Is Quantum Field Theory for the Gifted Amateur Effective for Self-Study?

[Hill]

I'd like to hear your professional opinion on and experience with using Quantum Field Theory for the Gifted Amateur by Tom Lancaster and Stephen J. Blundell as a self-study textbook. Thank you.

 

[PeroK]

It's pretty good, but the maths is still wild and woolly. I supplemented it by watching Tobias Osborne's lectures on YouTube. He had two courses on QFT.

The book has some nice insights.

The material is not dissimilar to David Tong's notes online.

https://www.damtp.cam.ac.uk/user/tong/qft.html

I also tried Zee's QFT in a Nutshell but he had lost me completely by page 5.

PS that's an amateur opinion, of course!

 

[Demystifier]

PS that's an amateur opinion, of course!

A gifted amateur opinion, I would add.

I like "QFT for the he Gifted Amateur" very much. It's not so good for developing technical computational skills, but it's fantastic for developing conceptual understanding. (And I'm a professional, btw.)

 

[vanhees71]

I think concerning the conceptional point of view, emphasizing the importance of locality and microcausality and Poincare symmetry, Coleman's lectures are better:

S. Coleman, Lectures of Sidney Coleman on Quantum Field
Theory, World Scientific Publishing Co. Pte. Ltd., Hackensack
(2018), https://doi.org/10.1142/9371

 

[Hill]

I also tried Zee's QFT in a Nutshell but he had lost me completely by page 5.

I tried several Zee's "in a nutshell" books. His writing style doesn't work for me.

 

[Hill]

I supplemented it by watching Tobias Osborne's lectures on YouTube.

Unfortunately, I cannot learn "by ear" because of a form of APD.

 

[vanhees71]

Usually the nutshells are too small for the addressed topics. There's no king's way to QFT!

 

[vanhees71]

Unfortunately, I cannot learn "by ear" because of a form of APD.

Hm, APD or not, you can't learn physics by just watching/attending lectures, but you have to do it yourself. Usually you start with attending a lecture or reading a book. That's just to get informed about the topic you want to learn, but then you have to practice by solving problems related to this topic for yourself. That's why it's good to have textbook with many problems (+solutions to check your work).

 

[Hill]

you have to do it yourself

This is my intent.

+solutions to check your work

Most / all textbooks don't provide these. I think the QFTftGA doesn't either.

 

[vanhees71]

Then the Homework section of this forum is a great opportunity!

 

[apostolosdt]

I think concerning the conceptional point of view, emphasizing the importance of locality and microcausality and Poincare symmetry, Coleman's lectures are better:

S. Coleman, Lectures of Sidney Coleman on Quantum Field
Theory, World Scientific Publishing Co. Pte. Ltd., Hackensack
(2018), https://doi.org/10.1142/9371

I should agree with vanhees71. Every time people ask for advice on what QFT textbook to use, Coleman Lectures always come up, and for a good reason. See them as some sort of "Feynman Lectures" on QFT. They are, perhaps, a bit more edited than Feynman's Red Books, but Coleman's students did a great job of keeping his spirit (and humor!). In my opinion, nothing can beat a verbatim lecture, edited or not edited. But the book is not for the light-hearted reader. Well, come to think of it, what QFT text is for light-hearted souls?

 

[vanhees71]

BTW, you get quite a part legally for free:

https://arxiv.org/abs/1110.5013v5

 

[Hill]

BTW, you get quite a part legally for free:

https://arxiv.org/abs/1110.5013v5

Thank you, downloaded. It does not have exercises though.

 

[haushofer]

I like the book, although it had some confusing aspects to me. But its refreshing style makes it different from most other textbooks and worth the effort.

 

[Hill]

I like the book a lot so far in spite of the aspects mentioned earlier. One confusing aspect to me is typos in some derivations, although finding them can be viewed as built-in exercises in addition to the ones at the end of each section.

One such typo has been pointed to in this thread. Here is, I believe, another:

Three terms have missing factors $e^{i\alpha}$, and the result should be rather $D_{\mu}\psi \, e^{i\alpha}$ to be non-trivial.

 

[Haborix]

The first of law of physics books is the conservation of exercises for the reader. Either your exercise is to derive it yourself or rederive it because of typos.

 

[vanhees71]

Well, it's reassuring for textbook/manuscript authors if the readers take the typos with good humor ;-)). I wish I could deliver typo-free texts though...

One should, however, mention that it's the crucial point of "gauging" a global symmetry to a gauge invariance that the gauge-covariant derivative transforms as the field for spacetime-dependent phase factors, i.e., you introduce the gauge field $A_{\mu}(x)$ as a connection to define the gauge-covariant derivative, i.e.,
$$\mathrm{D}_{\mu}=\partial_{\mu} + \mathrm{i} q A_{\mu}.$$
It's defined such that for
$$\psi'(x)=\exp[-\mathrm{i} q \alpha(x)]$$
the covariant derivative also fulfills the same transformation law,
$$\mathrm{D}_{\mu}' \psi'(x)=\exp[-\mathrm{i} q \alpha(x)] \mathrm{D}_{\mi} \psi(x),$$
and the question is how $A_{\mu}$ transforms to $A_{\mu}'$. Indeed
$$\mathrm{D}_{\mu}' \psi'(x)=(\partial_{\mu} + \mathrm{i} q A_{\mu}') [\exp(-\mathrm{i} q \alpha) \psi] = \exp(-\mathrm{i} q \alpha) [\partial_{\mu} +\mathrm{i} q (A_{\mu}'-\partial_{\mu} \alpha)] \psi$$
Since this should be
$$\exp[-\mathrm{i} q \alpha(x)] \mathrm{D}_{\mi} \psi(x) = \exp[-\mathrm{i} q \alpha(x)] (\partial_{\mu} +\mathrm{i} q A_{\mu}) \psi(x),$$
you get
$$A_{\mu}'-\partial_{\mu} \alpha=A_{\mu} \; \Rightarrow \; A_{\mu}'=A_{\mu} + \partial_{\mu} \alpha.$$

 

[Hill]

Well, it's reassuring for textbook/manuscript authors if the readers take the typos with good humor ;-)). I wish I could deliver typo-free texts though...

One should, however, mention that it's the crucial point of "gauging" a global symmetry to a gauge invariance that the gauge-covariant derivative transforms as the field for spacetime-dependent phase factors, i.e., you introduce the gauge field $A_{\mu}(x)$ as a connection to define the gauge-covariant derivative, i.e.,
$$\mathrm{D}_{\mu}=\partial_{\mu} + \mathrm{i} q A_{\mu}.$$
It's defined such that for
$$\psi'(x)=\exp[-\mathrm{i} q \alpha(x)]$$
the covariant derivative also fulfills the same transformation law,
$$\mathrm{D}_{\mu}' \psi'(x)=\exp[-\mathrm{i} q \alpha(x)] \mathrm{D}_{\mi} \psi(x),$$
and the question is how $A_{\mu}$ transforms to $A_{\mu}'$. Indeed
$$\mathrm{D}_{\mu}' \psi'(x)=(\partial_{\mu} + \mathrm{i} q A_{\mu}') [\exp(-\mathrm{i} q \alpha) \psi] = \exp(-\mathrm{i} q \alpha) [\partial_{\mu} +\mathrm{i} q (A_{\mu}'-\partial_{\mu} \alpha)] \psi$$
Since this should be
$$\exp[-\mathrm{i} q \alpha(x)] \mathrm{D}_{\mi} \psi(x) = \exp[-\mathrm{i} q \alpha(x)] (\partial_{\mu} +\mathrm{i} q A_{\mu}) \psi(x),$$
you get
$$A_{\mu}'-\partial_{\mu} \alpha=A_{\mu} \; \Rightarrow \; A_{\mu}'=A_{\mu} + \partial_{\mu} \alpha.$$

This is explained in the book, too. However, they have the $-q$ factor moved from the field $\psi$ to $A$:

The result, gauge invariance, is the same, of course.

 

[apostolosdt]

Well, it's reassuring for textbook/manuscript authors if the readers take the typos with good humor ;-)). I wish I could deliver typo-free texts though... [...]

Macmillan Publishing Co., at least up to the '80s, was known to publish typo-free books. I own only two from Macmillan, Birkhoff & Mac Lane's "Modern Algebra", and Leithold's "College Algebra". Not only without typos, but beautiful editions, too!

 

[vanhees71]

Well, at this time they may have had still a true lectorate at the publishing companies. Today you are completely on your own with proof-reading...