# Non relativistic limit for dirac propagator

1. Aug 13, 2009

### Sleuth

Hi everybody,
I don't know if this is the right section to ask for such a question but I have been dealing with this problem for a while and there's something I still cannot grasp...
Let us suppose that we have a dirac free particle with propagator (i'm sorry but i'm not able to obtain the feynman slash notation in TeX)

suppose $$p = (p_0, P)$$

$${i \over p/ - m} = i {p_0 \gamma_0 - P \cdot \gamma + m \over p_0^2 - P^2 -m^2}$$

My question is how can I perform a non relativistic expansion and recover the non relativistic propagator

$${1 \over E - {P^2 \over 2 m}}$$

with E as the kinetic energy?
I have a couple of ideas quite long to write down, but I cannot justify them completely so I'd accept some hint gladly :)

2. Aug 15, 2009

### Sleuth

I'm sorry if I am again over this question but would really appreciate some help.
Something I noticed is the putting
$$p_0 = E + M$$

$${p_0 \gamma_0 - P \cdot \gamma + M \over p_0^2 - P^2 -M^2 } &= {(1 + \gamma_0) M - P\cdot \gamma + E \gamma_0 \over E^2 + 2 E M - P^2} \\$$

$$&= { \left[ (1 + \gamma_0) M - P\cdot \gamma + E \gamma_0 \right] \over \left( E - (\sqrt{P^2 + M^2} - M) \right) \left( E + M + \sqrt{P^2 + M^2} \right) }$$

$$&= \left[ (1 + \gamma_0) M - P\cdot \gamma + E \gamma_0 \right] \\ \; {1 \over 2(E+M)}\left[ {1 \over E - (\sqrt{P^2 + M^2} - M) } + {1 \over E + M + \sqrt{P^2 + M^2} } \right]$$

Now the first one is obviously something like a propagator for a relativistic particle with the usual kinetic energy (a particle, say electron) while the other should be something entailing the presence in the theory of the antiparticle (say positron) with a negative kinetic energy. But now how do I separate the two contributes? can I simply delete the second propagator and go on with the first one? Would I be missing something?

notice that by projecting this propagator over the positive energy components with the use of the projector
$${1 + \gamma_0 \over 2}$$
on the left and on the right, the $$P \cdot \gamma$$ part is cancelled out and we are left with something really similar to the non relativistic propagator provided we expand the denominator in $${P^2 \over M^2}$$

3. Aug 15, 2009

### clem

I think you can use $${\slash p}+m=2m$$ in the numerator.
Factor the denominator into $$(E-\sqrt{p^2+m^2})(E+\sqrt{p^2+m^2})$$.
This has the NR limit 2m(p^2/2m).

4. Aug 16, 2009

### nrqed

You are on the right track.

Note that if you do the integral over $$p_o$$, you pick up two poles. The antiparticle pole is suppressed relative to the particle pole. So keep only the particle pole.

5. Aug 17, 2009

### Sleuth

Hi there! since your nickname, nrqed, I guess you're the right person to refer to for this kind of questions.
I'm just studying effective field theory, but I'm really not completely ok with the way they use to obtain the nrqed hamiltonian by studying the scattering amplitude.
I was looking for a way to obtain first of all the foldy-wouthuysen hamiltonian matching the dirac propagator

$${1 \over p/ - m} + {1 \over p/ - m} q A/ {1 \over p/ -m} \cdots$$

with a non realtivistic expansion
$${1 \over E - P^2/2M} + {1 \over E - P^2/2M} V {1 \over E - P^2/2M} + \cdots$$

with the potential V as the foldy wouthuysen potential.
By considering only the non interacting Dirac propagator, only the positive enrgy pole, I'm able to obtain the kinetic energy terms by putting E=0, but I have some problems with the fine structure terms, in particular I'm not able to obtain the complete darwin term, infact I obtain only the spin-orbit, the pure coulomb term and only a part of the darwin term which is
$${P \cdot Q \over 4 M^2}$$
so that a -P^2/8M^2 and -Q^2/8M^2 are missing.
I thought they could come by the negative energy pole but I'm not able to prove it.

6. Aug 18, 2009

### tiny-tim

p-slash

Hi!

Just thought I'd pop my head round the door and say it's "\not{p}": $\not{p}$

7. Aug 22, 2009

### nrqed

Hey! Sorry for the delay in replying, I am swamped these days.
It's been a long time since I have looked at all this. I will check it out.
A quick comment: the -P^2/8m^2 simply comes from the expansion of the $$\sqrt{m^2 + p^2}$$. I will try to get back to your question in the next few days.

Regards

8. Aug 23, 2009

### Sleuth

Re: p-slash

Hi! thank you :)
there schould be a way to move any character back or forward of a given amount of points but I've not been able to find it...

9. Aug 24, 2009

### tiny-tim

negative space

Hi Sleuth!

robphy found this …

\hspace{} with a positive distance inside will move the next character forward that distance, and with a negative distance inside will move it backwards …

eg $$\bigcap\hspace{-2.1ex}i$$

(a forum tag-search for "slash" would have got you this …)

10. Aug 24, 2009

### Sleuth

Yeah! it was just what I was looking for. My advisor, for my bachelor, more than one year ago told me about this feature but I was just learning teX in that period and I forgot it few hours later :P
thank you again :)
Sleuth