# Vacuum Wavefunctional

I am having this problem in understanding the vacuum wavefunctional in QFT. Hence this naive question.

I mean, if someone say vacuum wavefunctional, I can think of an element like wavefunction as in Quantum mechanics but now a function of fields. Its easier to solve for it, I think, in Schroedinger representation, as e.g Hatfield does in chapter 10 of his book QFT of point particles and strings. I seemingly understood that. Step by step. easy. But now I see that in path integral formalism, say in Euclidean FT, the vacuum state is defined as path integral over half of the total spacetime with some fixed boundary condition (B.C) $\psi(\tau=0,x,z)\psi_0(x,z)$ obeyed in the boundary of the half space and is given by $\mathbf{\Psi}[\psi_0]=\int_{fixed B.C}\mathcal{D}\psi e^{-W[\psi]}$ with $W$ the action. Isn't it reminiscent of the generating or partition function of an Euclidean field theory?

Anyway, I don't understand (may be also because I didn't find a reference, could you suggest one?) why this is and how to see it. An intuitive explanation will be very helpful. Thanks.

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The path integral over a "thick layer" of spacetime always produces the transition amplitudes $$\langle {\rm final}| U | {\rm initial}\rangle$$ where $U$ is the appropriate unitary evolution operator. This is already true in non-relativistic quantum mechanics where the equivalence between Feynman's path integral approach and the operator formalism is being shown most explicitly. The only difference in quantum field theory is that there are infinitely many degrees of freedom. It's like having infinitely many components of $x_i$ where the discrete index $i$ becomes continuous and is renamed as a point in space, $(x,y,z)$, and $x$ is replaced and renamed by fields $\phi$, so $x_i$ is replaced by $\phi(x)$.

It means that if we have a "thick layer" of spacetime given by time coordinate $t$ satisfying $$t_0 < t < t_1$$ then the path integral with boundary conditions $\phi_0$ and $\phi_1$ at the initial and final slice calculates the matrix element $$\langle \phi_1| U | \phi_0 \rangle$$ in a full analogy with (just an infinite-dimensional extension of) non-relativistic quantum mechanics. Just to be sure, the initial and final states above are really meant to be given by the wave functional $$\Psi[\phi(x,y,z)] = \Delta [\phi(x,y,z) - \phi_0(x,y,z)]$$ which holds for one of them (initial/final) while the other is obtained by replacing $0$ by $1$. So far, everything is completely isomorphic to the case of non-relativistic quantum mechanics except that the number of independent degrees of freedom $\hat x$, now called $\hat \phi$, is higher.

The only new thing in quantum field theory is that we also often need path integrals where the initial or final state is replaced by a semiinfinite line. When it's done so, we no longer specify a particular configuration on this initial slice or final slice because there's really no initial slice or final slice on this side (or on both sides, if the path integral integrates over configurations in the whole spacetime)

We don't specify the boundary conditions; in fact, when we follow the correct rules, the path integral immediately and automatically replaces the initial or final state (replaced by the semi-infinite line or half-spacetime) by the ground state $|0\rangle$ or $\langle 0|$, whichever is appropriate. Why is it so? It's because in the operator formalism, such a path integral still contains the evolution operator $$\exp(\hat H \cdot \Delta t / i\hbar)$$ over an infinite period $\Delta t$. In fact, the $i\epsilon$ and related rules in the path integral – the way how we treat the poles in the complex energy/momentum plane – really guarantee that $$\Delta t = \infty (1+i \epsilon)$$ where $\epsilon$ is an infinitesimal constant which is however greater than $1/\infty$ where $\infty$ is the positive real prefactor above. Consequently, the exponential (evolution operator) above contains the factor of $$\exp(-\infty \epsilon \hat H )$$ which is suppressing states in the relevant initial and/or final state according to their energy. Because $\infty\epsilon$ is still infinite, all the excited states are suppressed much more than the ground state and only the ground state survives.

That means that if we integrate in Feynman's path integral over all configurations in the whole spacetime, we automatically get the matrix elements in the vacuum $$\langle 0 | (\dots ) |0 \rangle .$$ Similarly, if we integrate over configurations in a semi-infinite spacetime and specify the boundary condition for the fields (resembling a classical field configuration) at the boundary, we obtain matrix elements like $$\langle 0 | (\dots) | \phi_0 \rangle$$ or the Hermitian conjugates of them where $\phi_0$ is the boundary condition. If the inserted operators $(\dots)$ are empty, just an identity operator, the expression above clearly reduces to $$\langle 0 | \phi_0 \rangle \equiv \Psi^*[\phi_0]$$ where the last identity is nothing else than an infinite-dimensional generalization of $$\langle \psi| x \rangle = \psi^*(x)$$ in non-relativistic quantum mechanics. We just have infinitely many $x$-like degrees of freedom in quantum field theory which is why wave functions are replaced by wave functionals.

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Ahh..Believe it or not, I mostly needed your line .."If the inserted operators (...) are empty, just an identity operator, the expression above clearly reduces to".. stupid me..Thanks a lot for this nice explanation. – user1349 Jan 25 '12 at 13:12
You're welcome. The stumbling block you mentioned may be atypical but the fact that there are often atypical stumbling blocks is pretty typical. ;-) – Luboš Motl Jan 25 '12 at 14:37