What stands for probability in QFT?

[fxdung]

When we apply creation operator in vacuum we certainly have one particle,similarly for annihilation operator.Then what is stand for chance(probability) in QFT?

 

[atyy]

Probability in QFT is still given using the Born rule. You can use the number operator as the observable in the Born rule, eg. section 4.6 http://hitoshi.berkeley.edu/221b/QFT.pdf. When you use that in the Born rule, you get the average number of particles that will be measured for the state.

There is an analogous formalism in the simple harmonic oscillator.
https://ocw.mit.edu/courses/physics/8-04-quantum-physics-i-spring-2016/video-lectures/part-2/number-operator-and-commutators/

 

[fxdung]

Can we know explicitly the form of /x1,x2,...xn> or /1p,2p,...,np> in QFT,then we can calculate the probability of that state?

 

[fxdung]

Classical EM field equal expectation: <n_k/E/n_k>, here <n_k/=/0_k>+/1_k>+...
What is similar expression for static EM field?(Because it seems to me <n_k/ for static EM field were /0_k> so corresponding expectation were 0)

 

[vanhees71]

Probability in QFT is still given using the Born rule. You can use the number operator as the observable in the Born rule, eg. section 4.6 http://hitoshi.berkeley.edu/221b/QFT.pdf. When you use that in the Born rule, you get the average number of particles that will be measured for the state.

There is an analogous formalism in the simple harmonic oscillator.
https://ocw.mit.edu/courses/physics/8-04-quantum-physics-i-spring-2016/video-lectures/part-2/number-operator-and-commutators/

The Born rule doesn't only give the average number of particles but even more information, namely with which probability you find $N$ particles, $N \in \mathbb{N}_0$. If you have an $N$-particle Fock state, you find with probability 1 $N$ particles (with an idealized detector of course).

 

[atyy]

Can we know explicitly the form of /x1,x2,...xn> or /1p,2p,...,np> in QFT,then we can calculate the probability of that state?

In condensed matter physics, there is non-relativistic QFT which is equivalent to the non-relativistic Schroedinger equation for many identical particles. So there the expression should have an explicit form as a wave function.
https://www.cond-mat.de/events/correl13/manuscripts/koch.pdf (see section 4 on second quantization)

 

[atyy]

Classical EM field equal expectation: <n_k/E/n_k>, here <n_k/=/0_k>+/1_k>+...
What is similar expression for static EM field?(Because it seems to me <n_k/ for static EM field were /0_k> so corresponding expectation were 0)

There is a discussion in Weinberg's Quantum Theory of Fields Vol 1 Section 13.6 "External Field Approximation": "It is intuitively obvious that a heavy charged particle like the nucleus of an atom acts like the source of a classical external field. In this section we will see how to justify this approximation, and will gain some idea of its limitations."