# What is a quantum field?

[2017-09-24 Sun]

As a mathematician trying to learn QFT, a question that always bothered me was "what is a quantum field?". In mathematics, we like to have a precise idea of what kind of objects we are working with (is it a set? is it a function? is it a topological space?). In QFT, the analytical difficulties in making the definition precise for even the simplest nontrivial theories are substantial; as a result, there are two standard attitudes to the question which are implicit in the way people write books about the subject:
• The first, completely reasonable, attitude is: "We will introduce the formalism and show you how to perform computations: that's what's important. What you do with the formalism in the privacy of your own room is your business."
• The second, also completely reasonable, attitude is: "We have thought long and hard about this, and the following is the cleanest and most beautiful axiomatisation of the subject we could find; we wish you the best of luck unpicking this."
Below, I will try and explain that the question does have a reasonable, intuitive answer, which you can explain without going too deeply into the analytical details. The moral is that there is actually no difference between quantisation in ordinary quantum mechanics and in QFT: there should be a Hilbert space of wavefunctions and some operators, the only difference is that the wavefunctions are defined on an infinite-dimensional space of field configurations (so we call them "wavefunctionals").

I will start by reviewing quantisation in ordinary quantum mechanics, before moving on to the simplest QFT (the Klein-Gordon field) and explaining the wavefunctional picture. Hopefully, once you have seen this intuition, the abstract frameworks built by books like Glimm-Jaffe or Streater-Wightman will make more sense.

## Review of quantisation

When we quantise, we start with a classical system and try to construct a quantum mechanical system which reduces to this classical system in the $$\hbar\to 0$$ limit. Of course, it's not clear that there is an answer, let alone a unique answer, or that there will be some prescriptive way to find the answer for an arbitrary classical system, even if it exists.

Nonetheless, let's review the basic set-up:

• A classical mechanical system comprises a phase space (symplectic manifold $$(M,\omega)$$) of positions and momenta; observable quantities are functions on this phase space. There is a distinguished function $$H$$ called the Hamiltonian which governs time evolution of the system. Explicitly, the Hamiltonian generates a Hamiltonian vector field $$X_H$$ on the phase space via Hamilton's equations $$\omega(X_H,-)=-dH$$ and integrating this vector field gives a flow $$\phi_H^t\colon M\to M$$ which is the time evolution of the system.
• A quantum mechanical system comprises a Hilbert space together with a collection of operators whose spectrum (eigenstates/eigenvalues) you want to compute. There is a distinguished operator $$H$$ called the Hamiltonian which governs time evolution of the system. In the Schrödinger picture, this time evolution works as follows. The Hamiltonian generates a one-parameter family of unitary transformations $$U_t$$ satisfying the Schrödinger equation $\frac{dU_t}{dt}=\frac{i}{\hbar}HU_t.$ (Unitarity of the transformation is equivalent to $$H$$ being Hermitian). For example, if $$H$$ is time-independent then $$U_t=e^{iHt/\hbar}$$. If the system is in the state $$\psi$$ at time $$0$$ then at time $$t$$ it will be in the state $$U_t\psi$$.

## Canonical quantisation of the free particle

The most basic classical system is a particle living in a 1-dimensional space with coordinate $$q$$. The phase space keeps track of the position $$q$$ but also the momentum $$p$$, so it is $$\mathbf{R}^2$$ with coordinates $$p,q$$ and symplectic form $$\omega=dp\wedge dq$$. We want to turn this into a quantum system by finding a suitable Hilbert space and turning functions of $$q,p$$ into operators on this Hilbert space.

We take our Hilbert space to be the space of square-integrable wavefunctions/ of $$q$$, written $$L^2(\mathbf{R})$$. The two most famous wavefunctions, whose physical interpretation is clear, actually don't live in this space (but can be closely approximated by wavefunctions in $$L^2$$):

• Fix a point $$q_0\in\mathbf{R}$$. The delta function $$\delta(q-q_0)$$ should represent something like a particle localised at the point $$q_0$$. Delta functions are not really functions at all, let alone square-integrable, but one can approximate a delta function arbitrarily closely by a strongly peaked Gaussian.
• Fix a frequency $$\omega$$. The function $$e^{i2\pi\omega q}$$ is a wave with pure frequency $$\omega$$. This should represent a particle (like a photon) whose momentum is $$\omega$$. To see why, recall:
• De Broglie's formula $$E=h\omega$$ ($$h=2\pi$$ in our units) relating the energy and frequency of matter waves,
• the fact that a light wave has energy equal to its momentum (up to a factor of $$c=1$$) since its energy-momentum vector lives on the null-cone in spacetime.
Again, this wavefunction is not square-integrable, but you could cut it off outside a large ball to get a reasonably good approximation.

We want these wavefunctions to be eigenfunctions of the operators $$\hat{q}$$ and $$\hat{p}$$ corresponding to position and momentum, with eigenvalues $$q_0$$ and $$p=2\pi\omega$$ respectively.

The wavefunction $$\delta(x-q_0)$$ is the unique $$q_0$$-eigenfunction of the operator $$(\hat{q}\psi)(q)=q\psi(q)$$. The wavefunction $$e^{ipx}$$ is the unique $$p$$-eigenfunction of the operator $$\hat{p}\psi=-i\partial_q\psi$$. The standard guess at a quantisation of this phase space is therefore to replace $$q$$ and $$p$$ by the operators $$\hat{q}$$ and $$\hat{p}$$ defined above, acting on the Hilbert space of $$L^2$$-functions of $$q$$.

If $$P$$ is a polynomial in $$p$$ and $$q$$ then we can try replacing $$P$$ by the corresponding polynomial in the operators $$\hat{p}$$ and $$\hat{q}$$. Alas, the operators $$\hat{p}$$ and $$\hat{q}$$ no longer commute, indeed, we have $[\hat{q},\hat{p}]=i,$ so there are many choices of how to order the operators in the polynomial $$P$$. For an example of how one quantises a particular quadratic Hamiltonian $$H=\frac{1}{2}p^2+\omega^2q^2$$, see my pre-QFT post on the simple harmonic oscillator.

## Quantum field theory

Let us now turn to the problem of finding a quantum system to replace the classical Klein-Gordon field. The Klein-Gordon field is a complex-valued function $$\phi(x)$$. The classical Hamiltonian associated to this field is $\frac{1}{2}\int d^3x\left(\pi^2+|\nabla\phi|^2+m^2\phi^2\right).$ The quantity $$\pi$$ is the analogue of "momentum" for a field: in the case of the Klein-Gordon field, it is just the time-derivative of $$\phi$$ (just as momentum is related to the time derivative of $$q$$).

When we quantised a free particle, our Hamiltonian was a function on the space of possible positions and momenta of the particle.

Now we are quantising a field, the Hamiltonian is a function on the space of possible configurations and momenta of the field.

When we quantised a free particle, we took as our Hilbert space the space of wavefunctions on $$\mathbf{R}^3$$, where $$\mathbf{R}^3$$ is the space of possible configurations (positions) of our particle.

Now we are quantising a field, we will take as our Hilbert space the space of wavefunctions on $$\mathcal{F}$$, the space of possible configurations of the field.

A configuration of the field is just a complex-valued function on $$\mathbf{R}^3$$. In other words, $$\mathcal{F}$$ is a suitable space of functions $$\phi(x)$$. A wavefunction will be a map $$\Psi\colon\mathcal{F}\to\mathbf{C}$$; something which eats a function $$\phi$$ and outputs a number, for example: $\Psi(\phi)=\int d^3x\phi(x).$ To distinguish the functions $$\phi\colon\mathbf{R}^3\to\mathbf{C}$$ from the wavefunctions $$\Psi\colon\mathcal{F}\to\mathbf{C}$$, we will call $$\Psi$$ a wavefunctional. The suffix -al denotes a function which eats functions and outputs numbers.

Just as a wavefunction in quantum mechanics describes some kind of superposition of particles at different points, the wavefunctional in QFT describes a superposition of field configurations.

I am not going to address the questions "which wavefunctionals do we allow?" or "what is the inner product making them into a Hilbert space?", nor will I talk about normalising wavefunctionals; therein lie some analytical issues. Instead, here are some examples of wavefunctionals.

• Given a function $$f\in\mathcal{F}$$, there is a delta-functional $$\delta(\phi-f)$$ concentrated at $$f$$ (which vanishes unless $$\phi\equiv f$$). This probably shouldn't be allowed in the rigorous theory, in the same way that the delta function $$\delta(q-q_0)$$ is not an $$L^2$$-function; nonetheless, it's a convenient storytelling device. You can imagine this as a QFT state where the field has a definite value $$f(x)$$ at each point $$x$$ (in the same way that the delta function $$\delta(q-q_0)$$ is a quantum mechanical state where the particle has definite position $$q_0$$).
• Given a function $$\lambda(x)$$, the wavefunctional $$\Psi(\phi)=\exp(i\int d^3x\lambda(x)\phi(x))$$ is going to be a QFT state where the field has definite "momentum" $$\pi=\lambda$$. This is analogous to the quantum mechanical "plane wave" state $$\psi(x)=e^{ipx}$$, which has momentum $$p$$.

Here are some examples of classical observables for fields and the corresponding QFT operators.

The function $$ev_x\colon\mathcal{F}\to\mathbf{C}$$ ("evaluation at $$x$$") is defined by $$ev_x\phi=\phi(x)$$. This is the field theory analogue of the position function $$q$$ for a free particle. What is the operator associated to $$ev_x$$ in QFT? When we quantised $$q$$ we used $$(\hat{q}\psi)(x)=x\psi(x)$$. Consider the operator $$\widehat{ev_x}$$ defined by $(\widehat{ev_x}\Psi)(\phi)=\phi(x)\Psi.$ The wavefunctional $$\Psi$$ is a $$\lambda$$-eigenstate of $$\widehat{ev_x}$$ if and only if $$\phi(x)\Psi(\phi)=\lambda\Psi(\phi)$$ for all $$\phi\in\mathcal{F}$$. This means that $$\Psi$$ must vanish on $$\phi$$ unless $$\phi(x)=\lambda$$. For example, we see that, if $$f\in\mathcal{F}$$ is a function, the delta-functional $$\Psi(\phi)=\delta(\phi-f)$$ is a simultaneous eigenstate for $$\widehat{ev_x}$$ for all $$x$$, with eigenvalues $$f(x)$$. This operator is usually called $$\phi(x)$$ in QFT textbooks.

There are also classical field observables $$ev_g(\phi)=\int d^3xg(x)\phi(x)$$ defined for any "smearing function" $$g$$, which averages the values of $$\phi$$ over the support of $$g$$, weighted by $$g$$. The corresponding operator should be $(\widehat{ev_g}\Psi)(\phi)=\left(\int d^3xg\phi\right)\Psi(\phi).$ For example, the operator $$\widehat{ev_x}$$ we already defined corresponds to $$g=\delta_x$$, the delta function at $$x$$. The map sending $$g$$ to the operator $$\widehat{ev_g}$$ is, I think what people mean when they talk about a quantum field as being an operator-valued distribution.

There should be an operator corresponding to the momentum $$\pi$$, and, by analogy with usual quantum mechanics, it should be $$-i\partial$$. However, now $$\partial$$ means differentiation in the space $$\mathcal{F}$$ of functions. Given a wavefunctional $$\Psi\colon\mathcal{F}\to\mathbf{C}$$ and functions $$f,\phi\in\mathcal{F}$$, we define the directional derivative $$(d_\phi\Psi)(f)$$ of $$\Psi$$ at $$\phi$$ in the $$f$$-direction to be $\left.\frac{d}{dt}\right|_{t=0}\Psi(\phi+tf).$ When $$f(x)=\delta(x-x_0)$$ is the delta function at $$x_0$$, this is often written as $\frac{\delta\Psi}{\delta\phi(x_0)}.$ The operator corresponding to field momentum at $$x$$ will be $$-i\frac{\delta}{\delta\phi(x)}$$. The state $\Psi(\phi)=e^{i\int d^3x\phi(x)\lambda(x)}$ is a simultaneous eigenstate for all these momentum operators, with eigenvalues $$\lambda(x)$$. To see this, note that \begin{align*} \lim_{t\to 0} \frac{e^{i\int d^3x(\phi(x)+t\delta(x))\lambda(x)}-e^{i\int d^3x\phi(x)\lambda(x)}}{t} &=e^{i\int d^3x\phi(x)\lambda(x)}\lim_{t\to 0}\frac{e^{it\int d^3x\delta(x)\lambda(x)}-1}{t}\\ &=e^{i\int d^3x\phi(x)\lambda(x)}i\lambda(x), \end{align*} so $-i\frac{\delta\Psi}{\delta\phi(x)}(\phi)=\lambda(x)\Psi(\phi).$

Finally, we turn to the most important operator in the theory of the Klein-Gordon field, the Hamiltonian.

We can rewrite the classical Hamiltonian $H=\frac{1}{2}\int d^3x\left(\pi^2+|\nabla\phi|^2+m^2\phi^2\right)$ as $H=\frac{1}{(2\pi)^3}\int d^3k a(k)^\dagger a(k),$ where \begin{align*} a(k)&=\int d^3x e^{ik\cdot x}\left(\omega_k\phi(x)-i\pi(x)\right)\\ a(k)^{\dagger}&=\int d^3x e^{-ik\cdot x}\left(\omega_k\phi(x)+i\pi(x)\right)\\ \omega_k&=\sqrt{|k|^2+m}. \end{align*}
We have $\int\frac{d^3k}{(2\pi)^3}a(k)^\dagger a(k)=\int\int\int\frac{d^3k}{(2\pi)^3}d^3xd^3ye^{ik\cdot(x-y)}\left(\omega_k\phi(x)-i\pi(x)\right)\left(\omega_k\phi(x)+i\pi(x)\right)$ Multiplying out gives four terms:
• The terms $$\int\int\int d^3kd^3xd^3e^{ik\cdot(x-y)}yi\pi(y)\omega_k\phi(x)$$ and $$-\int\int\int d^3kx^3xd^3ye^{ik\cdot(x-y)}i\pi(x)\omega_k\phi(y)$$ cancel. To see this, take the second term, switch $$x\leftrightarrow y$$ and change $$k\leftrightarrow -k$$. The integrand becomes the integrand in the first term, the volume form $$d^3kd^xd^3y$$ is invariant, so the terms cancel because one appears with a minus sign.
• The term $\int\int\int\frac{d^3k}{(2\pi)^3}d^3xd^3ye^{ik\cdot(x-y)}\pi(x)\pi(y),$ becomes $\int d^3x\pi(x)^2,$ using the identity $\delta(x-y)=\int\frac{d^3k}{(2\pi)^3}e^{ik\cdot(x-y)}.$
• The term $\int\int\int\frac{d^3k}{(2\pi)^3}d^3xd^3ye^{ik\cdot(x-y)}\left(|k|^2+m^2\right)\phi(x)\phi(y),$ becomes $\int d^3x\left(|\nabla\phi|^2+m^2\right)\phi^2,$ using the standard trick in Fourier theory of exchanging derivatives for factors of $$k$$.

When we quantise, we turn $$a(k)$$ and $$a(k)^\dagger$$ into operators $\widehat{a(k)}=\int d^3xe^{ik\cdot x}\left(\omega_k\phi(x)+\frac{\delta}{\delta\phi(x)}\right),\qquad \widehat{a(k)}^\dagger=\int d^3xe^{-ik\cdot x}\left(\omega_k\phi(x)-\frac{\delta}{\delta\phi(x)}\right).$ This notation has been chosen to suggest a formal similarity with the simple harmonic oscillator system, reviewed in another blog post. As with that system, there is a lowest-energy (or vacuum) state $$\Psi_0$$ which is annihilated by the operator $$\widehat{a(k)}$$, and therefore satisfies $\hat{H}\Psi_0=\int\frac{d^3k}{(2\pi)^3}a(k)^\dagger a(k)\Psi_0=0.$ One can generate other states by acting on the vacuum with $$\widehat{a(k)^\dagger}$$, $$\widehat{ev_x}$$ or combinations of these operators. The state $\widehat{a(k)}^\dagger\Psi_0$ has the interpretation of a single particle with momentum $$k$$. The state $$\widehat{ev_x}\Psi_0$$ has the interpretation of a single particle at $$x$$.

The wavefunctional $\Psi_0(\phi)=\exp\left(-\int\int\int\frac{d^3k}{(2\pi)^3}d^3xd^3y e^{ik\cdot(x-y)}\frac{\omega_k}{2}\phi(x)\phi(y)\right)$ is a vacuum state of the quantised Klein-Gordon theory (i.e. it is annihilated by $$\widehat{a(k)}$$ for all $$k$$).
We have \begin{align*} \frac{\delta\Psi_0}{\delta\phi(z)}(\phi)&= -\int\int\int\frac{d^3k}{(2\pi)^3}d^3x d^3y e^{ik\cdot(x-y)}\frac{\omega_k}{2} \left(\delta(x-z)\phi(y)+\phi(x)\delta(y-z)\right)\Psi_0(\phi)\\ &=-\omega_k\phi(z)\Psi_0(\phi), \end{align*} so if we apply $$\widehat{a(k)}$$ to $$\Psi_0$$, we obtain \begin{align*} \widehat{a(k)}\Psi_0&= \int d^3z e^{ik\cdot z}\left(\omega_k\phi(z)+\frac{\delta}{\delta\phi(z)}\right)\Psi_0\\ &=\int d^3z(\omega_k\phi(z)-\omega_k\phi(z))\\ &=0. \end{align*}

Note that this wavefunctional is even nicer if you write it in terms of the Fourier transform $$\tilde{\phi}(k)$$ of $$\phi(x)$$: $\Psi_0(\phi)=\exp\left(-\int\frac{d^3k}{(2\pi)^3}\frac{\omega_k}{2}\tilde{\phi}(k)^2\right),$ where we have used the fact that $$\tilde{\phi}(k)=\tilde{\phi}(-k)$$ since $$\phi$$ is real.

## Conclusion

The aim of all this was to show that, if you understand quantum mechanics, you can also understand quantum field theory (in principle!). The ideas are the same, but the wavefunctionals of QFT are on a different plane of difficulty: they are functions on the infinite-dimensional space of field configurations. To solve an eigenvalue problem in this context means solving an infinite-dimensional differential equation (like $$\widehat{a(k)}\Psi_0=0$$ in the final theorem).

For more on this, there is an excellent physics textbook which covers this material (thereby taking neither of the attitudes described in the first paragraph). Chapter 10 covers the material I described above:

• Hatfield "The quantum field theory of point particles and strings." Frontiers in Physics, Perseus, 1998.

The classic book by Glimm and Jaffe also covers this from a mathematical standpoint (and addresses issues which we glossed over regarding integration of wavefunctionals); Chapter 6 covers the material I described above:

• Glimm and Jaffe, "Quantum physics: a functional integral point of view", Springer Verlag, 1987.

For completeness, the other resources I have found most helpful for learning about QFT are:

• David Tong's lecture notes on QFT.
• Peskin and Schroeder, "An introduction to quantum field theory", Avalon, 1995.
• Ryder, "Quantum field theory", Cambridge University Press, Second Edition 1996.

Comments, corrections and contributions are very welcome; please drop me an email at j.d.evans at lancaster.ac.uk if you have something to share.

CC-BY-SA 4.0 Jonny Evans.