Probability of finding vacuum? Consider a real scalar quantum field $\varphi (x)$, interacting with a classical real scalar field $J(x)$ :
$$ \mathcal{L} = \frac{1}{2}(\partial \varphi)^2 - \frac{m^2}{2} \varphi^2 + \varphi J$$
Assuming the classical source is nonzero only in the interval $t_i < t < t_f$, the solution for $t>t_f$ is given by:
$$\varphi(x) = \int \frac{d^3 p}{(2 \pi)^3 \sqrt{2 \omega_p}} \Big[ \Big(  a_p +  \frac{i }{\sqrt{2 \omega_p}} \tilde{J}(p) \Big) e^{-ip \cdot x}  +   \Big(  a_p +  \frac{i }{\sqrt{2 \omega_p}} \tilde{J}(p) \Big)^\dagger e^{ip \cdot x}  \Big]$$
The claim is that the probability of finding the system in the vacuum state is given by
$$P(0) = e^{-\lambda}$$
where $\lambda = \Delta N$, the difference between $$\big< n \big| \hat{N} \big| n \big>$$ in the asymptotic future and asymptotic past (here simply $t_f$ and $t_i$).
I want to derive it myself but I don't understand the meaning of that probability, I don't understand what is meant by $P(0)$. What is the state vector $\left| n \right>$?
I'm new to QFT, but I understand non-relativistic QM. The terminology and formalism of QFT isn't that clear to me here.
 A: You can read off from your solution that the classical source has shifted the creation/annihilation operators. That is, the creation/annihilation operators in the asymptotic future are $a_{p,+}= a_{p,-} + \frac{\mathrm{i}}{\sqrt{2\omega_p}} \tilde{J}(p)$ and the same for the conjugate, where I wrote $a_{p,-}$ for the original modes $a_p$ of the field.
Now, the defining property of the vacuum with respect to one of these pairs of operators is
$$ a_{p,-}\lvert 0,-\rangle = 0 \quad\land\quad a_{p,+}\lvert 0,+\rangle = 0$$
and the probability you seek is the overlap $P_{-\to +} = \lvert\langle 0,-\vert 0,+\rangle\rvert^2$.
Hence, we must compute $\langle 0,-\vert 0,+\rangle$. To that end, observe that the future asymptotic vacuum is a "coherent" state for the past annihilation operators since it fulfills
$$ a_{p,-}\lvert 0,+ \rangle = -\frac{i}{\sqrt{2\omega_p}}\tilde{J}(p)\lvert 0,+ \rangle$$
by definition. For the quantum harmonic oscillator, we know that a coherent state with eigenvalue $\alpha$ can be written as $\mathrm{e}^{-\lvert\alpha\rvert^2/2}\mathrm{e}^{\alpha(t) a^\dagger(t)}\lvert 0 \rangle$ in the Heisenberg picture and as $N\mathrm{e}^{\alpha(t) a^\dagger}\lvert 0 \rangle$ in the Schrödinger picture. Generalizing this we find
$$ \lvert 0,+\rangle = \exp\left(-\frac{1}{2}\int\lvert \tilde{J}(p)\rvert^2\frac{\mathrm{d}^3 p}{2\omega_p}\right)\underbrace{\exp\left(-\mathrm{e}^{\mathrm{i}\omega_p t}\mathrm{i}\int \tilde{J}(p)a_{p,-}^\dagger\frac{\mathrm{d}^3 p}{\sqrt{2\omega_p}}\right)}_{=: A}\lvert 0,-\rangle$$
and in $\langle 0,-\vert 0,+\rangle$ we can now finally let $A$ act to the left, where all the $a_{p,-}$ just give 0, so the whole operator just gives 1, and what remains is
$$\langle 0,-\vert 0,+\rangle = \exp\left(-\frac{1}{2}\int\lvert \tilde{J}(p)\rvert^2\frac{\mathrm{d}^3 p}{2\omega_p}\right)$$
where the exponent of the r.h.s. is indeed half of the difference between the past and the future number operator.
