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I am quite confused about the Hermitian conjugate of the 4-vector momentum $p=(p^0, p^1, p^2, p3)$. The confusion mainly arises when deriving the Dirac adjoint and the charge current probability.

(1).Conder the adjoint Dirac equation:

start from the Dirac equation:

$p_u \gamma^u \phi = mc \phi$

Taking the Hermitian conjugate of the Dirac equation:

$(p_u)^+ \phi^+ {\gamma^u}^+ = mc \phi^+$

note that ${\gamma^u}^+ = \gamma^0 \gamma^u \gamma^0$ and $\bar{\phi} = \phi^+ \gamma^0$.

So now If $(p^u)^+=p^0u$:

$p_u\bar{\phi} \gamma^u=mc\bar{\phi}$

So we can obtain the Dirac adjoint equation:

$\bar{\phi}(\overleftarrow{p_u}\gamma^u-mc)=0$

while If ${p^u}^+ = -p^u$, we cannot get such the Dirac adjoint equation above.

(2). Cosider the charge current probability:

Still start from the Dirac equation:

$p_u {\gamma^u} {\phi}=mc{\phi}$

multiply on the left side by ${\bar{\phi}}$:

${\bar{\phi}}p_u {\gamma^u} {\phi}={\bar{\phi}}mc{\phi}$ -----eq(1)

Now again, taking the Hermitian conjugate of the Dirac equation:

${p_u}^+ {\phi}^+ {\gamma^u}^+ =mc{\phi}^+$ -->${p_u}^+ \bar{\phi} {\gamma^u} =mc\bar{\phi}$

Then multiplying on the right side by $\phi$

$({p_u}^+ \bar{\phi} {\gamma^u}){\phi} =mc\bar{\phi}{\phi}----eq(2)$

It seems that in order to arrive at the correct charge current probability $j_u=c\bar{\phi} {\gamma^u}{\phi}$, here we have to use ${p^u}^+=-{p^u}$ in eq(2) such that we can subtract eq(1) by eq(2) to cancel $mc{\bar{\phi}}{\phi}$ on the right-hand side and therefore reach the correct result.

Based on the derivation above, it seems there is some contradiction about ${p^u}^+$, which must be taken different values when deriving the Dirac adjoint equation and the charge current probability, i.e. ${p^u}^+=p^u$ for the Dirac adjoint equation; ${p^u}^+=-p^u$ for the charge current probability.

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  • $\begingroup$ @bistoal, try to understand the difference between Dirac adjoint and Hermitian adjoint. That will eventually help you to get the right road. $\endgroup$ – AMS Nov 19 '16 at 12:43
  • $\begingroup$ @AMS I know they are different. $\endgroup$ – bitsoal Nov 20 '16 at 1:47

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