Quantum Field Theory

  • We talk about the most important concepts of quantum field theory and why they are important
  • You’ll learn how to compute the probability that certain new particles are created if we smash two particles, for example electrons, together
  • You’ll understand how particles can be described in a field theory

Let‘s start with some good news: quantum field theory is easy, especially if you already know how quantum mechanics works.

Of course, not every computation in quantum field theory is easy, but understanding how the framework works in principle is.

Essential Concepts

Before we talk about how quantum field theory works, here is a summary of the most important concepts and why they are important.

This will help you in your further studies, because most books have no summary at the beginning that tells you what is important.

Why a field theory?

Field theories have the big advantage that they treat time and space on an equal footing, which is necessary because of special relativity.

In contrast, in a particle theory we always use the location as a function of time $q=q(t)$ to describe the particles in question. A field $\Phi$ is a function of space and time $\Phi(x,t)$.

The Lagrangian Formalism

It was invented originally as an alternative for Newton’s classical mechanics, but is now an essential part of the best theory of physics that we have.

Why? It turns out guessing equations that describe nature at the most fundamental level isn’t a very good method, because the human intuition fails on this scale. The Lagrangian formalism makes it possible to derive the correct equations systematically.

In simple terms the Lagrangian, unsurprisingly the most important thing in this formalism, is the object that we use to derive the fundamental equations.

We want equations that look the same for every observer, because otherwise our equations would be useless. Therefore, we have the strong condition that our Lagrangian must not only look the same for different observers, but must stay exactly the same. Otherwise we would get different equations.

This helps immensely and rules out a lot of possibilities.

Formulated differently: The Lagrangian formalism is the perfect framework to work with symmetries and symmetries are the best thing that we can use to derive the fundamental equations of nature instead of guessing.

The Quantum Formalism

The same ideas developed for ordinary quantum mechanics, can be used in quantum field theory.

We need something abstract that describes the physical system in question $| \Psi >$. This is called a ket. Exactly as in quantum mechanics we need something, called a bra $<\Phi|$, to denote a possible final state.

If we put these together we get the probability amplitude that our system changed from the initial state $| \Psi >$ into the final state $<\Phi|$:  $$\text{ amplitude } = <\Phi| | \Psi >.$$ The probability for this process is then the absolute value squared of this probability amplitude.

This is sometimes called the Born rule:$$ \text{ Probability } = | \text{ amplitude }|² $$

Noether’s Theorem

Using the Lagrangian formalism, we can derive this theorem, which tells us that for every symmetry we have a conserved quantity.

This is important, because in a field theory we get for rotational symmetry a conserved quantity that has two parts. One part is ordinary angular momentum, but the second part is something new, called Spin.


One of the biggest discoveries in the last century was that elementary particles have spin, which is some kind of internal angular momentum. This was discovered by the famous Stern-Gerlach experiment.

In abstract terms you can thing about spin as a label that tells us how particles behave in experiments, exactly as the mass or the electric charge. For example, a particle with electric charge behaves different than one without in experiments and the same is true for spin.

There are particles with spin $0$, particles with spin $\frac{1}{2}$ and particles with spin $1$. For each of these different particle types we have an equation describes their behaviour.

Field Equations and their Solutions

Using clever mathematics we can derive from the Lagrangians the corresponding equations of motion. It turns out there is not only one equation that describes all elementary particles, but there are different kinds of elementary particles and one equation of motion for each type.

In a field theory the equation of motion are called field equations and the most important examples are:

  • For spin $0$ we have the Klein-Gordon equation:\begin{equation} ( \partial _{\mu} \partial ^{\mu}+m^2)\Phi =0. \end{equation}
  •  For spin $\frac{1}{2}$ the Dirac equation$$(i\gamma_\mu \partial^\mu – m ) \Psi =0. $$
  •  For spin $\frac{1}{2}$ anti-particles$$(i \partial^\mu \bar{\Psi} \gamma_\mu + m \bar{\Psi} ) =0$$
  •  For spin $1$ the Proca equation$$ m^2 A^\rho = \frac{1}{2} \partial_\sigma ( \partial^\sigma A^\rho – \partial^\rho A^\sigma) $$
  • which for massless particles $m=0$ is commonly called inhomogeneous Maxwell equation$$0 = \frac{1}{2} \partial_\sigma ( \partial^\sigma A^\rho – \partial^\rho A^\sigma) $$

As noted above, these equations can be derived from the corresponding Lagrangians. For example, the Klein-Gordon equation follows from the Lagrangian

\begin{equation} \mathscr{L}= \frac{1}{2}( \partial _{\mu} \Phi \partial ^{\mu} \Phi -m^2 \Phi^2) \end{equation}

The Canonical Commutation Relation

This overcomplicated name denotes the most important equation of quantum field theory:
\begin{equation} \label{qftcomm}  [\Phi(x), \pi(y)]=\Phi(x) \pi(y) – \pi(y) \Phi(x) = i \delta(x-y) \end{equation}where $\delta(x-y)$ is the Dirac delta distribution and $\pi(y) = \frac{\partial \mathscr{L}}{\partial(\partial_0\Phi)}$ is the conjugate momentum, which is the conserved quantity that follows from Noether‘s theorem from invariance under translations of $\Phi(x)$ itself: $ \Phi(x) \rightarrow \Phi(x) + \epsilon$. For example, for spin $0$ fields the conjugate momentum is$$\pi(x) = \frac{\partial \mathscr{L}}{\partial(\partial_0\Phi(x))} = \frac{\partial }{\partial(\partial_0\Phi(x))} \frac{1}{2}( \partial _{\mu} \Phi(x) \partial ^{\mu} \Phi(x) -m \Phi^2(x))  $$ $$ = \partial_0\Phi(x) $$Take note that, for brevity, it‘s conventional to write $\Phi(x)$ instead of $\Phi(x,t)$ , which means we include $t$ in $x$: $x_0=t$, $x_1=x$, $x_2=y$ and $x_3=z$.There is a beautiful explanation for this equation, but this is enough stuff for another article.

Solutions of the Field Equations

The next step is, of course, solving these equations.The most general solution of the Klein-Gordon equation is\begin{equation}\label{KGsol} \Phi(x)= \int \mathrm{d }k^3 \frac{1}{(2\pi)^3 2\omega_k} \left( a(k){\mathrm{e }}^{ -i(k x)} + a^\dagger(k) {\mathrm{e }}^{ i(kx)}\right)\end{equation}The solutions for the other equations look similar. Equation \ref{qftcomm} tells us then that the field $\Phi(x)$ we use to describe particles (we will see in a moment how this works), can‘t be simply a function, but must be an operator. Ordinary functions and numbers commutate: For example $f(x)=3x$ and $g(y)= 7y^2 +3$ clearly commutate$$ [f(x) , g(x)]= f(x)g(x) – g(x) f(x) = 3x (7y^2 +3) -(7y^2 +3) 3x =0 $$Therefore if we look at the solution $a(k)$ and $a^\dagger(k)$ can‘t be numbers, but must be operators, because we can compute

$$ [a^\dagger(k), a(k)] = i \delta(x-y) .$$

Everything else in the solutions are just numbers. Quantum fields are operators that act, like every operator in quantum theory, on abstract states.

Can we understand what these states are here and what they do? Absolutely!

This is one of the most beautiful aspects of quantum field theory. To understand what a quantum field does we will focus on the operator parts of the fields, i.e. $a(k)$ and $a^\dagger(k)$.

The Hamiltonian

One thing that we understand is energy. We can compute, using the Lagrangian, the corresponding Hamiltonian, which represents the energy and is the conserved quantity that follows from invariance under time-translation, if we use Noether‘s theorem.

For example, for  spin $0$ fields we have\begin{equation} \label{hamil}   H= \frac{1}{2} \Big( \big(\partial_0 \Phi \big)^2 + ( \partial_i \Phi )^2 + m \Phi^2 \Big).  \end{equation}The fields are operators and therefore the Hamiltonian is an operator. For reasons explained above we call it the energy operator, which means if we act with the Hamiltonian on an abstract state $ | \Psi >$ that describes the system in question, we get the energy of the system:$$ H | \Psi > = E | \Psi > $$

Creation and Annihilation Operators

Do the operators $a(k)$ and $a^\dagger(k)$ have any effect on the energy of the system? That may seem like a strange question, but will make a lot of sense in a moment. To find an answer we must compute $H a(k) | \Psi >$.

Click here to see some Details how we can compute this!

The result is:

$$ \hat H \big( a(k’) | \Psi > \big) =  \Big( E – \omega_{k’} ) \big( a(k’) | \Psi > \big) $$

In words this means that our operator $a(k’)$ lowers the energy of the system by the amount $ \omega_{k’} $. Equivalently, we can compute that $a^\dagger(k’)$ raises the energy by the amount $ \omega_{k’} $.

Imagine we have a completely empty system, which means energy zero:

$$ H | 0> = 0 | 0> $$

If now our operators $a^\dagger(k’)$ acts on this completely empty system it suddenly has energy $ \omega_{k’} $ instead of zero:

$$ H a^\dagger(k’) | 0> = \omega_{k’} a^\dagger(k’) | 0> $$

Therefore $a^\dagger(k’) | 0>$ describes a new system, which isn‘t empty any more. We can act again with $a^\dagger(k’)$ on this system, which has then energy $2 \omega_{k’} $:

$$ H a^\dagger(k’) a^\dagger(k’) | 0> = 2\omega_{k’} a^\dagger(k’) a^\dagger(k’) | 0> $$

Recall that $a^\dagger(k’) and $a(k’) $ are the operator parts of our quantum fields. Here we learn that quantum fields create and annihilate particles!

$| 0>$ is an empty system

$a^\dagger(k’) | 0>$ is a system with one particle, with energy $\omega_{k’} $. Therefore we write $a^\dagger(k’) | 0> = | 1_{k‘}>$

$a^\dagger(k’) a^\dagger(k’) | 0>$ is a system with two particles, with energy $\omega_{k’} $ each. Therefore we write $a^\dagger(k’) a^\dagger(k’) | 0>= | 2_{k‘}>$.

Of course it is possible to create particles with different energy. For example if we consider $a^\dagger(k’‘)a^\dagger(k’) a^\dagger(k’) | 0>$, we have a system with two particles with energy $\omega_{k’} $ and one with energy $\omega_{k’‘} $. Therefore $a^\dagger(k’‘) a^\dagger(k’) a^\dagger(k’) | 0>= | 2_{k‘}, 1_{k‘‘}>$.

It‘s absolutely normal to have now thousands of question popping up in your head. You‘re in good company. For example Richard Feynman:

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I remember that when someone had started to teach me about creation and annihilation operators, that this operator creates an electron, I said, “how do you create an electron? It disagrees with the conservation of charge”, and in that way, I blocked my mind from learning a very practical scheme of calculation.


You‘ll find very good answers to these questions if you dive deeper into the concepts of quantum field theory. Conservation of charge, for example, is never violated. For the moment, the message to take away is: Quantum fields create and annihilate particles. We‘ll see in a moment where this leads us.

The Scattering Operator

Most parts of quantum field theory are about scattering processes. This means we want to use the framework to answer questions like: If we smash an electron and a positron together what are the odds that we detect an outgoing photon in our detectors afterwards.

In mathematical terms, using the usual Dirac notion, we can write this question as $$< \gamma(k‘)| \hat S | e_ -(p), e_+(p‘) >= ?$$Here $p$ and $p‘$ denote the momentum of the colliding electron and positron, $k$ the momentum of the outgoing photon and $ \hat S$ is some operator that describes the scattering process.

What should this operator do? We start with some state at an initial point in time and want to calculate the probability to find it at some later time in another state. Therefore, the scattering operator is simply the time-evolution operator that is used in quantum mechanics, too.

Time Evolution

The time-evolution in quantum mechanics is described by the Schrödinger equation:\begin{equation} \label{eq:schroedinger} i \partial_t | \Psi (t)> = H | \Psi (t)>, \end{equation}where $H$ denotes the Hamiltonian.

Click here to see some thoughts that make plausible why this equation holds in quantum field theory, too.

We can introduce formally an time-evolution operator that enables us to compute how our state transforms from the initial point in time $t_0$ to a later point in time:$$ S(t-t_0) | \Psi (t_0)> = | \Psi (t)>. $$

This means we simply write down what an operator that we call time-evolution operator must do. For notational brevity, let‘s use $t_0=0$, which is just a matter of choice.

$$ S(t) | \Psi (0)> = | \Psi (t)>. $$

We can put this into equation \ref{eq:schroedinger}, which yields

\begin{equation} i \partial_t S(t) | \Psi (0)> = H S(t) | \Psi (0)>. \end{equation}

This equation holds for arbitrary $ | \Psi (0)>$ and therefore

\begin{equation} i \partial_t S(t) = H S(t) \end{equation}

The general solution of this equation is

$$ S(t) = \mathrm{e}^{-i \int H dt},$$


\begin{align} & \quad i \partial_t S(t) | \Psi (0)> = H S(t) | \Psi (0)> \\
& \rightarrow i \partial_t \mathrm{e}^{-i \int H dt}| \Psi (0)> = H \mathrm{e}^{-i \int H dt} | \Psi (0)> \\
& \rightarrow H \mathrm{e}^{-i \int H dt}| \Psi (0)> = H \mathrm{e}^{-i \int H dt} | \Psi (0)>

We conclude: The mysterious scattering operator isn‘t mysterious at all.

The interaction picture: Please take note that there are some very important concepts we haven‘t talked about here like the interaction picture, but you can read about these in the books recommended at the end of this article. The interaction picture enables us to use everything we learn about free fields, which means fields without interactions, in an interaction theory. The behaviour of the free fields are exactly what is described by the equations like the Klein-Gordon or the Dirac equation above.

Gauge Theory and Interactions

Now we have the the scattering operator and all we need to compute $$< \gamma(k‘)| \hat S | e_ -(p), e_+(p‘) >$$ is $H$.

As noted above, we can compute the Hamiltonian $H$ from the corresponding Lagrangian and the Lagrangian is what defines our physical theory.

The derivation of the Lagrangian that describe the interactions of particles is called gauge theory. The name “gauge theory” does not make any sense, but is only kept for historic reasons. 

Gauge theory means we use abstract symmetries, called $U(1)$, $SU(2)$ and $SU(3)$ to derive the correct terms in Lagrangian that describe interactions.

This yields one Lagrangian (and therefore one Hamiltonian) for electromagnetic interactions, one for weak and one for strong interactions.  For example the Lagrangian, describing electromagnetic interactions is$$ H = \int d^3x \left( g A_\mu \bar \Psi \gamma^\mu \Psi \right) $$Here $A_\mu$ describes a spin $1$ field and $\Psi$ a spin $ \frac{1}{2}$ field.

The derivation of this Hamiltonian is really, really awesome, but unfortunately we can‘t talk about the details here and you‘re referred to the books at the end of this article.

If we want to consider all interactions at once we need to write these into one big Lagrangian and derive the corresponding Hamiltonian.

How Quantum Field Theory Works

We have now everything at hand to understand what is really going on in quantum field theory. To recapitulate:

  • We want to compute $< \gamma(k‘)| \hat S | e_ -(p), e_+(p‘) >$.
  • We derived $ S(t) = \mathrm{e}^{-i \int H dt}$.
  • We have $H = \int d^3x \left( g A_\mu \bar \Psi \gamma^\mu \Psi \right) $.
  • We know that quantum fields create and annihilate particles.

One key feature of quantum field theory is that we can‘t evaluate $ \mathrm{e}^{-i \int H dt}$ simply at once, but must write the exponential function as a series and evaluate the series term by term. Therefore

\hat{S} &= \underbrace{1}_{S^{(0)}} \underbrace{-i \int dt_1 H(t_1)}_{S^{(1)}} \\ & \quad \underbrace{- \frac{1}{2!} T \Bigg \{ \left( \int dt_1 H(t_1) \right) \left( \in dt_2 H(t_2) \right) \Bigg \}}_{S^{(2)}} + \ldots

This is possible because we have coupling constants $g$ in the Hamiltonian. The first term is proportional to $g$, the second to $g^2$, the third to $g^3$.

If the coupling constant is smaller than one, the higher order terms of the series expansion are less expansions and we get a good approximation of the solution of we focus on the first few terms.

Happily for electromagnetic and weak interactions the coupling constant is smaller than one!

The first term $S^{(0)}=1$ changes nothing and is only interesting for computations like $< e_ -(k), e_+(k‘) | \hat S | e_ -(p), e_+(p‘) >$. The next term of the series $S^{(1)}$ is more interesting. We now need the explicit form of the Hamiltonian, which we cited above.

$$ S_1 = -i \int dt_1 H(t_1) = -i \int dt \int d^3x \left( g A_\mu \bar \Psi \gamma^\mu \Psi \right) $$

Take note that this Hamiltonian takes only electromagnetic interactions into account. If we want to consider weak-interactions, too we need to use a longer and more complicated Hamiltonian.

For each field, we can use the solution of the equations cited at the beginning. Each solution consists of two terms (see Eq. \ref{KGsol}) and therefore this one term of the series is actually eight terms. Happily, the interpretation of these terms is straight-forward.

  • $\Phi= \Phi^+ \Phi^-$, where $ \Phi^+$ annihilates and $\Phi^-$ creates spin $0$ particles.
  • $\Psi= \Psi^+ \Psi^-$, where $ \Psi^+$ annihilates and $\Psi^-$ creates spin $\frac{1}{2}$ particles.
  • $\bar \Psi= \bar \Psi^+\bar \Psi^-$, where $ \bar \Psi^+$ annihilates and $\bar \Psi^-$ creates spin $\frac{1}{2}$ anti-particles.
  • $A_\mu= A_\mu^+A_\mu^-$, where $ A_\mu^+$ annihilates and $A_\mu^-$ creates spin $1$ particles.

We will focus here on only one of these terms and you‘ll understand in a moment why:

$$ < \gamma(k‘)| \hat S | e_ -(p), e_+(p‘) > \approx < \gamma(k‘)| ig \int d^4x A_\mu^- \bar \Psi^+ \gamma^\mu \Psi^+|e^-(p), e^+(p‘)>$$

What happens here? Well, $\Psi^+$ destroys the electron, $\bar \Psi^+$ destroys the positron, which leaves us with an empty system:

$$ < \gamma(k‘)| ig \int d^4x A_\mu^+ \bar \Psi^+ \gamma^\mu \Psi^+|e^-(p), e^+(p‘)> $$
$$ = < \gamma(k‘)| const \times ig \int d^4x A_\mu^- |0 > $$

Then $A_\mu^-$ creates a photon:

$$ < \gamma(k‘)| const \times ig \int d^4x A_\mu^- | 0> = < \gamma(k‘)| const’ \times ig int d^4x | \gamma(p+p‘) >$$

$ \gamma(p+p‘) >$ and $ < \gamma(k‘)| $ fit nicely together

$$ < \gamma(k‘)| \gamma(p+p‘) > = \delta(k-(p+p‘)) $$

and therefore we get something non-zero. The resulting number is exactly the probability amplitude we are interested in.

For every other term of the sum we would get zero, because for example:

$$ < \gamma(k‘)| ig \int d^4x A_\mu^+ \bar \Psi^+ \gamma^\mu \Psi^-|e^-(p), e^+(p‘)> = < \gamma(k‘)| const \times ig \int d^4x |e^-(p), e^-(p‘‘) , \gamma(p‘‘‘)>$$

and we have

$$ < \gamma(k‘)| |e^-(p), e^-(p‘‘) , \gamma(p‘‘‘)> =0,$$

because the states are orthogonal. This should be plausible, because assume we have just an electron $ |e^-(p)> and we want to know what the probability is to find it as an photon at the same moment in time, without any interaction:

$$ < \gamma(p)|| e^-(p)> =0 $$

This must be zero. For example, because of charge conservation.

I think you‘re ready now to dive deeper into these concepts. Here are a few comments on the books about quantum field theory that I like most:

Recommended Books

There are hundreds of books about quantum field theory, but only a few are really, really good.

More book recommendations can be found here.