### 物理代写|量子力学代写quantum mechanics代考|PHYSICS 7544

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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|量子力学代写quantum mechanics代考|Higher Dimensions

So far, for simplicity, we have worked in just one dimension where the Schrödinger equation reads
$$\left[-\frac{\hbar^{2}}{2 m} \frac{\partial^{2}}{\partial x^{2}}+V(x)\right] \Psi(x, t)=i \hbar \frac{\partial \Psi(x, t)}{\partial t}$$
Here the partial derivatives imply that the other variable in the set $(x, t)$ is to be kept constant. To increase the number of dimensions, we can simply follow our work on the wave equation and replace
$$\frac{\partial^{2}}{\partial x^{2}} \rightarrow \nabla^{2}$$
where $\nabla^{2}$ is the laplacian
\begin{aligned} \nabla^{2} &=\frac{\partial^{2}}{\partial x^{2}} & & ; \text { one dimension } \ &=\frac{\partial^{2}}{\partial x^{2}}+\frac{\partial^{2}}{\partial y^{2}} & & ; \text { two dimensions } \ &=\frac{\partial^{2}}{\partial x^{2}}+\frac{\partial^{2}}{\partial y^{2}}+\frac{\partial^{2}}{\partial z^{2}} & & ; \text { three dimensions } \end{aligned}
This is equivalent to writing the Schrödinger equation as
$$H \Psi(\vec{x}, t)=\left[\frac{\vec{p}^{2}}{2 m}+V(\vec{x})\right] \Psi(\vec{x}, t)=i \hbar \frac{\partial \Psi(\vec{x}, t)}{\partial t}$$

and expanding the momentum to read
$$p_{j}=\frac{\hbar}{i} \frac{\partial}{\partial x_{j}} \quad ; j=1,2, \cdots$$
where the index $j$ now labels the cartesian axes.
As one example, consider a particle of mass $m$ in a square twodimensional box with sides $L$. Here the boundary conditions are those of walls, and the eigenfunctions and eigenvalues are evidently
\begin{aligned} \psi_{n_{x}, n_{y}}(x, y) &=\left(\frac{2}{L}\right) \sin \left(\frac{n_{x} \pi x}{L}\right) \sin \left(\frac{n_{y} \pi y}{L}\right) \quad ;\left(n_{x}, n_{y}\right)=1,2,3, \cdots \ E_{n_{x}, n_{y}} &=\frac{\hbar^{2} \pi^{2}}{2 m L^{2}}\left(n_{x}^{2}+n_{y}^{2}\right) \end{aligned}
The general solution to the Schrödinger equation is correspondingly
$$\Psi(x, y, t)=\sum_{n_{x}} \sum_{n_{y}} c_{n_{x}, n_{y}} \psi_{n_{x}, n_{y}}(x, y) e^{-i E_{n_{x}, n_{y}} t / \hbar}$$

## 物理代写|量子力学代写quantum mechanics代考|Perturbation Theory

Suppose the hamiltonian has an additional small piece $\delta V(x)$, which makes the Schrödinger equation difficult to solve analytically
$$H \rightarrow H_{0}+\delta V(x)$$
We return to $\mathrm{Eq} .(3.6)$
$$E=\frac{\int d x \psi^{}(x) H \psi(x)}{\int d x|\psi(x)|^{2}}=\frac{\int d x \psi^{}(x)\left[H_{0}+\delta V(x)\right] \psi(x)}{\int d x|\psi(x)|^{2}}$$
Let us use the eigenfunction $\psi_{n}(x)$ of $H_{0}$ in this expression to obtain
$$E_{n}=E_{n}^{0}+\frac{\int d x \psi_{n}^{}(x)[\delta V(x)] \psi_{n}(x)}{\int d x\left|\psi_{n}(x)\right|^{2}}$$ This provides the first-order perturbation theory expression for the shift in the eigenvalue $$\delta E_{n}=\frac{\int d x \psi_{n}^{}(x)[\delta V(x)] \psi_{n}(x)}{\int d x\left|\psi_{n}(x)\right|^{2}} \quad ; \text { perturbation theory (3.46) }$$
The small shift in the eigenvalue is the integral of the perturbation over the eigenfunction.

As an example, suppose that with the particle in the box in Fig. $3.1$ there is a small, narrow potential step at the midpoint
$$\delta V(x)=\nu_{0} \quad ;\left|x-\frac{L}{2}\right|<l$$
where $l \ll L$. The eigenfunctions are $\sqrt{2 / L} \sin (n \pi x / L)$. For odd $n$, the magnitude of the sine is unity at the midpoint where $x=L / 2$. For even $n$, it vanishes there. 4 Hence, for $l \ll L$, one has
\begin{aligned} \delta E_{n} &=4 \nu_{0} \frac{l}{L} & & ; n=1,3,5, \cdots \ &=0 & & ; n=2,4,6, \cdots \end{aligned}

## 物理代写|量子力学代写quantum mechanics代考|Non-Degenerate Perturbation Theory

Let us make the analysis more general. We want to solve for the eigenfunctions and eigenvalues in the Schrödinger equation
$$H \psi(x)=\left[H_{0}+\delta V(x)\right] \psi(x)=E \psi(x)$$
Expand the wave function $\psi(x)$ in the complete set of eigenstates of $H_{0}$
\begin{aligned} \psi(x) &=\sum_{m} c_{m} \psi_{m}(x) \ H_{0} \psi_{m}(x) &=E_{m}^{0} \psi_{m}(x) \end{aligned}
Substitute this in the above equation
$$\sum_{m}\left(E-E_{m}^{0}\right) c_{m} \psi_{m}(x)=\delta V(x) \psi(x)$$
Now multiply by $\psi_{n}^{}(x)$ on the left, integrate over $x$, and use the orthonormality of the eigenfunctions $$c_{n}=\frac{1}{E-E_{n}^{0}} \int d x \psi_{n}^{}(x) \delta V(x) \psi(x)$$
We now make a rather unusual choice of norm for $\psi(x)$
$$c_{n}=1 \quad ; \text { choice of norm }$$
Let us discuss this:

• This choice is for a given $n$ :

## 物理代写|量子力学代写quantum mechanics代考|Higher Dimensions

[−⁇22米∂2∂X2+在(X)]Ψ(X,吨)=一世⁇∂Ψ(X,吨)∂吨

∂2∂X2→∇2

∇2=∂2∂X2; 一维  =∂2∂X2+∂2∂是2; 二维  =∂2∂X2+∂2∂是2+∂2∂和2; 三个维度

HΨ(X→,吨)=[p→22米+在(X→)]Ψ(X→,吨)=一世⁇∂Ψ(X→,吨)∂吨

pj=⁇一世∂∂Xj;j=1,2,⋯

ψnX,n是(X,是)=(2大号)罪⁡(nX圆周率X大号)罪⁡(n是圆周率是大号);(nX,n是)=1,2,3,⋯ 和nX,n是=⁇2圆周率22米大号2(nX2+n是2)

Ψ(X,是,吨)=∑nX∑n是CnX,n是ψnX,n是(X,是)和−一世和nX,n是吨/⁇

## 物理代写|量子力学代写quantum mechanics代考|Perturbation Theory

H→H0+d在(X)

d和n=∫dXψn(X)[d在(X)]ψn(X)∫dX|ψn(X)|2; 微扰理论 (3.46)

d在(X)=ν0;|X−大号2|<l

d和n=4ν0l大号;n=1,3,5,⋯ =0;n=2,4,6,⋯

## 物理代写|量子力学代写quantum mechanics代考|Non-Degenerate Perturbation Theory

Hψ(X)=[H0+d在(X)]ψ(X)=和ψ(X)

ψ(X)=∑米C米ψ米(X) H0ψ米(X)=和米0ψ米(X)

∑米(和−和米0)C米ψ米(X)=d在(X)ψ(X)

Cn=1和−和n0∫dXψn(X)d在(X)ψ(X)

Cn=1; 规范的选择

• 这个选择是给定的n :

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