1.4 Harmonic oscillator

We work out the path integral quantization of a harmonic oscillator, which is an example of systems for which the path integral may be evaluated exactly. We also introduce the zeta function regularization, which is a useful tool in many areas of theoretical physics.

1.4.1 Transition amplitude

The Lagrangian of a one-dimensional harmonic oscillator is

$$L=\frac{1}{2}m{\dot{x}}^2-\frac{1}{2}m{\omega }^2{x}^2.\text{\hspace{1em}(1.130)}$$

The transition amplitude is given by

$$⟨x_f,t_f|x_i,t_i⟩=\int \mathcal{D}x{e}^{iS[x(t)]},\text{\hspace{1em}(1.131)}$$

where $S[x(t)]={\int }_{t_i}^{t_f}L\mathrm{d}t$ is the action.

Let us expand $S[x]$ around its extremum $x_c(t)$ satisfying

$${\frac{\delta S[x]}{\delta x}|}_{x=x_c(t)}=0.\text{\hspace{1em}(1.132)}$$

Clearly $x_{\mathrm{c}}(t)$ is the classical path connecting $(x_i,t_i)$ and $(x_f,t_f)$ and satisfies the Euler-Lagrange equation

$${\ddot{x}}_{\mathrm{c}}+{\omega }^2{x}_{\mathrm{c}}=0.\text{\hspace{1em}(1.133)}$$

The solution of equation (1.133) satisfying $x_{\mathrm{c}}(t_i)=x_i$ and $x_{\mathrm{c}}(t_f)=x_f$ is easily obtained as

$$x_{\mathrm{c}}(t)=\frac{1}{\sin \omega T}\left[x_f\sin \omega \left(t-t_i\right)+x_i\sin \omega \left(t_f-t\right)\right]\text{\hspace{1em}(1.134)}$$

where $T=t_f-t_i$ . Substituting this solution into the action, we obtain (exercise)

$$\begin{array}{rl}& S_c\equiv S[x_c]\\ & =\frac{m\omega }{2\sin \omega T}\left[\left(x_f^2+x_i^2\right)\cos \omega T-2x_f{x}_i\right].\end{array}\text{\hspace{1em}(1.135)}$$

Now the expansion of $S[x]$ around $x=x_c$ takes the form

$$S\left[x_{\mathrm{c}}+y\right]=S\left[x_{\mathrm{c}}\right]+\frac{1}{2!}\int \mathrm{d}{t}_1\mathrm{d}{t}_{2}y\left(t_1\right)y\left(t_2\right)\frac{{\delta }^{2}S[x]}{\delta x\left(t_1\right)\delta x\left(t_2\right)}{|}_{x=x_{\mathrm{c}}}\text{\hspace{1em}(1.136)}$$

where $y(t)$ satisfies the boundary condition $y(t_i)=y(t_f)=0$ . Note that (1) the first-order term vanishes since $\delta S[x]/\delta x=0$ at $x=x_{\mathrm{c}}$ and (2) terms of order three and higher do not exist since the action is second order in $x$ . Therefore, this expansion is exact and this problem is exactly solvable as we see later.

By noting that

$$\begin{array}{rl}\frac{\delta }{\delta x\left(t_1\right)}{\int }_{t_i}^{t_f}\mathrm{d}t\left[\frac{1}{2}m\dot{x}(t{)}^2-\frac{1}{2}m{\omega }^{2}x(t{)}^2\right]& =-m\frac{{\mathrm{d}}^2}{\mathrm{d}{t}_1^2}x\left(t_1\right)-m{\omega }^{2}x\left(t_1\right)\\ & =-m\left(\frac{{\mathrm{d}}^2}{\mathrm{d}{t}_1^2}+{\omega }^2\right)x(t_1)\end{array}$$

and that

$$\frac{\delta x(t_1)}{\delta x(t_2)}=\delta (t_1-t_2)$$

we obtain the second-order functional derivative

$$\frac{{\delta }^{2}S[x]}{\delta x\left(t_1\right)\delta x\left(t_2\right)}=-m\left(\frac{{\mathrm{d}}^2}{\mathrm{d}{t}_1^2}+{\omega }^2\right)\delta \left(t_1-t_2\right).\text{\hspace{1em}(1.137)}$$

Substituting this into equation (1.136) we find that

$$\begin{array}{rl}S\left[x_{\mathrm{c}}+y\right]& =S\left[x_{\mathrm{c}}\right]-\frac{m}{2!}\int \mathrm{d}{t}_1\mathrm{d}{t}_{2}y\left(t_1\right)y\left(t_2\right)\left(\frac{{\mathrm{d}}^2}{\mathrm{d}{t}_1^2}+{\omega }^2\right)\delta \left(t_1-t_2\right)\\ & =S\left[x_{\mathrm{c}}\right]+\frac{m}{2}\int \mathrm{d}t\left({\dot{y}}^2-{\omega }^2{y}^2\right),\end{array}\text{\hspace{1em}(1.138)}$$

where the boundary condition $y(t_i)=y(t_f)=0$ has been taken into account.

Since $\mathcal{D}x$ is translationally invariant, ¹ we may replace $\mathcal{D}x$ by $\mathcal{D}y$ to obtain

$$⟨x_f,t_f\mid x_i,t_i⟩={\mathrm{e}}^{\mathrm{i}S[x_c]}{\int }_{y(t_i)=y(t_f)=0}\mathcal{D}y{\mathrm{e}}^{\mathrm{i}\frac{m}{2}{\int }_{t_i}^{t_f}\mathrm{d}t\left({\dot{y}}^2-{\omega }^2{y}^2\right)}.\text{\hspace{1em}(1.139)}$$

Let us evaluate the fluctuation part

$$I_f={\int }_{y(0)=y(T)=0}\mathcal{D}y{e}^{i\frac{m}{2}{\int }_0^{T}dt\left({\dot{y}}^2-{\omega }^2{y}^2\right)}\text{\hspace{1em}(1.140)}$$

where we have shifted the $t$ variable so that $t_i$ now becomes $t=0$ . We expand $y(t)$ as

$$y(t)=\sum _{n\in \mathbb{N}}{a}_n\sin \frac{n\pi t}{T}\text{\hspace{1em}(1.141)}$$

in conformity with the boundary condition. Substitution of this expansion into the integral in the exponent yields

$${\int }_0^T\mathrm{d}t({\dot{y}}^2-{\omega }^2{y}^2)=\frac{T}{2}\sum _{n\in \mathbb{N}}{a}_n^2\left[{\left(\frac{n\pi }{T}\right)}^2-{\omega }^2\right].$$

The Fourier transform from $y(t)$ to $\{a_n\}$ may be regarded as a change of variables in the integration. For this transformation to be well defined, the number of variables must be the same. Suppose the number of the time slice is $N+1$ , including $t=0$ and $t=T$ , for which there are $N-1$ independent $y_k$ . Correspondingly, we must put $a_n=0$ for $n>N-1$ . The Jacobian associated with this change of variables is

$$J_N=\det \frac{\partial y_k}{\partial a_n}=\det \left[\sin \left(\frac{n\pi t_k}{T}\right)\right]\text{\hspace{1em}(1.142)}$$

where $t_k$ is the $k$ th time step when $[0,T]$ is divided into $N$ infinitesimal steps.

This Jacobian can be evaluated most easily for a free particle. Since the transformation $\{y_k\}\to \{a_n\}$ is independent of the potential, the Jacobian should be identical for both cases. The probability amplitude for a free particle has been obtained in (1.104) leading to

$$⟨x_f,T|x_i,0⟩={\left(\frac{1}{2\pi \mathrm{i}T}\right)}^{1/2}\exp \left[\mathrm{i}\frac{m}{2T}(x_f-x_i{)}^2\right]={\left(\frac{1}{2\pi \mathrm{i}T}\right)}^{1/2}{\mathrm{e}}^{\mathrm{i}S[x_{\mathrm{c}}]}.\text{\hspace{1em}(1.143)}$$

This is written in terms of a path integral as

$${\mathrm{e}}^{\mathrm{i}S\left[x_{\mathrm{c}}\right]}{\int }_{y(0)=y(T)=0}\mathcal{D}y{\mathrm{e}}^{\mathrm{i}\frac{m}{2}{\int }_0^T\mathrm{d}t{\dot{y}}^2}.\text{\hspace{1em}(1.144)}$$

By comparing these two expressions and noting that

$$\frac{m}{2}{\int }_0^T\mathrm{d}t{\dot{y}}^2\to m\sum _{n=1}^N\frac{a_n^2{n}^2{\pi }^2}{4T}$$

we arrive at the equality

$$\begin{array}{rl}{\left(\frac{1}{2\pi \mathrm{i}T}\right)}^{1/2}& ={\int }_{y(0)=y(T)=0}\mathcal{D}y{\mathrm{e}}^{\mathrm{i}\frac{m}{2}{\int }_0^T\mathrm{d}t{\dot{y}}^2}\\ & =\underset{N\to \infty }{\lim }{J}_N{\left(\frac{1}{2\pi \mathrm{i}\epsilon }\right)}^{1/2}\int \mathrm{d}{a}_1\dots \mathrm{d}{a}_{N-1}\exp \left(\mathrm{i}m\sum _{n=1}^{N-1}\frac{a_n^2{\pi }^2{n}^2}{4T}\right).\end{array}$$

By carrying out the Gaussian integrals, it is found that

$$\begin{array}{rl}{\left(\frac{1}{2\pi \mathrm{i}T}\right)}^{1/2}& =\underset{N\to \infty }{\lim }{J}_N{\left(\frac{1}{2\pi \mathrm{i}\epsilon }\right)}^{N/2}\prod _{n=1}^{N-1}\frac{1}{n}{\left(\frac{4\pi \mathrm{i}T}{{\pi }^2}\right)}^{1/2}\\ & =\underset{N\to \infty }{\lim }{J}_N{\left(\frac{1}{2\pi \mathrm{i}\epsilon }\right)}^{N/2}\frac{1}{(N-1)!}{\left(\frac{4\pi \mathrm{i}T}{{\pi }^2}\right)}^{(N-1)/2}\end{array}$$

from which we finally obtain, for finite $N$ , that

$$J_N=N^{-N/2}{2}^{-(N-1)/2}{\pi }^{N-1}(N-1)!\text{\hspace{1em}(1.145)}$$

The Jacobian $J_N$ clearly diverges as $N\to \infty$ . This does not matter at all, however, since we are not interested in $J_N$ on its own but a combination with other (divergent) factors.

The transition amplitude of a harmonic oscillator is now given by

$$\begin{array}{rl}& ⟨x_f,T|x_i,0⟩=\underset{N\to \infty }{\lim }{J}_N{\left(\frac{1}{2\pi \mathrm{i}\epsilon }\right)}^{N/2}{\mathrm{e}}^{\mathrm{i}S[x_{\mathrm{c}}]}\\ & \times \int \mathrm{d}{a}_1\dots \mathrm{d}{a}_{N-1}\exp \left[\mathrm{i}\frac{mT}{4}\sum _{n=1}^{N-1}{a}_n^2\left\{{\left(\frac{n\pi }{T}\right)}^2-{\omega }^2\right\}\right].\end{array}\text{\hspace{1em}(1.146)}$$

The integrals over $a_n$ are simple Gaussian integrals and easily carried out to yield

$$\int \mathrm{d}{a}_n\exp \left[\frac{\mathrm{i}mT}{4}{a}_n^2\left\{{\left(\frac{n\pi }{T}\right)}^2-{\omega }^2\right\}\right]={\left(\frac{4\mathrm{i}T}{\pi n^2}\right)}^{1/2}{\left[1-{\left(\frac{\omega T}{n\pi }\right)}^2\right]}^{-1/2}.$$

By substituting this result into equation (1.146), we obtain

$$\begin{array}{rl}& ⟨x_f,t_f|x_i,t_i⟩=\underset{N\to \infty }{\lim }{J}_N{\left(\frac{N}{2\pi \mathrm{i}T}\right)}^{N/2}{\mathrm{e}}^{\mathrm{i}S[x_{\mathrm{c}}]}\\ & \times \prod _{k=1}^{N-1}\left[\frac{1}{k}{\left(\frac{4\mathrm{i}T}{\pi }\right)}^{1/2}\right]\prod _{n=1}^{N-1}{\left[1-{\left(\frac{\omega T}{n\pi }\right)}^2\right]}^{-1/2}\\ & ={\left(\frac{1}{2\pi \mathrm{i}T}\right)}^{1/2}{\mathrm{e}}^{\mathrm{i}S[x_{\mathrm{c}}]}\prod _{n=1}^{N-1}{\left[1-{\left(\frac{\omega T}{n\pi }\right)}^2\right]}^{-1/2}.\end{array}\text{\hspace{1em}(1.147)}$$

The infinite product over $n$ is well known and reduces to

$$\underset{N\to \infty }{\lim }\prod _{n=1}^N\left[1-{\left(\frac{\omega T}{n\pi }\right)}^2\right]=\frac{\sin \omega T}{\omega T}\text{\hspace{1em}(1.148)}$$

Note that the divergence of $J_N$ cancelled with the divergence of the other terms to yield a finite value. Finally we have shown that

$$\begin{array}{rl}⟨x_f,t_f|x_i,t_i⟩& ={\left(\frac{\omega }{2\pi \mathrm{i}\sin \omega T}\right)}^{1/2}{\mathrm{e}}^{\mathrm{i}S[x_{\mathrm{c}}]}\\ & ={\left(\frac{\omega }{2\pi \mathrm{i}\sin \omega T}\right)}^{1/2}\exp \left[\frac{\mathrm{i}\omega }{2\sin \omega T}\left\{\left(x_f^2+x_i^2\right)\cos \omega T-2x_i{x}_f\right\}\right].\end{array}\text{\hspace{1em}(1.149)}$$

1.4.2 Partition function

The partition function of a harmonic oscillator is easily obtained from the eigenvalue $E_n=(n+1/2)\omega$ ,

$$\mathrm{Tr}{\mathrm{e}}^{-\beta \hat{H}}=\sum _{n=0}^{\infty }{\mathrm{e}}^{-\beta (n+1/2)\omega }=\frac{1}{2\sinh (\beta \omega /2)}.\text{\hspace{1em}(1.150)}$$

The inverse temperature $\beta$ can be regarded as the imaginary time by putting $\mathrm{i}T=\beta$ . Then the partition function may be evaluated from the path integral point of view.

Method 1: The trace may be taken over $\{|x⟩\}$ to yield

$$\begin{array}{rl}& Z(\beta )=\int \mathrm{d}x⟨x|{\mathrm{e}}^{-\beta \hat{H}}|x⟩\\ & ={\left(\frac{\omega }{2\pi \mathrm{i}(-\mathrm{i}\sinh \beta \omega )}\right)}^{1/2}\\ & \times \int \mathrm{d}x\exp \mathrm{i}\left[\frac{\omega }{-2\mathrm{i}\sinh \beta \omega }(2x^2\cosh \beta \omega -2x^2)\right]\\ & ={\left(\frac{\omega }{2\pi \sinh \beta \omega }\right)}^{1/2}{\left[\frac{\pi }{\omega \tanh (\beta \omega /2)}\right]}^{1/2}\\ & =\frac{1}{2\sinh (\beta \omega /2)}\end{array}\text{\hspace{1em}(1.151)}$$

where use has been made of equation (1.149).

The following exercise serves as a preliminary to Method 2.

Exercise 1.5

(1) Let $A$ be a symmetric positive-definite $n\times n$ matrix. Show that

$$\int \mathrm{d}{x}_1\dots \mathrm{d}{x}_n\exp \left(-\sum _{i,j}{x}_i{A}_{ij}{x}_j\right)={\pi }^{n/2}(\det A{)}^{-1/2}={\pi }^{n/2}\prod _i{\lambda }_i^{-1/2}\text{\hspace{1em}(1.152)}$$

where ${\lambda }_i$ is the eigenvalue of $A$

(2) Let $A$ be a positive-definite $n\times n$ Hermite matrix. Show that

$$\int \mathrm{d}{z}_1\mathrm{d}{\bar{z}}_1\dots \mathrm{d}{z}_n\mathrm{d}{\bar{z}}_n\exp \left(-\sum _{i,j}{\bar{z}}_i{A}_{ij}{z}_j\right)={\pi }^n(\det A{)}^{-1}={\pi }^n\prod _i{\lambda }_i^{-1}.\text{\hspace{1em}(1.153)}$$

Method 2: We next obtain the partition function by evaluating the path integral over the fluctuations with the help of the functional determinant and the $\zeta$ -function regularization. We introduce the imaginary time $\tau =\mathrm{i}t$ and rewrite the path integral as

$$\begin{array}{rl}& {\int }_{y(0)=y(T)=0}\mathcal{D}y\exp \left[\frac{\mathrm{i}}{2}\int \mathrm{d}ty\left(-\frac{{\mathrm{d}}^2}{\mathrm{d}{t}^2}-{\omega }^2\right)y\right]\\ & \to {\int }_{y(0)=y(\beta )=0}\bar{\mathcal{D}}y\exp \left[-\frac{1}{2}\int d\tau y\left(-\frac{d^2}{d{\tau }^2}+{\omega }^2\right)y\right],\end{array}$$

where we noted the boundary condition $y(0)=y(\beta )=0$ . Here the bar on $\mathcal{D}$ implies the path integration measure with imaginary time.

Let $A$ be an $n\times n$ Hermitian matrix with positive-definite eigenvalues ${\lambda }_k$ ( $1\le k\le n$ ). Then for real variables $x_k$ , we obtain from exercise 1.5 that

$$\prod _{k=1}^n\left({\int }_{-\infty }^{\infty }\mathrm{d}{x}_k\right){\mathrm{e}}^{-\frac{1}{2}{\sum }_{p,q}{x}_p{A}_{pq}{x}_q}=\prod _{k=1}^n\frac{1}{\sqrt{{\lambda }_k}}=\frac{1}{\sqrt{\det A}}$$

where we neglected numerical factors. This is a generalization of the well-known Gaussian integral

$${\int }_{-\infty }^{\infty }dx{e}^{-\frac{1}{2}\lambda x^2}=\sqrt{\frac{2\pi }{\lambda }}$$

for $\lambda >0$ . We define the determinant of an operator $\mathcal{O}$ by the (properly regularized) infinite product of its eigenvalues ${\lambda }_k$ as $\mathrm{Det}\mathcal{O}={\prod }_k{\lambda }_k$ . Then the previous path integral is written as

$${\int }_{y(0)=y(\beta )=0}\bar{\mathcal{D}}y\exp \left[-\frac{1}{2}\int d\tau y\left(-\frac{d^2}{d{\tau }^2}+{\omega }^2\right)y\right]=\frac{1}{\sqrt{{\mathrm{Det}}_{\mathrm{D}}\left(-d^2/d{\tau }^2+{\omega }^2\right)}},\text{\hspace{1em}(1.154)}$$

where the subscript ‘D’ implies that the eigenvalues are evaluated with the Dirichlet boundary condition $y(0)=y(\beta )=0$ .

The general solution $y(\tau )$ satisfying the boundary condition is written as

$$y(\tau )=\frac{1}{\sqrt{\beta }}\sum _{n\in \mathbb{N}}{y}_n\sin \frac{n\pi \tau }{\beta }.\text{\hspace{1em}(1.155)}$$

Note that $y_n\in \mathbb{R}$ since $y(\tau )$ is a real function. Since the eigenvalue of the eigenfunction $\sin (n\pi \tau /\beta )$ is ${\lambda }_n=(n\pi /\beta {)}^2+{\omega }^2$ , the functional determinant is formally written as

$$\begin{array}{rl}{\mathrm{Det}}_{\mathrm{D}}\left(-\frac{{\mathrm{d}}^2}{\mathrm{d}{\tau }^2}+{\omega }^2\right)& =\prod _{n=1}^{\infty }{\lambda }_n=\prod _{n=1}^{\infty }\left[{\left(\frac{n\pi }{\beta }\right)}^2+{\omega }^2\right]\\ & =\prod _{n=1}^{\infty }{\left(\frac{n\pi }{\beta }\right)}^2\prod _{p=1}^{\infty }\left[1+{\left(\frac{\beta \omega }{p\pi }\right)}^2\right].\end{array}\text{\hspace{1em}(1.156)}$$

The first infinite product in the last line is written as

$${\mathrm{Det}}_{\mathrm{D}}\left(-\frac{{\mathrm{d}}^2}{\mathrm{d}{\tau }^2}\right).$$

We will evaluate this infinite product through the $\zeta$ -function regularization. Let $\mathcal{O}$ be an operator with positive-definite eigenvalues ${\lambda }_n$ . Then we have formally

$$\log \mathrm{Det}\mathcal{O}=\mathrm{Tr}\log \mathcal{O}=\sum _n\log {\lambda }_n.\text{\hspace{1em}(1.157)}$$

Now we define the spectral $\zeta$ -function as

$${\zeta }_{\mathcal{O}}(s)\equiv \sum _n\frac{1}{{\lambda }_n^s}.\text{\hspace{1em}(1.158)}$$

The RHS converges for sufficiently large $\mathrm{Re}s$ and ${\zeta }_{\mathcal{O}}(s)$ is analytic with respect to $s$ in this region. Moreover, it can be analytically continued to the whole $s$ -plane except at a possible finite number of points. By noting that

$${\frac{\mathrm{d}{\zeta }_{\mathcal{O}}(s)}{\mathrm{d}s}|}_{s=0}=-\sum _n\log {\lambda }_n$$

we arrive at the expression

$$\mathrm{Det}\mathcal{O}=\exp {\left[-\frac{\mathrm{d}{\zeta }_{\mathcal{O}}(s)}{\mathrm{d}s}\right|}_{s=0}].\text{\hspace{1em}(1.159)}$$

We replace $\mathcal{O}$ by $-{\mathrm{d}}^2/\mathrm{d}{\tau }^2$ in the case at hand to find

$${\zeta }_{-{\mathrm{d}}^2/\mathrm{d}{\tau }^2}(s)=\sum _{n\ge 1}{\left(\frac{n\pi }{\beta }\right)}^{-2s}={\left(\frac{\beta }{\pi }\right)}^{2s}\zeta (2s)\text{\hspace{1em}(1.160)}$$

where $\zeta (2s)$ is the celebrated Riemann $\zeta$ -function. It is analytic over the whole $s$ -plane except at the simple pole at $s=1$ . From the well-known values

$$\zeta (0)=-\frac{1}{2}{\zeta }^{\prime }(0)=-\frac{1}{2}\log (2\pi )\text{\hspace{1em}(1.161)}$$

we obtain

$${\zeta }_{-{\mathrm{d}}^2/\mathrm{d}{\tau }^2}^{\prime }(0)=2\log \left(\frac{\beta }{\pi }\right)\zeta (0)+2{\zeta }^{\prime }(0)=-\log (2\beta ).$$

We have finally shown that

$${\mathrm{Det}}_{\mathrm{D}}\left(-\frac{{\mathrm{d}}^2}{\mathrm{d}{\tau }^2}\right)={\mathrm{e}}^{\log (2\beta )}=2\beta \text{\hspace{1em}(1.162)}$$

and that

$${\mathrm{Det}}_{\mathrm{D}}\left(-\frac{{\mathrm{d}}^2}{\mathrm{d}{\tau }^2}+{\omega }^2\right)=2\beta \prod _{p=1}^{\infty }\left[1+{\left(\frac{\beta \omega }{p\pi }\right)}^2\right].\text{\hspace{1em}(1.163)}$$

The infinite product in this equation is well known but let us pretend that we are ignorant about this product.

The partition function is now expressed as

$${\mathrm{Tre}}^{-\beta H}={\left[2\beta \prod _{p=1}^{\infty }\left\{1+{\left(\frac{\beta \pi }{p\pi }\right)}^2\right\}\right]}^{-1/2}{\left[\frac{\pi }{\omega \tanh (\beta \omega /2)}\right]}^{1/2}.\text{\hspace{1em}(1.164)}$$

By comparing this with the result (1.151), we have proved the formula

$$\prod _{n=1}^{\infty }\left[1+{\left(\frac{\beta \omega }{n\pi }\right)}^2\right]=\frac{\pi }{\beta \omega }\sinh (\beta \omega )$$

namely

$$\prod _{n=1}^{\infty }\left(1+\frac{x^2}{n^2}\right)=\frac{\sinh (\pi x)}{\pi x}.\text{\hspace{1em}(1.165)}$$

What about the infinite product expansion of the cosh function? This is given by using the path integral with respect to the fermion, which we will work out in the next section.

Endnotes

Unplaced source note

We will use ‘det’ for the determinant of a finite dimensional matrix while ‘Det’ for the (formal) determinant of an operator throughout this book. Similarly, the trace of a finite-dimensional matrix is denoted ‘tr’ while that of an operator is denoted ‘Tr’.

来源与版本

正文：英文 · OCR 机器稿 · 待校对

核对状态：OCR 机器稿 · 待校对

原书 PDF

来源版本：second-edition-local

来源 PDF SHA-256：fcd24c3371d092a8c699ccf5cf0765462924b6c48412804231076e8084b1940d

OCR 来源 SHA-256：0b6324f01172a3548cd34382e82fd8bfae574602b6b2cfaa357ba96bc2c35478

OCR 产物 SHA-256：f1a774ec3b60244bc65d108d62c84e246c6fd34527df5584a43465d641cf48c8

Footnotes

1. Integrating over all possible paths $x(t)$ with $x(t_i)=x_i$ and $x(t_f)=x_f$ is equivalent to integrating over all possible paths $y(t)$ with $y(t_i)=y(t_f)=0$ , where $x(t)=x_c(t)+y(t)$ . ↩