Non relativistic limit for dirac propagator

[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 :)

 

[Sleuth]

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

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

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

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

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?

[edit]
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 canceled out and we are left with something really similar to the non relativistic propagator provided we expand the denominator in $${P^2 \over M^2}$$

 

[Meir Achuz]

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).

 

[nrqed]

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

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

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

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

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?

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.

 

[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 energy 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.

 

[tiny-tim]

p-slash

i'm sorry but I'm not able to obtain the feynman slash notation in TeX

Hi!

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

 

[nrqed]

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 energy 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.

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

 

[Sleuth]

Hi!

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

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...

 

[tiny-tim]

negative space

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...

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 …)

oh, robphy, that's great! …

you've invented negative space!
​

you've inserted a negative space "after" the d (using \hspace{-0.8ex}), and put a horizontal bar in it.

(btw, you can leave out "\newcommand" at the start, and "\dslash " at the end: $${d \hspace{-0.8ex}\rule[1.2ex]{0.8ex}{.1ex}}$$ )

Does CERN know about this? ​

 

[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