# Derivative with respect to ${\not}{p}$

When studying renormalization of QED in standard textbooks, we typically encounter derivatives with respect to ${\not}{p}=p^\mu \gamma_\mu$, i.e., $\partial/\partial{\not}p$. As far as I understand, this is just a symbolic notation to perform derivative operations over functions of symbols like ${\not}{p}$ itself or $p^2={\not}{p}^{\,2}$, as if ${\not}{p}$ were just some variable which could be replace by $x$, for instance. Clearly, there is something more going on because ${\not}{p}$ is a matrix, so a derivative with respect to it would be something like a derivative with respect to a matrix, which I cannot understand. As far as it concerns the conventional QED, that works fine -- we only need to perform this derivative on ${\not}{p}$ or $p^2$--, but if we depart from it to consider different models, we may encounter operations like $\frac{\partial\,p^\mu}{\partial\,{\not}{p}}$.

I wonder if someone has ever dealt with that, and what was the solution found. I have been trying to keep this symbolic notation to find the answer, but I always get stucked at some point or I end up finding something inconsistent -- for instance, I would expect the result to be Lorentz covariant, but I cannot find a result with such property. Conversely, I am trying to give up this notation and trying to make real sense of the operation, but with no success so far.

Edit: The basic context is set in the discussion of renormalization of QED. You have the renormalized electron's propagator written as $\frac{i}{{\not}{p}-m-\Sigma(p)}$. When you take your quantities on shell, i.e., at ${\not}{p}=m$ (where both sides of this equality are understood to be acting on a spinor $u(p)$), you require that the pole of this propagator is located at $p^2=m^2$ and the residue is $i$. This amounts to the requirement of $\Sigma({\not}{p}=m)=0$ and $\frac{\partial\,\Sigma}{\partial\,{\not}{p}}|_{{\not}{p}=m}=0$, once you assumed the symbolic notation of the derivative with respect to ${\not}{p}$. For the conventional QED, $\Sigma=\Sigma({\not}{p})$ and this symbolic derivative can be taken without a problem. Modified models based on QED can lead to a $\Sigma(p)$ that also contains $p^\mu$, for instance, making that derivative operation unclear.

-
Where, precisely, have you seen this notation? It would be helpful to have more specific context to interpret it properly. –  joshphysics Dec 13 '13 at 20:05
See Eq. (17.116) of Hatfield's "Quantum field theory of point particles and strings", or Eq. (7.25) of Peskin & Schroeder's "An introduction to quantum field theory", or the comment before Eq. (62.38) of Srednicki's "Quantum field theory", for instance. –  andrehgomes Dec 13 '13 at 20:25
{\not}p yields ${\not}p$, which looks prettier than \not{p} rendered as $\not{p}$ –  Christoph Dec 13 '13 at 21:55
I believe @joshphysics is asking for reference of $\frac{\partial\,p^\mu}{\partial\,{\not}{p}}$, but the examples you gave(at least the ones from P&S and Srednicki) are of the form $\frac{\partial f({\not}{p})}{\partial\,{\not}{p}}$ –  Jia Yiyang Dec 14 '13 at 1:38
I see now, sorry. Yes, that's correct, all given textbook examples are of the form $\frac{\partial\,f({\not}{p})}{\partial\,{\not}{p}}$. The case for $\frac{\partial\,p^{\mu}}{\partial\,{\not}{p}}$ is something particular of a nonconventional QED model I am working on. The nonconventional part gives contributions to the electron's self-energy $\Sigma$ that are proportional also to $p^{\mu}$ (there are new degrees of freedom coupling to $p^\mu$ so the expression remains covariant). –  andrehgomes Dec 14 '13 at 13:47

Here is my tentative answer. First of all, this answer is all based on the conventional QED, where we know the electron's self-energy is like $\Sigma=\Sigma({\not}{p})$. My original question deals with nonconventional QED where $\Sigma$ may also have momentum dependence on $p^\mu$ (which could be contracted with a background field of some kind, for instance, therefore resulting in an observer scalar quantity). Hopefully, my arguments may be extended to that nonconventional case.

Basically, I cannot find a consistent way of doing $\frac{\partial p^\mu}{\partial {\not}{p}}$, and I suspect there isn't one at all because $\frac{\partial}{\partial {\not}{p}}$ may not be a well defined operation but instead just a practical symbolic notation. Therefore, my approach is to avoid using $\frac{\partial}{\partial {\not}{p}}$ and instead use $\frac{\partial}{\partial p^2}$, which seems to be the meaningful operation.

As far as I could track, in the context mentioned above, the only instance where we need $\frac{\partial}{\partial {\not}{p}}$ is when imposing $\frac{\partial \Sigma}{\partial {\not}{p}}\Big|_{{\not}{p}=m}=0$. This requirement shows up when we ask the propagator $\frac{i}{{\not}{p}-m-\Sigma}$ to have residue at $i$ when ${\not}{p}=m$ once we already have the condition $\Sigma({\not}p=m)=0$ that ensures it's pole is located at $p^2=m^2$. The important point here is that the specific value $i$ arises after a series of loose symbolic operations for taking the residue:

$Residue = \displaystyle \lim_{{\not}{p}\rightarrow m} ({\not}p-m) \frac{i}{{\not}{p}-m-\Sigma}$,

but in the limit we would get $\frac{0}{0}$ because $\Sigma({\not}p=m)=0$. Thus, we use l'Hôpital's rule (by taking $\frac{\partial}{\partial {\not}{p}}$ on numerator and denominator) and in the end we see that imposing $\frac{\partial \Sigma}{\partial {\not}{p}}\Big|_{{\not}{p}=m}=0$ indeed fix the residue at $i$. This is motivated by the wish of having the renormalized propagator to look like the free one (where there is no $\Sigma$).

On the other hand, we can do the same thing in a meaningful way writing the propagator as $i({\not}{p}-m-\Sigma)^{-1} = i\frac{{\not}{p}+m+\Sigma}{p^2-(m+\Sigma)^2}$ and look for the residue,

$Residue = \displaystyle \lim_{p^2\rightarrow m^2} (p^2-m^2) i\frac{{\not}{p}+m+\Sigma}{p^2-(m+\Sigma)^2}$,

and again we would find $\frac{0}{0}$ and then use l'Hôpital, but by means of the $\frac{\partial}{\partial p^2}$ derivative instead. We find that imposing $\frac{\partial \Sigma}{\partial p^2}\Big|_{{\not}{p}=m}=0$ fix the residue at $2m i$. The extra $2m$ factor is indeed the same we would find for the free case ($\Sigma=0$).

The value of the residue is different in both cases first because the dimensionality of the derivatives are different. Second, the latter approach have a pole at $p^2=m^2$ and this makes precise sense, and the former symbolic approach mention before has the pole at ${\not}{p}=m$ in a very loose sense.

The first point is: in the end, both conditions are expected to give precisely the same result. Both conditions are always taken on shell (${\not}{p}=m$), therefore, given the correct dimensionality factor $2m$, derivatives with respect to $p^2$ should be equivalent to ones with respect to ${\not}{p}$. (One way of understand that is by looking at the effect of both derivatives on $p^2={\not}{p}^{\,2}$ after imposing ${\not}{p}=m$.)

The second point is: because $\Sigma$ depends on ${\not}{p}$ it is much easier to work with $\frac{\partial}{\partial {\not}{p}}$ instead of $\frac{\partial}{\partial p^2}$, also because to use the latter we need to know $\frac{\partial p^\mu}{\partial p^2}$, which is not immediately obvious. Once we know the symbolic notation is equivalent to the meaningful one, using $\frac{\partial}{\partial {\not}{p}}$ is more practical for the conventional QED, because all we need to derive depends on ${\not}{p}$ or $p^2={\not}{p}^{\,2}$.

In some model based on nonconventional QED, we may need to know the derivative of $p^\mu$ with respect to either ${\not}{p}$ or $p^2$. The former I cannot figure out (and I suspect it may not make sense at all), but I expect that figuring out $\frac{\partial p^\mu}{\partial p^2}$ should be enough. For the latter task I invoke Lorentz covariance for the result and write $\frac{\partial p^\mu}{\partial p^2}=a(p^2) p^\mu$. Multiplying both sides by $p_\mu$ and further differentiating by $p^2$ leads to the conclusion that $a=\frac{1}{2p^2}$ and, therefore, $\frac{\partial p^\mu}{\partial p^2}=\frac{p^\mu}{2p^2}$.

In the conventional QED I can check the consistency of this result. There, we know that the structure of $\Sigma$ is $\Sigma({\not}{p})=A(p^2){\not}{p}-B(p^2)m$. If we use this explicity structure when finding the conditions for the residue of the propagator to be $2mi$, by the meaningful approach, we would find

$\left[\frac{1}{2}A+m^2\left(\frac{\partial A}{\partial p^2}-\frac{\partial B}{\partial p^2}\right) \right]_{{\not}{p}=m}=0$,

and this results does not require any derivative of $p^\mu$ with respect to $p^2$. On the other hand, we already know this result should be equivalent to $\frac{\partial \Sigma}{\partial p^2}\Big|_{{\not}{p}=m}=0$ which, using the explicit form of $\Sigma$ does require derivation of $p^\mu$ with respect to $p^2$. Using $\frac{\partial p^\mu}{\partial p^2}=\frac{p^\mu}{2p^2}$ indeed gives the expected result, indicating that the result $\frac{\partial p^\mu}{\partial p^2}=\frac{p^\mu}{2p^2}$ is consistent so far.

The next step would be to generalize the discussion to consider $\Sigma$ with a more general form than the mentioned above (including, for instance, factors with $\gamma_5$ and $\sigma^{\mu\nu}$ contracted with $p^\mu$ and suitable background fields). I am able to outline some points:

• The result $\frac{\partial p^\mu}{\partial p^2}=\frac{p^\mu}{2p^2}$ remains the same.

• The condition for the residue using the symbolic notation seemly remains the same, $\frac{\partial \Sigma}{\partial {\not}{p}}\Big|_{{\not}{p}=m}=0$, but it may not be useful as long as we don't know $\frac{\partial p^\mu}{\partial {\not}{p}}$.

• It seems to be harder to find the analogous condition by the meaningful approach with $\frac{\partial}{\partial p^2}$ because we need to invert a more complicated $\Sigma$ structure, but nevertheless we would know how to actually apply the resulting condition, because we know how to do the derivative.

The way out, and I may be streching too far, I suspect may be a correspondence like $\left[\frac{\partial}{\partial {\not}{p}}(\dots)\right]_{{\not}{p}=m} = 2m \left[\frac{\partial}{\partial p^2}(\dots)\right]_{{\not}{p}=m}$.

-