Quantum Field Theory

This page is absolutely meant for people who are just getting started with particle physics. The theory of particle physics deeply relies on what we know as the Quantum Field Theory. You can take a look at a basic ground in understanding what QFT deals with and its significance in particle physics in my blog post The Standard Model and Beyond. However, it is basically a hand wavy way of introducing the subject. Unlike other posts on my page, this is going to be about the theoretical framework on which particle physics is actually based on. The post is still going to be very basic in terms of level of understanding, however it might require a bit of an expertise in basic linear algebra, group theory, intermediate level classical mechanics, special relativity and quantum mechanics.

The Lagrangian method to understand Quantum Fields

This is a very interesting and simple outline to a method in mathematics that shall help us in understanding Quantum Fields.
In classical physics when we describe the mechanics of a system, we often use what is called the 'Lagrangian'. It is just a function that summarizes you about the dynamics of the system, and allows you to solve what is actually going on. It is written by taking the Kinetic energy of a system, and subtracting from it the potential energy of the system: $L=T-V$ . In classical physics, generally $T$ , and $V$ are functions of position and velocity. Then, we can apply what we call the Euler Equation on the Lagrangian, to derive its equations of motion. In classical mechanics it is written as: $\boxed{\frac{d}{dt}\left(\frac{\partial L}{\partial\dot{x}}\right)-\frac{\partial L}{\partial x}=0}$ . These two equations, are the only ones we need to solve a system.

In case you are not already familiar with Lagrangian dynamics that I summarized here, you can refer to the notes and the example in here to understand what it actualy means: Notes, Pendulum Example. Now, when we come to the Standard Model, we will be dealing with fields and not solid objects as we saw previously. Moreover these are going to be Quantum Fields and not classical Fields. But it turns out there exists a similar kind of approach, a Lagrangian method, applicable to quantum fields as well. I would simply ask you to trust me and accept it without going into much of a detail.

But before getting into that let me try to answer a few questions that might have cropped into your mind.

What kinds of fields will we be talking about?
How can we have a Lagrangian for a field?
How can a field have Kinetic and Potential energy?

So, as we discussed before, the particles in the Standard Model are simply the excitation of their respective fields. These fields are present in all of space, even in empty space (having a value of 0). If energy is given to it to excite it, it can produce ripples, or disturbances from its zero value (the particle). The mss of the particle will come from some of the energy we give by Einstein's famous equation $E=mc^2$ . So, we are going to be interested in a lots of different fields, which break down into three main types:

Fermion Field: When excited will form a Spin-1/2 particle. Eg: The Electron field
Vector Field: When excited will form a Spin-1 particle (a Gauge Boson). Eg: The Photon field or the Electromagnetic field
Scalar Field: When excited will form a Spin-0 particle (Scalar Boson). Eg: The Higgs field

The first of these the Fermion field or better known as the Dirac Field. These are represented by $\psi$ . This describes the lepton and the quark fields that produce spin-1/2 particle upon excitation. The Dirac Equation, was the first equation to marry special relativity and quantum mechanics sensibly. The Dirac equation can be written down as: $\boxed{i\gamma^\mu\partial_\mu\psi-m\psi=0}$ I donot want you at this stage to get intimidated by this equation. I donot expect you to understand what this equation means at this stage, rather just want you to realize that here $\psi$ is a fermionic field, which upon excitation produces a fermion of mass $m$ . Quickly moving on to the Vector Field, that is the fields for Spin-1 bosons. If we say talk about electromagnetic field: denoted by $A$ . We can use the Maxwell's equationsas we shall see in a minute to derive the value of electric and magnetic field at a point from the information stored in $A$ , which we can measure. Clearly the $A$ field is not just a number that gives a value at any $(x,y,z)$ . It also has to describe the direction of this special kind of field at at any point, in space and time. It is known as the Electromagnetic 4 potential: with $A_\mu\equiv(A_0,\vec{A})$ . $A_0$ is the electric scalar potential, and $\vec{A}$ is the magnetic vector potential with 3-components along the coordinate axes. This four vector $A_\mu$ is all we need to describe all electromagnetic phenomena. It means that although the electromagnetic field $A$ is zero everywhere, its excitation produces the electromagnetic waves, or photons.
The Maxwell's equations to calculate the Electric and the Magnetic Fields: $\boxed{\vec{B}=\nabla\times\vec{A}}$ $\boxed{\vec{E}=\nabla\varphi-\frac{\partial\vec{A}}{\partial t}}$

Now, before we go on to discussing the next kind of field, it is important to understand what we mean by the term ‘Gauge’ when we say ‘Gauge bosons’ or ‘Gauge Fields’ or ‘Gauge Theory’. The values of the field that can be measured, $\vec{E}$ and $\vec{B}$ are not dependent on certain changes in the electromagnetic four potential $A_\mu$ . What it means is we can tweak the value of $A_\mu$ unders some constraints and still get the same answers for the measurables. To give you an example, say we have a scalar function $\alpha(t,x,y,z)$ , and we change $A_\mu \equiv (\varphi,\vec{A})$ as: $\varphi\rightarrow\varphi-\frac{\partial\alpha}{\partial t}$ $\vec{A}\rightarrow \vec{A} + \nabla a$
We will then have a totally new value of $A_\mu$ , but putting this into Maxwell’s equations that we saw previously, we derive the same value of $\vec{B}$ , and $\vec{E}$ . We say that the field $A_\mu$ is gauge invariant, or we can choose a different gauge (add or subtract different things and still get the same answer to the measurables!). Another word for this is Gauge Symmetry. Even though nothing changes, if we add or subtract some terms ins the equation, it turns out that some choices give us a cleaner and easier way to work out the calculations. So, a gauge theory is a type of field theory in which the Lagrangian is invariant for a continuous group of local transformations of the field. The word ‘local’ or ‘local transformation’ has a certain significance here. What the statement mean is for a field theory to be Gauge Theory, it must be invariant to changes in the underlying field which vary in some way fro point to point in space-time. The change is not ‘global’, neither in a haphazard way, rather the changes are different in different points in space and in time, but are predictive and in form of some ‘function’ of space-time.

The last kind of field we are going to look at is the Scalar Field, or the Higgs Field. For this we will use the symbol $\phi$ . There is no direction at any point, we just have a value at a particular point. So, now we have an idea about what kind of fields we are going to talk about. This brings us to the second question: How can we have a Lagrangian for a field? But how can we write anything down for kinectic energy or potential energy of $\psi$ , $A_\mu$ , or $\phi$

. A field stretches over all space, so we cannot write its energy down as we did for the case of classical systems. So we need to figure a way out! So we use what we call the Lagrangian density, denoted by $\mathcal{L}$ . So, we are considering here, the KE, or the PE per unit volume of a field (Although here volume refers to something non-trivial, its not the spatial volume, rather the volume of space-time). This shall bring us to the next question: How can a field have Potential Energy or a Kinetic Energy? Potential energy as we know is some kind of a ‘stored energy’. But how much energy is stored in a field? We propose that amount of energy stored in a field is proportional to the value of the field squared. I cannot give a proof of this at this stage, however can help motivating it from a simple electroststic example of parallel plate capacitor, where the energy per unit volume stored in between the two plates in form of electric field is given by: $\frac{\epsilon_0 E^2}{2}$ . So if in future we write a lagrangian with a part that looks like $\frac{\phi^2}{2}$ , it shouldn’t be a surprse. Coming to kinetic energy, it is weird that we are not talking about particles, or objects moving around here and there, rather it is a field. How can a field have KE? Not just that we should be able to write down KE per unit volume of space-time. Just like for an object KE has to do with how its position changes with time, so for a field KE is how the field changes with spacetime. So, changes in the value of the field in spacetime is what we will take as the kinetic energy of the field. So, the kinetic energy part in the lagrangian will have terms that somewhat looks like: $(\partial\phi/\partial t)^2$ - $(\partial\phi/\partial x)^2$ - $(\partial\phi/\partial y)^2$ - $(\partial\phi/\partial z)^2$ . I have not considered constants in this equation. This is certinly not a simple idea, atleast not as simple as it might seem. However, for our purposes, this works quite well. In order to condense the kinetic energy part by using the einstein summation notation: $(\partial_\mu\phi)^2$ , or $\partial^\mu\phi\partial_\mu\phi$ .

A Scalar Spin-0 Field

From the discussion above, it should not be a matter of surprise that I can write the Lagrangian density of a scalar field as $\boxed{\mathcal{L} = \frac{1}{2}(\partial_\mu\phi)^2-\frac{1}{2}\left(\frac{mc^2}{\hbar}\right)^2\phi^2}$ Now, in these kinds of equations the constants that we see here in the potential energy part might seem quite unintuitive. In units we use in particle physics, we genrally set $c=\hbar=1$ . Also, we write the mass of the excited particle as $\mu$ rather than m. So, our equation now looks like: $\boxed{\frac{1}{2}(\partial_\mu\phi)^2-\frac{1}{2}\mu^2\phi^2}$ . Its high time for me to appologize for my notations, the two $\mu$ in the above equation are two completely different things. The $\mu$ in the derivative is just an index taking values 0,1,2,3, whereas the $\mu$ in the potential energy term is what gives us the mass of the particle that is produced as a result of the excitation of the field.Now, its time for us to apply Euler equation. But here in case of QFT, it becomes: $\boxed{\partial_\mu\left(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\right)-\frac{\partial\mathcal{L}}{\partial\phi}=0}$ So, applying this to our simple scalar field lagrangian, we get: $\partial^2_\mu\phi - \mu^2\phi=0$ . This is what we can call the equation of motion of the field. This equation looks somewhat like the equation of motion of a SHO $\frac{d^2x}{dt^2}-\omega^2x=0$ , where $x\rightarrow\phi$ and $t\rightarrow x^\mu$ .

So, if in the equation, the sign of the second term is negative, as in here $\partial^2_\mu\phi - \mu^2\phi=0$ ,then the system is perfectly straight forward. The cup like potential is curved upwards, the field is in stable equilibrium. Any disturbance will make the field oscillate. There will be energy involved in the oscillation, as the field increases and decreases, and the disturbance (the particle), will have a mass $\mu$ , which is real. If for some reason the second term became zero,the mass term shall disappear, and the lagrangian would be $\mathcal{L}=\frac{1}{2}(\partial_\mu\phi)^2$ . In that case we will have a massless particle with no potential energy involved. It is a bit like a ball rolling on a flat surface. The field could still be excited, but as the field goes up and down, no potential energy would be involved! Now, if for some reason the mass $\mu$ becomes imaginary (we will see later), this shall make $\mu^2<0$ , so we will have a positive sign instead of a negative sign before the second term (the one containing $\phi^2$ term). It would be very strange, like a ball on the top of a hill.Somehow the field would be unstable. We will get back to this later! So, this is the basic of the Higgs Field Lagrangian.

But there is one more thing: the Higgs field is called a free field, because there are no kinds of charges, from which it emanates or ends. However the fact that it is self-sourcing means that it also interacts with itself, this is called Higgs self interaction. This will give an extra potential term to the field. It is not possible to give a classical analogy of this, so we will simply take this as an ansatz that the potential energy of the Lagrangian doesn’t just contain a $\phi^2$ term, but also a quartic term,a term with $\phi^4$ . This quartic term means that the field somehow interacts with itself. So our final take to the Lagrangian of a free scalar field looks somewhat like $\boxed{\mathcal{L}=\frac{1}{2}(\partial_\mu\phi)^2-\left( \frac{1}{2}\mu^2\phi^2+\frac{1}{4}\lambda\phi^4\right )}$ . The parameter $\lambda$ is dimensionless, called the quartic self coupling. It accounts for the strength of the Hggs self interaction, and affects the production rate of the Higgs particles pairs in experiments. The actual value of $\mu$ , and $\lambda$ are not predicted by the standard model. But as of now as the mass of Higgs has been calculated from experiments(~125GeV), the value of $\lambda$ can be calculated, and has been found to be around 0.12.

Free Spin-1/2 Fermion Field

Also known as the Dirac Field, the Lagrangian for this field looks somewhat like $\boxed{\mathcal{L}=i\bar{\psi}\gamma^\mu\partial_\mu\psi-m\bar{\psi}\psi}$ . This might look a bit intimidating, but we can try to understand this in a naive way. The first term has a 4-vector analogy of rate of change of field, which is some kind of a ‘velocity’, so it looks like the KE part of the lagrangian (Although, there is no squaring!). And the second part looks somewhat like the potential term in case of the scalar field, with $\bar{\psi}\psi$ instead of $\phi^2$ . The $\gamma^\mu$ needs a bit explanation. It is a special set of matrices. It is not very simple, but lets not get into that intricacy as of now. What is important is, if we apply to this the Euler Lagrange Equation, we get the Diract equation: $\boxed{i\gamma^\mu\partial_\mu\psi-m\psi=0}$ .

Spin-1 Vector Boson Field

The Lagrangin for this field is called a Proca Lagrangian. It looks somewhat like this $\boxed{\mathcal{L}=-\frac{1}{16\pi}(\partial^\mu A^\nu-\partial^\nu A^\mu)(\partial_\mu A_\nu-\partial_\nu A_\mu)+\frac{m^2}{8\pi}A^\nu A_\nu}$ . It looks awful!, but lets look some important stuff about it for now! Firstly, the equation has mass $m$ in it. Now, as we know for some gauge bosons such as photons have no mass, so there we simply set $m=0$ . This is just a more general case. Now, in the lagrangian if say for electromagnetic field, the two brackets, in the first term are called the Electromagnetic tensors, denoted by letter $F$ . So, the Proca Lagrangian can be simply written as: $\boxed{\mathcal{L}=-\frac{1}{16\pi}F^{\mu\nu}F_{\mu\nu}+\frac{m^2}{8\pi}A^\nu A_\nu}$ . Now, with this Lagrangian if we focus on the Electromagnetic Field, and choose a particulae gauge called the Lorentz gauge, which makes $m=0$ , then $\mathcal{L}=-\frac{1}{16\pi}F^{\mu\nu}F_{\mu\nu}$ . If we play around with this massless Lagrangian, to put it in the Euler’s equation, we will get the maxwell’s equations with no charge or current in them, because it is a free field.

Putting Together

Now, if we ask a question, what if we put together all these? I mean lets say two electrons, which are described by the Dirac field interact with the help of electromagnetic field. So, in that case the Lagrangian should have both $\psi$ and $A_\mu$ . This Lagrangian will be called the Lagrangian of Quantum Electrodynamics, which looks like: $\boxed{\mathcal{L}=i\bar{\psi}\gamma^\mu\partial_\mu\psi-e\bar{\psi}\gamma^\mu A_\mu\psi-m\bar{\psi}\psi-\frac{1}{4}F^{\mu\nu}F_{\mu\nu}}$ . I donot expect it to make a great deal of sense to you at the moment, but it woulb be great if you can identify the fact that the Lagrangian contains free fermionic terms, the free mass-less proca term, as well as a interaction term containing both $\psi$ and $A_\mu$ . The second term (the interaction term) in the lagrangian is what causes the attraction, or repulsion. The strength of this term is given by the multiplier $e$ , the electric charge.

Mass of the Gauge Bosons

At this point we might take up the statement I raised earlier, 'The Standard Model insists that the gauge bosons have to be described mathematically by field equations and Lagrangians for massless particles.' Why is that so? It has all to do with Gauge invariance. It applies to all the four gauge bosons. So, lets consider the free vector field $A_\mu$ that could have a mass. Now I transform this vector field by adding something to it, say a scalar function $\alpha(t,x,y,z)$ , which maynot be constant in space-time. If we find a derivative of this with respect to space-time, this shall also vary in space-time in a very general case. So we transform $A_\mu\rightarrow A_\mu+\partial_\mu\alpha$ . Now, the Proca Lagrangian was $\mathcal{L}=-\frac{1}{16\pi}F^2+\frac{m^2}{8\pi}A^\nu A_\nu$ . So, if I make the local transformation does, this Lagrangian change? If we take a look at just the mass term: $\mathcal{L}_{mass}=m^2A^2$ . If we transform $A\rightarrow A+\partial\alpha$ , then it becomes :

$\mathcal{L}_{mass}=m^2(A+\partial\alpha)(A+\partial\alpha)=m^2A^2+2m^2A\partial\alpha+m^2(\partial\alpha)^2$
But in order for it to be invariant the 2^nd, and the 3^rd terms should disappear. If we look at the 3^rd term, we see that it is just some kind of a kinetic energy term and has nothing to do with the mass of the disturbance. So, we are just left with the $\mathcal{L}_{mass}=m^2(A+\partial\alpha)(A+\partial\alpha)=m^2A^2+2m^2A\partial\alpha$ . The 2^nd term has to vanish for Gauge symmetry. The only way is to put the extra bit to be zero. This can be done in three ways:

$\partial\alpha=0$ : Which means adding nothing! Doesn't help.
$A=0$ : Which means underlying field itself vanishes, which cannot be the case either.
$m=0$ : which means mass of the excited particle of the field, the gauge boson is zero. It is the only sensible way out! So all gauge bosons, of spin-1 must be massless, and the forces that they describe must be long ranged.<\li> </ul>
But the point is ofcourse, they all donot have a zero mass.The $W^\pm$ and the $Z^0$ have quite large masses from experiments. Moreover, the weak forces are short ranged forces. Which means that in low energy situations, where the energy is small, they have to borrow large amount of energy out of vacuum in order to be created, so that the interaction can take place. And, so according to energy-time uncertainty, they have a very small decay width or lifetime. So this energy has to be paid back quickly and the particle cannot travel very far before it decays. Now, that we have discussed why the standard model insists that the gauge bosons have zero mass, we can start our discussion about why some of them have a large mass.

Spontaneous Symmetry Breaking and the Higgs Mechanism