### 物理代写|量子场论代写Quantum field theory代考|PHYSICS 3544

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• (Generalized) Linear Models 广义线性模型
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• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 物理代写|量子场论代写Quantum field theory代考|Path Integrals in Quantum and Statistical Mechanics

Already back in 1933 , Dirac asked himself whether the classical Lagrangian and action are as significant in quantum mechanics as they are in classical mechanics $[1,2]$. He observed that for simple systems, the probability amplitude
$$K\left(t, q^{\prime}, q\right)=\left\langle q^{\prime}\left|\mathrm{e}^{-\mathrm{i} \hat{A} t / h}\right| q\right\rangle$$
for the propagation from a point with coordinate $q$ to another point with coordinate $q^{\prime}$ in time $t$ is given by
$$K\left(t, q^{\prime}, q\right) \propto \mathrm{e}^{\mathrm{i} S\left[q_{\mathrm{cl}}\right] / h}$$
where $q_{\mathrm{cl}}$ denotes the classical trajectory from $q$ to $q^{\prime}$. In the exponent the action of this trajectory enters as a multiple of Planck’s reduced constant $h$. For a free particle with Lagrangian
$$L_{0}=\frac{m}{2} \dot{q}^{2}$$ the formula $(2.2)$ is verified easily: A free particle moves with constant velocity $\left(q^{\prime}-q\right) / t$ from $q$ to $q^{\prime}$ and the action of the classical trajectory is
$$S\left[q_{\mathrm{cl}}\right]=\int_{0}^{t} \mathrm{~d} s L_{0}\left[q_{\mathrm{cl}}(s)\right]=\frac{m}{2 t}\left(q^{\prime}-q\right)^{2}$$
The factor of proportionality in $(2.2)$ is then uniquely fixed by the condition $\mathrm{e}^{-\mathrm{i} \hat{H} t / \hbar} \longrightarrow 1$ for $t \rightarrow 0$ which in position space reads
$$\lim {t \rightarrow 0} K\left(t, q^{\prime}, q\right)=\delta\left(q^{\prime}, q\right)$$ Alternatively, it is fixed by the property $\mathrm{e}^{-\mathrm{i} \hat{H} t / h} \mathrm{e}^{-\mathrm{i} \hat{H} s / h}=\mathrm{e}^{-\mathrm{i} \hat{H}(t+s) / h}$ that takes the form $$\int \mathrm{d} u K\left(t, q^{\prime}, u\right) K(s, u, q)=K\left(t+s, q^{\prime}, q\right)$$ in position space. Thus, the correct free particle propagator on a line is given by $$K{0}\left(t, q^{\prime} \cdot q\right)=\left(\frac{m}{2 \pi \mathrm{i} \hbar t}\right)^{1 / 2} \mathrm{c}^{\mathrm{i} m\left(q^{\prime}-q\right)^{2} / 2 h t}$$
Similar results hold for the harmonic oscillator or systems for which $\langle\hat{q}(t)\rangle$ fulfills the classical equation of motion. For such systems $\left\langle V^{\prime}(\hat{q})\right\rangle=V^{\prime}(\langle\hat{q}\rangle)$ holds true. However, for general systems, the simple formula (2.2) must be extended, and it was Feynman who discovered this extension back in 1948. He realized that all paths from $q$ to $q^{\prime}$ (and not only the classical path) contribute to the propagator. This means that in quantum mechanics a particle can potentially move on any path $q(s)$ from the initial to the final destination,
$$q(0)=q \quad \text { and } \quad q(t)=q^{\prime}$$

## 物理代写|量子场论代写Quantum field theory代考|Recalling Quantum Mechanics

There are two well-established ways to quantize a classical system: canonical quantization and path integral quantization. For completeness and later use, we recall the main steps of canonical quantization both in Schrödinger’s wave mechanics and Heisenberg’s matrix mechanics.

A classical system is described by its coordinates $\left{q^{i}\right}$ and momenta $\left{p_{i}\right}$ on phase space $\Gamma$. An observable $O$ is a real-valued function on $\Gamma$. Examples are the coordinates on phase space and the energy $H(q, p)$. We assume that phase space comes along with a symplectic structure and has local coordinates with Poisson brackets
$$\left{q^{i}, p_{j}\right}=\delta_{j}^{i}$$
The brackets are extended to observables on through antisymmetry and the derivation rule ${O P, Q}=O{P, Q}+{O, Q} P$. The evolution in time of an observable is determined by
$$\dot{O}={O, H}, \quad \text { e.g. } \quad \dot{q}^{i}=\left{q^{i}, H\right} \quad \text { and } \quad \dot{p}{i}=\left{p{i}, H\right}$$
In the canonical quantization, functions on phase space are mapped to operators, and the Poisson brackets of two functions become commutators of the associated operators:
$$O(q, p) \rightarrow \hat{O}(\hat{q}, \hat{p}) \quad \text { and } \quad{O, P} \longrightarrow \frac{1}{\mathrm{i} \hbar}[\hat{O}, \hat{P}]$$

The time evolution of an (not explicitly time-dependent) observable is determined by Heisenberg’s equation
$$\frac{\mathrm{d} \hat{O}}{\mathrm{~d} t}=\frac{\mathrm{i}}{\hbar}[\hat{H}, \hat{O}]$$
In particular the phase space coordinates $\left(q^{l}, p_{i}\right)$ become operators with commutation relations $\left[\hat{q}^{i}, \hat{p}{j}\right]=\mathrm{i} \hbar \delta{j}^{i}$, and their time evolution is determined by
$$\frac{\mathrm{d} \hat{q}^{i}}{\mathrm{~d} t}=\frac{\mathrm{i}}{\hbar}\left[\hat{H}, \hat{q}^{i}\right] \quad \text { and } \quad \frac{\mathrm{d} \hat{p}{i}}{\mathrm{~d} t}=\frac{\mathrm{i}}{\hbar}\left[\hat{H}, \hat{p}{i}\right]$$
For a system of non-relativistic and spinless particles, the Hamiltonian reads
$$\hat{H}=\hat{H}{0}+\hat{V} \quad \text { with } \quad \hat{H}{0}=\frac{1}{2 m} \sum \hat{p}{i}^{2}$$ and one arrives at Heisenberg’s equations of motion $$\frac{\mathrm{d} \hat{q}^{i}}{\mathrm{~d} t}=\frac{\hat{p}{i}}{m} \quad \text { and } \quad \frac{\mathrm{d} \hat{p}{i}}{\mathrm{~d} t}=-\hat{V}{, i}$$
Observables are represented by Hermitian operators on a Hilbert space $\mathscr{H}$, whose elements characterize the states of the system:
$$\hat{O}(\hat{q}, \hat{p}): \mathcal{H} \longrightarrow \mathcal{H}$$
Consider a particle confined to an endless wire. Its Hilbert space is $\mathcal{H}=L_{2}(\mathbb{R})$, and its position and momentum operator are represented in position space as
$$(\hat{q} \psi)(q)=q \psi(q) \quad \text { and } \quad(\hat{p} \psi)(q)=\frac{\hbar}{i} \partial_{q} \psi(q)$$

## 物理代写|量子场论代写Quantum field theory代考|Feynman–Kac Formula

We shall derive Feynman’s path integral representation for the unitary time evolution operator $\exp (-\mathrm{i} \hat{H} t)$ as well as Kac’s path integral representation for the positive operator $\exp (-\hat{H} \tau)$. Thereby we shall utilize the product formula of Trotter. In case of matrices, this formula was already verified by Lie and has the form:
Theorem 2.1 (Lie’s Theorem) For two matrices $\mathrm{A}$ and $\mathrm{B}$
$$\mathrm{e}^{\mathrm{A}+\mathrm{B}}=\lim {n \rightarrow \infty}\left(\mathrm{e}^{\mathrm{A} / n} \mathrm{e}^{\mathrm{B} / n}\right)^{n}$$ To prove this theorem, we define for each $n$ the two matrices $\mathrm{S}{n}:=\exp (\mathrm{A} / n+\mathrm{B} / n)$ and $\mathrm{T}{n}:=\exp (\mathrm{A} / n) \exp (\mathrm{B} / n)$ and telescope the difference of their $n$ ‘th powers, $$\mathrm{S}{n}^{n}-\mathrm{T}{n}^{n}=\mathrm{S}{n}^{n-1}\left(\mathrm{~S}{n}-\mathrm{T}{n}\right)+\mathrm{S}{n}^{n-2}\left(\mathrm{~S}{n}-\mathrm{T}{n}\right) \mathrm{T}{n}+\cdots+\left(\mathrm{S}{n}-\mathrm{T}{n}\right) \mathrm{T}{n}^{n-1}$$ Now we choose any (sub-multiplicative) matrix norm, for example, the Frobenius norm. The triangle inequality together with $|X Y| \leq|X \mid| Y |$ imply the inequality $|\exp (X)| \leq \exp (|X|)$ such that $$\left|\mathrm{S}{n}\right|,\left|\mathrm{T}{n}\right| \leq a^{1 / n} \quad \text { with } \quad a=\mathrm{e}^{|\mathrm{A}|+|\mathrm{B}|}$$ Thus we conclude $$\left|\mathrm{S}{n}^{n}-\mathrm{T}{n}^{n}\right| \equiv\left|\mathrm{e}^{\mathrm{A}+B}-\left(\mathrm{e}^{\mathrm{A} / n} \mathrm{e}^{B / n}\right)^{n}\right| \leq n \times a^{(n-1) / n}\left|\mathrm{~S}{n}-\mathrm{T}{n}\right|$$ Finally, using $\mathrm{S}{n}-\mathrm{T}_{n}=-[\mathrm{A}, \mathrm{B}] / 2 n^{2}+O\left(1 / n^{3}\right)$, the product formula is verified for matrices. But the theorem also holds for self-adjoint operators.

Theorem $2.2$ (Trotter’s Theorem) If $\hat{A}$ and $\hat{B}$ are self-adjoint operators and $\hat{A}+$ $\hat{B}$ is essentially self-adjoint on the intersection $\mathscr{D}$ of their domains, then
$$\mathrm{e}^{-\mathrm{i} t(\hat{A}+\hat{B})}=s-\lim {n \rightarrow \infty}\left(\mathrm{e}^{-\mathrm{i} t \hat{A} / n} \mathrm{e}^{-\mathrm{i} t \hat{B} / n}\right)^{n}$$ If in addition $\hat{A}$ and $\hat{B}$ are bounded from below, then $$\mathrm{e}^{-\tau(\hat{A}+\hat{B})}=s-\lim {n \rightarrow \infty}\left(\mathrm{e}^{-\tau \hat{A} / n} \mathrm{e}^{-\tau \hat{B} / n}\right)^{n}$$
The operators need not be bounded and the convergence is with respect to the strong operator topology. For operators $\hat{A}{n}$ and $\hat{A}$ on a common domain $\mathscr{D}$ in the Hilbert space, we have s- $\lim {n \rightarrow \infty} \hat{A}{n}=\hat{A}$ iff $\left|\hat{A}{n} \psi-\hat{A} \psi\right| \rightarrow 0$ for all $\psi \in \mathscr{D}$. Formula (2.27) is used in quantum mechanics, and formula $(2.28)$ finds its application in statistical physics and the Euclidean formulation of quantum mechanics [16].

Let us assume that $\hat{H}$ can be written as $\hat{H}=\hat{H}{0}+\hat{V}$ and apply the product formula to the evolution kernel in (2.22). With $\varepsilon=t / n$ and $\hbar=1$, we obtain \begin{aligned} K\left(t, q^{\prime}, q\right) &=\lim {n \rightarrow \infty}\left\langle q^{\prime}\left|\left(\mathrm{e}^{-\mathrm{i} \varepsilon \hat{H}{0}} \mathrm{e}^{-\mathrm{i} \varepsilon \hat{V}}\right)^{n}\right| q\right\rangle \ &=\lim {n \rightarrow \infty} \int \mathrm{d} q_{1} \cdots \mathrm{d} q_{n-1} \prod_{j=0}^{j=n-1}\left|q_{j+1}\right| \mathrm{e}^{-\mathrm{i} \varepsilon \hat{H}{0}} \mathrm{e}^{-i \varepsilon \hat{V}}\left|q{j}\right\rangle \end{aligned}
where we repeatedly inserted the resolution of the identity $(2.21)$ and denoted the initial and final point by $q_{0}=q$ and $q_{n}=q^{\prime}$, respectively. The potential $\hat{V}$ is diagonal in position space such that
$$\left\langle q_{j+1}\left|\mathrm{e}^{-\mathrm{i} \varepsilon \hat{H}{0}} \mathrm{e}^{-\mathrm{i} \varepsilon \hat{V}}\right| q{j}\right\rangle=\left\langle q_{j+1}\left|\mathrm{e}^{-\mathrm{i} \varepsilon \hat{H}{0}}\right| q{j}\right\rangle \mathrm{e}^{-\mathrm{i} \varepsilon V\left(q_{j}\right)}$$

## 物理代写|量子场论代写Quantum field theory代考|Path Integrals in Quantum and Statistical Mechanics

ķ(吨,q′,q)=⟨q′|和−一世一个^吨/H|q⟩

ķ(吨,q′,q)∝和一世小号[qCl]/H

∫d在ķ(吨,q′,在)ķ(s,在,q)=ķ(吨+s,q′,q)在位置空间。因此，一条线上正确的自由粒子传播子由下式给出

ķ0(吨,q′⋅q)=(米2圆周率一世ℏ吨)1/2C一世米(q′−q)2/2H吨

q(0)=q 和 q(吨)=q′

## 物理代写|量子场论代写Quantum field theory代考|Recalling Quantum Mechanics

\left{q^{i}, p_{j}\right}=\delta_{j}^{i}\left{q^{i}, p_{j}\right}=\delta_{j}^{i}

○(q,p)→○^(q^,p^) 和 ○,磷⟶1一世ℏ[○^,磷^]

（不是明确的时间相关的）可观测的时间演化由海森堡方程确定

d○^ d吨=一世ℏ[H^,○^]

dq^一世 d吨=一世ℏ[H^,q^一世] 和 dp^一世 d吨=一世ℏ[H^,p^一世]

H^=H^0+在^ 和 H^0=12米∑p^一世2一个到达海森堡的运动方程

dq^一世 d吨=p^一世米 和 dp^一世 d吨=−在^,一世
Observables 由希尔伯特空间上的 Hermitian 算子表示H，其元素表征系统的状态：

○^(q^,p^):H⟶H

(q^ψ)(q)=qψ(q) 和 (p^ψ)(q)=ℏ一世∂qψ(q)

## 物理代写|量子场论代写Quantum field theory代考|Feynman–Kac Formula

Theorem 2.1 (Lie’s Theorem) 对于两个矩阵一个和乙

|小号n|,|吨n|≤一个1/n 和 一个=和|一个|+|乙|因此我们得出结论

|小号nn−吨nn|≡|和一个+乙−(和一个/n和乙/n)n|≤n×一个(n−1)/n| 小号n−吨n|最后，使用小号n−吨n=−[一个,乙]/2n2+○(1/n3)，乘积公式针对矩阵进行验证。但该定理也适用于自伴算子。

ķ(吨,q′,q)=林n→∞⟨q′|(和−一世eH^0和−一世e在^)n|q⟩ =林n→∞∫dq1⋯dqn−1∏j=0j=n−1|qj+1|和−一世eH^0和−一世e在^|qj⟩

⟨qj+1|和−一世eH^0和−一世e在^|qj⟩=⟨qj+1|和−一世eH^0|qj⟩和−一世e在(qj)

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