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Consider free complex scalar theory. Let's denote $a(p)^{\dagger}$ as the particle creation operator, and $b(p)^{\dagger}$ as the antiparticle creation operator. I know that an arbitrary one particle state is given by $$|1\rangle=\int{}d^3p\,g(p)a(p)^{\dagger}|0\rangle$$ where $g(p)$ is a sufficiently well behaved function. I wanna know how to compute the wave function of such a state, but I have never been told how to compute $|x\rangle$ in this context.

So, how do I compute the wave function? Would there be any difference in the way to compute $|x\rangle$ had I chosen a one antiparticle state?

Yossarian
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  • for a particle that is allowed to decay or give place to something else in general its position makes little sense... – Phoenix87 May 26 '15 at 12:21
  • There is not really a good position representation in relativistic QFT, and hence no good notion of "wavefunction", see this and this, for instance. If you are asking about the non-relativistic case (as the non-Lorentz invariant integration measure would imply), then I don't see what your problem would be, just Fourier transform $g(p)$. – ACuriousMind May 26 '15 at 12:24

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Supposing that $p$ denotes a momentum coordinate, the function $g(p)\in L^2(\mathbb{R}^3)$ is your one particle wavefunction in the momentum representation.

More precisely, the vector $\lvert 1\rangle=\Bigl(\lvert 1\rangle_0,\lvert 1\rangle_1(p_1),\dotsc,\lvert 1\rangle_n(p_1,p_2,\dotsc,p_n),\dotsc\Bigl)$ that you has written (where $\lvert 1\rangle_n(p_1,\dotsc,p_n)$ is the $n$-particle component in momentum representation) has the form $$\lvert 1_g\rangle=\Bigl(0,g(p),0,\dotsc,0,\dotsc\Bigl)\; ;$$ i.e. it is an infinite vector with zero components, apart from the second that represents a single particle with wavefunction $g(p)$.

As usual, you can recover the wavefunction in position representation by Fourier transform in $L^2$, i.e. $$g(x)=\frac{1}{(2\pi)^{3/2}}\int_{\mathbb{R}^3} e^{ip\cdot x}g(p)d^3p\; .$$

yuggib
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