# Complex momentum density

1. Feb 7, 2009

### blue2script

Hello!

Just a short question: I am currently calculating the momentum density of a free dirac wavefunction under boost. However, I get some complex density - can this happen or will it be a mistake on my side?

Blue2script

PS: In some forum there was a note that momentum density is derived from the real Lagrangian. What does that mean? I just took the momentum density given from the energy-momentum tensor of the normal Dirac Lagrangian.

2. Feb 7, 2009

### xepma

Not completely sure what the context is, but I think you want to get some distributionfunction of the momenta p, i.e. $\Psi(p)$. The function itself can be complex, since when you calculate what the chance is of some outcome you always use the absolute squared function, $|\Psi(p)|^2$. The argument p is, ofcoure, real-valued.

So yes, it is possible.

3. Feb 7, 2009

### blue2script

Thank you xepma for your answer! I guess I was just dump. I wanted to calculate the momentum density of a wavefunction that solves the dirac equation (in my context meaning $$\Psi\left(x\right)$$). From the free Dirac equation you can calculate the energy-momentum density $$T^{\mu\nu}$$. Integrating $$T^{00}$$ over the whole space gives you the energy E of the wavefunction, integrating the 0i components gives you the momentum, and so on. However, sure the momentum density can be complex, but the integral over all the space has to be real again. That's just what happens to me if I get my $$i$$'s right. So I was just blind and dump.

Anyway, thanks again!
With very best regards,
Blue2script

4. Feb 7, 2009

### blue2script

Hmmm... now I ran into another problem: how would you define the momentum density of a non-normalizable wavefunction? Consider for example the plane wave solutions of the free Dirac equation, $$\psi\left(x\right) = u\left(p\right) e^{ipx}$$ (in one dimension). They have momentum p and energy $$E = \sqrt{p^2 + m^2}$$. However, these modes are not normalizable since the density of these modes is just constant in space. How can one then define a suitable momentum density whose integral over the space yields $$p$$?

Blue2script

5. Feb 7, 2009

### blue2script

Ok, forget it again, think I got it. It is just the momentum density divided by the fermion density. Anyway, thanks for reading!

6. Feb 7, 2009

### RedX

Hi. You can show that the energy tensor should always be real. You just have to remember that $$\overline{u(p)}u(p)=a real number$$, and that $$\gamma_{\mu}p^{\mu}u(p)=\pm mu(p)$$. When you linearly expand the field in fourier-series, bilinears where p and p' different are sumed in conjugates, and where p and p' are the same are just real numbers.

Last edited: Feb 7, 2009