物理代写|量子场论代写Quantum field theory代考|PHYSICS 7013

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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• 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|Position State Space for a Particle

In this section we analyze how the previous machinery works to describe a very simple system, a massive point that can be located anywhere on the real line. This provides a first concrete example, and at the same time allows us to discuss the intricacies of considering infinite-dimensional state spaces. Almost nothing of what we explained in the finite-dimensional case will carry on exactly the same, but a suitable infinite-dimensional reinterpretation of the concepts will basically suffice.

The state space $\mathcal{H}$ is the space $L^{2}=L^{2}(\mathbb{R}, \mathrm{d} x)$ of complex-valued square-integrable functions ${ }^{23}$ on the real line. ${ }^{24}$ An element of $\mathcal{H}$ is thus a complex-valued function ${ }^{25} f$ on $\mathbb{R}$. The traditional terminology is to call this function the wave function. A wave function of norm 1 therefore describes the possible state of a massive point, which for simplicity we will call a particle.

The basic idea is that the position of a particle in state $f$ is not really determined, but that the function $|f|^{2}$ represents the probability density to find this particle at a given location. This statement will eventually appear as the proper interpretation of (2.6) in the present “continuous case”. To develop this idea, consider an interval $I$ of $\mathbb{R}$, and the operator $1_{I}$ defined by $1_{l}(f)(x)=f(x)$ if $x \in I$ and $1_{l}(f)(x)=0$ if $x \notin I$. This operator is bounded since $\left|1_{I}(f)\right| \leq|f|$. After we develop the right generalization of Hermitian operators in infinite dimensions, it will become apparent that this operator corresponds to an observable, and the average value of this observable on the state $f$ is
$$\left(f, 1_{I}(f)\right)=\int_{I} \mathrm{~d} x|f(x)|^{2},$$
which is the probability to find the particle in the set $I$. It is worth repeating the fundamental fact: When you actually measure whether the particle is in $I$ or not, you get a yes/no answer. But you are certain to find the particle in $I$ only if its state vector $f$ is an eigenvector of $1_{I}$

of eigenvalue 1, i.e. $f(x)=0$ for $x \notin I$, and you are certain not to find it in $I$ only when $f$ is an eigenvector of $1_{l}$ of eigenvalue 0 , i.e. $f(x)=0$ for $x \in I .^{26}$

In the present setting, the position of the particle is an observable so that it corresponds to a “Hermitian operator” $X$, which will be called the position operator. It is not difficult to guess what the operator $X$ should be. If indeed $|f|^{2}$ represents the probability density that the particle is at a given location, its average position is given by
$$\int \mathrm{d} x x|f(x)|^{2}=\int \mathrm{d} x f(x)^{*}(x f(x))$$

物理代写|量子场论代写Quantum field theory代考|Unitary Operators

We introduce now unitary operators between Hilbert spaces. It is a fundamental notion in at least two respects:

• Mathematically it provides (as explained in this section) a way to recognize whether two different models “are in fact the same”.
• Physically, countless processes are represented by unitary operators. Why this is the case is explained at the beginning of Section 2.10.

Definition 2.6.1 A linear operator $U$ between Hilbert spaces is called unitary if it is one-to-one, onto, and preserves the norm, $|U(x)|=|x|$.

Unitary transformations are in a sense the “natural class of isomorphisms between Hilbert spaces”. The “polarization identity” $|x+y|^{2}=|x|^{2}+|y|^{2}+2 \operatorname{Re}(x, y)$ shows that a unitary operator preserves the inner product, ${ }^{38}$
$$(U(x), U(y))=(x, y) .$$
It is almost obvious that the set $\mathcal{U}(\mathcal{H})$ of unitary operators on a Hilbert space $\mathcal{H}$ forms a group.

Next we reformulate the condition that an operator on a Hilbert space $\mathcal{H}$ is unitary using the notion of adjoint operator. As already noted, for a bounded operator $A$ one has $\mathcal{D}(A)=$ $\mathcal{D}\left(A^{\dagger}\right)=\mathcal{H}$ and
$$\forall x, y \in \mathcal{H} ;\left(A^{\dagger}(x), y\right)=(x, A(y)) \text {. }$$
A unitary operator on a Hilbert space is bounded, so its adjoint is defined everywhere. Furthermore by $(2.21)$ one has $\left(U^{\dagger} U(x), y\right)=(U(x), U(y))$ and $(2.20)$ is equivalent to

$U^{\dagger} U=1$. Consequently, an operator on a complex Hilbert space is unitary if and only if it is invertible and
$$U^{-1}=U^{\dagger}$$
Thus a unitary operator also satisfies $U U^{\dagger}=1$.
The following trivial fact is stressed because of its considerable importance.

物理代写|量子场论代写Quantum field theory代考|Momentum State Space for a Particle

To put the ideas of the previous section to use we go back to the model of a massive particle on the line which we studied in Section $2.5$. Let us consider $\mathcal{H}^{\prime}=L^{2}(\mathbb{R}, \mathrm{d} p /(2 \pi h))$, where the notation $\mathrm{d} p /(2 \pi \hbar)$ means that we include the factor $1 /(2 \pi \hbar)$ whenever we integrate in p. Consider the Fourier transform $U: f \mapsto \hat{f}$ of $(1.30)$ from $\mathcal{H}$ to $\mathcal{H}^{\prime}$. It is a unitary operator since it preserves the scalar product by (1.32) and since it has an inverse, the inverse Fourier transform $\varphi \mapsto \check{\varphi}$. The state which was represented by $f \in \mathcal{H}$ is now represented by $\varphi=\hat{f} \in \mathcal{H}^{\prime}$. As in Lemma 2.6.2 the Fourier transform $U$ transports an operator $A$ on $\mathcal{H}$ to the operator $A^{\prime}=U A U^{-1}$ on $\mathcal{H}^{\prime}$ given by $A^{\prime}(\varphi)=\widehat{A(\breve{\varphi})}$. Using (1.33) for $f=\breve{\varphi}$ yields (provided $\varphi$ is well behaved)
$$P^{\prime}(\varphi)(p)=p \varphi(p)$$
and, similarly,
$$X^{\prime}(\varphi)=\mathrm{i} \hbar \frac{\mathrm{d} \varphi}{\mathrm{d} p} .$$
(The plus sign here is not surprising since (2.19) implies that $\left[X^{\prime}, P^{\prime}\right]=\mathrm{i} \hbar 1$.) It is now the momentum operator which looks simple and the position operator which looks complicated. Just as in position state space $|f|^{2}$ represented the probability density of the location of the particle in state $f$, we can now argue that $|\varphi|^{2}$ represents the probability density of the momentum of the particle (when the basic measure is $\mathrm{d} p / 2 \pi h$ ). For this reason we will call this space $\mathcal{H}^{\prime}$ the momentum state space. Taking the Fourier transform is the standard way to analyze a wave. Thus it can be said that using momentum state space amounts to thinking of a particle as wave. This is the wave-particle duality. ${ }^{39}$

物理代写|量子场论代写Quantum field theory代考|Position State Space for a Particle

(F,1我(F))=∫我 dX|F(X)|2,

∫dXX|F(X)|2=∫dXF(X)∗(XF(X))

物理代写|量子场论代写Quantum field theory代考|Unitary Operators

• 从数学上讲，它提供了（如本节所述）一种识别两个不同模型是否“实际上相同”的方法。
• 在物理上，无数的过程由单一运算符表示。为什么会出现这种情况在 2.10 节的开头进行了解释。

(在(X),在(是))=(X,是).

∀X,是∈H;(一个†(X),是)=(X,一个(是)).

物理代写|量子场论代写Quantum field theory代考|Momentum State Space for a Particle

X′(披)=一世⁇d披dp.
（这里的加号并不奇怪，因为（2.19）意味着[X′,磷′]=一世⁇1.) 现在是动量算子看起来很简单，而位置算子看起来很复杂。就像在位置状态空间中一样|F|2表示状态中粒子位置的概率密度F，我们现在可以说|披|2表示粒子动量的概率密度（当基本度量为dp/2圆周率H）。出于这个原因，我们将这个空间称为H′动量状态空间。傅里叶变换是分析波的标准方法。因此可以说，使用动量态空间相当于将粒子视为波。这就是波粒二象性。39

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