This article is aimed at being a review as comprehensive as possible of the different resources (known to me) that are available for a beginner to self-learn QFT. My motivations for writing it are twofold; I would like to collect together a list of resources that I find/have found useful, and that may also be of use to other people. In addition, I was inspired to write this because of this blogpost. The problem with it is that it is now several years out of date, and some new resources have appeared in the intervening time. As well as this, I want to add some of my own takes to the matter (what I find hard, easy or confusing) in order to clarify it in my own mind.

I intend to be *as opinionated as possible*, but I will try and ignore the biases that are specific to my situation. Do you agree or disagree with my opinion? I would be happy if you contacted me and let me know! I encourage you to also check out the original blogpost. I am *specifically going to avoid* ‘difficult for beginners’ texts such as Weinberg. When you’re ready, you’ll find them.

I plan to continually update this page as more resources are suggested.

Quantum Field Theory (QFT), and by extension the standard model of particle physics, is the most accurate theory of nature that we currently have. It is well tested to extreme levels of precision, and describes all known natural phenomena (excluding gravity). Many undergraduates are excited to learn about quantum field theory due to its application at the frontiers of physics, but aren’t sure where to begin. At the simplest level, QFT is the unification of quantum mechanics and special relativity, so you should be comfortable with both. Since there are so many different books/lecture notes/YouTube courses available, this post was written to hopefully clarify the situation somewhat.

For a beginner, learning QFT can be a considerable challenge – sometimes even the prerequisites aren’t obvious. This list is absolutely not exhaustive, but is, in my opinion, the **bare minimum** requirements.

- You should know about poles, residues, and how to take a contour integral. Trust me, you’ll need it!

- You should be able to take a Fourier transform and an inverse Fourier transform. I have some notes on the absolute basics here (be aware that in QFT some slightly different conventions are standard). You should know that ‘position’ and ‘momentum’ are Fourier dual to each other (and know what that means). You may have seen this in a condensed matter course.

- You should know what a group is. You don’t have to go into depth, just read the Wikipedia page. This is important because symmetries in physics are expressed using the language of group theory. The
*Lorentz group*is the symmetry group of rotations and boosts in spacetime. Combining the Lorentz group with spacetime translations is known as the*Poincare group*. Eventually you will need to know about representations of groups as linear transformations (to know how different*fields*transform under the action of a group), but you can pick this up when you need it.

You need to know about Lagrangians and Hamiltonians. Knowledge of Noether’s theorem would be a bonus.

You should be familiar with the particles of the standard model and their interactions (at least be aware of them). You should have seen a Feynman diagram before (but you don’t have to know what it *means*, QFT will explain that). It would help to be vaguely aware of conservation laws (charge, lepton number, baryon number etc.)

This is extremely important – you should have a good foundation in special relativity, be able to perform calculations involving four-vectors, and know the difference between an ‘up’ index and a ‘down’ index. I strongly recommend Taylor & Wheeler if you want to review the subject. If you want to understand more of the mathematical structure behind it ((co)-tangent spaces, tensor fields etc.) then I **strongly** recommend the Heraeus Winter School on Gravity and Light (this is definitely overkill for learning QFT though, and most working physicists don’t have to be familiar with it).

A course at the level of Griffiths or Shankar will do just fine.

This is the standard text on the subject at the first-year graduate level (and beyond). I find it quite terse and hard to follow at times, but everything you need is there. Tong’s notes are essentially an expanded version of the first few chapters of this book. Perhaps best used as a second look at topics, this book is very calculation-oriented, so it’s great for dipping in and out of for specific topics.

My favourite QFT textbook. A lot of fun to read, and some excellent explanations throughout. I find it very friendly and engaging, but it doesn’t go anywhere near as deep as Peskin & Schroeder in terms of calculations. I see it as a perfect first course in QFT, and as one reviewer suggested that “grad students will curl up and read this by the fire” – perhaps not, but I do indeed keep it beside my bed!

This book is a true gift to humanity! It has clear, well thought out explanations of many many topics, and beautiful illustrations. It may seem a little hokey, but this textbook is not to be underestimated. It lets you know exactly what you’re expected to already know, and what you’re not expected to, and has some incredible insights dotted around. Whenever I find some aspect of QFT confusing, I find myself continually returning to this book for its insights. This book coupled with Peskin and Schroeder would make for a winning combination in my opinion. Another one to keep by the bedside.

This book takes things very very slowly, and goes to great pains to explain every little detail, which, if left unexplained, might turn out to be a source of confusion for later. I’m not a huge fan of its style, but if you want to see everything spelled out explicitly, then I highly recommend you check out this book. However, if you’re a good student, you might get bored quickly. I never managed to make it past 100 pages when I read it linearly, but later on I found it useful to dip in and see how Klauber explained this or that point.

I only briefly used this book, but the style and level of exposition is absolutely solid. There’s plenty here to get your teeth into, and it seems to be a good mixture of detailed explanations, but letting you fend for yourself. I plan to check out more of Greiner’s books in the future because I liked the style so much.

The title says it all, this is a book of problems (with solutions). Importantly the solutions are **detailed**, with all steps given. In preparing for exams, I found this book *extremely useful*. There’s a chapter per major topic, and a little summary of the topic at the start of the chapter (telling you what you should remember from what you have studied). The questions are at a good level for the first few chapters of Peskin and Schroeder, but would supplement any introductory course well. My only complaint is that there’s not a version of the book covering advanced quantum field theory, because I would buy it in a second (covering path integrals, the standard model, renormalisation group etc.). Radovanovic, get on it!

Tong is a hero among undergraduates and postgraduates alike for his somewhat incredible arsenal of lecture notes for different physics courses. His notes on QFT are legend for being clear and concise, with some great worked examples. Video recordings of his lectures can also be found here. However, I prefer to recommend Tobias Osborne’s lectures if you want video recordings.

These are at a similar level to David Tong’s course, if slightly more mathematical. However, the broad strokes of the course are exactly the same. There are very high quality video lectures, and notes available here, typed up by A.V. St. John. He also has an advanced course available for when you’re ready, again with typed notes.

Warren Siegel has put out an enormous ebook (over 1000 pages!) on quantum field theory, taking an orthogonal approach to many other texts. So far I have not read further than the contents page, but I plan to read it in the future. It is definitely worth checking out, as I feel that reading this (if you can) would provide a ridiculously thorough background in QFT.

Did you agree with my opinion? Am I being pointlessly dogmatic? Do you have a suggestion for something to add to the list? Contact me if any or all of the above are true, as I would love to hear about it!

]]>A brief overview of Fourier analysis (Fourier series and transform).

A function is periodic with period \(2\pi\) if

\[f(x) = f(x + 2\pi).\]\(f(x)\) need only be given on the half-open interval \(x \in [0, 2\pi)\) for it to be specified everywhere.

Fourier’s theorem states that any such \(f(x)\) which is *well behaved* can be written in each one of the three following (equivalent) forms:

The different sets of coefficients \((a_n, b_n), \ (A_n, \phi_n), \ c_n\) can be converted between easily via trigonometric relations.

We are going to make use of the integrals

\[\begin{aligned} I_{n,m} &= \int_0^{2\pi } \mathrm{cos} (nx) \ \mathrm{cos} (mx) \ dx = \delta_{n,m}\pi \\ J_{n,m} &= \int_0^{2\pi } \mathrm{cos} (nx) \ \mathrm{sin} (mx) \ dx = 0 \\ K_{n,m} &= \int_0^{2\pi } \mathrm{sin} (nx) \ \mathrm{sin} (mx) \ dx = \delta_{n,m}\pi. \end{aligned}\]If we integrate \(f(x)\) over its range, \(\int_0^{2\pi} f(x) \ dx\), we see that since the average value of \(\mathrm{sin}(nx)\) and \(\mathrm{cos}(nx)\) is zero. The integral reduces to \(\int_0^{2\pi} \frac{a_0}{2} \ dx\), so we find

\[a_0 = \frac{1}{\pi} \int_0^{2\pi} f(x) \ dx.\]Note: we can integrate over \(2\pi\) at any point in the function’s range, so this is perfectly equivalent to

\[\frac{a_0}{2} = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(x) \ dx.\]In other words, \(\frac{a_0}{2}\) is the *average value* of \(f(x)\).

For the \(a_n\) coefficients, multiply \(f(x)\) by \(\cos (mx)\) and integrate over the range:

\[\begin{aligned} \int_0^{2\pi} \cos(mx) f(x) \ dx &= \int_0^{2\pi} \sum_{n=1}^{\infty} a_n \cos(nx) \cos(mx) \ dx \\ &= \sum_{n=1}^{\infty}\int_0^{2\pi} a_n \cos(nx) \cos(mx) \ dx \\ &= a_m \int_0^{2\pi} \cos^2(mx) \ dx = \pi a_m, \end{aligned}\]finally, relabelling \(m\) to \(n\) gives

\[\fcolorbox{white}{indigo}{$ \begin{aligned} a_n = \frac{1}{\pi} \int_0^{2\pi} \cos(nx) f(x) \ dx. \end{aligned} $}\]Exactly the same argument holds for the \(b_n\), so similarly

\[\fcolorbox{white}{indigo}{$ \begin{aligned} b_n = \frac{1}{\pi} \int_0^{2\pi} \sin(nx) f(x) \ dx. \end{aligned} $ }\]Given the function

\[f(x)=\begin{cases} -1 & -\pi \leq x < 0\\ \phantom{-} 1 & \phantom{-} 0 \leq x < \pi \end{cases}\]We see that since it is *odd*, \(a_n\) must be zero, and we find the \(b_n\) to be \(b_n = \frac{2}{n\pi} \left( 1 - (-1)^n \right)\)

So the Fourier series representation of \(f(x)\) is

\[f(x) = \sum_{n=1}^{\infty} \frac{2}{n\pi} \left( 1 - (-1)^n \right) \sin(nx).\]Given the function

\[f(x)=\begin{cases} -x & -\pi \leq x < 0\\ \phantom{-} x & \phantom{-} 0 \leq x < \pi \end{cases}\]We see that since it is *even*, \(b_n\) must be zero, and we find the \(a_n\) to be
\(a_n = \frac{2}{n^2 \pi} \left( (-1)^n - 1 \right)\)

So the Fourier series representation of \(f(x)\) is

\[f(x) = \sum_{n=1}^{\infty} \frac{2}{n^2 \pi} \left( (-1)^n - 1 \right) \cos(nx) .\]Suppose \(f(x) = f(x + T),\) that is \(f(x)\) is periodic with some arbitrary period. Then we would write

\[f(x) = \frac{a_0}{2} + \sum_{n = 1}^{\infty} \left[ a_n \mathrm{cos}\left(\frac{2\pi nx}{T} \right) + b_n \mathrm{sin}\left(\frac{2\pi nx}{T} \right) \right]\]And the formulae for the coefficients would become

\[\begin{aligned} a_0 &= \frac{2}{T} \int_{-\frac{T}{2}}^{\frac{T}{2}} f(x) \ dx \\ a_n &= \frac{2}{T} \int_{-\frac{T}{2}}^{\frac{T}{2}} \cos \left(\frac{2\pi nx}{T} \right) f(x) \ dx \\ b_n &= \frac{2}{T} \int_{-\frac{T}{2}}^{\frac{T}{2}} \sin \left(\frac{2\pi nx}{T} \right) f(x) \ dx. \\ \end{aligned}\]This can be simplified by rewriting in terms of angular frequency \(T = \frac{2\pi}{\omega}\)

giving

\[\fcolorbox{white}{indigo}{ $ \begin{aligned} f(x) &= \frac{a_0}{2} + \sum_{n = 1}^{\infty} \left[ a_n \mathrm{cos}(n \omega x) + b_n \mathrm{sin}(n \omega x) \right]\\ a_n &= \frac{2}{T} \int_{-\frac{T}{2}}^{\frac{T}{2}} \cos \left(n \omega x \right) f(x) \ dx \\ b_n &= \frac{2}{T} \int_{-\frac{T}{2}}^{\frac{T}{2}} \sin \left(n \omega x \right) f(x) \ dx. \\ \end{aligned}$ }\]Earlier, we noted that the Fourier series can be written

\[f(x) = \sum_{n=-\infty}^{\infty} c_n e^{inx}.\]Using the fact that \(\int_{-\pi}^{\pi} e^{-imx} e^{inx} \ dx = 2\pi \delta_{n,m}\) and integrating over the series representation, we extract the coefficients

\[c_n = \frac{1}{2\pi} \int_{-\pi}^{\pi} e^{-inx} f(x) \ dx.\]For a period of \(2\pi L\), the series becomes

\[\begin{aligned} f(x) &= \sum_{n=-\infty}^{\infty} c_n e^{\frac{inx}{L}}\\ c_n &= \frac{1}{2\pi L} \int_{-\pi L}^{\pi L} e^{\frac{-inx}{L}} f(x) \ dx.\\ \end{aligned}\]Now we define \(\hat{f}\left( \frac{n}{L}\right) = c_n,\) that is, a function that is only defined at the points \(\frac{n}{L}\). The series now looks like

\[\begin{aligned} f(x) &= \sum_{n=-\infty}^{\infty}\hat{f}\left( \frac{n}{L}\right) e^{\frac{inx}{L}}\\ \hat{f}\left( \frac{n}{L}\right) &= \frac{1}{2\pi L} \int_{-\pi L}^{\pi L} e^{\frac{-inx}{L}} f(x) \ dx.\\ \end{aligned}\]Further, if we define \(\widetilde{f}\left( \frac{n}{L} \right) = \sqrt{2\pi} L \hat{f}\left( \frac{n}{L}\right)\) we get

\[\begin{aligned} f(x) &= \frac{1}{\sqrt{2\pi}} \sum_{n=-\infty}^{\infty}\widetilde{f}\left( \frac{n}{L}\right) e^{\frac{inx}{L}} \frac{1}{L}\\ \widetilde{f}\left( \frac{n}{L} \right) &= \frac{1}{\sqrt{2\pi}} \int_{-\pi L}^{\pi L} e^{\frac{-inx}{L}} f(x) \ dx.\\ \end{aligned}\]Rewriting \(\frac{n}{L}\) as \(k\) and taking \(L \rightarrow \infty\) finally gives us

\[\fcolorbox{white}{indigo}{ $ \begin{aligned} f(x) &= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{ikx}\widetilde{f}(k) \ dx\\ \widetilde{f}(k) &= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-ikx} f(x) \ dx.\\ \end{aligned}$ }\]Suppose we have \(f(x) = e^{-\alpha x^2}.\) The Fourier transform of this is

\[\begin{aligned} \widetilde{f}(k) &= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-ikx} e^{-\alpha x^2} \ dx\\ &= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-\alpha \left( x + \frac{ik}{2\alpha} \right)^2 - \frac{k^2}{4\alpha}} \ dx.\\ \end{aligned}\]Shifting this by the substitution \(y = x + \frac{ik}{2\alpha}\) gives

\[\begin{aligned} \widetilde{f}(k) &= \frac{1}{\sqrt{2\pi}} e^{-\frac{k^2}{4\alpha}} \int_{-\infty}^{\infty} e^{-\alpha y^2} \ dy\\ &= \frac{1}{\sqrt{2\pi}} e^{-\frac{k^2}{4\alpha}} \sqrt{\frac{\pi}{\alpha}}\\ &= \frac{1}{\sqrt{2\alpha}} e^{-\frac{k^2}{4\alpha}}. \end{aligned}\]Another Gaussian, but stretched.

Given the function

\[f(x)= \begin{cases} \frac{1}{A} & -A \leq x \leq A\\ 0 & \phantom{--..} |x| > A\\ \end{cases}\]its Fourier transform is

\[\begin{aligned} \widetilde{f}(k) &= \frac{1}{\sqrt{2\pi}} \int_{-A}^{A} \frac{1}{A} e^{-ikx} \ dx \\ &= \frac{1}{A\sqrt{2\pi}} \left[ \frac{e^{ikA} - e^{-ikA}}{ik} \right]\\ &= \sqrt{\frac{2}{\pi}} \frac{\sin(kA)}{kA}\\ &= \sqrt{\frac{2}{\pi}} \mathrm{sinc}(kA)\\ \end{aligned}\]A convolution is an integral transform that takes two functions and composes them, forming a new function. The definition is

\[(f \star g)(x) = \int_{-\infty}^{\infty} f(x')g(x-x') \ dx'.\]The Fourier transform of the convolution of two functions is \(\sqrt{2\pi}\) times the product of the Fourier transforms.

\[\begin{aligned} \mathcal{F}\{f \star g\} &= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-ikx} \left[ \int_{-\infty}^{\infty} f(x') g(x-x') \ dx' \right] dx\\ &= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} f(x') \int_{-\infty}^{\infty} g(x-x') e^{-ikx} \ dx' dx\\ \end{aligned}\]Setting \(y = x - x'\) gives

\[\begin{aligned} \mathcal{F}\{f \star g\} &= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} f(x') \ dx' \int_{-\infty}^{\infty} e^{-ik(x'+y)} g(y) \ dy\\ &= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-ikx'} f(x') \ dx' \int_{-\infty}^{\infty} e^{-iky} g(y) \ dy\\ &= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-ikx'} f(x') \ dx' \int_{-\infty}^{\infty} e^{-iky} g(y) \ dy\\ &= \sqrt{2\pi} \widetilde{f}(k) \widetilde{g}(k). \end{aligned}\]The basic dimensions in physics are length, mass and time, which are commonly expressed in the SI system of units as the metre, the kilogram and the second.

In particle physics, the typical scale of quantities is lengths of \(10^{-15}\mathrm{m}\), masses of \(10^{-27}\mathrm{kg}\) and durations of \(10^{-7}\mathrm{s}\)

As in any branch of physics, we should adopt a system of units that is appropriate for the scale of quantities involved. We adopt a system called *natural units*, where the fundamental constants \(c\) and \(\hbar\) are set to unity:

SI Units | Natural Units | ||
---|---|---|---|

\(c\) | Speed of light | \(3\times 10^8\mathrm{ms}^{-1}\) | 1 |

\(\hbar\) | Reduced Planck constant | \(1.05\times 10^{-34}\mathrm{Js}\) | 1 |

That is to say, \(c\) and \(\hbar\) are *used* as the base units of velocity and action (or angular momentum) respectively.

One further quirk is that when we write quantities in natural units, there is no need to write in the units of \(c\) or \(\hbar\), because they are defined to be unity, so they are suppressed.

Natural units are very neat in particle physics, because many of the formulae we write down contain factors of \(\hbar\) and \(c\), and so become neater and more transparent (and easier to remember for the exam!) when we use natural units. Contrast

\[\left(\beta mc^2 + c \sum_{n \mathop =1}^{3}\alpha_n p_n\right) \psi (x,t) = i \hbar \frac{\partial\psi(x,t) }{\partial t}\]with

\[(i{\cancel{\partial}} - m) \psi = 0.\](Ok, the Feynman slash tidies it up a little too… That’s a topic for a future post though)

There’s another, more physical reason for using natural units though. The only reason we need the speed of light in formulae is because we don’t measure space and time in the same units. It is properly seen as the *conversion factor* between distances and times (seconds and metres, for example). But relativity tells us that space and time are unified into one object – *spacetime*. If we treat time and space on the same footing, we should measure distances and durations in **the same units**. What’s the speed of light in metres of distance divided by metres of time? One! (for more on this, check out Spacetime Physics).

We can still specify one more unit, which we choose to be the unit of energy. We take the electronvolt (eV), which is defined as the kinetic energy an electron picks up falling through a potential of one volt.

\[1\mathrm{eV} = 1.6\times 10^{-19}\mathrm{J}.\]Different authors may use MeV or even GeV. Don’t panic! It’s just scaling by a power of ten.

In natural units, then, *all* quantities have dimensions of a power of energy, since the whole system is specified in terms of \(\hbar\), \(c\) and \(\mathrm{eV}\).

From known relations, we can show that if we have a quantity of some certain dimensions

\[[\mathrm{quantity}] = \mathbf{M}^\alpha \mathbf{L}^\beta \mathbf{T}^\gamma,\]then this can be written in terms of \(\hbar\), \(c\) and energy as

\[\left( \frac{\mathbf{E}}{c^2} \right) ^\alpha \left(\frac{c \hbar}{\mathbf{E}} \right)^\beta \left(\frac{\hbar}{\mathbf{E}}\right)^\gamma\]and since in NU \(\hbar\) and \(c\) are set to unity, we have that in the NU system, the quantity has dimensions of

\[\mathbf{E}^{\alpha - \beta - \gamma}.\]For example, if I have a duration (dimensions of time), then in NU it will have units of \(\mathrm{eV}^{-1}\).

As an example, let’s convert a length of \(l_{NU} = 1\mathrm{eV}^{-1}\) back to SI units. We need to put back in the factors of \(\hbar\) and \(c\) as follows:

\[\left[ \hbar^\alpha \right] \left[ c^\beta \right] \left[ \mathrm{eV}^{-1} \right] = \mathbf{L}.\]We end up with a set of equations for \(\alpha\) and \(\beta\), and find that

\[\begin{aligned} l_{SI} &= \hbar c\ l_{NU}\\ &= 1.97\times 10^{-7} \mathrm{m}. \end{aligned}\]Remember to make sure to use the right units for the numerical values of \(\hbar\) and \(c\), and happy physicsing!

(though I recommend the \(1^{st}\) edition)

]]>I figured it was probably time to get a personal blog, so here it is.

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