# Can we regard field operator $\Psi (x)$ as $a_{x}^{\dagger }$ ,$a_{x}$?

In real scalar CG-field, do we have $a_{x}^{\dagger }$ and $a_{x}$ operators?

Because we have $a_{p}^{\dagger }$ and $a_{p}$ , also the relation $\Psi (x)=\int dp\, \, a^{\dagger }e^{-ipx-i\omega t}+ae^{ipx+i\omega t}$

Since
$a_{p}^{\dagger }|0 \rangle=|p\rangle$ meaning a state with one particle of momentum p

Then I wish to have something similar,

$a_{x}^{\dagger }|0\rangle=|x\rangle$ meaning a state with one particle of location x

$\Psi (x) |0 \rangle=\int dp\, \, e^{-ipx}|p\rangle=|x\rangle$ meaning a state with one particle localized in x

$|x,x',x''\rangle$ meaning a state with three particles localized in x x' x''

So I think $\Psi (x)$ is $a_{x}^{\dagger }$, because $a_{x}^{\dagger }|0\rangle=|x\rangle$,

or I am wrong $\Psi (x) |0 \rangle=\int dp\, \, e^{-ipx}|p\rangle=|\Psi (x)\rangle$ which is different from $|x\rangle$

$\Psi (x) =\Psi^{\dagger } (x)$ implies $a_{x}^{\dagger }=a_{x}$ , then $a_{x}^{\dagger }|0\rangle=|x\rangle=a_{x}|0\rangle=0$ wrong!

Are there well defined $a_{x}^{\dagger }$ and $a_{x}$?

• In solid state tight binding model, we often write Hamiltonian as sum of $b_{i}^{\dagger }a_{j}+a_{j}^{\dagger }b_{i}$, these are the creation operators of real space. Well in QFT, it is integration over $\Psi (x)$. – wwwjjj Nov 4 '14 at 1:33

You can't really have free field operators $a_x$ and $a^\dagger_x$ because the excitations at nearby locations are coupled because of the derivative in the lagrangian density.

When you exploit translational symmetry in the theory and fourier transform, you go to a basis where modes with momentum $k$ do not couple to modes with momentum $k^\prime$. So you can write free-field operators as above.

If you insisted on constructing creation/annihilation operators at each point, they would couple nearby points and the Lagrangian/Hamiltonian would look like something in your comment.

If you're okay with that, then you're right in identifying $a_x \equiv \Psi(x)$

No you cannot. The issue is that in QFT we reduce spacetime coordinates from observables to mere parameters of our fields.

In non-relativistic quantum mechanics, you have time $t$ as a parameter governing the evolution of your states. Note that there is no operator that corresponds to time!

To implement relativity, we need to deal with space and time at the same level. This means to degrade space to a mere parameter in our theory. Our operators (like your $\Psi(x)$) now not only depend on time (which the operators in non-relativistic QM do in the Heisenberg picture), but they also depend on space. A quantization procedure like in regular QM is no longer available.

Note that the quantities that appear in the commutators that you impose in the quatization procedure are not $\vec x, \vec p$, but $\Psi, \Pi$, where $\Pi$ is the conjugate momentum to the field $\Psi$ and not the momentum operator that appears e.g. in the Poincare algebra.