### Connection classical and Keldysh path integrals for the noninteracting case

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After I had been telling him many times how good path integrals were, my friend in biophysics sent me a related article he had found [Hanggi1989]. I am not sure if he has delved into the matter, but I looked at the paper and thought it a very clear introduction to classical path integrals. Having in mind the intriguing statement that "the free classical and quantum theories are governed by the same dynamics for same initial conditions'' [Berges2007], I figured that it was worth a shot trying to put two and two together and see how the classical path integral relates to the Keldysh path integral.

We will start by looking into the derivation of the one-dimensional classical path integral from the Langevin equation. The multivariate case follows immediately. Finally we will show that the simplest kind of Keldysh path integral is apparently equivalent to a two-dimensional classical stochastic process.

Langevin equation

Since a straightforward quantum-classical correspondence only works for noninteracting systems, let us specialize right away to a linear law of motion (corresponding to a quadratic path integral). Observe that in the theory of irreversible processes, the classical equation of motion, which is turned into the Langevin equation by adding a noise term, is routinely referred to as the ``phenomenological law'' or the ``macroscopic law''. With an eye to the final results, we also restrict to additive, delta-correlated Gaussian noise. The equation then reads

\begin{align}\label{eq:langevin}

\dot{x}(t) = \alpha x(t) + \xi(t),

\end{align}

where \(\langle \xi(t) \rangle = 0\) and \(\langle \xi(t)\xi(t') \rangle = D\delta(t-t')\). Much of the subtlety of stochastic processes lies in the question of how this equation should be

\begin{align}

\frac{x_n-x_{n-1}}{\varepsilon} = \alpha\cdot \frac{1}{2}(x_n + x_{n-1}) + \xi_n.

\end{align}

For the purpose of this note, however, we shall rather follow Kamenev [kamenev2011field], who uses

\begin{align}

\frac{x_n-x_{n-1}}{\varepsilon} = \alpha x_{n-1} + \xi_{n-1},

\end{align}

which apparently corresponds to the

Derivation of the path integral

What is the path integral? Why do we want it? Well, the noise in the Langevin equation introduces uncertainty into an otherwise deterministic law of motion. Still we would like to know the probability for a particle to pass from a point \((x_0, t_0)\) to a later point \((x, t > t_0)\). This transition probability \(R(x|x_0)\) is related to the full probability density along the discretized path according to the rule of marginalization,

\begin{align}

R(x|x_0) = \int \prod_{n=1}^{N-1}\mathrm{d}x_n\; p(x_N = x, x_{N-1}, ..., x_1|x_0).

\end{align}

How can we find \(p(x_N, x_{N-1}, ..., x_1|x_0)\)? Probability theory tells us that two densities \(p(x)\) and \(p(\xi)\) are related via substitution ([VanKampen2007], p. 18)

\begin{align}

p(x)=p(\xi)\left|\frac{\partial \xi}{\partial x}\right|,

\end{align}

where the determinant of the Jacobian comes in on the right-hand side. Therefore, we can find the probability density of the \(x\) from that of the noise,

\begin{align}

p(x_N, x_{N-1}, ..., x_1|x_0) = p(\xi_N, \xi_{N-1}, ..., \xi_1),

\end{align}

where we have chosen a regularization such that the Jacobian determinant becomes unity (as in Chp. 4 of Kamenev, and unlike as in Hänggi's derivation, where the expansion of the determinant is a central point).

For the noise, we assume that we know its

\begin{align}

p(\xi_N, \xi_{N-1}, ..., \xi_1) \equiv \int \mathrm{d}z_1 \cdots \mathrm{d}z_N \; \chi(\varepsilon z_1, ..., \varepsilon z_N) \exp{\left( i\sum_{n=1}^{N} \varepsilon z_n \xi_n \right)},

\end{align}

where I have neglected some (potentially important) details. In the continuum limit, we then find for the transition probability

\begin{align}

R(x|x_0) = \lim_{N\to\infty}\int \prod_{n=1}^{N-1}\mathrm{d}x_n\; \int \prod_{n=1}^{N}\mathrm{d}z_n \; \chi[z] \exp{\left( i\int_{t_0}^{t} \mathrm{d}\bar{t}\; z(\bar{t}) \xi(\bar{t}) \right)}.

\end{align}

Substituting for \(\xi(t)\) from the very first equation, this becomes

\begin{align}

R(x|x_0) = \int \mathcal{D}x \mathcal{D}z \; \chi[z] \exp{\left( i\int_{t_0}^{t} \mathrm{d}\bar{t}\; z \left(\dot{x} - \alpha x \right) \right)},

\end{align}

which is the final path integral. Important note: it can very well be that the order of the last two steps matters and should be reversed, i.e. that first one should substitute for \(\xi\) and only then take the continuum limit. This is a question one has to answer.

For delta-correlated Gaussian noise the characteristic function reads

\begin{align}

\chi[z] = \exp{\left( -\frac{D}{2}\int\mathrm{d}\bar{t}\; z^2(\bar{t}) \right)},

\end{align}

and we come out with \(R(x|x_0) = \int \mathcal{D}x \mathcal{D}z \; e^{i S[x, z]}\),

\begin{align}

S[x,z] = \int \mathrm{d}\bar{t}\;\left\{ z \left(\dot{x} - \alpha x \right) + iDz^2/2 \right\}.

\end{align}

Multivariate case

The corresponding multivariate path integral is a straightforward generalization,

\begin{align}

S[\boldsymbol{x},\boldsymbol{z}] = \int \mathrm{d}\bar{t}\;\sum_{k}\left\{ z_k \left(\dot{x}_k - f_k(\boldsymbol{x}) \right) + iDz^2_k/2 \right\},

\end{align}

with the \(f_k(\boldsymbol{x})\) for now being restricted to linear functions of their arguments.

The free Keldysh action for a bosonic field with frequency \(\omega\) and (infinitesimal) damping \(\eta\) is

\begin{align}

\int \mathcal{D}[\phi_{\mathrm{c}}^{*}, \phi_{\mathrm{c}}^{}] \mathcal{D}[\phi_{\mathrm{q}}^{*}, \phi_{\mathrm{q}}^{}]e^{iS[\phi_{\mathrm{c}}^{*}, \phi_{\mathrm{c}}^{}, \phi_{\mathrm{q}}^{*}, \phi_{\mathrm{q}}^{}]},

\end{align}

with

\begin{align}

S[\phi_{\mathrm{c}}^{*}, \phi_{\mathrm{c}}^{}, \phi_{\mathrm{q}}^{*}, \phi_{\mathrm{q}}^{}] = \phi_{\mathrm{q}}^{*}(i\partial_t + i\eta) \phi_{\mathrm{c}}^{} - \phi_{\mathrm{q}}^{}(i\partial_t + i\eta) \phi_{\mathrm{c}}^{*} - \omega(\phi_{\mathrm{q}}^{*}\phi_{\mathrm{c}}^{} + \phi_{\mathrm{c}}^{*} \phi_{\mathrm{q}}^{}) + 2i\eta \phi_{\mathrm{q}}^{*}\phi_{\mathrm{q}}^{}.

\end{align}

Note that the free action cannot be defined without an (infinitesimal) damping rate.

If we set \(\phi_{\mathrm{q}}^{} = i(z_1 + i z_2)\), \(\phi_{\mathrm{c}}^{} = (x_1 + ix_2)/2\), we come out with

\begin{align}

S[x_1, x_2, z_1, z_2] = z_1 \dot{x}_1 + z_2 \dot{x}_2 +\omega (z_2 x_1 - z_1 x_2 ) + \eta(z_1 x_1 + z_2 x_2) + 2i\eta (z_1^2 + z_2^2).

\end{align}

This is equivalent to the above classical path integral for \(D=4\eta\) and

\begin{align}

\boldsymbol{f}(\boldsymbol{x}) =

\begin{pmatrix}

-\eta x_1 + \omega x_2 \\ -\eta x_2 - \omega x_1

\end{pmatrix}.

\end{align}

The corresponding macroscopic law in Newtonian form is

\begin{align}

\begin{split}

\ddot{x}_1 = -\eta\dot{x}_1 -\omega^2 x_1 - \eta\omega x_2,\\

\ddot{x}_2 = -\eta\dot{x}_2 -\omega^2 x_2 + \eta\omega x_1.

\end{split}

\end{align}

From all of this we learn chiefly two things: 1) free quantum noise is seemingly equivalent to

The open questions are, among others: What is the physical meaning of the Langevin equation corresponding to the macroscopic law that we have found? How does the quantum-classical correspondence break down for interacting theories?

[1] Hänggi, P. (1989). “Path integral solutions for non-Markovian processes”. Zeit. Phys. B 75(2): 275–281.

[2] Berges, J. and T. Gasenzer (2007). “Quantum versus classical statistical dynamics of an ultracold Bose gas”. Phys. Rev. A 76(3).

[3] Kamenev, A. (2011). “Field Theory of Non-Equilibrium Systems”. Cambridge University Press.

[4] van Kampen, N. G. (2007). “Stochastic Processes in Physics and Chemistry”. (3ed).

We will start by looking into the derivation of the one-dimensional classical path integral from the Langevin equation. The multivariate case follows immediately. Finally we will show that the simplest kind of Keldysh path integral is apparently equivalent to a two-dimensional classical stochastic process.

**Classical case**Langevin equation

Since a straightforward quantum-classical correspondence only works for noninteracting systems, let us specialize right away to a linear law of motion (corresponding to a quadratic path integral). Observe that in the theory of irreversible processes, the classical equation of motion, which is turned into the Langevin equation by adding a noise term, is routinely referred to as the ``phenomenological law'' or the ``macroscopic law''. With an eye to the final results, we also restrict to additive, delta-correlated Gaussian noise. The equation then reads

\begin{align}\label{eq:langevin}

\dot{x}(t) = \alpha x(t) + \xi(t),

\end{align}

where \(\langle \xi(t) \rangle = 0\) and \(\langle \xi(t)\xi(t') \rangle = D\delta(t-t')\). Much of the subtlety of stochastic processes lies in the question of how this equation should be

*discretized*. Let \(\varepsilon = \frac{t-t_0}{N}\), \(t_n = t_0 +n \varepsilon \), and \(x_n=x(t_n)\). Hänggi for example writes\begin{align}

\frac{x_n-x_{n-1}}{\varepsilon} = \alpha\cdot \frac{1}{2}(x_n + x_{n-1}) + \xi_n.

\end{align}

For the purpose of this note, however, we shall rather follow Kamenev [kamenev2011field], who uses

\begin{align}

\frac{x_n-x_{n-1}}{\varepsilon} = \alpha x_{n-1} + \xi_{n-1},

\end{align}

which apparently corresponds to the

*Ito convention*. There is much to be learned here, but not now.Derivation of the path integral

What is the path integral? Why do we want it? Well, the noise in the Langevin equation introduces uncertainty into an otherwise deterministic law of motion. Still we would like to know the probability for a particle to pass from a point \((x_0, t_0)\) to a later point \((x, t > t_0)\). This transition probability \(R(x|x_0)\) is related to the full probability density along the discretized path according to the rule of marginalization,

\begin{align}

R(x|x_0) = \int \prod_{n=1}^{N-1}\mathrm{d}x_n\; p(x_N = x, x_{N-1}, ..., x_1|x_0).

\end{align}

How can we find \(p(x_N, x_{N-1}, ..., x_1|x_0)\)? Probability theory tells us that two densities \(p(x)\) and \(p(\xi)\) are related via substitution ([VanKampen2007], p. 18)

\begin{align}

p(x)=p(\xi)\left|\frac{\partial \xi}{\partial x}\right|,

\end{align}

where the determinant of the Jacobian comes in on the right-hand side. Therefore, we can find the probability density of the \(x\) from that of the noise,

\begin{align}

p(x_N, x_{N-1}, ..., x_1|x_0) = p(\xi_N, \xi_{N-1}, ..., \xi_1),

\end{align}

where we have chosen a regularization such that the Jacobian determinant becomes unity (as in Chp. 4 of Kamenev, and unlike as in Hänggi's derivation, where the expansion of the determinant is a central point).

For the noise, we assume that we know its

*characteristic function*, which is the Fourier transform of the distribution\begin{align}

p(\xi_N, \xi_{N-1}, ..., \xi_1) \equiv \int \mathrm{d}z_1 \cdots \mathrm{d}z_N \; \chi(\varepsilon z_1, ..., \varepsilon z_N) \exp{\left( i\sum_{n=1}^{N} \varepsilon z_n \xi_n \right)},

\end{align}

where I have neglected some (potentially important) details. In the continuum limit, we then find for the transition probability

\begin{align}

R(x|x_0) = \lim_{N\to\infty}\int \prod_{n=1}^{N-1}\mathrm{d}x_n\; \int \prod_{n=1}^{N}\mathrm{d}z_n \; \chi[z] \exp{\left( i\int_{t_0}^{t} \mathrm{d}\bar{t}\; z(\bar{t}) \xi(\bar{t}) \right)}.

\end{align}

Substituting for \(\xi(t)\) from the very first equation, this becomes

\begin{align}

R(x|x_0) = \int \mathcal{D}x \mathcal{D}z \; \chi[z] \exp{\left( i\int_{t_0}^{t} \mathrm{d}\bar{t}\; z \left(\dot{x} - \alpha x \right) \right)},

\end{align}

which is the final path integral. Important note: it can very well be that the order of the last two steps matters and should be reversed, i.e. that first one should substitute for \(\xi\) and only then take the continuum limit. This is a question one has to answer.

For delta-correlated Gaussian noise the characteristic function reads

\begin{align}

\chi[z] = \exp{\left( -\frac{D}{2}\int\mathrm{d}\bar{t}\; z^2(\bar{t}) \right)},

\end{align}

and we come out with \(R(x|x_0) = \int \mathcal{D}x \mathcal{D}z \; e^{i S[x, z]}\),

\begin{align}

S[x,z] = \int \mathrm{d}\bar{t}\;\left\{ z \left(\dot{x} - \alpha x \right) + iDz^2/2 \right\}.

\end{align}

Multivariate case

The corresponding multivariate path integral is a straightforward generalization,

\begin{align}

S[\boldsymbol{x},\boldsymbol{z}] = \int \mathrm{d}\bar{t}\;\sum_{k}\left\{ z_k \left(\dot{x}_k - f_k(\boldsymbol{x}) \right) + iDz^2_k/2 \right\},

\end{align}

with the \(f_k(\boldsymbol{x})\) for now being restricted to linear functions of their arguments.

**Keldysh action**The free Keldysh action for a bosonic field with frequency \(\omega\) and (infinitesimal) damping \(\eta\) is

\begin{align}

\int \mathcal{D}[\phi_{\mathrm{c}}^{*}, \phi_{\mathrm{c}}^{}] \mathcal{D}[\phi_{\mathrm{q}}^{*}, \phi_{\mathrm{q}}^{}]e^{iS[\phi_{\mathrm{c}}^{*}, \phi_{\mathrm{c}}^{}, \phi_{\mathrm{q}}^{*}, \phi_{\mathrm{q}}^{}]},

\end{align}

with

\begin{align}

S[\phi_{\mathrm{c}}^{*}, \phi_{\mathrm{c}}^{}, \phi_{\mathrm{q}}^{*}, \phi_{\mathrm{q}}^{}] = \phi_{\mathrm{q}}^{*}(i\partial_t + i\eta) \phi_{\mathrm{c}}^{} - \phi_{\mathrm{q}}^{}(i\partial_t + i\eta) \phi_{\mathrm{c}}^{*} - \omega(\phi_{\mathrm{q}}^{*}\phi_{\mathrm{c}}^{} + \phi_{\mathrm{c}}^{*} \phi_{\mathrm{q}}^{}) + 2i\eta \phi_{\mathrm{q}}^{*}\phi_{\mathrm{q}}^{}.

\end{align}

Note that the free action cannot be defined without an (infinitesimal) damping rate.

If we set \(\phi_{\mathrm{q}}^{} = i(z_1 + i z_2)\), \(\phi_{\mathrm{c}}^{} = (x_1 + ix_2)/2\), we come out with

\begin{align}

S[x_1, x_2, z_1, z_2] = z_1 \dot{x}_1 + z_2 \dot{x}_2 +\omega (z_2 x_1 - z_1 x_2 ) + \eta(z_1 x_1 + z_2 x_2) + 2i\eta (z_1^2 + z_2^2).

\end{align}

This is equivalent to the above classical path integral for \(D=4\eta\) and

\begin{align}

\boldsymbol{f}(\boldsymbol{x}) =

\begin{pmatrix}

-\eta x_1 + \omega x_2 \\ -\eta x_2 - \omega x_1

\end{pmatrix}.

\end{align}

The corresponding macroscopic law in Newtonian form is

\begin{align}

\begin{split}

\ddot{x}_1 = -\eta\dot{x}_1 -\omega^2 x_1 - \eta\omega x_2,\\

\ddot{x}_2 = -\eta\dot{x}_2 -\omega^2 x_2 + \eta\omega x_1.

\end{split}

\end{align}

From all of this we learn chiefly two things: 1) free quantum noise is seemingly equivalent to

*two*classical uncorrelated Gaussian noises, and 2) the "quantum'' field \(\phi_{\mathrm{q}}^{}\) belongs to the characteristic function of the noise.The open questions are, among others: What is the physical meaning of the Langevin equation corresponding to the macroscopic law that we have found? How does the quantum-classical correspondence break down for interacting theories?

**References**[1] Hänggi, P. (1989). “Path integral solutions for non-Markovian processes”. Zeit. Phys. B 75(2): 275–281.

[2] Berges, J. and T. Gasenzer (2007). “Quantum versus classical statistical dynamics of an ultracold Bose gas”. Phys. Rev. A 76(3).

[3] Kamenev, A. (2011). “Field Theory of Non-Equilibrium Systems”. Cambridge University Press.

[4] van Kampen, N. G. (2007). “Stochastic Processes in Physics and Chemistry”. (3ed).

Yes, the point was mainly to gain more understanding for the Keldysh formalism in path-integral representation by linking it to classical stochastic processes.

What do you mean by "the nature of these... differ[s] immensely"? And regarding the "known fact" that bosonic actions have classical counterparts - how is that to be understood? As far as I can see, this is certainly true for quadratic actions (which is the basis for the post here), but not for interacting ones, for instance according to [2]. In the path-integral Keldysh formalism, this is especially obvious from the different combinations of \( ( \phi_c^*\phi_q\phi_q\phi_q + \mathrm{c.c.} ) \) vs. \( ( \phi_q^*\phi_c\phi_c\phi_c + \mathrm{c.c.} ) \), where the former is the classically absent vertex.

Concerning rotating the fields, your statement is not true - it's not a rotation (which would presumably

What do you mean by "the nature of these... differ[s] immensely"? And regarding the "known fact" that bosonic actions have classical counterparts - how is that to be understood? As far as I can see, this is certainly true for quadratic actions (which is the basis for the post here), but not for interacting ones, for instance according to [2]. In the path-integral Keldysh formalism, this is especially obvious from the different combinations of \( ( \phi_c^*\phi_q\phi_q\phi_q + \mathrm{c.c.} ) \) vs. \( ( \phi_q^*\phi_c\phi_c\phi_c + \mathrm{c.c.} ) \), where the former is the classically absent vertex.

Concerning rotating the fields, your statement is not true - it's not a rotation (which would presumably

*mix*classical and quantum components), but simply a reparametrization in terms of real and imaginay parts... The q-field is a vector of two noises, whereas the classical field holds the (two) quantities of interest!
written
4 months ago by
Tim

“The nature of these classical and quantum Lagrangians differ immensely”: This is the main part of my comment. I mean, concisely put, they refer to very different

“Bosonic actions have classical counterparts”: This can be proven, as far as I remember. A proof can be found — if I'm not mistaken — in Greiner's “Field Quantization” book.

“rotating the fields”: This is a minor point; I do not insist on the word “rotation”. ‘Rotation’ is often misused in place of ‘unitary transformation’.

*physical*quantities, although they may be*formally*the same. I have to explain this properly though. Just that two equations look similar, does _not_ mean that they refer to the same physical quantities — we have actually, a lot of such similarities, since (simply) you can categorize the equations-of-motion as in math (elliptic, hyperbolic, ...).“Bosonic actions have classical counterparts”: This can be proven, as far as I remember. A proof can be found — if I'm not mistaken — in Greiner's “Field Quantization” book.

“rotating the fields”: This is a minor point; I do not insist on the word “rotation”. ‘Rotation’ is often misused in place of ‘unitary transformation’.

written
4 months ago by
AlQuemist

Concerning the ''rotation": I did not refer to your wording - I disagreed with your statement that the "classical/quantum" labels are irrelevant. Here again, wording does not matter, but the \(q\) fields

*are*different in nature from the \(c\) fields...
written
4 months ago by
Tim

That alleged difference between c- and q- fields is still unclear to me. Could you kindly explain the difference between the two and why one is called “classic”, the other “quantum”?

written
4 months ago by
AlQuemist

Please login to add an answer/comment or follow this question.

formal similarityof the quantum and classical Lagrangians; isn't it?I do not see what else one can read from this derivation. Essentially, it is a known fact that

bosonicLagrangians (in contrast to fermionic ones) have classical counterparts. Yet, the nature of these classical and quantum Lagrangians differ immensely.Btw, it seems that labelings like "classical"/"quantum" for \( \phi_c / \phi_q \) are not that relevant, since in your derivation, one has to rotate them to arrive at a "classical" picture.