Annihilation operator of a particle at $\bf x$ for canonically qauntized Klein Gordon field

I am now reading the David Tong's lecture notes on quantum field theory.

http://www.damtp.cam.ac.uk/user/tong/qft/two.pdf

And I have some questions on whether there is some well-defined particle annihilation operator at position $\bf x$.

I already know that we can interpret $\phi ({\bf x}) \left| {0} \right\rangle$ as a single-particle state at position $\bf x$. However, it seems that we can not say that $\phi ({\bf x})$ is the creation operator for a particle at $\bf x$. This is because it is a real scalar field which satisfies $\phi ^{\dagger} ({\bf x}) = \phi ({\bf x})$. Therefore, if we try to interpret $\phi ({\bf x})$ as the particle creation operator then we will totally get lost of the corresponding particle annihilation operator at $\bf x$.

Therefore I am confused, is there any particle annihilation operator at position $\bf x$ in this rather simple real scalar field theory?

I would be grateful for any suggestion! Thanks!

• In spite of what Tong is saying, we cannot interpret $\phi(\boldsymbol x)|0\rangle$ as a single-particle state at position $\boldsymbol x$. That's a lie to the children that doesn't hold up. – AccidentalFourierTransform Apr 27 '18 at 15:28
• Oh really? Then what is the proper interpretation of the state $\phi ({\bf x}) \left| {0} \right\rangle$? Thanks! Btw in Peskin's book it is stated that "We will therefore put forward the same interpretation and claim that the operator $\phi ({\bf x})$ acting on the vacuum creates a particle at position $\bf x$" – Kuan-Sen Lin Apr 27 '18 at 15:44
• No interpretation whatsoever. It's just a mathematical tool. Broadly speaking, the only object with a physical interpretation is the $S$-matrix. This object can be calculated using $\phi(\boldsymbol x)$; but $\phi(\boldsymbol x)$ itself has no meaning. – AccidentalFourierTransform Apr 27 '18 at 15:51
• @AccidentalFourierTransform In a free quantum field theory, everything is properly defined, and it makes perfect sense to interpret the elements of the $n$-particle sectors of the Fock space as wavefunctions describing $n$ particles. In fact there is a very explicit characterization of the dynamics (or action of the Poincaré group if you prefer) in such case, that fits very well with such interpretation. – yuggib Apr 27 '18 at 16:42
• For interacting theories that are non-Fock, such as the few ones we know how to deal with mathematically in low dimensions, then the Fock vectors and free fields do not have much physical meaning (since the canonical observables are not represented in Fock space). The interacting fields and vacuum state should instead be considered. – yuggib Apr 27 '18 at 16:45

The Fock representation of the free field operator (valued distribution) $\varphi(x)$ can be written as the sum $\frac{1}{\sqrt{2}}(a^*(x)+a(x))$, where $a(x)$ and $a^*(x)$ are the annihilation and creation operator( valued distribution)s on the Fock space.
Since the acting on the vacuum $\Omega$, $a(x)$ satisfies $a(x)\Omega=0$, it then follows that $\varphi(x)\Omega= \frac{1}{\sqrt{2}}a^*(x)\Omega$, that is a Fock space vector (valued distribution) belonging to the sector with only one particle ("localized" at position $x$).
Alternatively, one could also define the field momentum operator (valued distribution) $\pi(x)$, satisfying the usual canonical commutation relation with $\varphi(x)$: $[\varphi(x),\pi(y)]=i\delta(x-y)$. Then $a=\frac{1}{\sqrt{2}}(\varphi + i\pi)$ and $a^*=\frac{1}{\sqrt{2}}(\varphi-i\pi)$.
• Sorry, but how does this answer the question "is there any particle annihilation operator at position $\boldsymbol x$ in this rather simple real scalar field theory?" You only described how $a^*(x)$ creates a particle at $x$ (which OP already knows), but you didn't say anything about annihilating something at $x$. – AccidentalFourierTransform Apr 27 '18 at 16:54