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}


\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:

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

  2. 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?

  3. If the answer to 1 is yes? How does it work and did I make a mistake in my assumption?


2 Answers 2


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 $t\geq t_0$


There is already a correct answer from Ryder Rude. In this answer we provide more information and a proof.

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

  2. The free 2-point function/Greens function $$ \langle T[\hat{\psi}(x)\hat{\psi}^{\dagger}(x^{\prime})] \rangle^{\rm free} ~=~\hbar G(x\!-\!x^{\prime}) \tag{3} $$ 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.

  3. 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 negative frequency, 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. This answers OP's question.

  4. In the corresponding second-quantized operator formalism with CCRs $$ \begin{align} [\hat{a}_{\bf k}, \hat{a}^{\dagger}_{{\bf k}^{\prime}}] ~=~&(2\pi)^3\hbar{\bf 1}~\delta^3({\bf k}\!-\!{\bf k}^{\prime}), \cr [\hat{\psi}({\bf x},t),\hat{\psi}^{\dagger}({\bf x}^{\prime},t)] ~=~& \hbar{\bf 1}~\delta^3({\bf x}\!-\!{\bf x}^{\prime}), \end{align}\tag{8}$$ (and other CCRs vanishing), the Fourier expansion of the operator field $$ \hat{\psi}(x)~=~\left.\int \! \frac{\mathrm{d}^3{\bf k}}{(2\pi)^3}\hat{a}_{\bf k}e^{ik\cdot x}\right|_{k^0=\frac{1}{2m_2}{\bf k}^2}, \tag{9} $$ only contains a particle annihilation operator mode with positive frequency in order to correctly reproduce the retarded Greens function (3) & (6). In particular, there is no antiparticle creation operator mode with negative frequency in the $\hat{\psi}$ Fourier expansion (9), in contrast to a complex relativistic field.

    Concerning the CCR (8), see also 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.

  • $\begingroup$ Why do you say that the pole of the Fourier transformed Green's function is related to the existence of antiparticles? $\endgroup$
    – Ryder Rude
    Nov 25, 2022 at 12:43
  • $\begingroup$ I do get that the second pole leads to the backward time theta function term in the propagator of QFT. But does this thing have a physical interpretation in terms of anti-particles, or is it just a popular saying that this is an anti particle? $\endgroup$
    – Ryder Rude
    Nov 25, 2022 at 12:46
  • $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic
    Aug 23, 2023 at 10:51

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