Dirac equation: 4 solution vectors

Reference: References: Robert D. Klauber, Student Friendly Quantum Field Theory, (Sandtrove Press, 2013) – Chapter 4, Problem 4.5.

The Dirac equation in relativistic quantum mechanics can be written as

\displaystyle  \left(i\gamma^{\mu}\partial_{\mu}-mI\right)\left|\psi\right\rangle =0 \ \ \ \ \ (1)

When written out in its matrix components, this equation is actually 4 differential equations.

\displaystyle   \left(i\partial_{0}-m\right)\left|\psi\right\rangle _{1}+i\partial_{3}\left|\psi\right\rangle _{3}+\left(i\partial_{1}+\partial_{2}\right)\left|\psi\right\rangle _{4} \displaystyle  = \displaystyle  0\ \ \ \ \ (2)
\displaystyle  \left(i\partial_{0}-m\right)\left|\psi\right\rangle _{2}+\left(i\partial_{1}-\partial_{2}\right)\left|\psi\right\rangle _{3}-i\partial_{3}\left|\psi\right\rangle _{4} \displaystyle  = \displaystyle  0\ \ \ \ \ (3)
\displaystyle  -i\partial_{3}\left|\psi\right\rangle _{1}-\left(i\partial_{1}+\partial_{2}\right)\left|\psi\right\rangle _{2}-\left(i\partial_{0}+m\right)\left|\psi\right\rangle _{3} \displaystyle  = \displaystyle  0\ \ \ \ \ (4)
\displaystyle  -i\left(\partial_{1}+i\partial_{2}\right)\left|\psi\right\rangle _{1}+i\partial_{3}\left|\psi\right\rangle _{2}-\left(i\partial_{0}+m\right)\left|\psi\right\rangle _{4} \displaystyle  = \displaystyle  0 \ \ \ \ \ (5)

Remember that {\left|\psi\right\rangle } is a 4-d column vector in spinor space rather than a single function, so that the subscript index {j} in {\left|\psi\right\rangle _{j}} indicates which component in spinor space we’re dealing with. These equations have four solutions denoted by {\left|\psi^{\left(n\right)}\right\rangle } for {n=1,2,3,4}. Note that each {\left|\psi^{\left(n\right)}\right\rangle } is a full 4-component vector in spinor space; that is, the superscript {\left(n\right)} indicates which complete vector we’re dealing with. Thus {\left|\psi^{\left(n\right)}\right\rangle _{j}} is the {j}th component of the {n}th vector.

We can write the 4 PDEs as a matrix eigenvalue equation by moving the terms involving {m} to the RHS and factoring out an {i} from the terms remaining on the LHS:

\displaystyle  i\left[\begin{array}{cccc} \partial_{0} & 0 & \partial_{3} & \partial_{1}-i\partial_{2}\\ 0 & \partial_{0} & \partial_{1}+i\partial_{2} & -\partial_{3}\\ -\partial_{3} & -\partial_{1}+i\partial_{2} & -\partial_{0} & 0\\ -\partial_{1}-i\partial_{2} & \partial_{3} & 0 & -\partial_{0} \end{array}\right]\left[\begin{array}{c} \psi_{1}\\ \psi_{2}\\ \psi_{3}\\ \psi_{4} \end{array}\right]=m\left[\begin{array}{c} \psi_{1}\\ \psi_{2}\\ \psi_{3}\\ \psi_{4} \end{array}\right] \ \ \ \ \ (6)

We’ll now look at the four solutions {\left|\psi^{\left(n\right)}\right\rangle } and verify that they satisfy 6. First, we have

\displaystyle  \left|\psi^{\left(1\right)}\right\rangle =\sqrt{\frac{E+m}{2m}}\left[\begin{array}{c} 1\\ 0\\ \frac{p^{3}}{E+m}\\ \frac{p^{1}+ip^{2}}{E+m} \end{array}\right]e^{-ipx}\equiv u_{1}e^{-ipx} \ \ \ \ \ (7)

where {u_{1}} is defined by this equation as the constant {\sqrt{\frac{E+m}{2m}}} multiplied by the 4-d spinor factor. Remember that {px} is a 4-vector product:

\displaystyle  px=p^{\mu}x_{\mu}=Et-\mathbf{p}\cdot\mathbf{x} \ \ \ \ \ (8)

The derivatives in 6 are all with respect to spacetime variables, so act only on {e^{-ipx}}; the spinor components are constants with respect to these derivatives. The first row in 6 is therefore

\displaystyle   i\sqrt{\frac{E+m}{2m}}e^{-ipx}\left[-iE+0+\frac{p^{3}}{E+m}\left(ip^{3}\right)+\frac{p^{1}+ip^{2}}{E+m}\left(ip^{1}+p^{2}\right)\right] \displaystyle  =
\displaystyle  -\sqrt{\frac{E+m}{2m}}e^{-ipx}\left[-E+\frac{\mathbf{p}^{2}}{E+m}\right] \ \ \ \ \ (9) \displaystyle  =
\displaystyle  -\sqrt{\frac{E+m}{2m}}e^{-ipx}\left[-E+\frac{E^{2}-m^{2}}{E+m}\right] \displaystyle  =\ \ \ \ \ (10)
\displaystyle  -\sqrt{\frac{E+m}{2m}}e^{-ipx}\left[-E+\frac{\left(E+m\right)\left(E-m\right)}{E+m}\right] \displaystyle  = \displaystyle  \sqrt{\frac{E+m}{2m}}e^{-ipx}m\ \ \ \ \ (11)
\displaystyle  \displaystyle  = \displaystyle  m\psi_{1} \ \ \ \ \ (12)

Thus the first row of 6 is verified. The other 3 rows can be verified similarly. For row 2:

\displaystyle  i\sqrt{\frac{E+m}{2m}}e^{-ipx}\left[0+0+\frac{p^{3}}{E+m}\left(ip^{1}-p^{2}\right)+\frac{p^{1}+ip^{2}}{E+m}\left(-ip^{3}\right)\right]=0=m\psi_{2} \ \ \ \ \ (13)

For row 3:

\displaystyle   i\sqrt{\frac{E+m}{2m}}e^{-ipx}\left[-ip^{3}+0+\frac{p^{3}}{E+m}\left(iE\right)+0\right] \displaystyle  = \displaystyle  i\sqrt{\frac{E+m}{2m}}e^{-ipx}\left[\frac{-ip^{3}\left(E+m\right)+ip^{3}E}{E+m}\right]\ \ \ \ \ (14)
\displaystyle  \displaystyle  = \displaystyle  \sqrt{\frac{E+m}{2m}}e^{-ipx}m\frac{p^{3}}{E+m}\ \ \ \ \ (15)
\displaystyle  \displaystyle  = \displaystyle  m\psi_{3} \ \ \ \ \ (16)

And for row 4:

\displaystyle   i\sqrt{\frac{E+m}{2m}}e^{-ipx}\left[\left(-ip^{1}+p^{2}\right)+0+0+\frac{p^{1}+ip^{2}}{E+m}iE\right] \displaystyle  =
\displaystyle  \sqrt{\frac{E+m}{2m}}e^{-ipx}\left[\left(p^{1}+ip^{2}\right)-\frac{p^{1}+ip^{2}}{E+m}E\right] \ \ \ \ \ (17) \displaystyle  =
\displaystyle  \sqrt{\frac{E+m}{2m}}e^{-ipx}\left[\frac{\left(p^{1}+ip^{2}\right)\left(E+m\right)-\left(p^{1}+ip^{2}\right)E}{E+m}\right] \displaystyle  =\ \ \ \ \ (18)
\displaystyle  \sqrt{\frac{E+m}{2m}}e^{-ipx}\frac{p^{1}+ip^{2}}{E+m}m \displaystyle  = \displaystyle  m\psi_{4} \ \ \ \ \ (19)

The other 3 solutions are

\displaystyle   \left|\psi^{\left(2\right)}\right\rangle \displaystyle  = \displaystyle  \sqrt{\frac{E+m}{2m}}\left[\begin{array}{c} 0\\ 1\\ \frac{p^{1}-ip^{2}}{E+m}\\ -\frac{p^{3}}{E+m} \end{array}\right]e^{-ipx}\equiv u_{2}e^{-ipx}\ \ \ \ \ (20)
\displaystyle  \left|\psi^{\left(3\right)}\right\rangle \displaystyle  = \displaystyle  \sqrt{\frac{E+m}{2m}}\left[\begin{array}{c} \frac{p^{3}}{E+m}\\ \frac{p^{1}+ip^{2}}{E+m}\\ 1\\ 0 \end{array}\right]e^{ipx}\equiv v_{2}e^{ipx}\ \ \ \ \ (21)
\displaystyle  \left|\psi^{\left(4\right)}\right\rangle \displaystyle  = \displaystyle  \sqrt{\frac{E+m}{2m}}\left[\begin{array}{c} \frac{p^{1}-ip^{2}}{E+m}\\ -\frac{p^{3}}{E+m}\\ 0\\ 1 \end{array}\right]e^{ipx}\equiv v_{1}e^{ipx} \ \ \ \ \ (22)

If you really want to, you can verify that these 3 vectors satisfy 6 by grinding through the calculations as above. One point worth noting is that the constant {\sqrt{\frac{E+m}{2m}}} that multiplies all the solutions could be any other constant and still satisfy 6 (since the constant just cancels off both sides). It’s chosen to be {\sqrt{\frac{E+m}{2m}}} to make later calculations easier.