Probability Density and Current of Dirac Equation

In summary: Perhaps because those are solutions to the Dirac equation?In summary, the author is trying to determine the probability density and current of the Dirac equation by comparison to the general continuity equation. He has attempted to do this by multiplying one term by 'i' and the other by '-i', but he can only get it to work if he does this in the order of matrix transpositions. He is confused about the meaning of the hermitian conjugate of the Dirac equation and wonders if someone could help him understand it. He also suggests that the Klein Gordon solutions might be easier to understand because they are solutions to the Dirac equation.
  • #1
Sekonda
207
0
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
 
Physics news on Phys.org
  • #2
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
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
Have you considered that taking the hermitian conjugate is not only taking the complex conjugate but also the transposition?
 
  • #5
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
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
Sekonda said:
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. [itex] i \dfrac{d\Psi}{dt} = -i \alpha \cdot (\nabla \Psi) + \beta m \Psi[/itex]
  2. [itex] -i \dfrac{d\Psi^\dagger}{dt} = +i (\nabla \Psi^\dagger \cdot \alpha) + \Psi^\dagger \beta m[/itex]
Taking the conjugate reverses the order of matrices. So if you multiply the top equation on the left by [itex]-i \Psi^\dagger[/itex] and multiply the bottom equation on the right by [itex]+i \Psi[/itex] and add them, you get:
[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
Thanks Stevendaryl!

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

Thanks again,
SK
 
  • #9
stevendaryl said:
The complicated thing is correctly taking the Hermitian conjugate of the Dirac equation. What I think is true is this:

  1. [itex] i \dfrac{d\Psi}{dt} = -i \alpha \cdot (\nabla \Psi) + \beta m \Psi[/itex]
  2. [itex] -i \dfrac{d\Psi^\dagger}{dt} = +i (\nabla \Psi^\dagger \cdot \alpha) + \Psi^\dagger \beta m[/itex]
Taking the conjugate reverses the order of matrices. So if you multiply the top equation on the left by [itex]-i \Psi^\dagger[/itex] and multiply the bottom equation on the right by [itex]+i \Psi[/itex] and add them, you get:
[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.

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.
 

1. What is the Dirac equation?

The Dirac equation is a mathematical equation that describes the behavior of spin-1/2 particles, such as electrons, in quantum mechanics. It was developed by physicist Paul Dirac in 1928 as an extension of the Schrödinger equation to account for the relativistic effects of high-speed particles.

2. What is probability density in the context of the Dirac equation?

In the Dirac equation, probability density refers to the likelihood of finding a particle at a specific position in space at a given time. It is represented by the square of the wavefunction, which is a solution to the Dirac equation. The higher the probability density, the greater the chance of finding the particle in that location.

3. How is current related to the Dirac equation?

In the Dirac equation, current refers to the flow of probability density, or the movement of a particle. It is represented by the probability current density, which is a vector quantity that describes the direction and magnitude of the flow of probability density at a given point in space and time.

4. What is the significance of the probability density and current in the Dirac equation?

The probability density and current in the Dirac equation are important because they provide a way to describe the behavior of particles in quantum mechanics. They allow us to make predictions about the behavior of particles and understand their properties, such as spin and charge.

5. How is the Dirac equation used in practical applications?

The Dirac equation has many practical applications in fields such as particle physics, solid-state physics, and quantum computing. It is used to describe the behavior of particles in particle accelerators, semiconductors, and quantum computers. It also played a crucial role in the development of the Standard Model of particle physics, which describes the fundamental building blocks of matter and their interactions.

Similar threads

  • Quantum Physics
Replies
9
Views
787
Replies
9
Views
350
Replies
17
Views
1K
Replies
12
Views
2K
  • Quantum Physics
Replies
31
Views
2K
  • Quantum Physics
Replies
5
Views
329
Replies
1
Views
629
  • Quantum Physics
Replies
9
Views
1K
Replies
4
Views
1K
  • Quantum Physics
Replies
9
Views
981
Back
Top