High Energy Possible typo in Peskin & Schroeder's QFT Textbook (p. 666)?

[murillo137]

Hi everyone!

I'm going through Peskin & Schroeder's Chapter 19 (Perturbation Theory Anomalies) and it seems to be that equation 19.74 in page 666 has a minus sign missing on the RHS. Namely, I think the correct equation should read
\begin{align}
(i\not\!\! D)^2 = -D^2 - \frac{e}{2}\sigma^{\mu\nu}F_{\mu\nu}
\end{align}
This depends on the convention for the covariant derivative. For Chapter 19, the convention seems to be $D_{\mu} = \partial_{\mu} + ieA_{\mu}$, with the plus sign, as established in the first line of p. 652, at least for the 2D case which is discussed there. It's also the convention for Chapter 4, when QED is introduced. I don't seem to find any point in Chapter 19 where they switch to a different convention.
We have then:
\begin{align}
(i\not\!\! D)^2 &= - \gamma^{\mu}\gamma^{\nu}(D_{\mu}D_{\nu}) \\
(i\not\!\! D)^2 &= - \frac{1}{2}\{\gamma^{\mu},\gamma^{\nu}\}(D_{\mu}D_{\nu}) - \frac{1}{2}[\gamma^{\mu},\gamma^{\nu}](D_{\mu}D_{\nu})\\
(i\not\!\! D)^2 &= -\frac{1}{2}(2g^{\mu\nu})(D_{\mu}D_{\nu}) - \frac{1}{4}[\gamma^{\mu},\gamma^{\nu}][D_{\mu},D_{\nu}]\\
(i\not\!\! D)^2 &= -D^2 + \frac{i}{2}\sigma^{\mu\nu}[D_{\mu},D_{\nu}],
\end{align}
where $\sigma^{\mu\nu} = \frac{i}{2}[\gamma^{\mu},\gamma^{\nu}]$. Then, we have
\begin{align}
[D_{\mu},D_{\nu}] &= [\partial_{\mu} + ieA_{\mu}, \partial_{\nu} + ieA_{\nu}] = [\partial_{\mu}, ieA_{\nu}] - [\partial_{\nu}, ieA_{\mu}]\\
[D_{\mu},D_{\nu}] &= ie(\partial_{\mu}A_{\nu} - \partial_{\nu}A_{\mu}) = ieF_{\mu\nu},
\end{align}
such that $(i\not\!\! D)^2 = -D^2 - \frac{e}{2}\sigma^{\mu\nu}F_{\mu\nu}$.

I looked it up in the errata for the textbook, and there is no mention of this anywhere. Can someone confirm this? Or is there some issue in my derivation?

 

[billtodd]

I think D slash should be without the $\imath=sqrt(-1)$ inside the parantheses.
And then the problem with minus gets cancelled.

I wonder how come I didn't encountered this problem...

since I finished reading the book a few years ago and also read the solution manual, really hard to grasp QFT.

 

[renormalize]

I think D slash should be without the $\imath=sqrt(-1)$ inside the parantheses.
And then the problem with minus gets cancelled.

Can you explain how that will change the relative sign between the terms $D^2$ and $\frac{e}{2}\sigma^{\mu\nu}F_{\mu\nu}$?

 

[billtodd]

He has $\slash{D}^2=D^2+...$ where .... is the the term with F and without the minus sign.
Or so I think, does it make sense, nothing in QFT made sense....

 

[renormalize]

He has $\slash{D}^2=D^2+...$ where .... is the the term with F and without the minus sign.
Or so I think, does it make sense, nothing in QFT made sense....

Here is Peskin & Schroder pg. 666 eq.(19.74):$$(i\not\!\! D)^2 = -D^2 + \frac{e}{2}\sigma^{\mu\nu}F_{\mu\nu}$$and this is what @murillo137 derives in post #1:$$(i\not\!\! D)^2 = -D^2 - \frac{e}{2}\sigma^{\mu\nu}F_{\mu\nu}$$How does your comment help reconcile the difference between these equations?

 

[billtodd]

Here is Peskin & Schroder pg. 666 eq.(19.74):$$(i\not\!\! D)^2 = -D^2 + \frac{e}{2}\sigma^{\mu\nu}F_{\mu\nu}$$and this is what @murillo137 derives in post #1:$$(i\not\!\! D)^2 = -D^2 - \frac{e}{2}\sigma^{\mu\nu}F_{\mu\nu}$$How does your comment help reconcile the difference between these equations?

Which edition of the book does he use?
The latest one.
I don't have the book at my disposal, tomorrow I will look at it.

Indeed at least one someone did a mistake. (besides me).

 

[billtodd]

Hi everyone!

I'm going through Peskin & Schroeder's Chapter 19 (Perturbation Theory Anomalies) and it seems to be that equation 19.74 in page 666 has a minus sign missing on the RHS. Namely, I think the correct equation should read
\begin{align}
(i\not\!\! D)^2 = -D^2 - \frac{e}{2}\sigma^{\mu\nu}F_{\mu\nu}
\end{align}
This depends on the convention for the covariant derivative. For Chapter 19, the convention seems to be $D_{\mu} = \partial_{\mu} + ieA_{\mu}$, with the plus sign, as established in the first line of p. 652, at least for the 2D case which is discussed there. It's also the convention for Chapter 4, when QED is introduced. I don't seem to find any point in Chapter 19 where they switch to a different convention.
We have then:
\begin{align}
(i\not\!\! D)^2 &= - \gamma^{\mu}\gamma^{\nu}(D_{\mu}D_{\nu}) \\
(i\not\!\! D)^2 &= - \frac{1}{2}\{\gamma^{\mu},\gamma^{\nu}\}(D_{\mu}D_{\nu}) - \frac{1}{2}[\gamma^{\mu},\gamma^{\nu}](D_{\mu}D_{\nu})\\
(i\not\!\! D)^2 &= -\frac{1}{2}(2g^{\mu\nu})(D_{\mu}D_{\nu}) - \frac{1}{4}[\gamma^{\mu},\gamma^{\nu}][D_{\mu},D_{\nu}]\\
(i\not\!\! D)^2 &= -D^2 + \frac{i}{2}\sigma^{\mu\nu}[D_{\mu},D_{\nu}],
\end{align}
where $\sigma^{\mu\nu} = \frac{i}{2}[\gamma^{\mu},\gamma^{\nu}]$. Then, we have
\begin{align}
[D_{\mu},D_{\nu}] &= [\partial_{\mu} + ieA_{\mu}, \partial_{\nu} + ieA_{\nu}] = [\partial_{\mu}, ieA_{\nu}] - [\partial_{\nu}, ieA_{\mu}]\\
[D_{\mu},D_{\nu}] &= ie(\partial_{\mu}A_{\nu} - \partial_{\nu}A_{\mu}) = ieF_{\mu\nu},
\end{align}
such that $(i\not\!\! D)^2 = -D^2 - \frac{e}{2}\sigma^{\mu\nu}F_{\mu\nu}$.

I looked it up in the errata for the textbook, and there is no mention of this anywhere. Can someone confirm this? Or is there some issue in my derivation?

First thing first.
They write before deriving eq (19.74) that: "...according to eq (16.107), in our present conventions, this equation reads (what you wrote without the minus sign)".
Now equation (16.107) reads as follows the LHS is iDslash squared and the RHS is:$-D^2+2(1/2 F^b_{\rho\sigma}S^{\rho\sigma})t^b$
Now the S is the generator of Lorentz transformation in the spinor representation.

perhaps @vanhees71 can help you more than I here.

 

[mpeskin]

Murillo137 is correct. Equation (19.74) is copied from eq. (16.107) of Peskin and Schroeder. But, in Chapter 16, D_mu = (del_mu - ig A_mu), the P&S convention for gauge theories, but in Chapter 19, D_mu = (del_mu + ieA_mu), the P&S convention for QED. So, in copying the equation, P&S should have changed the sign. Note that this sign cancels out of eq. (19.77)-(19.79), but eq. (19.80) is off by a minus sign. (Thanks to Alon Brook-Ray for alerting me to this question.) -- Michael Peskin