- #1

DoobleD

- 259

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- TL;DR 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.

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.