equation:

\displaystyle i\hbar\frac{\partial}{\partial t}\psi_{a}\left(x\right)=\left[-i\hbar c\left(\alpha^{j}\right)_{ab}\partial_{j}+mc^{2}\left(\beta\right)_{ab}\right]\psi_{b}\left(x\right) \ \ \ \ \ (1)

Here, {\psi} is now a vector in spin space with components {\psi_{a}}. The objects {\beta} and {\alpha^{j}} (for {j=1,2,3}) are square matrices (where the subscript {ab} indicates the component of the matrix being considered), also in spin space, and repeated indices are summed over spatial coordinates only. [We won’t worry about how Dirac arrived at this equation for now; we’ll just accept it and see where it leads.]

To make this equation formally equivalent to the Schrödinger equation, the hamiltonian operator {H} on the RHS must now be a matrix. We can also use the definition of the momentum operator {P_{j}=-i\hbar\partial_{j}} to get

\displaystyle H_{ab}=cP_{j}\left(\alpha^{j}\right)_{ab}+mc^{2}\left(\beta\right)_{ab} \ \ \ \ \ (2)

 

This might not look much like the relativistic energy:

\displaystyle E=\sqrt{p^{2}c^{2}+m^{2}c^{4}} \ \ \ \ \ (3)

but if we square 2 (remembering that matrix products need not commute), we have

\displaystyle \left(H^{2}\right)_{ab}=c^{2}P_{j}P_{k}\left(\alpha^{j}\alpha^{k}\right)_{ab}+mc^{3}P_{j}\left(\alpha^{j}\beta+\beta\alpha^{j}\right)_{ab}+m^{2}c^{4}\left(\beta^{2}\right)_{ab} \ \ \ \ \ (4)

 

We can define the anticommutator as

\displaystyle \left\{ A,B\right\} \equiv AB+BA \ \ \ \ \ (5)

We can write the first term on the RHS of 4 as

\displaystyle c^{2}P_{j}P_{k}\left(\alpha^{j}\alpha^{k}\right)_{ab}=\frac{1}{2}c^{2}P_{j}P_{k}\left\{ \alpha^{j},\alpha^{k}\right\} \ \ \ \ \ (6)

so we get

\displaystyle \left(H^{2}\right)_{ab}=\frac{1}{2}c^{2}P_{j}P_{k}\left\{ \alpha^{j},\alpha^{k}\right\} +mc^{3}P_{j}\left\{ \alpha^{j},\beta\right\} +m^{2}c^{4}\left(\beta^{2}\right)_{ab} \ \ \ \ \ (7)

 

In order to make this equal to {E^{2}}, we need the matrices {\alpha^{j}} and {\beta} to satisfy the conditions:

\displaystyle \left\{ \alpha^{j},\alpha^{k}\right\} \displaystyle = \displaystyle 2\delta^{jk}\delta_{ab}\ \ \ \ \ (8)
\displaystyle \left\{ \alpha^{j},\beta\right\} \displaystyle = \displaystyle 0\ \ \ \ \ (9)
\displaystyle \left(\beta^{2}\right)_{ab} \displaystyle = \displaystyle \delta_{ab} \ \ \ \ \ (10)

The first condition requires the anticommutator of {\alpha^{j}} and {\alpha^{k}} to be zero unless {j=k}, in which case the anticommutator gives the identity matrix. Remember that the superscripts {j} and {k} specify which matrix we’re talking about, while the subscripts {ab} indicate the component of the matrix. The conditions aren’t derived; rather they are imposed to make the energy come out right. With these conditions, we have

\displaystyle \left(H^{2}\right)_{ab}=\left(\mathbf{P}^{2}c^{2}+m^{2}c^{4}\right)\delta_{ab} \ \ \ \ \ (11)

which gives the correct operator for the square of the energy.

The question arises as to what these matrices {\alpha^{j}} and {\beta} are. One candidate is the set of three Pauli spin matrices

\displaystyle \sigma_{x} \displaystyle = \displaystyle \left[\begin{array}{cc} 0 & 1\\ 1 & 0 \end{array}\right]\ \ \ \ \ (12)
\displaystyle \sigma_{y} \displaystyle = \displaystyle \left[\begin{array}{cc} 0 & -i\\ i & 0 \end{array}\right]\ \ \ \ \ (13)
\displaystyle \sigma_{z} \displaystyle = \displaystyle \left[\begin{array}{cc} 1 & 0\\ 0 & -1 \end{array}\right] \ \ \ \ \ (14)

By direct calculation, we see that {\left\{ \sigma^{i},\sigma^{j}\right\} =2\delta^{ij}}. For example

\displaystyle \left\{ \sigma_{x},\sigma_{y}\right\} \displaystyle = \displaystyle \left[\begin{array}{cc} 0 & 1\\ 1 & 0 \end{array}\right]\left[\begin{array}{cc} 0 & -i\\ i & 0 \end{array}\right]+\left[\begin{array}{cc} 0 & -i\\ i & 0 \end{array}\right]\left[\begin{array}{cc} 0 & 1\\ 1 & 0 \end{array}\right]\ \ \ \ \ (15)
\displaystyle \displaystyle = \displaystyle \left[\begin{array}{cc} i & 0\\ 0 & -i \end{array}\right]+\left[\begin{array}{cc} -i & 0\\ 0 & i \end{array}\right]\ \ \ \ \ (16)
\displaystyle \displaystyle = \displaystyle 0\ \ \ \ \ (17)
\displaystyle \left\{ \sigma_{x},\sigma_{x}\right\} \displaystyle = \displaystyle 2\sigma_{x}^{2}\ \ \ \ \ (18)
\displaystyle \displaystyle = \displaystyle 2\left[\begin{array}{cc} 1 & 0\\ 0 & 1 \end{array}\right] \ \ \ \ \ (19)

and so on. However, in order to satisfy 9, we need to find a single matrix that anticommutes with all 3 spin matrices. We get

\displaystyle \left\{ \sigma_{x},\beta\right\} \displaystyle = \displaystyle \left[\begin{array}{cc} \beta_{21} & \beta_{22}\\ \beta_{11} & \beta_{12} \end{array}\right]+\left[\begin{array}{cc} \beta_{12} & \beta_{11}\\ \beta_{22} & \beta_{21} \end{array}\right]\ \ \ \ \ (20)
\displaystyle \displaystyle = \displaystyle \left[\begin{array}{cc} 0 & 0\\ 0 & 0 \end{array}\right] \ \ \ \ \ (21)

This gives

\displaystyle \beta_{12} \displaystyle = \displaystyle -\beta_{21}\equiv\gamma\ \ \ \ \ (22)
\displaystyle \beta_{11} \displaystyle = \displaystyle -\beta_{22}\equiv\epsilon \ \ \ \ \ (23)

We then get

\displaystyle \left\{ \sigma_{z},\beta\right\} \displaystyle = \displaystyle \left[\begin{array}{cc} 1 & 0\\ 0 & -1 \end{array}\right]\left[\begin{array}{cc} \epsilon & \gamma\\ -\gamma & -\epsilon \end{array}\right]+\left[\begin{array}{cc} \epsilon & \gamma\\ -\gamma & -\epsilon \end{array}\right]\left[\begin{array}{cc} 1 & 0\\ 0 & -1 \end{array}\right]\ \ \ \ \ (24)
\displaystyle \displaystyle = \displaystyle \left[\begin{array}{cc} \epsilon & \gamma\\ \gamma & \epsilon \end{array}\right]+\left[\begin{array}{cc} \epsilon & -\gamma\\ -\gamma & \epsilon \end{array}\right]\ \ \ \ \ (25)
\displaystyle \displaystyle = \displaystyle \left[\begin{array}{cc} 0 & 0\\ 0 & 0 \end{array}\right] \ \ \ \ \ (26)

This gives

\displaystyle \epsilon=0 \ \ \ \ \ (27)

So finally

\displaystyle \left\{ \sigma_{y},\beta\right\} \displaystyle = \displaystyle \left[\begin{array}{cc} 0 & -i\\ i & 0 \end{array}\right]\left[\begin{array}{cc} 0 & \gamma\\ -\gamma & 0 \end{array}\right]+\left[\begin{array}{cc} 0 & \gamma\\ -\gamma & 0 \end{array}\right]\left[\begin{array}{cc} 0 & -i\\ i & 0 \end{array}\right]\ \ \ \ \ (28)
\displaystyle \displaystyle = \displaystyle \left[\begin{array}{cc} i\gamma & 0\\ 0 & i\gamma \end{array}\right]+\left[\begin{array}{cc} i\gamma & 0\\ 0 & i\gamma \end{array}\right]\ \ \ \ \ (29)
\displaystyle \displaystyle = \displaystyle \left[\begin{array}{cc} 0 & 0\\ 0 & 0 \end{array}\right] \ \ \ \ \ (30)

So {\gamma=0} resulting in {\left(\beta\right)_{ab}=0}. Thus there is no non-zero matrix {\beta} that anticommutes with all 3 of the Pauli spin matrices.

So what can we say about the Dirac matrices? From 10, we see that the eigenvalues of {\beta^{2}=I} are all 1, so the eigenvalues of {\beta} must be {\pm1}.

The trace (sum of the diagonal elements) of a matrix is equal to the sum of its eigenvalues (theorem from matrix algebra). To find the trace of {\beta}, we can use the anticommutators 8 and 9, together with another theorem from matrix algebra which states that {\mbox{tr}\left(AB\right)=\mbox{tr}\left(BA\right)} for any square matrices {A} and {B} of the same order.

\displaystyle \mbox{tr}\left(\alpha_{1}^{2}\beta\right) \displaystyle = \displaystyle \mbox{tr}\left(\alpha_{1}\left(\alpha_{1}\beta\right)\right)\ \ \ \ \ (31)
\displaystyle \displaystyle = \displaystyle \mbox{tr}\left(\left(\alpha_{1}\beta\right)\alpha_{1}\right) \ \ \ \ \ (32)

However, from 9, {\alpha_{1}\beta=-\beta\alpha_{1}} and from 8, {\alpha_{1}^{2}=I} (the identity matrix), so

\displaystyle \mbox{tr}\left(\alpha_{1}^{2}\beta\right) \displaystyle = \displaystyle \mbox{tr}\left(\beta\right)\ \ \ \ \ (33)
\displaystyle \displaystyle = \displaystyle \mbox{tr}\left(\left(\alpha_{1}\beta\right)\alpha_{1}\right)\ \ \ \ \ (34)
\displaystyle \displaystyle = \displaystyle -\mbox{tr}\left(\left(\beta\alpha_{1}\right)\alpha_{1}\right)\ \ \ \ \ (35)
\displaystyle \displaystyle = \displaystyle -\mbox{tr}\left(\beta\alpha_{1}^{2}\right)\ \ \ \ \ (36)
\displaystyle \displaystyle = \displaystyle -\mbox{tr}\left(\beta\right) \ \ \ \ \ (37)

Hence {\mbox{tr}\left(\beta\right)=-\mbox{tr}\left(\beta\right)=0}, so {\beta} must have an equal number of {+1} and {-1} eigenvalues. In other words, {\beta} must be even dimensional, so the smallest size is {4\times4}.

We can also find the trace of {\alpha^{j}} by starting with {\mbox{tr}\left(\alpha^{j}\beta^{2}\right)} and following through the same steps as above (using {\beta^{2}=I}) to show that {\mbox{tr}\left(\alpha^{j}\right)=-\mbox{tr}\left(\alpha^{j}\right)=0}.

8 thoughts on “Dirac equation

  1. Pingback: Dirac equation: matrix properties | Physics pages

  2. Pingback: Dirac equation: the gamma matrices | Physics pages

  3. Pingback: Dirac equation as four coupled differential equations | Physics pages

  4. Pingback: Dirac equation: 4 solution vectors | Physics pages

  5. Pingback: Dirac equation: inner products of spinors | Physics pages

  6. Pingback: Dirac equation: orthogonality of solutions | Physics pages

  7. Pingback: Dirac equation: adjoint solutions | Physics pages

  8. Pingback: Adjoint Dirac equation | Physics pages

Leave a Reply

Your email address will not be published. Required fields are marked *