Schwartz 第14章 · 来源与作者路径

逐字来源、作者主张与候选译文分别保留。工程核验不批准 OCR、译文或数学正确性。

概念与作者路径

§14 路径积分的引入

作者引入路径积分这一不同于先前正则表述的方法。

条件与约定：引入不是正式定义；后续量子力学推导与场论推广分开。

逐字作者来源与版本

So far, we have studied quantum field theory using the canonical quantization approach, which is based on creation and annihilation operators. There is a completely different way to do quantum field theory called the path integral formulation. It says

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式(14.6) 一维高斯积分结果

平方配方和积分变换后的高斯结果。

条件与约定：复参数延拓与收敛条件不由此记录补充证明。

逐字作者来源与版本

$$
\int_ {- \infty} ^ {\infty} d p e ^ {- \frac {1}{2} a p ^ {2} + J p} = \sqrt {\frac {2 \pi}{a}} e ^ {\frac {J ^ {2}}{2 a}}. \tag {14.6}
$$

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后文使用：用式(14.6)计算短时间矩阵元。

式(14.10) 时间切片

在小时间片上分解振幅。

条件与约定：平滑时间依赖及无穷小区间近似。

逐字作者来源与版本

If instead we only assume  $\tilde{H}(t)$  is a smooth function of  $t$ , then we can only solve for the matrix element this way for infinitesimal time intervals. So, let us break this down into  $n$  small time intervals  $\delta t$  and define  $t_j = t_i + j\delta t$  and  $t_n = t_f$ . Then,

$$
\langle f | i \rangle = \int d x _ {n} \dots d x _ {1} \left\langle x _ {f} \right| e ^ {- i H \left(t _ {f}\right) \delta t} \left| x _ {n} \right\rangle \left\langle x _ {n} \right| \dots \left| x _ {2} \right\rangle \left\langle x _ {2} \right| e ^ {- i H \left(t _ {2}\right) \delta t} \left| x _ {1} \right\rangle \left\langle x _ {1} \right| e ^ {- i H \left(t _ {1}\right) \delta t} \left| x _ {i} \right\rangle . \tag {14.10}
$$

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后文使用：式(14.14) 路径积分前的时间片乘积。

用式(14.6)计算短时间矩阵元

作者明确代入高斯公式得到短时间矩阵元。

条件与约定：非相对论二次动量哈密顿量与时间片近似。

逐字作者来源与版本

Now we can use the Gaussian integral in Eq. (14.6),  $\int dp\exp \left(-\frac{1}{2} ap^2 +Jp\right) = \sqrt{\frac{2\pi}{a}}\exp \left(\frac{J^2}{2a}\right)$ , with  $a = i\frac{\delta t}{m}$  and  $J = i(x_{j + 1} - x_j)$  to get

$$
\langle x _ {j + 1} | e ^ {- i H \delta t} | x _ {j} \rangle = N e ^ {- i V (x _ {j}, t _ {j}) \delta t} e ^ {i \frac {m}{2} \delta t \frac {(x _ {j + 1} - x _ {j}) ^ {2}}{(\delta t) ^ {2}}} = N e ^ {i L (x, \dot {x}) \delta t}, \tag {14.12}
$$

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使用前置陈述：式(14.6) 一维高斯积分结果。作者有明确编号回指。

证据：作者编号回指

Now we can use the Gaussian integral in Eq. (14.6),  $\int dp\exp \left(-\frac{1}{2} ap^2 +Jp\right) = \sqrt{\frac{2\pi}{a}}\exp \left(\frac{J^2}{2a}\right)$ , with  $a = i\frac{\delta t}{m}$  and  $J = i(x_{j + 1} - x_j)$  to get

后文使用：式(14.14) 路径积分前的时间片乘积。

§14.2.2 高斯积分实现勒让德变换

高斯积分把哈密顿量表述转为拉格朗日量表述。

条件与约定：局部身份同时涉及拉格朗日量与勒让德变换；本绑定仅取后者，不能推广到所有受约束系统。

逐字作者来源与版本

where  $N$  is an  $x$ - and  $t$ -independent normalization constant, which we will justify ignoring later, and

$$
L (x, \dot {x}) = \frac {1}{2} m \dot {x} ^ {2} - V (x, t) \tag {14.13}
$$

is the Lagrangian. We see that the Gaussian integral effected a Legendre transform to go from  $H(x,p)$  to  $L(x,\dot{x})$ .

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式(14.14) 路径积分前的时间片乘积

作者把(14.12)代入(14.10)，得到拉格朗日量指数乘积。

条件与约定：依赖已求得的短时间矩阵元；还未取连续极限。

逐字作者来源与版本

Using Eq. (14.12), each term in Eq. (14.10) becomes just a number and the product reduces to

$$
\langle f | i \rangle = N ^ {n} \int d x _ {n} \dots d x _ {1} e ^ {i L \left(x _ {n}, \dot {x} _ {n}\right) \delta t} \dots e ^ {i L \left(x _ {1}, \dot {x} _ {1}\right) \delta t}. \tag {14.14}
$$

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使用前置陈述：用式(14.6)计算短时间矩阵元。作者有明确编号回指。

证据：作者编号回指

Using Eq. (14.12), each term in Eq. (14.10) becomes just a number and the product reduces to

使用前置陈述：式(14.10) 时间切片。作者有明确编号回指。

证据：作者编号回指

Using Eq. (14.12), each term in Eq. (14.10) becomes just a number and the product reduces to

§14.2.2 路径积分表达式 (14.15)

时间片极限得到固定端点路径求和与作用量指数。

条件与约定：$\hslash =1$；端点固定；归一化 N 的形式发散及消去按作者保留。

逐字作者来源与版本

Finally, taking the limit  $\delta t\to 0$ , the exponentials combine into an integral over  $dt$  and we get

$$
\langle f | i \rangle = N \int_ {x \left(t _ {i}\right) = x _ {i}} ^ {x \left(t _ {f}\right) = x _ {f}} \mathcal {D} x (t) e ^ {i S [ x ]}, \tag {14.15}
$$

where  $\mathcal{D}x$  means sum over all paths  $x(t)$  with the correct boundary conditions and the action is  $S[x] = \int dt\mathcal{L}[x(t),\dot{x} (t)]$ . Note that  $N$  has been redefined and is now formally infinite, but it will drop out of any physical quantities, as we will see in the path integral case.

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§14.2.3 薛定谔绘景场

场算符取代量子力学的位置算符；每个空间点对应算符。

条件与约定：是薛定谔绘景在场论中的使用，不是有限维态空间假设。

逐字作者来源与版本

The field theory derivation is very similar, but the set of intermediate states is more complicated. We will start by calculating the vacuum matrix element  $\langle 0; t_f | 0; t_i \rangle$ . In quantum mechanics we broke the amplitude down into integrals over  $|x\rangle \langle x|$  for intermediate times where the states  $|x\rangle$  are eigenstates of the  $\hat{x}$  operator. In field theory, the equivalents of  $\hat{x}$  are the Schrödinger picture fields  $\hat{\phi}(\vec{x})$ , which at any time  $t$  can be written as

$$
\hat {\phi} (\vec {x}) = \int \frac {d ^ {3} p}{(2 \pi) ^ {3}} \frac {1}{\sqrt {2 \omega_ {p}}} \left(a _ {p} e ^ {i \vec {p} \vec {x}} + a _ {p} ^ {\dagger} e ^ {- i \vec {p} \vec {x}}\right). \tag {14.16}
$$

Each field comprises an infinite number of operators, one at each point  $\vec{x}$ . We put the hat on  $\phi$  to remind you that it is an operator.

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§14.2.4 经典极限与稳相法

恢复 ℏ 后，作者用稳相法说明经典极限由作用量驻值配置主导。

条件与约定：形式路径积分与 $\hslash o0$；这里不新增严密测度存在性或渐近误差证明。

逐字作者来源与版本

As a first check on the path integral, we can take the classical limit. To do that, we need to put back  $\hbar$ , which can be done by dimensional analysis. Since  $\hbar$  has dimensions of action, it appears as

$$
\langle 0; t _ {f} | 0; t _ {i} \rangle = N \int \mathcal {D} \Phi (\vec {x}, t) e ^ {\frac {i}{\hbar} S [ \Phi ]}. \tag {14.31}
$$

Using the method of stationary phase we see that, in the limit  $\hbar \rightarrow 0$ , this integral is dominated by the value of  $\Phi$  for which  $S[\Phi]$  has an extremum. But  $\delta S = 0$  is precisely the condition that determines the Euler-Lagrange equations which a classical field satisfies. Therefore, the only configuration that contributes in the classical limit is the classical solution to the equations of motion.

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§14.6 格拉斯曼代数

由反对易生成元及复数系数组合得到格拉斯曼代数。

条件与约定：Schwartz 此处采用复系数；Woit 30.1 的定义采用实数域；不消去基域差别。

符号：$het{a}_{i}het{a}_j=-het{a}_{j}het{a}_i$

逐字作者来源与版本

A Grassmann algebra is a set of objects  $\mathcal{G}$  that are generated by a basis  $\{\theta_i\}$ . These  $\theta_i$  are Grassmann numbers, which anticommute with each other,  $\theta_i\theta_j = -\theta_j\theta_i$ , add commutatively,  $\theta_i + \theta_j = \theta_j + \theta_i$ , and can be multiplied by complex numbers,  $a\theta \in \mathcal{G}$  for  $\theta \in \mathcal{G}$  and  $a \in \mathbb{C}$ . The algebra must also have an element 0 so that  $\theta_i + 0 = \theta_i$ .

For one  $\theta$ , the most general element of the algebra is

$$
g = a + b \theta , \quad a, b \in \mathbb {C}, \tag {14.85}
$$

since  $\theta^2 = 0$ . For two  $\theta$ 's, the most general element is

$$
g = A + B \theta_ {1} + C \theta_ {2} + F \theta_ {1} \theta_ {2}, \tag {14.86}
$$

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§14.6 分次、外代数类比与费米子场

作者讨论奇偶分次及与微分形式外代数的类比，再给出费米子场生成元。

条件与约定：原文称 odd-graded subalgebra 随即又说其不封闭；原文保留，术语冲突待审，不据此断言奇部是子代数。

逐字作者来源与版本

and so on. Elements of the algebra that have an even number of  $\theta_{i}$  commute with all elements of the algebra, so they compose the even-graded or bosonic subalgebra. Similarly, the odd-graded or fermionic subalgebra has an odd number of  $\theta_{i}$  and anticommutes within itself (but commutes with the bosonic subalgebra). The fermionic subalgebra is not closed, since  $\theta_{1}\theta_{2}$  is bosonic.

Sometimes it is helpful to compare what we will do with fermions to an example of a Grassmann algebra that you might already be familiar with: the exterior algebra of differential forms. Two forms  $A$  and  $B$  form a Grassmann algebra with the product usually denoted with a wedge, so that  $A \wedge B = -B \wedge A$ . So, for example,  $\mathbf{dx}$  and  $\mathbf{dy}$  would generate a two-dimensional Grassmann algebra.

In physics our Grassmann numbers will be  $\theta_{1} = \psi (x_{1}),\theta_{2} = \psi (x_{2}),\ldots$  , so we will have an infinite number of them. Then quantities such as the Lagrangian are (bosonic) elements of  $\mathcal{G}$ . To get regular numbers out, we need to integrate over  $\mathcal{D}\psi$ . So we need to figure out a consistent way to define such integrals.

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