Interpretation of probability density in QFT

[Silviu]

Hello! I am a bit confused about the interpretation of probability density in QFT. Let's say we have the Klein-Gordon equation. I understand that this is the field equation for a spin-0 charged particle. So if we find a solution $\phi(x)$ of the Klein-Gordon equation, as far as I understand this can be interpreted as a probability density (i.e. where can you find the particle with a given probability, given by $\| \phi(x)^2 \|dx$), but at the same time it can be interpreted as a creation operator so when you apply $\phi(x)$ on the vacuum state $\|0>$, you create a particle at the position x (please let me know if I said something wrong up to know - I might have understood something wrong). As a first question, do these 2 interpretation hold only for a free particle, or even for Klein-Gordon equation with an interaction term? And I am a bit confused about how do these 2 interpretation can take place at the same time and when do you pick one over the other. The first one assume that the position of the particle is spread in space with different probabilities at each point while the second one assumes that the particle is localized at the point where it is created, but in the end, mathematically, you have a single solution to the Klein-Gordon equation. So, can someone explain to me more in depth the meaning of these 2 interpretations and the differences between them? Thank you!

 

[Demystifier]

First, Klein-Gordon equation is not a good example because $|\phi(x)|^2$ cannot be interpreted as a probability density in the usual sense. To make a better sense of your question, it is better to consider non-relativistic QFT, i.e. second-quantized Schrödinger equation.

Second, within non-relativistic QFT, one should distinguish field operator $\hat{\phi}(x)$ from the wave function $\psi(x)$. The object $\hat{\phi}(x)|0\rangle$ is a 1-particle state at position $x$. The quantity $|\psi(x)|^2$ is the probability density. For the relation between $\hat{\phi}(x)$ and $\psi(x)$, and for other details, see
https://arxiv.org/abs/quant-ph/0609163
Secs. 7 and 8.

 

[Demystifier]

For more advanced aspects, I also recommend a very recent paper "Localized States in Quantum Field Theories" by M. Pavšić:
https://arxiv.org/abs/1705.02774

 

[Silviu]

First, Klein-Gordon equation is not a good example because $|\phi(x)|^2$ cannot be interpreted as a probability density in the usual sense. To make a better sense of your question, it is better to consider non-relativistic QFT, i.e. second-quantized Schrödinger equation.

Second, within non-relativistic QFT, one should distinguish field operator $\hat{\phi}(x)$ from the wave function $\psi(x)$. The object $\hat{\phi}(x)|0\rangle$ is a 1-particle state at position $x$. The quantity $|\psi(x)|^2$ is the probability density. For the relation between $\hat{\phi}(x)$ and $\psi(x)$, and for other details, see
https://arxiv.org/abs/quant-ph/0609163
Secs. 7 and 8.

Hello! Thank you so so much for this paper. It really helped me a lot and also got me to understand the second quantization. If you know any other paper at this level, to prepare me for QFT it would be great!