# Spatial position and 3 momentum relation between particle states in canonical quantization (QFT)

In the beginning of the chapter on LSZ reduction, in Srednicki's book, he says that the operator $$a_1 ^{\dagger}:=\int d^3k\: \text{exp}\Big[-(\textbf{k}-\textbf{k}_{1})^2/4 \sigma^2\Big] a^{\dagger}(\textbf{k})$$ creates a particle that has a three momentum that is localised near $$\textbf{k}_1$$ and in position space it is localised around the origin. I get the momentum part, I can use the momentum operator given by $$\textbf{P}=\int \frac{d^3k}{2 \omega} \textbf{k}\:a^{\dagger}a$$ and show that $$\textbf{P}a_1 ^{\dagger}$$ acting on the vacuum state is equal to $$\int d^3k\: \textbf{k}\: \text{exp}\Big[-(\textbf{k}-\textbf{k}_{1})^2/4 \sigma^2\Big] a^{\dagger}(\textbf{k})$$ acting on the vacuum state. But I am having trouble understanding how $$a_1 ^{\dagger}|0>$$ represents a particle localised about the origin in position space. I would think that one can define the creation and annihilation operators in position space in terms of Fourier transforms $$a(\textbf{x})=\int d^3k\: \text{exp}[i \textbf{p.x}] a^{\dagger}(\textbf{k})$$ and then maybe we define a position operator $$\textbf{X}$$ and then use it to check the statement or is there some easier way of seeing it and is the above method flawed? Another lead maybe from the expression of the negative energy frequency part of $$\phi(x)$$ where the $$x$$ dependence is of the form $$e^{-ik.x}$$

• – SRS Apr 6 '19 at 16:46
• Virtual duplicate. The state created by φ(x) is almost localized near the origin. This is a well-known peculiarity of QFT, and, if you think about it, you'd never probe coordinates; in the lab, you essentially work in momentum space! – Cosmas Zachos Apr 6 '19 at 20:31
• – Cosmas Zachos Apr 7 '19 at 20:03
• The source of my confusion I think is this localisation, if I take $\sigma$ goes to zero I get a delta function and the regular $a^{\dagger}(\textbf{$k_1$})$. Is the localisation in some sense a reflection of the uncertainty of probing the momentum, that an exact particle with momentum $\textbf{$k_{1}$}$ isn't what we observe? I'll also go through the posts and try to figure out how the above expression translates to localisation around the origin – Soumil Apr 7 '19 at 20:21

The answer is really a computation. It doesn't make sense to talk about the position of a state; there's no operator $$X^\mu$$ to probe it. At best you can insert local operators and see what happens. In this case, the simplest thing you can look at is the one-point function $$f(x) = \langle 0 | \phi(t=0,x) a_1^\dagger | 0 \rangle.$$ Here $$f(x)$$ is a completely explicit function that you can compute. I'm too lazy to do it, but by rotation invariance the function $$f$$ can only depend on the dimensionless variables $$k_1 \cdot x$$ and $$\sigma^2|x|^2$$. If you carefully do the computation, you'll most likely find that $$f(x) \; \sim \; e^{i k_1 \cdot x - \sigma^2 |x|^2}.$$ So $$a_1^\dagger$$ creates a one-particle state, and by probing with $$\phi$$ you conclude that it's unlikely to find the particle outside of $$|x| \gtrsim 1/\sigma$$.
• Hi Cosmas. You're right, it's not really a one-point function in a strange state (because you'd need to sandwich it on the LHS with the same state $\langle 0 | a_1$) but not a two-point function either. At best this is "a matrix element of the operator $\phi(x)$". – Hans Moleman Apr 7 '19 at 22:12
The Fourier transform of a Gaussian is another Gaussian, so $$a_1^{\dagger}$$ can also be written as an integral of the position creation operator $$a^{\dagger}(x)$$ with a Gaussian. Thus it is localized in position in exactly the same way you would interpret it being localized in momentum.