2 Free quantised scalar field

Most discussions of QM describe the wavefunction as dynamic and time dependent and the operators acting on them as static. This is called the Schrödinger picture which describes wavefunctions as manifestly time dependent through the Schrödinger equation (2)

The Hamiton operator $\hat{H}_{S}$ itself does not depend on time.³³ 3 Note that for a brief period we will use a hat to indicate that an object is an operator Starting from some initial state $|{\Psi_{S}(0)}\rangle$, the time evolution of the state is described by

The operator $\hat{U}(t)$ fulfils the differential equation

Please keep in mind that this is a differential equation for an operator rather than a function. To ensure that $\langle{\Psi}|{\Psi}\rangle$ remains unchanged, we need the operator $\hat{U}$ to be unitary, i.e. $\hat{U}^{-1}=\hat{U}^{\dagger}$.

This picture is not ideal to derive a QFT in because the language we have developed for field theories explicitly talks about time dependence. Instead, we will use the Heisenberg picture where operators depend on time while states do not. To transform between the two pictures, we will use a unitary operator. The state of our system as it was at $t=0$ obviously does not depend on time which makes it our new state vector

To make use of this new picture, we also need to consider how an arbitrary operator $\hat{O}_{S}$ transforms. It is not difficult to see that

leaves observables invariant. Crucially for what we are about to attempt, this transformation also leaves the commutation relation between $\hat{q}$ and $\hat{p}$ invariant

This new operator fulfils the EoM if $\hat{O}$ has explicit time dependence through the Heisenberg equation assuming $\hat{O}_{H}$ has no explicit time dependence

This is very similar to the classical equivalent Poisson bracket

Suggested Exercise

Show that (68) indeed leaves expectation values invariant and that the EoM is follows from the Schrödinger equation.

From now on we will always assume the Heisenberg picture and drop in the subscript $H$.

One of the first things we have learned about QM was that the position and momentum operators do not commute, i.e. $[\hat{q},\hat{p}]=i$. Extending this for multiple particles, we can define the operator for the position (or momentum) of the $i$-th particle as $\hat{q}_{i}$ ($\hat{p}_{i}$). In position space, these just take the form

The commutation relation is now extended to

since operators acting on different degrees of freedom will always commute. These are referred to as canonical commutation relations.

In the continuum limit, we have locations in space $\vec{x}$ instead indices $j$, field operators $\hat{\phi}(\vec{x},t)$ instead of coordinate operators $\hat{q}_{i}$, and conjugate momenta operator $\hat{\pi}(\vec{x},t)$ instead of momentum operators $\hat{p}_{i}$. We can now write the CCRs at equal time $t$

Note that these are in the Heisenberg picture as they depend on time! While this looks like it might not be Lorentz invariant (after all, we are treating time and space very different here), the formalism we are developing is invariant which will become clearer later.

We will get a better idea of what the operator $\hat{\phi}$ actually does once we write down a solution that satisfies both $\mathcal{L}$ and the CCRs. However, it is helpful to get an idea early on so it is a bit less abstract. Consider the vacuum state $|{0}\rangle$, i.e. a completely empty universe devoid of any particles (we will formalise this later as well but this suffices for now), as our initial state. We will see shortly that applying $\hat{\phi}(\vec{x},t)$ on the vacuum will create a new particle at position $\vec{x}$.

We can verify the EoM $i\partial_{t}\hat{\phi}=[\hat{\phi},\hat{H}]$ and $i\partial_{t}\hat{\pi}=[\hat{\pi},\hat{H}]$ follow from the CCRs. For this, we use that the Hamiltonian is (55)

where we have used that for the KG field $\pi=\dot{\phi}$ which translates to operators $\hat{\pi}=\dot{\hat{\phi}}$. We now calculate commutators

Here the derivative in the $\vec{\nabla}_{y}$ operator is w.r.t. $\vec{y}$ and therefore it commutes with $\hat{\phi}(\vec{x},t)$. Therefore,

which is fulfilling the EoM as $\hat{\pi}=\dot{\hat{\phi}}$.

Suggested Exercise

Show that

$\displaystyle[\hat{\pi}(\vec{x},t),\hat{H}]=i\Big{(}\nabla^{2}\hat{\phi}(\vec{x},t)-m^{2}\hat{\phi}(\vec{x},t)\Big{)}\,.$

(79)

Together, we can deduce the KG equation as

In (63) we have seen a solution $\phi(\vec{x},t)$ to the KG wave equation. To turn this into a solution for the field operator $\hat{\phi}$, we promote the coefficients $a$ and $a^{*}$ to operators, i.e. $a_{\vec{k}}\to\hat{a}(\vec{k})$ and $a_{\vec{k}}^{*}\to\hat{a}^{\dagger}(\vec{k})$. Explicitly, we have for $\hat{\phi}$ and $\hat{\pi}$ (now written in terms of the four-vector $x$ rather than $\vec{x}$ and $t$)

where we still have $k^{0}=E_{\vec{k}}$.

It is easy to be overwhelmed by the number of symbols in the equation above so let us focus on the relevant terms to schematically write

These equations should look very familiar as they are (up to the $\vec{k}$ dependence), the decomposition of $\hat{x}$ and $\hat{p}$ in terms of ladder operators $\hat{a}$ and $\hat{a}^{\dagger}$ of the quantum harmonic oscillator. We can find the CCRs for the $\hat{a}$ and $\hat{a}^{\dagger}$ operators as

Suggested Exercise

Show that these follow from the CCRs of $\hat{\phi}$ and $\hat{\pi}$. To do this, inverse-Fourier transform (81) to show that

$\displaystyle\begin{split}\hat{a}(\vec{p})&=\frac{1}{\sqrt{2E_{\vec{p}}}}\int{\rm d}^{3}x\Big{[}E_{\vec{p}}\hat{\phi}(x)+i\hat{\pi}(x)\Big{]}{\rm e}^{ip\cdot x}\,,\\ \hat{a}^{\dagger}(\vec{p})&=\frac{1}{\sqrt{2E_{\vec{p}}}}\int{\rm d}^{3}x\Big{[}E_{\vec{p}}\hat{\phi}(x)-i\hat{\pi}(x)\Big{]}{\rm e}^{-ip\cdot x}\,.\end{split}$

(84)

Now you can calculate the commutators and reduce them to the CCRs.

The harmonic oscillator became a lot simpler once we wrote the Hamiltonian $\hat{H}$ in terms of the ladder operators. We can do something similar here with one exception

Let us swap $\vec{p}\to-\vec{p}$ and $\vec{q}\to-\vec{q}$ in the terms that have a positive exponential. Note that this only changes the spatial components, i.e. $p\cdot x=E_{\vec{p}}\cdot t-\vec{p}\cdot\vec{x}\to E_{\vec{p}}\cdot t+\vec{p}\cdot x=2E_{\vec{p}}t-p\cdot x$.

We can use that

to set $\vec{q}=-\vec{p}$ and $E_{\vec{q}}=E_{\vec{p}}$

where we have used $\vec{p}\cdot\vec{p}+m^{2}=E_{\vec{p}}^{2}$ since this on-shell condition was always assumed. We now need to multiply out the ladder operators to find

Swapping the integration variable $\vec{p}\to-\vec{p}$ in the second term and commuting the $\hat{a}$ to the right of the $\hat{a}^{\dagger}$

This is again very similar to the harmonic oscillator where we ended up setting $[\hat{a},\hat{a}^{\dagger}]=1$ and $\hat{H}=\omega(\hat{a}^{\dagger}\hat{a}+1/2)$. Here we have a problem though as the commutator evaluates to $\delta(0)$ which is not defined.

What does it mean for $\hat{H}$ to be infinite like this? In practice, nothing. Experiments can only ever measure the difference between energies so that this term will always cancel. However, this is rather unsatisfactory as it does not really solve the underlying problem that our result is infinite, nor does it help us work with the expression. It is quite unwieldy to have infinities like this that are not regulated somehow as it is all too easy to accidentally have the infinity appear “for real” if we for example had actually used the commutation relation rather than just writing the commutator. We need a formal and rigorous way of handling these terms from the beginning.

Lattice regularisation

The simplest way to formalise the removal of this to put our system in a box. Rather than having an infinitely large universe which allows for uncountably infinitely many momenta, we have a finite volume of length $L$ (for example with periodic boundary conditions). This means only discrete momenta are allowed

$\displaystyle p^{i}=\frac{2\pi}{L}n^{i}\qquad\text{with}\qquad n^{i}=0,\pm 1,\pm 2,...$

(93)

and we translate

	$\displaystyle\int\frac{{\rm d}^{3}p}{(2\pi)^{3}}$	$\displaystyle\to L^{-3}\sum_{\vec{n}}$		(94)
	$\displaystyle\delta^{(3)}(\vec{p}-\vec{q})$	$\displaystyle\to\bigg{(}\frac{L}{2\pi}\bigg{)}^{3}\delta_{\vec{p},\vec{q}}\,.$		(95)

Now we have $(2\pi)^{3}\delta(0)=L^{3}$ which is perfectly regular as long as $L$ is finite. In the limit $L\to\infty$, $\hat{H}$ still divergences but we can now define exactly how we want to subtract the $\delta(0)$ problem. Unfortunately, we are still not done because

$\displaystyle\hat{H}=L^{-3}\sum_{\vec{n}}E_{\vec{p}}\Big{(}\hat{a}^{\dagger}(\vec{p})\hat{a}(\vec{p})+\frac{1}{2}\underbrace{\big{[}\hat{a}(\vec{p}),\hat{a}^{\dagger}(\vec{p})\big{]}}_{(2\pi)^{3}\delta(0)}\Big{)}=L^{-3}\sum_{\vec{n}}E_{\vec{p}}\ \hat{a}^{\dagger}(\vec{p})\hat{a}(\vec{p})+\sum_{\vec{n}}\frac{E_{\vec{p}}}{2}\,.$

(96)

Even though the $L^{3}$ has cancelled, we still have a sum over all possible momenta that will diverge since $E_{\vec{p}}\sim|\vec{p}|\sim\vec{n}$. There are multiple ways of regularising this sum such as using Ramanujan summation (which is common in string theory and would assign $\sum_{n}n=-1/12$) or lattice regularisation (which defines a largest possible momentum or equivalently a smallest lattice spacing). Whatever way we end up choosing, we can modify our original $\mathcal{L}$ to avoid this problem as a function of the regulator.

This is our first encounter with regularisation (which makes the problem explicit) and renormalisation (which removes the problem at the level of $\mathcal{L}$).

Another more elegant way to deal with this problem is to backtrace a bit. Fundamentally, our problem arises from the commutator of $\hat{a}$ and $\hat{a}^{\dagger}$ which in turn arises from the $\hat{\phi}$ field operator. In our classical field theory, $a$ and $a^{*}$ were numbers so their order did not matter. Similarly, when we derived the Hamiltonian $\mathcal{H}$ we wrote $\phi^{2}$ because $\phi$ is a number. During quantisation, we translated $\phi^{2}\to(\hat{\phi})^{2}$ which lead to the commutator when we decided to move $\hat{a}$ to the right of $\hat{a}^{\dagger}$. The trick to avoid this is to always ensure that $\hat{a}$ is to the right of $\hat{a}^{\dagger}$ by using normal ordering. In a normal-ordered expression, which we indicate with colons, the operators are defined such that $\hat{a}^{\dagger}$ always comes before $\hat{a}$ when we go from a classical to a quantum theory. For example

instead of the unordered result

Both of these have the same classical limit but are different operators. However, only the normal-ordered one $:\hat{\phi}^{2}:$ avoids the problem of the $\delta(0)$ if we define $\hat{H}$ as

We can now finally answer the question raised earlier about the interpretation of $\hat{\phi}$ by using the ladder operators $\hat{a}$ and $\hat{a}^{\dagger}$. Similar to the discussion of the harmonic oscillator, we begin by defining a vacuum state $|{0}\rangle$ such that $\hat{a}$ destroys it

and it is properly normalised, i.e.

The (normal-ordered) energy of this state is just zero since it is immediately destroyed by $\hat{a}$. This is the reason that $|{0}\rangle$ is called the vacuum state and why the problematic contribution of the $\delta(0)$ to $\hat{H}$ we encountered earlier is called vacuum energy.

Now that we know what the vacuum is, what happens if we let $\hat{a}^{\dagger}$ act upon it? What is the interpretation of the new state

This state is clearly orthogonal

A good and Lorentz-invariant choice of the normalisation is $C_{p}=\sqrt{2E_{\vec{p}}}$ because then

which is Lorentz invariant.

Lorentz invariance and integrals

Now is a good time to look at this term a bit closer. We will very often see integrals of the form

$\displaystyle\int\frac{{\rm d}^{3}p}{(2\pi)^{3}\,2E_{\vec{p}}}f(p)\,,$

(105)

with $E_{\vec{p}}=\sqrt{\vec{p}^{2}+m^{2}}$. Even though it does not appear to be Lorentz-invariant, it is. This means that if $f(p)$ is Lorentz-invariant, so is the integral. Because it is so very common, the measure it is often abbreviated as ${\rm d}\Phi_{p}$, i.e.

$\displaystyle\int\frac{{\rm d}^{3}p}{(2\pi)^{3}\,2E_{\vec{p}}}=\int{\rm d}\Phi_{p}$

(106)

To show this, let us begin by adding an additional integration over $p^{0}=E_{\vec{p}}$ which is forced on-shell by a delta function

$\displaystyle\int{\rm d}\Phi=\int\frac{{\rm d}^{4}p}{(2\pi)^{4}}\frac{(2\pi)\delta(p^{0}-E_{\vec{p}})}{2E_{\vec{p}}}\,.$

(107)

We can use the identity

$\displaystyle\delta(\alpha x)=\frac{\delta(x)}{|\alpha|}\,,$

(108)

with $\alpha=2E_{\vec{p}}$ to write

$\displaystyle\frac{(2\pi)\delta(p^{0}-E_{\vec{p}})}{2E_{\vec{p}}}=(2\pi)\delta\Big{(}\big{(}2E_{\vec{p}}\big{)}\big{(}p^{0}-E_{\vec{p}}\big{)}\Big{)}=\theta(p^{0})\ (2\pi)\delta\Big{(}\big{(}p^{0}+E_{\vec{p}}\big{)}\big{(}p^{0}-E_{\vec{p}}\big{)}\Big{)}\,.$

(109)

Here we have used that, as long as $p^{0}>0$ (as enforced by the Heaviside function), $p^{0}+E_{\vec{p}}=2E_{\vec{p}}$. Expanding the argument and using the definition of $E_{\vec{p}}$

$\displaystyle\int{\rm d}\Phi=\int\frac{{\rm d}^{4}p}{(2\pi)^{4}}\theta(p^{0})\ (2\pi)\delta\big{(}p^{2}-m^{2}\big{)}\,.$

(110)

For proper, orthochronous Lorentz transformations, i.e. those that do not change the sign of $p^{0}$, this is manifestly Lorentz-invariant.

Note that we can create multiple particles as well. If we apply $\hat{a}^{\dagger}(\vec{p})$ and $\hat{a}^{\dagger}(\vec{q})$ we get

Since the two creation operators commute, $|{\vec{p},\vec{q}}\rangle=|{\vec{q},\vec{p}}\rangle$. This allows us to conclude that the particles described by the KG equation follow Bose-Einstein statistics, i.e. do not pick up a sign under permutation The spin-statistics theorem, for which no simple proof is known, states that particles following the Bose-Einstein statistics (i.e. bosons) have integer spin while those that follow the Fermi-Dirac statistics (i.e. fermions for which $|{\vec{p},\vec{q}}\rangle=-|{\vec{q},\vec{p}}\rangle$) have half-integer spin.

The first physics question we can ask of our theory is the amplitude of a particle travelling from $y$ to $x$. This is an important point from a causality perspective as well: a particle created at $y$ should only be able to reach $x$ if the distance between them is time-like, i.e. $(x-y)^{2}>0$.

(117) is our first example of a vacuum expectation value (vev) which is an object of the form

for some (potentially complicated) operator $\hat{O}$. The term is more commonly used for the vev of a single field $v=\langle{0}|\hat{\phi}|{0}\rangle$. The only field in nature with a non-zero is the Higgs field which has $v=246\,{\rm GeV}$ and is the reason for masses of the $W$ and $Z$ bosons.