Equivalence of second-quantized Schrödinger field and its first-quantized path integral formulation

Question

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

$$\mathcal{L} = \psi^{*}(x)\left(i\hbar\partial_t +\frac{\hbar^2}{2m}\Delta -V(x)\right)\psi(x)\tag{46.1}$$

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\psi^{*} D \psi \exp \left(i S[\psi^{*}, \psi]/\hbar \right) \end{equation}

With

\begin{equation} S[\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 \psi^{*} D \psi \exp(i S[\psi^{*}, \psi]/\hbar)\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?

Answer 1

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 $t\geq t_0$

Answer 2

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

1. The 1st-quantized Hamiltonian operator is $$ \hat{H}_1~=~-\frac{\hbar^2}{2m_1}\hat{\nabla}^2+V_1(\hat{\bf x}),\tag{1}$$ while the 2nd-quantized Hamiltonian operator is$^1$ $$\begin{align}\hat{H}_2 ~=~&\int\!d^3{\bf x}~ \hat{\cal H}_2 ({\bf x}),\cr \hat{\cal H}_2({\bf x})~=~&\frac{1}{2m_2}|\nabla\hat{\psi}_2({\bf x})|^2+V_2({\bf x})|\hat{\psi}_2({\bf x})|^2.\end{align}\tag{2}$$

2. In the 2nd-quantized Schrödinger field theory, the presence of the Feynman $i\epsilon$-prescription in the quadratic action$^2$ $$\begin{align} S~=~&\int \! \mathrm{d}^4x \left(i\psi_2^{\ast}\dot{\psi}_2-\frac{1}{2m_2} |\nabla\psi_2|^2 +(i\epsilon-V_2)|\psi_2|^2 \right)\cr ~\stackrel{V_2=0}{=}&\int \! \frac{\mathrm{d}^4k}{(2\pi)^4} \widetilde{\psi_2}^{\ast} \left( k^0 -\frac{1}{2m_2}{\bf k}^2 +i\epsilon \right)\widetilde{\psi_2} ,\end{align}\tag{3}$$ ensures that the path integral$^3$ $$Z~=~\int\! {\cal D}\frac{\psi_2}{\sqrt{\hbar}}{\cal D}\frac{\psi_2^{\ast}}{\sqrt{\hbar}} ~\exp\left(\frac{i}{\hbar} S\right),\tag{4}$$ is convergent.

3. The 2-point function/Greens function $$\begin{align} \langle T[\hat{\psi}_2(x)\hat{\psi}_2^{\dagger}(x^{\prime})] \rangle ~=~&\hbar G(x,x^{\prime}),\cr \langle T[\hat{\psi}_2(x)\hat{\psi}_2^{\dagger}(x^{\prime})] \rangle^{\rm free} ~=~&\hbar G_0(x\!-\!x^{\prime}), \end{align}\tag{5} $$ is the inverse $$\begin{align} \left(i\partial_0 + \frac{1}{2m_2} \nabla^2+i\epsilon -V_2\right) G(x,x^{\prime}) ~=~&i\delta^4(x\!-\!x^{\prime}), \cr \left(i\partial_0 + \frac{1}{2m_2} \nabla^2+i\epsilon \right) G_0(x\!-\!x^{\prime}) ~=~&i\delta^4(x\!-\!x^{\prime}), \end{align}\tag{6}$$ $$\widetilde{G}_0(k) ~\stackrel{(6)}{=}~\frac{i}{k^0 -\frac{1}{2m_2}{\bf k}^2 +i\epsilon},\tag{7}$$ of the differential operator in the $S$ action (3), cf. my Phys.SE answer here.

4. Now we want to derive the Greens function (5) from its Fourier transform (7). Notice that the $k^0$-pole in the Fourier transformed Greens function (7) 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 $$ \begin{align} G_0(x\!-\!x^{\prime})~=~~&\int \! \frac{\mathrm{d}^4k}{(2\pi)^4}\widetilde{G}_0(k)e^{ik\cdot (x-x^{\prime})}\cr ~\stackrel{(7)+(9)}{=}&\theta(x^0\!-\!x^{\prime 0})K_0(x\!-\!x^{\prime}), \end{align}\tag{8}$$ where $$ \begin{align} K_0(x\!-\!x^{\prime}) ~=~~~~&\left.\int \! \frac{\mathrm{d}^3{\bf k}}{(2\pi)^3}e^{ik\cdot (x-x^{\prime})}\right|_{k^0=\frac{1}{2m_2}{\bf k}^2}\cr ~\stackrel{\text{Gauss. int.}}{=}&\left(\frac{m_2}{2\pi i (x^0\!-\!x^{\prime 0})}\right)^{3/2} \exp\left\{ \frac{im_2}{2}\frac{({\bf x}\!-\!{\bf x}^{\prime})^2}{x^0\!-\!x^{\prime 0}}\right\}\cr ~=~~~~&\langle x| x^{\prime} \rangle^{\rm free} \end{align}\tag{9}$$ happens to be the free kernel/path integral from the first-quantized formalism. More generally $$ G(x,x^{\prime})~=~ \theta(x^0\!-\!x^{\prime 0})K(x,x^{\prime}),\qquad K(x,x^{\prime})~=~\langle x| x^{\prime} \rangle, \tag{10}$$ cf. e.g. this related Phys.SE post. This answers OP's question.

5. In the corresponding 2nd-quantized operator formalism with CCRs $$ [\hat{\psi}_2({\bf x},t),\hat{\psi}_2^{\dagger}({\bf x}^{\prime},t)] ~=~ \hbar{\bf 1}~\delta^3({\bf x}\!-\!{\bf x}^{\prime}), \tag{11} $$ (and other CCRs vanishing). If the potential $V_2$ is ${\bf x}$-independent, we can introduce modes $$ [\hat{a}_{\bf k}, \hat{a}^{\dagger}_{{\bf k}^{\prime}}] ~=~ (2\pi)^3\hbar{\bf 1}~\delta^3({\bf k}\!-\!{\bf k}^{\prime}),\tag{12} $$ (and other CCRs vanishing). For $V_2=0$ the Fourier expansion of the operator field $$ \hat{\psi}_2(x)~=~\left.\int \! \frac{\mathrm{d}^3{\bf k}}{(2\pi)^3}\hat{a}_{\bf k}e^{ik\cdot x}\right|_{k^0=\frac{1}{2m_2}{\bf k}^2} \tag{13} $$ only contains a particle annihilation operator mode with positive frequency in order to correctly reproduce the retarded Greens function (5) & (8). In particular, there is no antiparticle creation operator mode with negative frequency in the $\hat{\psi}_2$ Fourier expansion (13), in contrast to a complex relativistic field.

   Concerning the CCR (11), see also this related Phys.SE post.

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$^1$ A 2nd-quantized Hamiltonian operator (or action) could in principle also have higher-order interaction terms in $\hat{\psi}_2^{\dagger}$ & $\hat{\psi}_2$ associated with creation and annihilation of particles.

$^2$ 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 $(-,+,+,+)$.

$^3$ Concerning the correct handling of Planck's constant $\hbar$ in the 2nd-quantized vs. 1st-quantized theory in 3+1D, $$ \begin{align} \psi_2~=~& \sqrt{\hbar}\psi_1, \qquad [\psi_1]~=~ \text{Length}^{-3/2},\cr m_2~=~&\frac{m_1}{\hbar},\qquad [m_1]~=~ \text{Mass}, \qquad [m_2]~=~ \frac{\text{Time}}{\text{Length}^2},\cr V_2~=~&\frac{V_1}{\hbar}, \qquad [V_1]~=~ \text{Energy}, \qquad [V_2]~=~ \text{Time}^{-1}, \end{align} \tag{14}$$ cf. e.g. this related Phys.SE post.