# How to write localised wave function of a particular shape in Quantum Field theory?

We know how to write localised wave function for Quantum Mechanics . Say for a free particle of mass $$m$$ in non-relativistic Quantum Mechanics , we can have a Gaussian wave packet,

$$\Psi(x, 0 ) = \small \bigg[\dfrac{1}{{2 \pi \sigma^2}}\bigg]^{ \dfrac{1}{4}} \; \;\Large{e}^{ \small{-\big[ \dfrac{x-a}{2\sigma} \big]^{\Large2}}} \Large{e^{\large{i k x}}}$$

which describes a wave packet localised in time $$t = 0$$ at position $$x = a$$ with spread $$\sigma$$. Obeying Schrodinger evolution equation it spreads as time increases. After solving we get;

$$\Psi(x,t ) = \small \Bigg[\dfrac{1}{{2 \pi \sigma^2 {\big(1+ \dfrac{it}{\tau}} \big )}}\Bigg]^{ \dfrac{1}{4}} \; \;\Large{e}^{ \small{-\Bigg[ \dfrac{x-a-2k\sigma^2t}{2\sigma \sqrt{1+ \dfrac{it}{\tau}}} \Bigg]^{\Large2}}} \Large{e^{\large{i k( x-k \sigma^2t)}}}.$$

Can we have analogous localised state in Quantum Field Theory? How can we define them say in Free Scalar Field Theory of mass $$m$$? And how does it evolve?

This is more subtle than you might think. The simple final answer is shown at the end, in equation (6). The rest of this post explains why its interpretation is subtle.

The question The concept of particle in QFT is related, but this new question is more specific because it focuses on the idea of a localized wavefunction.

## What defines "location" in QFT?

Consider a free scalar field, and work in one-dimensional space for simplicity. The equal-time canonical commutation relations are $$\big[\phi(x,t),\dot\phi(y,t)\big]=i\delta(x-y) \tag{1}$$ and $$\big[\phi(x,t),\phi(y,t)\big]=0 \hskip2cm \big[\dot\phi(x,t),\dot\phi(y,t)\big]=0, \tag{2}$$ and the equation of motion is $$\ddot\phi(x,t)-\nabla^2\phi(x,t)+m^2\phi(x,t)=0 \tag{3}$$ where $$\nabla$$ is the derivative with respect to $$x$$.

By definition, the field operator $$\phi(x,t)$$ is localized at $$x$$ at time $$t$$. This defines the relationship between observables and regions of spacetime, which is central to the question.

## In relativistic QFT, particles can only be approximately localized

The familiar concept of "particle" combines two logically distinct attributes: particles are countable, and a particle has a location. In relativistic QFT, the first attribute remains meaningful in the case of a free scalar field, but the second attribute is only approximately meaningful. This is because in relativistic QFT, the vacuum state — which has no particles by definition — is entangled with respect to location: the connected correlation function $$\langle 0|\phi(x,t)\phi(y,t)|0\rangle- \langle 0|\phi(x,t)|0\rangle\, \langle 0|\phi(y,t)|0\rangle \tag{4}$$ is non-zero even for $$x\neq y$$. For this reason, any attempt to precisely define the location of a particle is doomed from the start.

However, a particle can still be approximately localized. Write $$\phi(x,t)=\phi^+(x,t)+\phi^-(x,t)$$ where the superscripts $$\pm$$ denote the positive/negative-frequency parts of $$\phi(x,t)$$, respectively. We can use $$\phi^\pm(x,t)$$ as the creation/annihilation operators of a particle that is approximately localized at $$x$$ at time $$t$$. The localization is only approximate because the operators $$\phi^\pm(x,t)$$ do not commute with $$\phi(y,t)$$ when $$x\neq y$$, so applying $$\phi^\pm(x,t)$$ to a state unavoidably affects observables at other points $$y$$ at the same time. These effects fall of exponentially with distance $$|x-y|$$, with characteristic scale $$\sim 1/m$$ determined by the particle's mass. That's a very small scale for a typical particle (the Compton wavelength), so the approximation is pretty good for most macroscopic purposes and becomes exact in the non-relativistic limit.

## Creating a particle with a given "wavefunction"

For any complex-valued function $$\Psi(x)$$, the operator $$\int dx\ \Psi(x)\phi^-(x,t) \tag{5}$$ and creates a particle with "wavefunction" $$\Psi(x)$$, but only in a loose sense. Even if $$\Psi(x)$$ is zero everywhere outside a region $$R$$, observables outside of $$R$$ are still sensitive to the particle's presence, as explained above. With that caveat in mind, the state $$\int dx\ \Psi(x)\phi^-(x,t)|0\rangle \tag{6}$$ answers the question.$$^\dagger$$ In the non-relativistic limit, where $$\phi^-(x,t)$$ becomes strictly localized, the wavefunction $$\Psi(x)$$ regains its familiar meaning.

$$^\dagger$$ With the usual conventions, the negative-frequency part of $$\phi$$ acts as a creation operator, as explained in Why is photon annihilation associated with the POSITIVE frequency component of the electric field?