# Meaning of Peskin equation (7.25)

1. Dec 28, 2011

### kof9595995

Equation (7.25)
$$(\displaystyle{\not}p - m)(1 - {\left. {\frac{{d\Sigma }}{{d\displaystyle{\not}p}}} \right|_{\displaystyle{\not}p = m}}) + O({(\displaystyle{\not}p - m)^2})$$
Formally it looks like a Taylor expansion of $\displaystyle{\not}p-m_{0}-\Sigma(\displaystyle{\not}p)$. However it involves a differentiation of a matrix, and what's worse is, he lets $\displaystyle{\not}p=m$, which is impossible because $\displaystyle{\not}p$ is always off-diagonal(peskin uses weyl representaion), while m is diagonal.
The best I can make of this $\displaystyle{\not}p=m$ is that this is just a formal replacement, but then Taylor expansion loses its meaning of "polynomial approximation around the neighbourhood of a point", since $\displaystyle{\not}p$ can never really approach m.

2. Dec 30, 2011

### strangerep

Isn't it just a matter of writing something like $p\!\!\!\!/ - \xi m$, where $\xi$ is some gamma matrix, and then re-expressing the formulas as a limit as $\xi\to 1$? I.e., take the limit at the end of the mass-shift calculation?

IOW, find a convenient way to perform the limit as $p^2\to m^2$, which doesn't have gamma matrix headaches.

3. Dec 30, 2011

### kof9595995

Sorry I can't follow you. What's the meaning of $\xi m$, and how can you let $\xi\to 1$ if $\xi$ is always off diagonal(as you said $\xi$ is one of the gamma matrices)

But I think I am having "gamma matrix headaches" with Peskin's way of doing it.

4. Dec 30, 2011

### kof9595995

Ok, I still don't understand strangerep's argument, but now I find a way, though involved, of understanding this. The idea is to prove that Peskin's formal manipulation actually gives the same result with an ordinary derivation:
The goal is to derive the residue at $p^2=m^2$. Let's start with
$$\frac{1}{\displaystyle{\not}p-m_{0}-\Sigma(\displaystyle{\not}p)}......(*)$$
It's not hard to convince oneself
$$\Sigma\equiv f(p^2) \displaystyle{\not} p -g(p^2)m_0......(1)$$
Thus the pole must be at the solution of
$$m-m_0- [f(m^2) \displaystyle{\not} p -g(m^2)m_0]=0$$
Tidy up a bit:
$$[1- f(m^2)]m-[1-g(m^2)]m_0=0............(2)$$

Now let's see what peskin's formal manipulation gives us:
Eqn (7.26) in the book:
$$Z^{-1}_2=1-\left.\frac{d\Sigma}{d\displaystyle{\not}p}\right|_{\displaystyle{\not}p = m}......(7.26)$$
Sub (1) into (7.26) we get
$$Z^{-1}_2=1-[2p^2f'(p^2)+f(p^2)-2m_0\displaystyle{\not}pg'(p^2)]_{\displaystyle{\not}p=m}=1-f(m^2)-2m^2f'(m^2)+2m_0mg'(m^2)......(3)$$
Notice that a rule is to take $p^2=\displaystyle{\not}p\displaystyle{\not}p$

Now let's see what's the ordinary way of getting the residue:
when $p^2\to m^2$, the propagator behaves like
$$\frac{Z_2(\displaystyle{\not}p+m)}{p^2- m^2}......(4)$$
So the natural way of solving the problem is to reduce (*) to the form of (4). Sub (1) into (*)
$$\frac{1}{\displaystyle{\not}p-m_{0}-\Sigma(\displaystyle{\not}p)}=\frac{1}{[1-f(p^2)]\displaystyle{\not}p-[1-g(p^2)]m_{0}}=\frac{[1-f(p^2)]\displaystyle{\not}p+[1-g(p^2)]m_{0}}{[1-f(p^2)]^2p^2-[1-g(p^2)]^2m^2_{0}}.......(5)$$
When $p^2\to m^2$, the nominator of (5) will be $$[1-f(m^2)](\displaystyle{\not}p+m)......(6)$$, where (2) is used.
The denominator of (5) will of course tend to 0, to get the first order dependence on $p^2- m^2$ we calculate
$$\mathop {\lim }\limits_{{p^2} \to {m^2}} \frac{[1-f(p^2)]^2p^2-[1-g(p^2)]^2m^2_{0}}{p^2- m^2}=2m^2[f(m^2)-1]f'(m^2)+[1-f(m^2)]^2-2m^2_0[g(m^2)-1]g'(m^2) .......(7)$$
by L'hospital's rule.
Combine (6) and (7) we conclude
$$Z_2=\frac{1-f(m^2)}{2m^2[f(m^2)-1]f'(m^2)+[1-f(m^2)]^2-2m^2_0[g(m^2)-1]g'(m^2)}=\frac{1}{1-f(m^2)-2m^2f'(m^2)+2m_0mg'(m^2)}......(8)$$
where (2) is used again.

And finally we can see (8) is exactly the same with (3). This is a bit tedious and I'm suspecting I've made a big detour to get the conclusion.

5. Dec 30, 2011

### strangerep

Neither do I. :-(

More later...

6. Dec 30, 2011

### strangerep

Let's avoid talking about poles for a moment, and just focus on the Dirac equation:
$$\def\Sp{p\!\!\!\!/} (i\Sp - m)\psi ~=~ 0 ~=~ (i\Sp - m_0 - \delta m)\psi ~.$$
The "m" in such an equation is identified with the physical electron mass. (The business about poles only occurs when we take the formal inverse of the operator on the LHS, i.e., a Green's fn, a.k.a. propagator).

Back to P&S...

They find a propagator [eq(7.23)], i.e.,
$$\frac{i}{\Sp -m_0 - \Sigma(p)}$$
which (in my interpretation) means that
$$\Big(i\Sp - m_0 - \Sigma(p) \Big) \psi ~=~ 0$$
Since $\Sigma(p)$ is a scalar, it can be written in the form
$$A \Sp + B$$
where A and B are functions of $m_0$ and $p^2$.
Substituting, we get
$$0 ~=~ \Big(i\Sp - m_0 - i\Sigma(p) \Big) \psi ~=~ (1-A) \Big( \Sp - \frac{m_0+B}{1-A} \Big) \psi$$
implying (for $A\ne 1$)
$$\Big( \Sp - \frac{m_0+B}{1-A} \Big) \psi ~=~ 0$$
so we interpret the physical mass to be
$$m ~=~ \frac{m_0 + B}{1-A} ~~~\Rightarrow ~ \delta m = m - m_0 = \frac{A m_0 + B}{1-A}$$
Now, in the approximation $\Sigma \approx \Sigma_2$, we see that both A and B are proportional to $\alpha$. [See eq(7.19).] Hence the denominator doesn't contribute if we're only evaluating $\delta m$ to $O(\alpha)$. Therefore,
$$\delta m ~\approx~ Am_0 + B$$
which just happens to be $\Sigma(p)$ evaluated by substituting $\Sp\to m_0$.

[Probably, I've screwed up some factors of $i$ in the above.]

I need to think a bit more about how/whether this still works for higher orders in $\alpha$. It's probably ok because $\Sp$ commutes with everything else in these calculations, and can thus be taken as simply another commutative variables. (There's a whole theory about the calculus of commutative B*-algebras that generalizes ordinary complex analysis, Cauchy integrals, and all that. It probably applies here.)
Maybe I'll say a bit more later.
[Edit: make that "tomorrow". Bedtime now.]

Last edited: Dec 31, 2011