Equivalence of second-quantized Schrödinger field and its first-quantized path integral formulation I was recently learning about the the second quantization of the Schrödinger field, and naturally got interested in how it aligns with the field theoretic path integral. So just as a short introduction.
What one can do is coming from the Lagrangian
$$\mathcal{L} = \bar{\psi}(x)\left(i\hbar\partial_t +\frac{\hbar^2}{2m}\Delta -V(x)\right)\psi(x)\tag{46.1}$$
and from here  do a quite regular canonical quantization. This is for example described in chapter 46 of the book "Quantum Mechanics" by Leonard I. Schiff. As it can be downloaded from this source: http://www.fulviofrisone.com/attachments/article/480/Schiff%20L.I.%20Quantum%20mechanics%20(MGH,%201949)(T)(417s).pdf
On the other hand, unrelated to this, there is a derivation of a configuration space path integral for a Schrödinger Wavefunction. Technically, one comes from the desire to construct an integration kernel s.t.
$$\psi(q', t')=\int dq K(q', q, t-t') \psi(q, t).$$
After applying some cool tricks, one comes to the conlcusion that:
$$K(q, q', t-t')= \int_{w(t)=q}^{w(t')=q'} Dw e^{iS[w]/\hbar}$$
Where $$S[w] = \int_t^{t'} \frac{m \dot{w}^2}{2}-V(w)$$ is the classical action. So this got me curious. If first I do canonical quantization and then retrieve this formula for the path integral with that Kernel? The answer is yes, and what you find is that
$$\langle \psi^\dagger(q', t') \psi(q, t) \rangle = K(q, q', t-t').$$
Then I asked myself: Can I retrieve this path integral from the second-quantized field-theoretic path integral?
So from
\begin{equation}
Z=\int D\bar{\psi} D \psi \exp \left(i S[\bar{\psi}, \psi]/\hbar \right)
\end{equation}
With
\begin{equation}
S[\bar{\psi}, \psi] = \int d^4 x \mathcal{L}.
\end{equation}
So my first thought was that this should equivalently be the 2-point function $G(q', t', q, t)$
\begin{equation}
G(q', t', q, t) = \int D \bar{\psi} D \psi \exp(i S[\bar{\psi}, \psi]/\hbar) \bar{\psi}(x', t') \psi(x, t).
\end{equation}
And now, since the integral we are looking at seems to be Gaussian, what we get is that
\begin{equation}
G = \frac{1}{i \hbar\partial_t + \frac{\hbar^2}{2m}\Delta-V(x)}.
\end{equation}
(Of course understood in a distributional sense.)
Now, this seems to be different, then what we had before, on the one hand, we have that
$K(q', q, t'-t)$ fulfills the schröedinger equation in both argument, i.e
$$i\hbar\partial_t K(q', t', q, t) = H K(q', t', q, t).$$ As a result in a distributional sense, we should find:
$$G^{-1} K = 0.$$
As a consequence, we don't have $$K = G.$$
So I guess I have the following questions:

*

*Is it indeed possible to derive the first version of the path integral from the field-theoretic one (the second)?


*If the answer to question 1 is no, doesn't that then mean that the path integral and the second quantization approach give different results?


*If the answer to 1 is yes? How does it work and did I make a mistake in my assumption?
 A: What you got from the QFT path integral is indeed not the Kernel. It's instead the Kernel times the theta function, $K\theta (t-t_0) $
First, observe that $K\theta (t-t_0)$ can only be used to evolve the Schrodinger wave, but only forward in time from the time $t_0$. This is the non-relativistic equivalent of the time-ordering stuff from relativistic QFT.
Second, $K\theta (t-t_0)$ is the Green's function of $i\frac{\partial }{\partial t}-H$.
So you did obtain the result of the path integral, by only for $t\geq t_0$
A: There is already a correct answer from Ryder Rude. In this answer we provide more information and a proof.

*

*In the second-quantized Schrödinger field theory, the presence of the Feynman $i\epsilon$-prescription in the free quadratic action$^1$
$$\begin{align}
S_2~=~&\int \! \mathrm{d}^4x \left(i\psi^{\ast}\dot{\psi}-\frac{1}{2m_2} |\nabla\psi|^2 +i\epsilon|\psi|^2  \right)\cr
~=~&\int \! \frac{\mathrm{d}^4k}{(2\pi)^4} \widetilde{\psi}^{\ast} \left( k^0 -\frac{1}{2m_2}{\bf k}^2 +i\epsilon \right)\widetilde{\psi}
,\end{align}\tag{1}$$
ensures that the path integral$^2$
$$Z~=~\int\! {\cal D}\frac{\psi}{\sqrt{\hbar}}{\cal D}\frac{\psi^{\ast}}{\sqrt{\hbar}} ~\exp\left(\frac{i}{\hbar} S\right),\tag{2}$$
is convergent.


*The free 2-point function/Greens function
$$ \langle T[\psi(x)\psi^{\ast}(x^{\prime})] \rangle^{\rm free} 
~=~\hbar G(x\!-\!x^{\prime}) \tag{3} $$
is the inverse
$$ \left(i\partial_0 + \frac{1}{2m_2} \nabla^2+i\epsilon \right) G(x\!-\!x^{\prime}) 
~=~i\delta^4(x\!-\!x^{\prime}),\tag{4}$$
$$\widetilde{G}(k)
~\stackrel{(4)}{=}~\frac{i}{k^0 -\frac{1}{2m_2}{\bf k}^2 +i\epsilon},\tag{5}$$
of the differential operator in the $S_2$ action (1), cf. my Phys.SE answer here.


*Now we want to derive the Greens function (3) from its Fourier transform (5). Notice that the $k^0$-pole in the Fourier transformed Greens function (5) is just below the positive ${\rm Re}(k^0)$ axis. This means that there is no antiparticle, and when we close the contour in the complex $k^0$-plane, there is only a non-zero residue for positive times $x^0\!-\!x^{\prime 0}>0$. This implies that $G$ will be the retarded Greens function. We calculate
$$ \begin{align} G(x\!-\!x^{\prime})~=~~&\int \! \frac{\mathrm{d}^4k}{(2\pi)^4}\widetilde{G}(k)e^{ik\cdot (x-x^{\prime})}\cr
~\stackrel{(5)+(7)}{=}&\theta(x^0\!-\!x^{\prime 0})K(x\!-\!x^{\prime}),
\end{align}\tag{6}$$
where
$$ \begin{align}  K(x\!-\!x^{\prime})~=~~~~&\left.\int \! \frac{\mathrm{d}^3{\bf k}}{(2\pi)^3}e^{ik\cdot (x-x^{\prime})}\right|_{k^0=\frac{1}{2m_2}{\bf k}^2}\cr
~\stackrel{\text{Gauss. int.}}{=}&\left(\frac{m_2}{2\pi i (x^0\!-\!x^{\prime 0})}\right)^{3/2}
\exp\left\{ 
\frac{im_2}{2}\frac{({\bf x}\!-\!{\bf x}^{\prime})^2}{x^0\!-\!x^{\prime 0}}\right\}\cr
~=~~~~&\langle x| x^{\prime} \rangle^{\rm free}
\end{align}\tag{7}$$
happens to be the free kernel/path integral
from the first-quantized formalism, cf. e.g. this related Phys.SE post.
--
$^1$ We put for simplicity $c=1$ to take advantage of relativistic notation, such as, $x^0=ct$ and $\omega=ck^0$. Since the theory is non-relativistic, it is in principle possible to remove all $c$-dependence from the notation. The Minkowski signature convention is $(-,+,+,+)$.
$^2$ Concerning the correct handling of Planck's constant $\hbar$ in the second-quantized theory, see e.g. this related Phys.SE post. Note that the parameter $m_2$ has dimension $[T]/[L]^2$, and it is replaced with $m/\hbar$ in the first-quantized theory.
