Explicit form of the wavefunctional

Question

In quantum mechanics, one in principle can write down an explicit form of the corresponding wave-function. For example, $V_i$ for the $i$-th level of quantum harmonic oscillator.

In QFT, the Hilbert space which the quantum fields act on is not the space of wave-functions, but rather of functionals on the space of field configurations. Is it possible to explicitly write down those functionals and the corresponding field configurations for $0, 1, 2, \ldots$ excitations (particles) of the field?

And I don't mean things like $a^\dagger_p a^\dagger_q|\mathrm{vacuum}\rangle$, but something in coordinates, like in the case of the quantum mechanical wave-function.

Answer 1

The correct analogy with the QM wave function can be understood from the path integral formulation. Given a theory with an action $S[\varphi]$ the path integral is denoted as $$ Z = \int [\mathrm{d}\varphi]\,e^{iS[\varphi]}\,. $$ A wave functional is, as the name says, a functional of the configuration variables of your QFT, which obviously are fields. In QM if you had, say, $N$ coordinates, your wave function would be $$ \psi(q_1,\ldots,q_N)\,. $$ QFT, on the other hand, can be though of as if there were infinitely many coordinates ($N\to\infty$). Those coordinates describe your field configuration at a certain time slice, which we can take to be $x^0 = 0$ for simplicity. Therefore the wave functional is something like $$ \Psi[\phi_0(\vec{x})]\,. $$ You can imagine discretizing your space, at which point the $q_i$'s would be the values of $\varphi$ at the point $x = (0,\vec{x}_i)$.

In the path integral language a field configuration can be thought of as a boundary condition for the integral at a certain time slice. We can therefore define a wave functional as the following path integral $$ \Psi[\phi_0(\vec{x})] = \int[\mathrm{d}\varphi]_{\varphi(0,\vec{x})\equiv\phi_0(\vec{x})}\,e^{iS[\varphi]}\,. $$ Where, as you can see, the integration is restricted only to those configurations which have $\varphi(0,\vec{x}) \equiv \phi_0(\vec{x})$. Now, this would actually the vacuum state because we are not inserting anything in the path integral. A more general state can be obtained by putting operators such as $$ \Psi_{1,\ldots,n}[\phi_0(\vec{x})] = \int[\mathrm{d}\varphi]_{\varphi(0,\vec{x})\equiv\phi_0(\vec{x})}\,\mathcal{O}_1(x_1)\cdots\mathcal{O}_n(x_n)\,e^{iS[\varphi]}\,. $$

Answer 2

Yes, you can generally map between second quantization picture (as in QFT) to first quantization (each particle has a coordinate and you get the wave functions). For more than one particle, the QFT version is guaranteed to contain the correct symmetrization or anti-symmetrization of the wave-function.

The wave-function for a two-particle state, for example, is given by $$ \Phi(r_1, r_2) = \langle r_1, r_2 | a^{\dagger}_p a^{\dagger}_q | {\rm{vac}}\rangle = \langle {\rm{vac}}| a(r_2)a(r_1) a^{\dagger}_p a^{\dagger}_q | {\rm{vac}}\rangle$$ where $a(r)$ is the field annihilating a particle at position $r$, which you can get from a Fourier-transform of $a_p$