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Is there a standard method for calculating the wavefunctional $\Psi[\phi] = \langle \phi | \psi \rangle$ for a given state $|\psi\rangle$, where $|\phi\rangle$ are field eigenstates? We can take a free scalar field for definiteness if it helps.

In particular, what are the wavefunctionals of one-particle states $|\mathbf{p}\rangle$? Can we get the multi-particle states through some sort of tensor product?

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The method by which we calculate wavefunctionals is actually remarkably similar to the methods by which we calculate the position space representations of states in standard quantum mechanics. The details of this will depend heavily on your exact theory, but for simplicity I will assume we are working with a free scalar field theory.

As always, we start with the Hamiltonian for our theory. In momentum space, this is given by

$$H=\frac{1}{2}\int\frac{\mathrm{d}^3\textbf{p}}{(2\pi)^3}\left(\hat{\pi}^2+\omega_{\textbf{p}}^2\hat{\phi}^2\right),$$

where $\pi$ is the momentum conjugate to the field $\phi$, and $\omega_{\textbf{p}}=\sqrt{\textbf{p}^2+m^2}$. In the "coordinate" representation (that is, when acting on wavefunctionals $\Psi[\phi]$), the field and momentum operators act as

$$\hat{\phi}(\textbf{p})\Psi[\phi]=\phi(\textbf{p})\Psi[\phi]$$ $$\hat{\pi}(\textbf{p})\Psi[\phi]=-i\frac{\delta}{\delta\phi(\textbf{p})}\Psi[\phi],$$

in perfect analogy with one-dimensional quantum mechanics.

Now, just like in standard quantum mechanics, we define

$$a(\textbf{p})=\sqrt{\frac{\omega_{\textbf{p}}}{2}}\left(\hat{\phi}(\textbf{p})+\frac{i}{\omega_{\textbf{p}}}\hat{\pi}(\textbf{p})\right),$$

for which the Hamiltonian takes the form

$$H=\int\frac{\mathrm{d}^3\textbf{p}}{(2\pi)^3}\,\omega_{\textbf{p}}a^{\dagger}(\textbf{p})a(\textbf{p})+E_0,$$

where $E_0$ is the ground-state energy.

So far, this is all rather standard. Now, let us recall that the ground state of the theory is the state $|0\rangle$ such that $a(\textbf{p})|0\rangle=0$ for all $\textbf{p}$. This implies that

$$a(\textbf{p})\Psi_0[\phi]=\sqrt{\frac{\omega_{\textbf{p}}}{2}}\left(\phi(\textbf{p})+\frac{1}{\omega_{\textbf{p}}}\frac{\delta}{\delta\phi(\textbf{p})}\right)\Psi_0[\phi]=0,$$

where $\Psi_0[\phi]=\langle\phi|0\rangle$. It is easily checked that the state

$$\Psi_0[\phi]=\exp\left(-\frac{1}{2}\int\mathrm{d}^3\textbf{p}\,\omega_{\textbf{p}}\,\phi(\textbf{p})^2\right)$$

satisfies this requirement.

Now that we have found the ground state, finding single-particle states is rather simple. We simply note that single particle states are made by acting on the ground state with the creation operator. This gives

$$\Psi_{\textbf{k}}[\phi]=a^{\dagger}(\textbf{k})\Psi_{0}[\phi]\propto\phi(\textbf{k})\exp\left(-\frac{1}{2}\int\mathrm{d}^3\textbf{p}\,\omega_{\textbf{p}}\,\phi(\textbf{p})^2\right).$$

Multiparticle states are constructed similarly, with repeated application of $a^{\dagger}$. For instance, we have

$$\Psi_{\textbf{k},\textbf{q}}[\phi]=a^{\dagger}(\textbf{q})\Psi_{\textbf{k}}[\phi]=\left[2\phi(\textbf{k})\phi(\textbf{q})-\frac{1}{\omega_{\textbf{q}}}\delta(\textbf{k}-\textbf{q})\right]\exp\left(-\frac{1}{2}\int\mathrm{d}^3\textbf{p}\,\omega_{\textbf{p}}\,\phi(\textbf{p})^2\right).$$

This demonstrates explicitly that one cannot simply write a multiparticle state as a tensor product of two single particle states. Indeed, multiparticle states are more complicated than that.

I hope this helps!

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