# I How to get the wavefunction of a single particle in QFT?

#### DoobleD

Summary
In QFT, can I create a single particle wavefunction by doing $\Psi(x,t)\left|0\right\rangle$?
Hi folks,

I'm trying to get a grasp on some of the basic concepts of QFT. Specifically, I'm trying to picture what are the actual fields of QFT and how they relate to wavefunctions. There are already many helpful posts about those concepts, here and in other places, but some points are fuzzy for me.

So it seems that in QFT:
- the fields are operator fields, and more specifically, quantum oscillator operator fields;
- the ground state is called the "vacuum state" and is devoid of particles (appart from virtual particles with short lifetimes due to the time/energy HUP), unlike in QM where states describe at least 1 particle.

I'm fine with the above. I've also read that the operator fields contain creation and annihiliation operators and this can be used to create a particle from the vacuum state, like so ($\Psi$ being my operator field) : $\Psi(x,t)\left|0\right\rangle$.

My (probably naïve) questions are then:
- is the result of $\Psi(x,t)\left|0\right\rangle$ a wavefunction $\Phi(x,t)$ (ignoring the energy eigenvalue factor) for the newly created single particle localized around the choosed $x$ position?
- is there a wavefunction for the vacuum state? I'd be tempted to say it's $\Phi(x,t) = 0$, meaning there's a 0 probability of finding a particle anywhere, but then $\Psi(x,t)\left|0\right\rangle$ would be $0$ too.

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#### vanhees71

Gold Member
No, wave functions do not make sense in QFT since QFT describes not a situation, where you have a fixed number of particles but you can create and destroy them in interactions, and that's why it's the natural description for collisions at relativistic energies (despite all the much more formal impossibilities of a first-quantization description of interacting relativistic particles).

A true single-particle state is something like
$$|\Phi,t \rangle =\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \Phi(\vec{x}) \hat{\psi}^{(+) \dagger}(t,\vec{x})|0 \rangle,$$
where $\Phi$ is some square-integrable function and $\hat{\psi}^{(+)$ is the positive-frequency part in the mode decomposition of free (sic!) fields.

There's no particle interpretation for states that are not asymptotically free states.

A very little known formalism that comes in an analogy way close to a "wave-mechanics formulation" is the wave-functional formalism of QFT, which you can find, e.g., in

B. Hatfield, Quantum Field Theory of Point Particles and Strings, Addison-Wesley, Reading, Massachusetts (1992).

• DarMM

#### DoobleD

Thank you for answering. This is not super clear to me but I'll try to work on it.