### 物理代写|量子场论代写Quantum field theory代考|PHYS3101

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## 物理代写|量子场论代写Quantum field theory代考|A First Contact with Creation and Annihilation Operators

High-energy interacting particles create other particles, and a relativistic theory must consider multiparticle systems, where the number of particles may vary.

Let us describe what is probably the simplest example of a multiparticle system. ${ }^{67}$ The “particles” are as simple as possible. ${ }^{68}$

Consider a separable Hilbert space with an orthonormal basis $\left(e_{n}\right){n \geq 0}$. The idea is that the state of the system is described by $e{n}$ when the system consists of $n$ particles. The important structure consists of the operators $a$ and $a^{\dagger}$ defined on the domain
$$\mathcal{D}=\left{\sum_{n \geq 0} \alpha_{n} e_{n} ; \sum_{n \geq 0} n\left|\alpha_{n}\right|^{2}<\infty\right}$$
by
$$a\left(e_{n}\right)=\sqrt{n} e_{n-1} ; a^{\dagger}\left(e_{n}\right)=\sqrt{n+1} e_{n+1} .$$
The definition of $a\left(e_{n}\right)$ is to be understood as $a\left(e_{0}\right)=0$ when $n=0$. The reason for the factors $\sqrt{n}$ and $\sqrt{n+1}$ is not intuitive, and will become clear only gradually.
The notation is consistent, since for each $n, m$,
$$\left(e_{n}, a\left(e_{m}\right)\right)=\sqrt{m} \delta_{n}^{m-1}=\sqrt{m} \delta_{n+1}^{m}=\left(a^{\dagger}\left(e_{n}\right), e_{m}\right)$$
where $\delta_{n}^{m}$ is the Kronecker symbol (equal to 1 if $n=m$ and to 0 otherwise).
Exercise 2.17.1 Prove that $a^{\dagger}$ is the adjoint of $a$. Prove in particular that if $|(y, a(x))| \leq$ $C|x|$ for $x \in \mathcal{D}$ then $y \in \mathcal{D}$.

Exercise 2.17.2 Prove that for each $\lambda \in \mathbb{C}$ the operator $a$ has an eigenvector with eigenvalue $\lambda$. Can this happen for a symmetric operator?
It should be obvious from $(2.88)$ that
$$a^{\dagger} a\left(e_{n}\right)=n e_{n} ; a a^{\dagger}\left(e_{n}\right)=(n+1) e_{n} .$$
Let us then consider the self-adjoint operator ${ }^{69}$
$$N:=a^{\dagger} a .$$
Thus $N\left(e_{n}\right)=n e_{n}$. Since a system in state $e_{n}$ has $n$ particles, the observable corresponding to this operator is “the number of particles”. The operator $N$ is therefore called the number operator.
As another consequence of $(2.90)$,
$$\left[a, a^{\dagger}\right]\left(e_{n}\right)=e_{n t}$$

## 物理代写|量子场论代写Quantum field theory代考|The Harmonic Oscillator

The fundamental structure outlined in the previous section is connected to an equally fundamental system, the harmonic oscillator. A classical one-dimensional harmonic oscillator of angular frequency ${ }^{73} \omega$ consists of a point of mass $m$ on the real line which is pulled back to the origin with a force $m \omega^{2}$ times the distance to the origin. The quantum version of this system is the space $\mathcal{H}=L^{2}(\mathbb{R})$ with Hamiltonian
$$H:=\frac{1}{2 m}\left(P^{2}+\omega^{2} m^{2} X^{2}\right),$$
where $P$ and $X$ are respectively the momentum and the position operators of Section $2.5$. This is the Hamiltonian $(2.79)$ in the case where $V(x)=m \omega^{2} x^{2} / 2$. That this formula provides a quantized version of the classical harmonic oscillator is not obvious at all. ${ }^{74}$ We will explain in Section $6.6$ the systematic procedure of “canonical quantization” to discover formulas such as (2.77) or (2.95). This procedure is by no means a proof of anything, and the resulting formulas are justified only by the fact that they provide a fruitful model. So there is little harm to accept at this stage that the formula (2.95) is indeed fundamental. We have not proved yet that this formula defines a self-adjoint operator, but this is a consequence of the analysis below.

Exercise 2.18.1 Prove that a symmetric operator which admits an orthonormal basis of eigenvectors is self-adjoint. Hint: If we denote by $\left(e_{n}\right)$ an orthonormal basis of eigenvectors, and $\lambda_{n}$ the eigenvalue of $e_{n}$, the natural domain $\mathcal{D}$ of the operator is
$$\left{x=\sum_{n} x_{n} e_{n} ; \sum_{n}\left(1+\left|\lambda_{n}\right|^{2}\right)\left|x_{n}\right|^{2}<\infty\right}$$
The program for this section is first to find a basis of eigenvectors for the Hamiltonian (2.95), and then to examine how some classical quantities transform under quantization.

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

The present section is standard material, but our presentation attempts to balance rigor and readability.

Principle 6 If the states of two systems $\mathcal{S}{1}$ and $\mathcal{S}{2}$ are represented by the unitary rays in two Hilbert spaces $\mathcal{H}{1}$ and $\mathcal{H}{2}$ respectively, the appropriate Hilbert space to represent the system consisting of the union of $\mathcal{S}{1}$ and $\mathcal{S}{2}$ is the tensor product $\mathcal{H}{1} \otimes \mathcal{H}{2}$.

Our first task is to describe this space. ${ }^{2}$ A mathematician would love to see an “intrinsic” definition of this tensor product, a definition that does not use bases or a special representation of these Hilbert spaces. This can be done elegantly as in e.g. Dimock’s book [23]. We shall not enjoy this piece of abstraction and we shall go the ugly way.

If $\left(e_{n}\right){n \geq 1}$ is an orthonormal basis of $\mathcal{H}{1}$ and $\left(f_{n}\right){n \geq 1}$ is an orthonormal basis of $\mathcal{H}{2}$ then the vectors $e_{n} \otimes f_{m}$ constitute an orthonormal basis of $\mathcal{H}{1} \otimes \mathcal{H}{2}$, which is thus the set of vectors of the type $\sum_{n, m \geq 1} a_{n, m} e_{n} \otimes f_{m}$ where the complex numbers $a_{n, m}$ satisfy $\sum_{n, m \geq 1}\left|a_{n, m}\right|^{2}<\infty$. Here the quantity $e_{n} \otimes f_{m}$ is just a notation, which is motivated by the fact that for $x=\sum_{n \geq 1} \alpha_{n} e_{n} \in \mathcal{H}{1}$ and $y=\sum{n \geq 1} \beta_{n} f_{n} \in \mathcal{H}{2}$ one defines $x \otimes y \in$ $\mathcal{H}{1} \otimes \mathcal{H}{2}$ by $$x \otimes y=\sum{m, n \geq 1} \alpha_{n} \beta_{m} e_{n} \otimes f_{m} .$$
When either $\mathcal{H}{1}$ or $\mathcal{H}{2}$, or both, are finite-dimensional, the definition is modified in the obvious manner.

Exercise 3.1.1 When $\mathcal{H}{1}$ and $\mathcal{H}{2}$ are finite-dimensional, what is the dimension of $\mathcal{H}{1} \otimes \mathcal{H}{2}$ ? How does it compare with the dimension of the usual product $\mathcal{H}{1} \times \mathcal{H}{2}$ ?
When either $\mathcal{H}{1}$ or $\mathcal{H}{2}$ is infinite-dimensional, $\mathcal{H}{1} \otimes \mathcal{H}{2}$ is an infinite-dimensional Hilbert space. The important structure is the bilinear form from $\mathcal{H}{1} \times \mathcal{H}{2}$ into this space given by (3.1).

Recalling that $(x, y)$ denotes the inner product in a Hilbert space we observe the formula
$$\left(x \otimes y, x^{\prime} \otimes y^{\prime}\right)=\left(x, x^{\prime}\right)\left(y, y^{\prime}\right),$$
which is a straightforward consequence of the fact that the basis $e_{n} \otimes f_{m}$ is orthonormal. $^{3}$
The problem with our definition of the tensor product is that one is supposed to check that “it does not depend on the choice of the orthonormal basis”, a tedious task that joins similar tasks under the carpet. ${ }^{4}$ The good news is that all the identifications one may wish for are true. If both $\mathcal{H}{1}$ and $\mathcal{H}{2}$ are the space of square-integrable functions on $\mathbb{R}^{3}$, then $\mathcal{H}{1} \otimes \mathcal{H}{2}$ is the space of square-integrable functions on $\mathbb{R}^{6} .5$ This fits very well with the Dirac formalism: If $|x\rangle$ denotes the Dirac function at $\boldsymbol{x}$, (so that these generalized vectors provide a generalized basis of $\mathcal{H}{1}$, and similarly for $|\boldsymbol{y}\rangle$, then $|\boldsymbol{x}\rangle|\boldsymbol{y}\rangle$ denotes the Dirac function at the point $(\boldsymbol{x}, \boldsymbol{y}) \in \mathbb{R}^{6}$, and these generalized vectors provide a generalized basis of $\mathcal{H}{1} \otimes \mathcal{H}{2}$. Furthermore, if $f \in \mathcal{H}{1}$ and $g \in \mathcal{H}_{2}$ then $f \otimes g$ identifies with the function $(\boldsymbol{x}, \boldsymbol{y}) \mapsto$ $f(x) g(y)$ on $\mathbb{R}^{6}$.

## 物理代写|量子场论代写Quantum field theory代考|A First Contact with Creation and Annihilation Operators

\mathcal{D}=\left{\sum_{n \geq 0} \alpha_{n}e_{n} ; \sum_{n \geq 0} n\left|\alpha_{n}\right|^{2}<\infty\right}\mathcal{D}=\left{\sum_{n \geq 0} \alpha_{n}e_{n} ; \sum_{n \geq 0} n\left|\alpha_{n}\right|^{2}<\infty\right}

(和n,一个(和米))=米dn米−1=米dn+1米=(一个†(和n),和米)

ñ:=一个†一个.

[一个,一个†](和n)=和n吨

## 物理代写|量子场论代写Quantum field theory代考|The Harmonic Oscillator

H:=12米(磷2+ω2米2X2),

\left{x=\sum_{n} x_{n} e_{n} ; \sum_{n}\left(1+\left|\lambda_{n}\right|^{2}\right)\left|x_{n}\right|^{2}<\infty\right}\left{x=\sum_{n} x_{n} e_{n} ; \sum_{n}\left(1+\left|\lambda_{n}\right|^{2}\right)\left|x_{n}\right|^{2}<\infty\right}

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

X⊗是=∑米,n≥1一个nb米和n⊗F米.

(X⊗是,X′⊗是′)=(X,X′)(是,是′),

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