- #1
MelvinSmith
- 5
- 0
Hello,
I have a question concerning the current in the Dirac equation and its corresponding operator. One can construct a current density that is
[itex]\textbf{j}^{i} = \psi^{\dagger}\gamma^{i}\psi[/itex]
If I want to have the current, I will have to integrate:
[itex]I = \oint \textbf{j} \cdot \textbf{n} \, dA [/itex]
With Gauss' theorem this should be:
[itex]I = \int \text{div} \textbf{j} \, d^{3}x = \int \text{div} (\psi^{\dagger}\gamma \psi) \, d^{3}x [/itex]
Okay, but now I've found in the literature, that the current operator is [itex]\hat{\textbf{j}} = \hat{\gamma} [/itex].
But when I calculate an average value with this operator, this is
[itex]I = \left\langle \psi | \hat{\gamma} | \psi \right\rangle = \int \psi^{\dagger}\gamma \psi d^{3}x [/itex]
without this divergence.
What's right? And why is one point wrong?
Thank you all for comments,
Melvin
I have a question concerning the current in the Dirac equation and its corresponding operator. One can construct a current density that is
[itex]\textbf{j}^{i} = \psi^{\dagger}\gamma^{i}\psi[/itex]
If I want to have the current, I will have to integrate:
[itex]I = \oint \textbf{j} \cdot \textbf{n} \, dA [/itex]
With Gauss' theorem this should be:
[itex]I = \int \text{div} \textbf{j} \, d^{3}x = \int \text{div} (\psi^{\dagger}\gamma \psi) \, d^{3}x [/itex]
Okay, but now I've found in the literature, that the current operator is [itex]\hat{\textbf{j}} = \hat{\gamma} [/itex].
But when I calculate an average value with this operator, this is
[itex]I = \left\langle \psi | \hat{\gamma} | \psi \right\rangle = \int \psi^{\dagger}\gamma \psi d^{3}x [/itex]
without this divergence.
What's right? And why is one point wrong?
Thank you all for comments,
Melvin