Preparing States using path integral in QFT I had some confusion about the idea of cutting the path integral  to define states in quantum field theory. There are two versions which I have seen:

*

*We do the path integral with an unspecified 'final' b.c. This would define a state at the cut.


*We do the path integral with both b.c specified and then integrate over the initial condition. Does this method give a state or a wavefunctional at the cut?
Is it just that (1) produces a ket state while (2) will give us a bra state and thus, the conjugate of the wavefunctional we get from (1) ?
 A: Consider a scalar field $\phi$ for simplicity. I'll write $[\phi]_I$ for the set of field variables whose time-argument is in the interval $I$, and I'll write $[\phi]_t$ for the set of field variables whose time-argument is equal to $t$.
The expression
$$
\tilde\Psi_1[\phi]_{\tilde t}
\equiv
\int [d\phi]_{(\tilde t,t]}\ \exp\left(iS[\phi]_{[\tilde t,t]}\right)\Psi_1[\phi]_{t}
\tag{1a}
$$
represents the state $\tilde\Psi_1$ at time $\tilde t$ obtained by taking the state $\Psi_1$ at time $t$ and evolves it forward to time $\tilde t>t$. Notice the subtle difference in the intervals: the action depends on all of the field variables in the closed interval $[\tilde t,t]$, which includes times $\tilde t$ and $t$, but the integral is only over the field variables in the half-open interval $(\tilde t,t]$, which doesn't include time $\tilde t$. We could also write (1a) as
$$
 |\tilde\Psi_1\rangle \equiv U(\tilde t-t)|\Psi_1\rangle.
\tag{1b}
$$
I assume that the expression (1a) is what option 1 means in the question.
The expression
$$
\langle\Psi_2|U(\tilde t-t)|\Psi_1\rangle
=
\int [d\phi]_{[\tilde t,t]}\ \Psi_2^*[\phi]_{\tilde t}
  \exp\left(iS[\phi]_{[\tilde t,t]}\right)\Psi_1[\phi]_{t}
\tag{2}
$$
represents the inner product of the state (1) with the state $\Psi_2$. Now the integral is also over the field variables at the final time $\tilde t$. This is almost option 2 in the question, except without integrating over the initial condition. I don't know what the question means by "integrate over the initial condition," because in order to integrate over initial states $\Psi_1$, we would need to specify some kind of distribution of initial states, which the question does not specify.
