# Explicit expansion of the term $\overline{\psi}(i \gamma_\mu \partial^\mu-m) \psi$

Explicit expansion of the term $$\overline{\psi}(i \gamma_\mu \partial^\mu-m) \psi$$

In QED, one finds the first part of the Lagrangian density to be $$\mathcal{L}=\overline{\psi}(i \gamma_\mu \partial^\mu-m) \psi +\dots$$

I am interested in expanding the term.

1. Am I correct to define $$\psi$$ as:

$$\psi=\pmatrix{R_0+i C_0\\R_1+iC_1\\R_2+iC_2\\R_3+iC_3}$$

where $$R_0,R_1,R_2,R_3,C_0,C_1,C_2,C_3\in \mathbb{R}$$.

1. Am I correct to define $$\overline{\psi}=\psi^\dagger\gamma_0$$ as

\begin{align} \psi^\dagger&=\pmatrix{R_0-i C_0 && R_1-iC_1 && R_2-iC_2&&R_3-iC_3}\pmatrix{0&0&0&-i\\0&0&i&0\\0&-i&0&0\\i&0&0&0}\\ &=\pmatrix{i(R_0-i C_0) && -i(R_1-iC_1) && i(R_2-iC_2) && -i(R_3-iC_3)} \end{align}

1. Am I correct to calculate the term $$\overline{\psi}m\psi$$ as

\begin{align} \overline{\psi}m\psi&=m\pmatrix{i(R_0-i C_0) && -i(R_1-iC_1) && i(R_2-iC_2) && -i(R_3-iC_3)}\pmatrix{R_0+i C_0\\R_1+iC_1\\R_2+iC_2\\R_3+iC_3}\\ &={{m (C_0^2 + C_1^2 + C_2^2 + C_3^2 + R_0^2 + R_1^2 + R_2^2 + R_3^2)}} \end{align}

1. For the other term $$\overline{\psi}(i \gamma_\mu \partial^\mu) \psi$$, am I correct to write:

$$\psi^\dagger \gamma_0 \gamma_0 \partial \psi + \psi^\dagger \gamma_0 \gamma_1 \partial \psi + \psi^\dagger \gamma_0 \gamma_2 \partial \psi+\psi^\dagger \gamma_0 \gamma_3 \partial \psi$$

1. Then for each of the terms in (4), as:

$$\psi^\dagger \gamma_0 \gamma_0 \partial \psi = \psi^\dagger \partial \psi \\ = (dC_0 - i dR_0) (C_0 + i R_0) + (dC_1 - i dR_1) (C_1 + i R_1) + (dC_2 - i dR_2) (C_2 + i R_2) + (dC_3 - i dR_3) (C_3 + i R_3)$$

then

$$\psi^\dagger \gamma_0 \gamma_1 \partial \psi \\ = -(dC_3 - i dR_3) (C_0 + i R_0) - (dC_2 - i dR_2) (C_1 + i R_1) - (dC_1 - i dR_1) (C_2 + i R_2) - (dC_0 - i dR_0) (C_3 + i R_3)$$

then

$$\psi^\dagger \gamma_0 \gamma_2 \partial \psi \\ = (dC_0 - i dR_0) (C_0 + i R_0) + (dC_1 - i dR_1) (C_1 + i R_1) - (dC_2 - i dR_2) (C_2 + i R_2) - (dC_3 - i dR_3) (C_3 + i R_3)$$

then

$$\psi^\dagger \gamma_0 \gamma_2 \partial \psi \\ = -(dC_2 - i dR_2) (C_0 + i R_0) + (dC_3 - i dR_3) (C_1 + i R_1) - (dC_0 - i dR_0) (C_2 + i R_2) + (dC_1 - i dR_1) (C_3 + i R_3)$$

1. Finally, adding everything together I get:

$$2 C0 dC0 - C2 dC0 - C3 dC0 + 2 C1 dC1 - C2 dC1 + C3 dC1 - C0 dC2 - C1 dC2 - C0 dC3 + C1 dC3 - 2 I C0 dR0 + I C2 dR0 + I C3 dR0 - 2 I C1 dR1 + I C2 dR1 - I C3 dR1 + I C0 dR2 + I C1 dR2 + I C0 dR3 - I C1 dR3 + C0^2 m + C1^2 m + C2^2 m + C3^2 m + 2 I dC0 R0 - I dC2 R0 - I dC3 R0 + 2 dR0 R0 - dR2 R0 - dR3 R0 + m R0^2 + 2 I dC1 R1 - I dC2 R1 + I dC3 R1 + 2 dR1 R1 - dR2 R1 + dR3 R1 + m R1^2 - I dC0 R2 - I dC1 R2 - dR0 R2 - dR1 R2 + m R2^2 - I dC0 R3 + I dC1 R3 - dR0 R3 + dR1 R3 + m R3^2$$

edit:

A. Is this correct?

$$\partial^t \psi=\pmatrix{\frac{\partial}{\partial t}R_0+i \frac{\partial}{\partial t}C_0\\\frac{\partial}{\partial t}R_1+i\frac{\partial}{\partial t}C_1\\\frac{\partial}{\partial t}R_2+i\frac{\partial}{\partial t}C_2\\\frac{\partial}{\partial t}R_3+i\frac{\partial}{\partial t}C_3}$$

B.

Are the quantities $$R_0,R_1,R_2,R_3,C_0,C_1,C_2,C_3$$ then not the correct entries of matrix, but instead the entries should be functions $$f:\mathbb{R}^4\to\mathbb{C}$$:

$$\psi=\pmatrix{ \psi_0[t,x,y,z] \\ \psi_1[t,x,y,z] \\ \psi_2[t,x,y,z] \\ \psi_3[t,x,y,z]}$$

• Paragraph 4 is wrong. You’ve lost the index on the partial derivative. I didn’t check the rest. Jun 8 '20 at 17:06
• What is your motivation behind this expansion? Jun 8 '20 at 17:39
• @JohnnyLongsom Same reason taking a TV apart and putting it back together makes it easy to understand how it works. Jun 8 '20 at 17:43
• @G.Smith What does index means on the partial derivative (after I distributed the sum) ? Jun 8 '20 at 17:47
• Regarding B, it’s OK to define $R_0$ and $C_0$ as real functions that are the real and imaginary parts of the complex function $\psi_0$, etc., but you have to let $\psi_0$, $R_0$, and $C_0$ be functions of $t$, $x$, $y$, and $z$. Jun 8 '20 at 18:56

I'll go through it step by step. First note that we mean $$\bar\psi (i\gamma^\mu \partial_\mu - m )\psi = \bar\psi (i\gamma^\mu \partial_\mu - m\mathbb{I} )\psi$$ where $$\mathbb I$$ is the $$4\times 4$$ identity matrix. Let us calculate first the operator in the middle.
$$\gamma^\mu \partial_\mu$$ is a matrix, after performing the contraction. We find:
$$\gamma^\mu \partial_\mu = \begin{pmatrix} \partial_0 & 0 & \partial_3 & \partial_1 - i\partial_2\\ 0& \partial_0 & \partial_1 + i\partial_2 & - \partial_3\\ -\partial_3& -\partial_1 + i\partial_2 & - \partial_0 &0 \\ -\partial_1 - i\partial_2 & \partial_3 & 0 & -\partial_0 \end{pmatrix}$$
by treating the $$\partial_\mu$$ as each component being a scalar, times the matrices which are the components of $$\gamma^\mu$$. Explicitly we have taken $$\sum_{i=0}^3 \gamma^i \partial_i$$. Now we include the mass term, giving,
$$(i\gamma^\mu \partial_\mu - m\mathbb{I} ) = \begin{pmatrix} i\partial_0 -m& 0 & i\partial_3 & i\partial_1 +\partial_2\\ 0& i\partial_0 -m& i\partial_1 -\partial_2 & - i\partial_3\\ -i\partial_3& -i\partial_1 -\partial_2 & - i\partial_0 -m &0 \\ -i\partial_1 +\partial_2 & i\partial_3 & 0 & -i\partial_0 -m \end{pmatrix}.$$
Now we have $$\psi = \begin{pmatrix} a\\ b\\ c\\ d \end{pmatrix}$$ where each component is complex. We have that $$\psi^\dagger = \begin{pmatrix} a^* &b^* & c^* & d^* \end{pmatrix}$$ and $$\bar \psi = \psi^\dagger \gamma^0$$. Thus, to write the Dirac equation explicitly, first act on $$\psi$$ with the messy matrix we computed. Then you act on the resultant column with $$\gamma^0$$. Finally, you take the dot product with $$\psi^\dagger$$.