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claim: $a^{\dagger}$= $\int d^{3}kf_{1}(k)a^{\dagger}(k)$ Creates a state with Localized momentum $k_{1}$and localized position near origin; where $f_{1}(k)$ $\propto exp[\frac{-(k-k_{1})}{4\sigma^{2}}]$

If $|0\rangle $ is a vacuum state in free(klein-gordon) field theory then

1) $a(k)|0\rangle$$=0$

2) $a^{\dagger}(k)|0\rangle = |k\rangle$

3) $\langle 0|0\rangle = 1$

4) $\langle k_1|k\rangle = (2\pi)^{3} 2\omega\delta^3(k - k_1)$

How should i prove it? Should i try to expand $a(k)$ etc in terms of $\Phi(\bf{x})$ fields and try to find position and momentum eigenvalues?

That way i have to know the meaning of $\Phi(\bf{x})|0\rangle$ etc.

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up vote 1 down vote accepted

Think about this in the contex of ordinary (fixed-particle-number) quantum mechanics. Indeed, $|k\rangle=a^\dagger(k)|0\rangle$ is a state with one particle with momentum $k$. Such states can be dealt by ordinary quantum mechanics, where your integral just becomes a wavefunction of Gaussian wavepacket.

In a more formal way, the momentum distribution is easily obtainable by the fact that $|k\rangle$ are defined to be the eigenstates with given momentum, so the $|f_1|^2$ is pretty much the answer by the general principles of QM (you state is a superposition of staes with definite momentum, so you just need to take absolute squares of the coefficients), if you forget about normalisation questions (your 4. is not exactly the delta function, you have to renormalise the states in order to extract the right probability density for $k$). I guess that the exact answer is $dp=(2\pi)^{3}2\omega|f_1|^2dk$ (dp is the probability).

The question about coordinate representation is more subtle -- you have to make additional definitions. As I said in the beginning, it is natural to define that a state with one particle at given position by analogy with one-particle QM (at least we have to be consistent with it). So you define $|x\rangle\propto\int dk\exp(-ikx)|k\rangle$ to be the discussed state. This is sketchy, you have to worry about the normalisation of $|k\rangle$ states -- it will introduce additional factor, which will make things explicitly covariant (my integral works when normalisation in 4 is pure delta-function, you have to rescale). Normalise the states as you like. Then $a^\dagger(x)|0\rangle=|x\rangle$ defines an operator that creates a particle at position $x$ and now you can write for the particle density something like $a^\dagger(x)a(x)$ and so on.

In order to find the probability density for $x$, you have to expand your state in $|x\rangle$'s and procced analogous to the momentum. Write $|\psi\rangle=a^\dagger|0\rangle=\sum_x|x\rangle\langle x|\psi\rangle$. I write it like sum to hide the normalisation. $\langle x|\psi\rangle$ is easy to obtain if you substitute the definitions. As with $k$, $|\langle x|\psi\rangle|^2$ is pretty much the answer, up to some normalisation.

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Thanks for the answer. – Aftnix May 1 '13 at 14:49

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