### 物理代写|量子场论代写Quantum field theory代考|PHYS4040

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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|One-parameter Unitary Groups and Stone’s Theorem

A (strongly continuous) one-parameter unitary group is simply a (strongly continuous) unitary representation of $\mathbb{R}$, that is a map which associates to $t \in \mathbb{R}$ a unitary operator $U(t)$ on Hilbert space $\mathcal{H}$ in such a manner that
$$U(s) U(t)=U(s+t),$$
and (the continuity condition)
$$\forall x, y \in \mathcal{H}, \lim _{t \rightarrow 0}(x, U(t)(y))=(x, y) .$$

The archetypical example is the operator $U(t)$ on $L^{2}(\mathbb{R})$ given for $f \in L^{2}$ and $w \in \mathbb{R}$ by
$$U(t)(f)(w)=\exp (\mathrm{i} t w / h) f(w) .$$
This is simply the operator “multiplication by the function exp(it $\cdot / \hbar) . “$ Another example is the operator $V(t)$ on $L^{2}(\mathbb{R})$ given for $f \in L^{2}$ and $w \in \mathbb{R}$ by
$$V(t)(f)(w)=f(w+t) .$$
In both cases it is a nice exercise of elementary analysis to prove that these operators are strongly continuous. These one-parameter groups are closely related by the Fourier transform. Indeed,
$$\widehat{V(t)(f)}=U(t) \hat{f} .$$
Exercise 2.14.1 Make sure you understand every detail of the proof of the important formula (2.55)

Theorem 2.14.2 (Stone’s theorem) There is a one-to-one correspondence between the strongly continuous one-parameter unitary groups on a Hilbert space $\mathcal{H}$ and the self-adjoint operators on $\mathcal{H}$. Given the unitary group $U$, the corresponding self-adjoint operator $A$ is called the infinitesimal generator of U. It is defined by the formula ${ }^{58}$
$$A(x)=\lim _{t \rightarrow 0} \frac{h}{i}(U(t)(x)-x) .$$
and its domain $\mathcal{D}=\mathcal{D}(A)$ is the set of $x$ for which the previous limit exists.

## 物理代写|量子场论代写Quantum field theory代考|Time-evolution

Consider a physical system, the state of which is described by a vector in $\mathcal{H}$.
Principle 5 If the system does not change with time (not in the sense that it does not evolve, but in the sense that it is not subjected to variable external influences), its evolution between time $t_{0}$ and time $t_{1}$ is described by a unitary operator ${ }^{62} U\left(t_{1}, t_{0}\right)$.

This operator depends only on $t_{1}-t_{0}$, reflecting the fact that the laws of physics are believed not to change with time ${ }^{63}$

Please observe the notation: the evolution $U\left(t_{1}, t_{0}\right)$ is from $t_{0}$ to $t_{1}$. The reason for this notation is that the evolution of the system from $t_{0}$ to $t_{1}$ and then from $t_{1}$ to $t_{2}$ is represented by $U\left(t_{2}, t_{1}\right) U\left(t_{1}, t_{0}\right)$, which also represents the same evolution as $U\left(t_{2}, t_{0}\right)$ so that as in (2.45) these two operators should differ only by a phase, $U\left(t_{2}, t_{1}\right) U\left(t_{1}, t_{0}\right)=c U\left(t_{2}, t_{0}\right)$ for some $c$ of modulus 1. Thus $U\left(t_{2}-t_{1}, 0\right) U\left(t_{1}-t_{0}, 0\right)=c U\left(t_{2}-t_{0}, 0\right)$. This means that $U(t):=$ $U(t, 0)$ should be a projective representation of $\mathbb{R}$ in $\mathcal{H}$, and on physical grounds it should be continuous in some sense. As shown in Section $2.13$ this projective representation arises from a true representation, so we can as well assume that it already is a true representation, and Stone’s theorem describes these. Therefore, there exists a self-adjoint operator $H$ on $\mathcal{H}$ such that
$$U(t)=\exp (-\mathrm{i} t H / \hbar) .$$
Probably it is worth making explicit the following fundamental point:
The time-evolution of a quantum system is entirely deterministic.
The minus sign in (2.75) is conventional. The reason for this convention will appear in Section 9.7. Since $h$ has the dimension of an energy times a time, $H$ has the dimension of an energy. It is called the Hamiltonian of the system. Although this is certainly not obvious ${ }^{64}$
The Hamiltonian should be thought of as representing the energy of the system.
A consequence of (2.75) is that if $\psi$ belongs to the domain of $H$, then, by the second part of $(2.64), \psi(t):=U(t)(\psi)$ satisfies the equation
$$\psi^{\prime}(t)=-\frac{\mathrm{i}}{h} H(\psi(t)) \text {. }$$

## 物理代写|量子场论代写Quantum field theory代考|Schrödinger and Heisenberg Pictures

The previous description of an evolving state $\psi(t)$ and of time-independent operators is called the Schrödinger picture.

We have seen that we may be able to improve matters by re-shuffling the state space using a unitary transformation. A fundamental idea is to try this using a time-dependent unitary transformation $V(t)$, replacing the state $\psi$ by $V(t)(\psi)$, and replacing the operator $A$ by $A(t)=V(t) A V(t)^{-1}$. Here we use a first simple implementation, with $V(t)=$ $\exp ($ it $H / \hbar)=U(t)^{-1}$. This is called the Heisenberg picture. In the Heisenberg picture the state $\psi(t)$ is replaced by $V(t)(\psi(t))=U(t)^{-1} U(t)(\psi)=\psi$, so that states do not change with time. On the other hand, an operator $A$ is replaced by the operator
$$A(t)=U(t)^{-1} A U(t)=\exp (\mathrm{i} t H / \hbar) A \exp (-\mathrm{i} t H / \hbar) .$$
Suppose that at time $t=0$ the system is in state $\psi$. Then, in the Schrödinger picture, the average value of $A$ at time $t$ is given by
$$(\psi(t), A \psi(t))=(U(t) \psi, A U(t) \psi)=\left(\psi, U(t)^{-1} A U(t) \psi\right)=(\psi, A(t) \psi),$$
where the last expression is the same quantity in the Heisenberg picture. Thus these two pictures are fortunately consistent with each other.

We may wonder why there is anything to gain by moving from the Schrödinger picture, where the simpler objects, the states, evolve, but where the complicated objects, the operators, are constant, to the Heisenberg picture where the simpler objects are constant, but the complicated ones evolve. One reason is that while it is correct in principle that the states, which are simply vectors of $\mathcal{H}$ are simpler objects than operators, it will often happen that the operators of interest have a simple description, while the states have a very complicated one. Another reason will be given at the end of the present section.

## 物理代写|量子场论代写Quantum field theory代考|One-parameter Unitary Groups and Stone’s Theorem

∀X,是∈H,林吨→0(X,在(吨)(是))=(X,是).

## 物理代写|量子场论代写Quantum field theory代考|Time-evolution

(2.75) 中的减号是常规的。这种约定的原因将出现在第 9.7 节中。自从H具有能量乘以时间的维度，H具有能量的维度。它被称为系统的哈密顿量。虽然这肯定不是很明显64

(2.75) 的结果是，如果ψ属于的领域H，然后，由第二部分(2.64),ψ(吨):=在(吨)(ψ)满足方程

ψ′(吨)=−一世HH(ψ(吨)).

## 物理代写|量子场论代写Quantum field theory代考|Schrödinger and Heisenberg Pictures

(ψ(吨),一个ψ(吨))=(在(吨)ψ,一个在(吨)ψ)=(ψ,在(吨)−1一个在(吨)ψ)=(ψ,一个(吨)ψ),

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## MATLAB代写

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