# Connection between quantum field and the wavefunction

The general question "What is a quantum field?" has been asked here before, but I'm looking for specific help in trying to iron out the details of my own personal interpretation and understanding.

In regular quantum mechanics, I noticed that a state $\left|\psi \right>$ can be created by introducing an operator $\Psi^{\dagger}$ (creation of the state) acting on the vacuum $\left|0\right>$ such that $\Psi^{\dagger}\left|0\right>=\left|\psi\right>$. This operator can be represented in position space as:

$$\Psi^{\dagger}=\int{dx}\,\psi(x){a_{x}}^{\dagger}$$

Where ${a_x}^{\dagger}$ ($a_x$) is creation (annihilation) of a particle at position $x$, satisfying the commutation relation $\left[a_x, {a_y}^{\dagger}\right]=\delta(x-y)$, and where $\psi(x)$ is the wavefunction we all know and love.

This representation isn't new, and gives rise to all the same physics and calculations, but it suggests that instead of the wavefunction $\psi(x)$ just being a probability amplitude of where to find the particle, it can also be thought of as a probability amplitude for a particle to be created at position $x$. It just so happens that when there is only one particle this amplitude is normalized to $1$.

Just as an example to give a flavor for the calculations in this reformulation, we can ask "Given our state $\left|\psi\right>$, what is the expected number of particles at positions x?" (The number operator being $N_x={a_x}^{\dagger}a_x$). This gives us:

$$\left<\psi\right|N_x\left|\psi\right>=\left<0\right|\Psi a_x^{\dagger}a_x\Psi^{\dagger}\left|0\right>=\left<0\right|\int{dy}\int{dz}\, \psi^*(y)\psi(z)a_ya_x^{\dagger}a_xa_z^{\dagger}\left|0\right>$$

$$=\left<0\right|\int{dy}\int{dz}\, \psi^*(y)\psi(z)\delta(y-x)\delta(x-z)\left|0\right>=\left|\psi(x)\right|^2$$

Which is the probability density. The "number of particles" you'd expect to find at this position is just this (times $dx$).

(You can also show other expected results in this formulation, such as $[X, P]=i\hbar$, by using $P=\int{dp} \, a_p^{\dagger}a_pp$, $X=\int{dx}\, a_x^{\dagger}a_xx$ and the relation $\left[a_x, a_p^{\dagger}\right]=(2\pi\hbar)^{-1/2}\exp(ixp/\hbar)$. Dynamics can also be found from using $i\hbar\frac{\partial \Psi^{\dagger}}{\partial t}=\left[H, \Psi^{\dagger}\right]$)

Now my attempt at extending this interpretation to a field goes like this:

Let $\Phi(x)$ be the superposition of the creation and annihilation operators of a state localized at position $x$.

$$\Phi(x)=\Psi_x+\Psi_x^{\dagger}$$

Where, using the fact that a localized state should be given by the wavefunction $\psi_x(y)=\sqrt{\delta(x-y)}$ (forget about the phase factor and the fact this is ill defined for now, I can still do calculations by rewriting the delta as a square or Gaussian and taking limits only later):

$$\Psi_x^{\dagger}=\int{dy}\, \psi_x(y)a_y^{\dagger}=\int{dy}\sqrt{\delta(x-y)}a_y^{\dagger}$$

Or equivalently in momentum space as (with $\hbar=1$):

$$\Psi_x^{\dagger}=\int{dp}\, \sqrt{\kappa(p)}e^{-ixp}a_p^{\dagger}$$

Where $\sqrt{\kappa(p)}e^{-ixp}$ is the Fourier transform of $\sqrt{\delta(y-x)}$, and $\kappa(p)$ is the limit of normalized Gaussian with standard deviation going to infinity. (Basically a tight distribution in position space and therefore large spread in momentum.)

This allows us to write $\Phi(x)$ as:

$$\Phi(x)=\int{dp}\, \sqrt{\kappa(p)}(e^{-ixp}a_p^{\dagger}+e^{ixp}a_p)$$

Which looks very similar to the fields used in Peskin and Schroeder as well as many other places, but with some differences that I can't figure out.

First of all $\Phi(x)$ is unit-less. Creation and annihilation operators generally have units of one over the root of their argument, so the operator creating a "state" should have units of one over that "state", which isn't a unit. You can also see that $a_p$ and $a_p^{\dagger}$ have units of $p^{-1/2}$, and a normalized Gaussian has units of one over it's argument, so $\sqrt{\kappa(p)}$ should have units of $p^{-1/2}$, which all together cancels with the $dp$ and makes $\Phi(x)$ unit-less. This will hold for higher dimensions too, since the integral will be over $d^dp$ and we'll have $\sqrt{\kappa^d(p)}$ and $a_p, a_p^{\dagger}$ have dimensions of $p^{-d/2}$.

However, in the standard formulation the field has units of $p$:

$$\phi(x)=\int\frac{d^3p}{\sqrt{2E_p}}(e^{-ixp}a_p^{\dagger}+e^{ixp}a_p)$$

I was hoping someone could help me find the connection or conversion between this and my formulation.