Probability Density and Current of Dirac Equationby Sekonda Tags: current, density, dirac, equation, probability 

#1
Feb413, 05:48 AM

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 [tex]i\frac{\partial \psi}{\partial t}=(i\underline{\alpha}\cdot\underline{\nabla}+\beta m)\psi[/tex] 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 [tex]\psi(Dirac)^{\dagger}+\psi^{\dagger}(Dirac)[/tex] 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. [tex]\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}[/tex] Any help is appreciated! Thanks, SK 



#2
Feb413, 11:55 AM

Sci Advisor
HW Helper
P: 11,866

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.




#3
Feb413, 03:04 PM

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.




#4
Feb413, 03:12 PM

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?




#5
Feb413, 05:43 PM

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! 



#6
Feb413, 06:00 PM

P: 123

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




#7
Feb413, 10:37 PM

P: 1,657

[itex]\Psi^\dagger \dfrac{d\Psi}{dt} + \dfrac{d\Psi^\dagger}{dt}\Psi=  \Psi^\dagger \alpha \cdot (\nabla \Psi)  (\nabla \Psi^\dagger) \cdot \alpha \Psi[/itex] (the terms involving [itex]\beta[/itex] cancel). You can rewrite this as (I think): [itex]\dfrac{d}{dt} (\Psi^\dagger \Psi) =  \nabla \cdot (\Psi^\dagger \alpha \Psi)[/itex] 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. 



#8
Feb513, 10:18 AM

P: 209

Thanks Stevendaryl!
This is exactly what I wanted, I can see what was incorrect now  brilliant! Thanks again, SK 



#9
Feb513, 12:09 PM

P: 1,657




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