# Ground State Functional and Vacuum-Vacuum Transition Amplitude

In Path Integral formalism, the vacuum-vacuum transition amplitude is defined to be (the functional integration is over all field configurations in the whole spacetime; $\Phi_{\vec{x}}(\tau)$ is the field configuration at a given time slice $\tau$):

$<0,+\infty|0,-\infty> = \int D[\Phi_\vec{x}(\tau)] e^{i S [\Phi(x)]}$ $(1)$

It doesn't make sense to me. Or, at the very least, I expect the answer to be different (I use the Heisenberg picture, where the operator is evolved with time, so the field $|\Phi_{\vec{x}},\tau>$ eigenstates of eigenconfiguration $\Phi_\vec{x}$ is also time-dependence). Indeed, adding the whole $\hat{1}$ ($|\Phi_\vec{x},\tau><\Phi_\vec{x},\tau|$) at every time slice ($\tau$): $<0,+\infty|0,-\infty> = \int d\Phi_{f,\vec{x}} d \Phi_{i,\vec{x}} <0,+\infty|\Phi_{f,\vec{x}},+\infty><\Phi_{i,\vec{x}},-\infty|0,-\infty>\int D[\Phi(\tau \neq \pm \infty)] e^{i S[\Phi]}$

Note that $<A,t|B,t>=<A,t=0|B,t=0>=<A|B>$, hence:

$= \int d\Phi_{f,\vec{x}} d \Phi_{i,\vec{x}} <0|\Phi_{f,\vec{x}}><\Phi_{i,\vec{x}}|0>\int D[\Phi(\tau \neq \pm \infty)] e^{i S[\Phi]}$

$= \int d\Phi_{f,\vec{x}} d \Phi_{i,\vec{x}} \Psi_0[\Phi_{f,\vec{x}}] \Psi_0[\Phi_{i,\vec{x}}] ^*\int D[\Phi(\tau \neq \pm \infty)] e^{i S[\Phi]}$ $(2)$

If only $\Psi_0[\Phi_{f,\vec{x}}] \Psi_0[\Phi_{i,\vec{x}}]$ is the same for any boundary of integration, then the answer of $(2)$ is physically the same as $(1)$, however in general it's not the case, I believe.

Indeed, the ground state wavefunctional is complicated, and it's hard to consider $\Psi_0[\Phi_{f,\vec{x}}] \Psi_0[\Phi_{i,\vec{x}}]$ as a constant or something like that. So, how one can find the wavefunctional $\Psi_0[\Phi_\vec{x}]$ ($<\Phi_\vec{x}|0>$)? Solving the functional Schordinger equation for the lowest value $E_0$ of the energy spectrum $E$, as far as I know (the answer for some special cases, such as non-interacting theories, can be found in literatures):

$\hat{H}(\hat{\Phi},\nabla \hat{\Phi},\hat{\Pi}) |\Psi> = E |\Psi> \Rightarrow H(\Phi,\nabla \Phi, -i\frac{\delta}{\delta \Phi}) \Psi[\Phi] = E \Psi[\Phi]$

My question is: How $(1)$ is derived in QFT as vacuum-vacuum transition amplitude?

I don't see how this equation can be right. There's a trick to derive it in Euclidean QFT, but seriously I don't buy it. However, I do agree that the equation can be used for lattice QFT, since the calculations must be done on Euclidean time.

• The answer is in section 9.2 of the first volume of the book by S. Weinberg. Basically, the wave function of the vacuum at past (or future) infinity will be a kind of a Gaussian (the fields approach the free ones at infinity), see eq. 9.2.9 in Weinberg, except that the kernel will be given by Henkel functions. – TwoBs May 31 '14 at 6:55
• It's not like you can always make sense of a non-interacting theory at far distance limit. For example, quark fields in QCD aren't free far away. – user109798 Jun 1 '14 at 5:46
• There is no such a problem if you work instead with fields associated to asymptotic states. Just change variables (but Of course the action would be a mess). In any case, in the scattering at high energy, the free field is a still good approximation even for quarks and gluons thanks to asymptotic freedom. – TwoBs Jun 1 '14 at 8:22