2
$\begingroup$

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 $f:\mathbb{R}^4\to\mathbb{C}$:

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

$\endgroup$
8
  • $\begingroup$ Paragraph 4 is wrong. You’ve lost the index on the partial derivative. I didn’t check the rest. $\endgroup$
    – G. Smith
    Jun 8 '20 at 17:06
  • 2
    $\begingroup$ What is your motivation behind this expansion? $\endgroup$ Jun 8 '20 at 17:39
  • $\begingroup$ @JohnnyLongsom Same reason taking a TV apart and putting it back together makes it easy to understand how it works. $\endgroup$
    – Anon21
    Jun 8 '20 at 17:43
  • $\begingroup$ @G.Smith What does index means on the partial derivative (after I distributed the sum) ? $\endgroup$
    – Anon21
    Jun 8 '20 at 17:47
  • 1
    $\begingroup$ 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 $z$. $\endgroup$
    – G. Smith
    Jun 8 '20 at 18:56
3
$\begingroup$

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 $\psi^\dagger$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.