# What kets represent on QFT?

In Quantum Mechanics kets are used to represent states of a system. This is indeed well written in the first postulate of Quantum Mechanics which states that to describe a quantum system we use a Hilbert Space whose elements are kets, each representing a state of the system.

In that setting, observables, which are physical quantities associated to a system that can be measured are represented by hermitian linear operators on the said Hilbert space. In that case, one observable is one operator which acts upon kets.

Now, in Quantum Field Theory things change. A Quantum Field is a field of operators. But as far as I know, the system itself is the field. So things get messy: the system we want to study is a field (like the Electromagnetic Field or the electron field, or whatever), then to represent the system we have a Quantum Field, which is one operator-valued function. But now, if the system is described by operators, what the kets really are?

I mean, suppose $\Psi$ is a Quantum Field, in that case $\Psi(\mathbf{r})$ is one observable. If that's true, $\Psi(\mathbf{r})$ acts upon kets of the Hilbert Space. But now, those kets represent exactly what, if the system itself is now being represented by $\Psi$?

• In quantum field theory, we use kets to represent states of a system. Operators (e.g. $\Psi$ above) are used to describe evolutions of a system or some kind of measurements. This is same as in quantum mechanics. So for your question, the answer is that kets in QFT describe states of a system. – Minkyu Jun 12 '16 at 2:57
• I think you may have the wrong impression. QFT is just QM applied to fields. True the field is indexed by the space time position so a field represents a collection of operators. Kets still represent the states, e.g. there is one particle with momentum $k$ is represented by $|k>$. – Virgo Jun 12 '16 at 3:01

## 2 Answers

They kets represent a configuration of the field (and the terminology is usually upgraded from "Hilbert space" to "Fock space"). $\lvert 0 1 0 \cdots 0\rangle$ is that there is 1 excitation (a particle) in some place specified by the second place in that vector.

To explain, let's take a simple example: A particle in a box. Normally, in undergrad quantum you solve the Schrödinger equation:

$$E_1 \psi_1(x) = - \partial_x^2 \psi_1(x), \quad \psi(0)=0=\psi(L)$$

Here $\lvert \psi_1\rangle$ is a well-defined object in single particle quantum mechanics (i.e. $\psi_1(x) = \langle x | \psi_1 \rangle$). If I solve this eigenvalue equation for the first one: $E_1 = \pi/L$ then I get the wave function $\psi_1(x) = \sqrt{\frac2L}\sin(\pi x/L)$. But I want more particles and it is convenient to think of $\psi_1(x)$ as a field now. So the way I do this is that I think of a larger "Fock" space that also takes into account number of particles.

If we say these are bosons, then we use boson ladder operators to create a particle. So, say that $a_x^\dagger$ creates a boson exactly at position $x$ in our system. Well, the $\psi(x)$ in our infinite square well is not at a single position, it is spread out over the whole box. So we begin with a state $\lvert{0}\rangle$ that has no particles, and to create the state associated with our solution to Schrödinger's equation we write

$$\lvert{1 0 0 \cdots}\rangle = \int_0^L dx \,\psi_1(x) a_x^\dagger \lvert{0}\rangle$$

The notation $\lvert{1 0 0 \cdots}\rangle$ is now that I have one particle in the first energy level. In fact, the boson operator that creates a particle at any energy $n\pi/L$ is just

$$a_n^\dagger = \int_0^L dx \,\psi_n(x) a_x^\dagger$$

where $\psi_n(x)$ has energy $n\pi/L$. You can easily verify that these are still well-defined boson ladder operators and $[a_n,a_m^\dagger]=0$ if $n\neq m$ (the magic of orthogonality!).

So it is in this sense that the kets are now just "field configurations": In the way I wrote the ket above for the second quantized square well $\lvert 1 2 5 0 0 \cdots \rangle$ means 1 particle in $\psi_1$, 2 particles in $\psi_2$ and 5 particles in $\psi_3$. People like to write this ket as

$$\lvert 1 2 5 0 0 \cdots \rangle = a_1^\dagger (a_2^\dagger)^2 (a_3^\dagger)^5 \lvert \vec{0}\rangle,$$

where the $a$'s are my field operator. Those are what ket's in quantum field theory mean.

The indices inside my ket represent a particular basis for my boson field operators $a_n$ (the energy basis). I could also write a similar quantity in position space basis with $a_x^\dagger$ and those represent the field in quantum field theory $\Psi(\mathbf r)$.

But how does the Hamiltonian change? Well, consider this object:

$$\mathcal H = \int dx \, a_x^\dagger (-\partial_x^2) a_x.$$

Now apply this to our ket $\lvert{10\cdots}\rangle$. After some boson algebra we get

$$\mathcal H \lvert{10\cdots}\rangle = \int_0^L dx \,(-\partial_x^2\psi_1(x)) a_x^\dagger \lvert{0}\rangle = E_1 \int_0^L dx \,\psi_1(x) a_x^\dagger \lvert{0}\rangle = E_1 \lvert{10\cdots}\rangle.$$

We have found our Hamiltonian for the second quantized space of kets! Now, people like to work with the field instead of with the kets, so people will use this Hamiltonian $\mathcal H$ with the field operator $a_x$ to obtain its full evolution with the Heisenberg picture.

I mean, suppose Ψ is a Quantum Field, in that case Ψ(r) is one observable.

Ψ(r) is the operator field, not an observable. The observable is what appears after the operation on the ground state.

Take the electron field. It is described mathematically over all space time, but obviously all of space time is not filled with electrons!

If that's true, Ψ(r) acts upon kets of the Hilbert Space.

A free electron, a plane wave in first quantization, in second quantization is represented by the creation operators of electrons , their function following, creating and annihilating , what in first quantization is the direction of the observable electron wavepacket. And do not forget that always there is an integral running in order to get the probability of a specific observation, even of a free particle.

But now, those kets represent exactly what,

They are the ground state wavefunction of the first quantization solution of the boundary value and potential problem at hand.

In the mathematical analysis of the other answer by qgp07 , the zeroes in the long kets are the ground state wave function of the problem at hand .

if the system itself is now being represented by Ψ?

It is a bad choice to represent the mathematical operator field by Ψ, it confuses the field with the wavefunction.

Contemplating a Feynman diagram in space can help.

The incoming lines come in space as the mathematically correct sequence of creation and annihilation of the particle the line represents. At the interaction points, what are potentials in first quantization, become vertex functions and propagator functions under the integral which will give the probability of interaction at that (x,y,z,t). It is energy/momentum space which is the most useful predictive function of Feynman diagrams, and the logic is the same.