Probability Density and Current of Dirac Equation

[Sekonda]

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

 

[dextercioby]

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.

 

[Sekonda]

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.

 

[kevinferreira]

Have you considered that taking the hermitian conjugate is not only taking the complex conjugate but also the transposition?

 

[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!

 

[kevinferreira]

Also, $\psi^{\dagger}\psi$ is a number, while $\psi\psi^{\dagger}$ is a matrix, so I don't really quite get the whole thing.

 

[stevendaryl]

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.

 

[Sekonda]

Thanks Stevendaryl!

This is exactly what I wanted, I can see what was incorrect now - brilliant!

Thanks again,
SK

 

[stevendaryl]

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.

My last comment was stupid. If you multiply probability current by the electron charge, you get the charge current. It's that simple. For some reason, though, that's not the case with the solutions of the Klein Gordon equation.