# Probability Density and Current of Dirac Equation

by Sekonda
Tags: current, density, dirac, equation, probability
 P: 209 Hey, I'm trying to determine the probability density and current of the Dirac equation by comparison to the general continuity equation. The form of the Dirac equation I have is $$i\frac{\partial \psi}{\partial t}=(-i\underline{\alpha}\cdot\underline{\nabla}+\beta m)\psi$$ According to my notes I am supposed to determine the following sum to make the relevant comparisons to the continuity equation and therefore determine the probability density/current $$\psi(Dirac)^{\dagger}+\psi^{\dagger}(Dirac)$$ Where 'Dirac' refers to the above equation. However I have tried this and I can only get it to work if I multiply one term by 'i' and the other by '-i' in the above. $$\psi(i\frac{\partial \psi}{\partial t})^{\dagger}+\psi^{\dagger}(i\frac{\partial \psi}{\partial t})=-i\psi\frac{\partial \psi^{*}}{\partial t}+i\psi^{*}\frac{\partial \psi}{\partial t}\neq i\frac{\partial (\psi^{*}\psi)}{\partial t}$$ Any help is appreciated! Thanks, SK
 Sci Advisor HW Helper P: 11,915 Your notes should use the gamma matrices. It's the modern treatment on the Dirac equation/field and make the special relativity invariance easier to see.
 P: 209 Perhaps they should but I'm reckoning these are introduced later, so considering we don't 'know' these yet this appears to be the simplest way of demonstrating the probability density/current of the Dirac equation. I'm confused though, I perhaps maybe taking the adjoint of the Dirac equation incorrectly.
 P: 123 Probability Density and Current of Dirac Equation Have you considered that taking the hermitian conjugate is not only taking the complex conjugate but also the transposition?
 P: 209 I think so, taking the adjoint of ψ doesn't bring out a minus sign does it? With regards to the RHS of the Dirac equation I think β is diagonal and so the transposition doesn't affect it, though I'm a bit confused as how I'd go about doing the hermitian conjugate on the dot product of the alpha matrix with the ∇... I wouldn't be surprised though if this transposition is the issue, I'll keep looking at it!
 P: 123 Also, $\psi^{\dagger}\psi$ is a number, while $\psi\psi^{\dagger}$ is a matrix, so I don't really quite get the whole thing.
P: 2,078
 Quote by Sekonda I think so, taking the adjoint of ψ doesn't bring out a minus sign does it? With regards to the RHS of the Dirac equation I think β is diagonal and so the transposition doesn't affect it, though I'm a bit confused as how I'd go about doing the hermitian conjugate on the dot product of the alpha matrix with the ∇... I wouldn't be surprised though if this transposition is the issue, I'll keep looking at it!
The complicated thing is correctly taking the Hermitian conjugate of the Dirac equation. What I think is true is this:
1. $i \dfrac{d\Psi}{dt} = -i \alpha \cdot (\nabla \Psi) + \beta m \Psi$
2. $-i \dfrac{d\Psi^\dagger}{dt} = +i (\nabla \Psi^\dagger \cdot \alpha) + \Psi^\dagger \beta m$
Taking the conjugate reverses the order of matrices. So if you multiply the top equation on the left by $-i \Psi^\dagger$ and multiply the bottom equation on the right by $+i \Psi$ and add them, you get:
$\Psi^\dagger \dfrac{d\Psi}{dt} + \dfrac{d\Psi^\dagger}{dt}\Psi= - \Psi^\dagger \alpha \cdot (\nabla \Psi) - (\nabla \Psi^\dagger) \cdot \alpha \Psi$

(the terms involving $\beta$ cancel). You can rewrite this as (I think):

$\dfrac{d}{dt} (\Psi^\dagger \Psi) = - \nabla \cdot (\Psi^\dagger \alpha \Psi)$

This can be rearranged as a continuity equation for probability.

There's a different continuity equation for electric charge, but I've forgotten what that is.
 P: 209 Thanks Stevendaryl! This is exactly what I wanted, I can see what was incorrect now - brilliant! Thanks again, SK
 Quote by stevendaryl The complicated thing is correctly taking the Hermitian conjugate of the Dirac equation. What I think is true is this: $i \dfrac{d\Psi}{dt} = -i \alpha \cdot (\nabla \Psi) + \beta m \Psi$ $-i \dfrac{d\Psi^\dagger}{dt} = +i (\nabla \Psi^\dagger \cdot \alpha) + \Psi^\dagger \beta m$ Taking the conjugate reverses the order of matrices. So if you multiply the top equation on the left by $-i \Psi^\dagger$ and multiply the bottom equation on the right by $+i \Psi$ and add them, you get: $\Psi^\dagger \dfrac{d\Psi}{dt} + \dfrac{d\Psi^\dagger}{dt}\Psi= - \Psi^\dagger \alpha \cdot (\nabla \Psi) - (\nabla \Psi^\dagger) \cdot \alpha \Psi$ (the terms involving $\beta$ cancel). You can rewrite this as (I think): $\dfrac{d}{dt} (\Psi^\dagger \Psi) = - \nabla \cdot (\Psi^\dagger \alpha \Psi)$ This can be rearranged as a continuity equation for probability. There's a different continuity equation for electric charge, but I've forgotten what that is.