Path Integral Series: 3. Path Integral

By 苏剑林 | June 02, 2016

The path integral is a description of quantum mechanics that originated with the physicist Feynman [5]. It is a type of functional integral and has become the mainstream form of modern quantum theory. In recent years, interest in it has increased, especially in its applications outside the quantum field, and several works have appeared, such as [7]. However, few people in China are familiar with path integrals, and many students majoring in quantum physics may not have even heard of them.

From a mathematical perspective, path integrals are a method for finding the Green's function of partial differential equations. We know that in the study of partial differential equations, if the corresponding Green's function can be found, it is of great help to the research. Usually, Green's functions are not easy to solve. However, constructing a path integral only requires the Green's function over an infinitesimal time interval, so it is quite simple in both form and concept.

This chapter contains no new material but aims to provide a concise and direct introduction to path integrals starting from the random walk problem, demonstrating how to transform parabolic partial differential equation problems into the path integral form.

From Point Probabilities to Path Probabilities

In the study of random walks in the previous chapter, we concluded that the probability density of starting from $x_0$ and arriving at $x_n$ after time $t$ is:

$$ \frac{1}{\sqrt{2\pi \alpha t}}\exp\left(-\frac{(x_n-x_0)^2}{2\alpha t}\right). \tag{22} $$

This is the probability from a certain point at one time to another point at another time. In mathematics, we call this the propagator or Green's function of the diffusion equation (21).

Now divide the time $T$ into $n$ equal parts, each of length $\Delta t = \frac{T}{n}$. At time $i\Delta t$, the position of the particle is $x_k$. The probability density for the particle to move from $x_k$ to $x_{k+1}$ is:

$$ \frac{1}{\sqrt{2\pi \alpha \Delta t}}\exp\left(-\frac{(x_{k+1}-x_k)^2}{2\alpha \Delta t}\right), \tag{23} $$

Therefore, the probability density for the particle to sequentially pass through $x_1, x_2, \dots, x_{n-1}, x_n$ is:

$$ \left(\frac{1}{\sqrt{2\pi \alpha \Delta t}}\right)^n\exp\left(-\frac{(x_1-x_0)^2+(x_2-x_1)^2+\dots+(x_n-x_{n-1})^2}{2\alpha \Delta t}\right), \tag{24} $$

Temporarily omitting the preceding factors and taking the limit as $\Delta t \to 0$, we consider the points $x_0, x_1, x_2, \dots, x_{n-1}, x_n$ as defining a path $x(t)$ from $(x_0, 0)$ to $(x_n, T)$, where:

$$ \frac{(x_1-x_0)^2+(x_2-x_1)^2+\dots+(x_n-x_{n-1})^2}{2\alpha \Delta t}=\frac{1}{2\alpha}\sum_{k=0}^{n-1} \left(\frac{x_{k+1}-x_k}{\Delta t}\right)^2\Delta t, \tag{25} $$

As $\Delta t \to 0$, we consider $\frac{x_{k+1}-x_k}{\Delta t}$ to be equal to the derivative $\dot{x}(t_k)$ of $x(t)$ at $t_k$. Thus, the above expression is exactly the discretized form of the integral $\frac{1}{2\alpha}\int\dot{x}^2dt$. In summary, we find that the probability of the particle following the path $x=x(t)$ is proportional to:

$$ P[x(t)] = \exp\left(-\frac{1}{2\alpha }\int\dot{x}^2dt\right). \tag{26} $$

This provides the probability of the particle following the path $x(t)$, which is a functional of $x(t)$. This is the path probability distribution for Brownian motion. Specifically, if $\alpha=1$, it is called standard Brownian motion.

Note:
If $W_t$ is a stochastic process satisfying the following conditions, it is called a Brownian motion:
1. $W_0 = 0$;
2. $\{W_t, t \geq 0\}$ is a process with stationary independent increments;
3. $\forall 0 \leq s \leq t, W_t - W_s \sim N(0, \sigma^2(t-s))$.
When $\sigma=1$, it is called standard Brownian motion.

Summing Over Paths

We have obtained the expression for the probability $P[x(t)]$ of a single path. Thus, the probability of going from $(x_0, 0)$ to $(x_n, T)$ should be the sum of the probabilities of all paths from $(x_0, 0)$ to $(x_n, T)$. In other words, we must traverse and sum all paths between the two points.

Traversing is achieved by discretizing the path. As shown in Figure 1, we still divide the time $T$. For any path, we can approximate it with a broken line $x_0, x_1, x_2, \dots, x_{n-2}, x_{n-1}, x_n$. Therefore, to traverse all paths $x(t)$, we only need to traverse all $x_1, x_2, \dots, x_{n-2}, x_{n-1}$ (imagine "flicking" each point $x_1, x_2, \dots, x_{n-1}$ up and down).

(Discretizing a path, then traversing)

If we use $P(x_0, 0; x_n, T)$ to denote the probability of reaching $(x_n, T)$ from $(x_0, 0)$, then (for simplicity, we omit the constant factors, which we can restore in actual problems; for now, we want to clarify the concept):

$$ \begin{aligned}&P(x_0,0;x_n,T)\\ =&\lim_{n\to\infty}\int_{-\infty}^{\infty} \exp\left(-\frac{1}{2\alpha}\sum_{k=0}^{n-1} \left(\frac{x_{k+1}-x_k}{\Delta t}\right)^2\Delta t\right)dx_1 dx_2\dots dx_{n-2}dx_{n-1}.\end{aligned} \tag{27} $$

The integration is performed $n-1$ times, then the limit $n \to \infty$ is taken. We denote the above expression simply as:

$$ P(x_0,0;x_n,T)=\int_{x_0}^{x_n} P[x(t)]\mathscr{D}x(t)=\int_{x_0}^{x_n} \exp\left(-\frac{1}{2\alpha }\int_0^T\dot{x}^2dt\right)\mathscr{D}x(t), \tag{28} $$

This is called the path integral (or functional integral) of the functional $P[x(t)]$, which is an infinite-dimensional integral.

Path Integral for Parabolic Equations

This section aims to convert the parabolic equation (19) into a path integral representation. From equation (27), we know that the key to constructing a path integral is to find the propagator (Green's function) for an infinitesimal time interval.

First, consider the case where $V=V(x)$, i.e., it is independent of time. Let $H=\frac{\alpha^2}{2}\frac{\partial^2}{\partial x^2}+ V$. We can write equation (19) briefly as:

$$ \alpha\frac{\partial \phi}{\partial t} = H\phi, \tag{29} $$

Since $H$ is independent of $t$, the solution to the above equation can be formally written as:

$$ \phi(x,t)=\exp\left(\frac{1}{\alpha}t H\right)\phi(x,0)=\exp\left[\frac{1}{\alpha}t\left(\frac{\alpha^2}{2}\frac{\partial^2}{\partial x^2}+ V(x)\right)\right]\phi(x,0). \tag{30} $$

The above equation is an exact solution, where $\exp\left(\frac{1}{\alpha}t H\right)$ should be understood as an operator series:

$$ \exp\left(\frac{1}{\alpha}t H\right)=\sum_{k=0}^{\infty} \frac{1}{k!}\frac{t^k}{\alpha^k}H^k. \tag{31} $$

Now we are only interested in the result for an infinitesimal interval, taking $t=\epsilon \to 0$:

$$ \phi(x,\epsilon)=\exp\left(\frac{\alpha}{2}\epsilon \frac{\partial^2}{\partial x^2}+\frac{1}{\alpha} \epsilon V(x)\right)\phi(x,0). \tag{32} $$

Note that $\frac{\partial^2}{\partial x^2}$ and $V(x)$ are generally non-commutative, so in general:

$$ \exp\left(\frac{\alpha}{2}\epsilon \frac{\partial^2}{\partial x^2}+\frac{1}{\alpha} \epsilon V(x)\right)\neq \exp\left(\frac{\alpha}{2}\epsilon \frac{\partial^2}{\partial x^2}\right)\exp\left(\frac{1}{\alpha} \epsilon V(x)\right), \tag{33} $$

However, in the first-order approximation, the two are equal (meaning the difference between them is second-order infinitesimal). So we have:

$$ \phi(x,\epsilon)=\exp\left(\frac{1}{\alpha} \epsilon V(x)\right)\exp\left(\frac{\alpha}{2}\epsilon \frac{\partial^2}{\partial x^2}\right)\phi(x,0). \tag{34} $$

Notice that:

$$ \hat{\phi}(x,\epsilon) = \exp\left(\frac{\alpha}{2}\epsilon \frac{\partial^2}{\partial x^2}\right)\phi(x,0) \tag{35} $$

is exactly the formal solution to the diffusion equation $\frac{\partial \hat{\phi}}{\partial t}=\frac{\alpha }{2}\frac{\partial^2 \hat{\phi}}{\partial x^2}$, for which we already have the general solution from mathematical physics tutorials:

$$ \hat{\phi}(x,\epsilon)=\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi\alpha\epsilon}}\exp\left(-\frac{1}{\alpha} \frac{(x-x_0)^2}{2\epsilon}\right)\phi(x_0,0)dx_0, \tag{36} $$

Thus equation (34) equals:

$$ \begin{aligned}\phi(x,\epsilon)=&\exp\left(\frac{1}{\alpha} \epsilon V(x)\right)\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi\alpha\epsilon}}\exp\left(-\frac{1}{\alpha} \frac{(x-x_0)^2}{2\epsilon}\right)\phi(x_0,0)dx_0\\ =&\frac{1}{\sqrt{2\pi\alpha\epsilon}}\int_{-\infty}^{\infty}\exp\left[-\frac{1}{\alpha} \left(\frac{1}{2}\frac{(x-x_0)^2}{\epsilon^2}-V(x)\right)\epsilon\right]\phi(x_0,0)dx_0\end{aligned}. \tag{37} $$

Since the above is for an infinitesimal time interval, $\frac{x-x_0}{\epsilon}$ can be regarded as the first derivative $\dot{x}$. Therefore, the Green's function for an infinitesimal time is (omitting the front factor):

$$ \exp\left[-\frac{1}{\alpha} \left(\frac{1}{2}\dot{x}^2-V(x)\right)\epsilon\right], \tag{38} $$

From this, we can successively find:

$$ \begin{aligned}&K(x_0,0;x_n,T)\\ =&\lim_{n\to\infty}\int_{-\infty}^{\infty} \exp\left\{-\frac{1}{2\alpha}\sum_{k=0}^{n-1} \left[\left(\frac{x_{k+1}-x_k}{\Delta t}\right)^2-V(x_k)\right]\Delta t\right\}dx_1 \dots dx_{n-1}.\end{aligned} \tag{39} $$

Since $\phi$ is not a probability in the true sense, $K$ is used here to represent the propagator of $\phi$. The above equation implies that the path functional for a specific path $x(t)$ is:

$$ K[x(t)] = \exp\left[-\frac{1}{\alpha} \int_{t_a}^{t_b} \left(\frac{1}{2}\dot{x}^2-V(x)\right)dt\right], \tag{40} $$

Thus, the probability from point $(x_a, t_a)$ to point $(x_b, t_b)$ can be expressed as a path integral between the two points:

$$ \begin{aligned}P(x_b,t_b;x_a,t_a)=&\int_{x_a}^{x_b} P[x(t)]\mathscr{D}x(t) \\ =&\int_{x_a}^{x_b}\exp\left[-\frac{1}{\alpha} \int_{t_a}^{t_b} \left(\frac{1}{2}\dot{x}^2-V(x)\right)dt\right]\mathscr{D}x(t).\end{aligned} \tag{41} $$

The derivation above is based on the special case where $V$ is independent of time $t$, but it can be applied to the time-dependent case with simple modifications, which will not be repeated here.

From Path Integrals to Partial Differential Equations

Starting from the path functional (40) and performing the path integral, one can conversely derive the partial differential equation (19). The specific process can be found in Quantum Mechanics and Path Integrals. Since they can be derived from each other, it means they are equivalent: given a partial differential equation of form (19), one can immediately write the corresponding path functional (40), and vice versa.

Some Calculation Examples

Path integrals are quite simple in concept, but their calculation is very complex. Very few solutions can be calculated exactly; in many cases, only approximations are used. This section provides some results without proof. these results are from works by physicist Feynman, such as Quantum Mechanics and Path Integrals and Statistical Mechanics: A Set of Lectures. This is simply to show the reader that effective calculation schemes exist for many path integral problems.

The Most Probable Path

We have already encountered some path integrals of the form:

$$ \int\exp\left(-\frac{1}{\alpha}S[x(t)]\right)\mathscr{D}x(t). \tag{42} $$

Where $S[x(t)]$ is a functional of $x(t)$, which in physical terms is called the "action". Its meaning is clear: it superimposes the effects of all paths. A natural question to consider is: which path contributes the most? If this effect is a probability, then this question is equivalent to asking which path has the highest probability.

Obviously, if we want $\exp\left(-\frac{1}{\alpha}S[x(t)]\right)$ to be as large as possible, we want $S[x(t)]$ to be as small as possible. However, this is not sufficient. The sufficient condition is that $S[x(t)]$ must be as stable as possible near $x(t)$ so that the contribution of $x(t)$ can be superimposed stably. Here, "stable" means the first variation of $S[x(t)]$ is zero. From this, we see that the most probable path problem in path integrals naturally leads to a "variational principle," indicating a close connection between the variational principle and path integrals.

For what we discussed earlier:

$$ S[x(t)]=\int_{t_a}^{t_b}\left[\frac{1}{2}\dot{x}^2-V(x,t)\right]dt, \tag{43} $$

$\delta S[x(t)]=0$ gives:

$$ \ddot{x}=-\frac{\partial V}{\partial x}. \tag{44} $$

With boundary conditions $x(t_a)=x_a, x(t_b)=x_b$. Let the solution be $x_{cl}(t)$. Substituting the solution back into $S[x(t)]$ yields $S_{cl}$, which is a function of $t_a, t_b, x_a, x_b$. These notations are helpful when solving quadratic problems.

For the $V(x,t)$ corresponding to a random walk, we have:

$$ \ddot{x}=\frac{1}{2}\alpha\frac{\partial^2 p}{\partial x^2}+p\frac{\partial p}{\partial x}+\alpha \frac{\partial p}{\partial t}. \tag{45} $$

Quadratic Actions

For any quadratic action, the path integral can be solved exactly. The answer is:

$$ P(b,a)=\left(\frac{1}{2\pi \hbar}\right)^{D/2} \sqrt{-\det\left(\frac{\partial^2 S_{cl}}{\partial x_a \partial x_b}\right)}\exp\left(-\frac{1}{\hbar}S_{cl}\right), \tag{46} $$

where $D$ is the dimension of the space, and $\det\left(\frac{\partial^2 S_{cl}}{\partial x_a \partial x_b}\right)$ is known as the van Vleck-Pauli-Morette determinant. The proof can be found in [6].

Perturbation Expansion

For path integrals with actions that cannot be calculated exactly, there is a perturbation expansion formula:

$$ P(b,a)=P_0(b,a)+\left(-\frac{1}{\hbar}\right)\int P(b,c)V(c)P(c,a) d\tau_c+\left(-\frac{1}{\hbar}\right)^2\int P(b,d)V(d)P(d,c)V(c)P(c,a) d\tau_c d\tau_d. \tag{47} $$

Specific meanings of the notation and the derivation can be found in Feynman's work [5].


< Path Integral Series: 2. Random Walk Model | Path Integral Series: 4. Stochastic Differential Equations >