Second quantization particle current

[daudaudaudau]

Hi. I'm reading an article which writes the following

"... and the well-known expression for the charge current is"
<br /> j=-\frac{ie}{m}\int dr\psi^\dagger (r)[\nabla-ieA(r)]\psi(r)<br />

Why does it have an integral sign? And when you define it this way, you integrate out the r-dependence, don't you?

 

[tom.stoer]

Hi. I'm reading an article which writes the following

"... and the well-known expression for the charge current is"
<br /> j=-\frac{ie}{m}\int dr\psi^\dagger (r)[\nabla-ieA(r)]\psi(r)<br />

Why does it have an integral sign? And when you define it this way, you integrate out the r-dependence, don't you?

Strange.

I guess it's not the el.-mag current from QED which does not contain A but carries a four-vector index. Instead it seems to be the probability current for the ordinary Schrödinger equation (with a vector potential) for a particle with mass m. But even in that case there is no integral

 

[daudaudaudau]

I've attached a picture from the particular part of the article (Thermal transport for many-body tight-binding models by Vanderbilt and King-Smith).

 

[tom.stoer]

OK

The author decided to call the integral a current (referring to the integrand as the current density).

 

[daudaudaudau]

But still he is taking a volume integral of a charge density operator. What is this?

 

[Demystifier]

But still he is taking a volume integral of a charge density operator. What is this?

For any extensive physical quantity Q, you can say that it is equal to the volume integral of the DENSITY of Q.

 

[daudaudaudau]

But when would you ever take the volume integral of the current density? The current density is a vector, and it has units A/m^2. What you would normally do is integrate it over a surface to get the current through the surface.

 

[DrDu]

Probably he had something like
<br /> j(r&#039;)=-\frac{ie}{2m}\int dr\psi^\dagger (r) \{\nabla_{r&#039;}-ieA(r&#039;),\delta(r-r&#039;)\}\psi(r)<br />
in mind.

 

[Demystifier]

But when would you ever take the volume integral of the current density? The current density is a vector, and it has units A/m^2. What you would normally do is integrate it over a surface to get the current through the surface.

Assume that you have a fluid, each part of which has some momentum density. How would you calculate the momentum of the fluid as a whole?

 

[daudaudaudau]

Assume that you have a fluid, each part of which has some momentum density. How would you calculate the momentum of the fluid as a whole?

Yes that sounds sensible, I have just never heard of currents defined in this way before. What should I google to find out more about this?

 

[Demystifier]

Yes that sounds sensible, I have just never heard of currents defined in this way before. What should I google to find out more about this?

I think you are right that such global vector quantities are usually not CALLED currents, so googling probably would not help.

In fact, in curved space the integral of a vector field is not even well defined. In other words, in curved space you can associate a vector to a point, but you cannot associate a vector to a whole region of space.

 

[Demystifier]

What you would normally do is integrate it over a surface to get the current through the surface.

Actually, normally you would obtain a FLUX, which is a scalar, not a vector. See also my post above.