Bonus Material

No-Nonsense Quantum Field Theory

- A great discussion of general wave properties and dispersion can be found here. Moreover, a great book on classical waves is The Physics of Waves by Georgi.
- A more detailed discussion of symmetry breaking and the Higgs mechanism is available here and here.
- To learn more about the philosophical aspects of the renormalization group and related topics, a great starting point is Renormalization Group Methods by Williams.
- If you want to learn more about spinors, try Spinors for everyone by Coddens and An introduction to spinors by Steane.

1. I am kind of struggling with upper and lower indexes of vectors as it looks for me the notation is opposite to what I could see everywhere. That is, normally vectors are using upper indexes and co-vectors are using lower indexes, so when lower indexes are used, the space coordinates are taken with negative sign, but in this book it is quite opposite. Just wondering why?

2. In (3.11) formula, did not we need to take a complex conjugation when writing the left vector? Then the result would be +1, not -1

1. This is just a convention. We also have the freedom to choose between the “West coast metric” (+,−,−,−) and the “East coast metric” (−,+,+,+).

2. The normalization condition in Eq. 3.9 is without complex conjugation. The Minkowski product of two four-vectors only involves transposition and the Minkowksi metric.

Hope this helps and let me know if you have any further questions!

Hello Jakob, thank you for writing such a fantastic textbook. I wish I had all of your no-nonsense series available during my undergrad years. I have one comment, on p.87 you mention parity transformation as transformation which mirror coordinate axes. Obviously, mathematically you mean (x,y,z) –> (-x,-y,-z). To the reader, this might create some confusion, specially with regards to the word mirror and the figure on p.87. Usually when we say mirror (or mirror reflection/transformation), we tend to picture a reflection about the x-z plane, thus mathematically (x,y,z) –> (x,-y,z). Also, it would be helpful if you could add a… Read more »

Thanks Fahim! I will add a comment in the next edition of the book.

Hi Jakob,

I have problems understanding the following:

In chapter four I consider that the global shift (Eq. (4.49)) of the scalar field should be a symmetry of the Klein-Gordon Lagrangian ((Eq. (5.2)), that is, the shift phi—>phi ‘=phi-ie should leave the Klein-Gordon Lagrangian unchanged, but a direct replacement in the Lagrangian does not show this symmetry explicitly, i.e., dL is nonzero. I ask this question because this symmetry is used in Chapter 8 (Eq. 8.2 ) in the context of scalar fields.

Best regards, Jakob. Thank you for your clear and simple explanations. I have a question. On page 50 you imagine a clock attached to an object that moves arbitrarily. Doesn’t the acceleration that this clock undergoes alter its proper time register in any way?

Strictly speaking, this image is only valid for infinitesimal time intervals. During an infinitesimal time interval, the velocity is constant and hence there are no problems due to acceleration.

Jakob,

Congratulations for writing books for true beginners. Established authors might take exception to your comments about them wanting only to show off their knowledge, but I think you’re correct here.

I was wondering: when you find typos, do you correct them in subsequent book printings, or are you waiting for a future edition?

Thank you.

Thanks Felix!

I update my books regularly. So instead of collecting the typos and then doing a big update (Edition 2), I do many small updates (Version 1.1, Version 1.2) etc.

Best,

Jakob

Hi Jakob

I just began reading your book.

Could you possibly clarify a confusion for me.

You wrote on page 24, “an elementary excitation of the electron field is what we call the electron.”.

What bothers me is this: Say, an electron in orbit around an atom gets excited. This means the excited electron was part of the electron field.

Is this electron field local to that atom or non-local to all possible existing electrons everywhere?

Hi,

there is just one electron field and all electrons are understood as excitations of this electron field.

Hi, Jakob. Great book (at least the 200 pages I’ve been through so far!)

I’ve got a question about the start of section 5.2: your gamma matrices don’t seem to satisfy the Clifford algebra. Using {mu, nu} = {0, 0} works out fine in your example, but for {1,1} and {2,2} the sign seems to be incorrect, i.e., g1*g1 and g2*g2 both give a positive identity matrix instead of negative, while {3,3} does seem to work correctly. Also several of the “off diagonal” terms don’t work out to a zero matrix.

Thanks again for a great book!

Hi Adam, Thanks for your feedback. There is indeed a typo in the gamma matrices. The correct matrices read $$ \gamma^1 = \begin{pmatrix} 0 & 0 &0 &1 \\ 0 & 0 &1 &0 \\ 0 & -1 &0 &0 \\ -1 & 0 &0 &0 \end{pmatrix} , \gamma^2 = \begin{pmatrix} 0 & 0 &0 &-i \\ 0 & 0 &i &0 \\ 0 & i &0 &0 \\ -i & 0 &0 &0 \end{pmatrix}\, $$ $$ \gamma^3 = \begin{pmatrix} 0 & 0 &1 &0 \\ 0 & 0 &0 &-1 \\ -1 & 0 &0 &0 \\ 0 &… Read more »

In 8.30 should not we get a sqrt(2) coefficient for |2k> ?

The same question is for page 299 where for the formula |E2>=a+|E1> I’d rather expect sqrt(2)*|E2> = a+|E1>

I see some consistency here but fail to understand if it’s a problem or not.

Hello Jakob, it’s Fahim again. My question is regarding the multiparticle states. In eq.(10.19) on p.414 you defined the 2-pion state created from vacuum by acting a^{\dagger}_1 and a^{\dagger}_2. Since they are bosons the creation operators commute, so we have an equally valid indistinguishable state with a^{\dagger}_1 acting before a^{\dagger}_2, in contrast to eqn.(10.19). Should we not add a prefactor 1 / \sqrt{2!} in eq.(10.19) to account for this? Or is it required only when we work with ‘occupation-number’ representation? How are the two notations related?