### 物理代写|量子力学代写quantum mechanics代考|PHYS3034

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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|Schr¨odinger Equation

Let us try and extend the Schrödinger equation to describe a non-relativistic particle of mass $m$ moving in a real potential $V(x)$. An evident approach is to just appeal to our classical mechanics arguments and extend the hamiltonian by
$$\frac{p^{2}}{2 m} \rightarrow \frac{p^{2}}{2 m}+V(x)$$
Let us see what happens to our previous quantum mechanics arguments if we work with the following hamiltonian
$$H(p, x)=\frac{p^{2}}{2 m}+V(x) \quad ; \text { hamiltonian }$$
We will continue to write the momentum in the Schrödinger equation as
$$p=\frac{\hbar}{i} \frac{\partial}{\partial x} \quad ; \text { momentum }$$
The hamiltonian is still hermitian, since a real potential is hermitian
$$\int d x \psi^{}(x) V \psi(x)=\int d x[V \psi(x)]^{} \psi(x) \quad ; \text { hermitian }$$
The separated solutions are then again stationary states
$$\Psi(x, t)=\psi(x) e^{-i E t / \hbar}$$
where $E$ is the real energy
$$\frac{\int d x \psi^{*}(x) H \psi(x)}{\int d x|\psi(x)|^{2}}=E$$

## 物理代写|量子力学代写quantum mechanics代考|Particle in a Box

Before investigating the general boundary conditions, let us first consider another simple physical situation where the potential is repulsive and grows very large. The potential then effectively presents a wall to the particle where the wave function must vanish. If a particle moves in one dimension along the $x$-axis and is in a box of length $L$, the boundary conditions become (see Fig. 3.1)
$$\psi(0)=\psi(L)=0 \quad ; \text { particle in box }$$
The energy eigenstates in this case are
\begin{aligned} \psi_{n}(x) &=\sqrt{\frac{2}{L}} \sin k_{n} x & \ k_{n} &=\frac{n \pi}{L} \quad ; n=1,2,3, \cdots \end{aligned}

The corresponding energy eigenvalues are
$$E_{n}=\frac{\left(\hbar k_{n}\right)^{2}}{2 m}=\frac{(\hbar \pi n)^{2}}{2 m L^{2}}$$
The energy eigenstates are no longer also eigenstates of momentum, since now the particle is bouncing off the walls; however, the momentum operator is still hermitian since the boundary term on the r.h.s. of Eq. (2.36) still vanishes
$$\left[\psi_{m}^{*}(x) \psi_{n}(x)\right]_{0}^{L}=0$$
We show the first four eigenfunctions and corresponding probability densities in Figs. $3.2$ and 3.3. If one has some way of repeatedly observing the location of the particle in these stationary states, then one will indeed observe the spatial distribution in Fig. 3.3. This is a real, quite amazing, consequence of quantum mechanics!

## 物理代写|量子力学代写quantum mechanics代考|Boundary Conditions

The separated Schrödinger equation is a second-order differential equation in space. With no additional input, the evident boundary condition is to ask that the physically acceptable solutions, and their first derivatives, be continuous
$$\psi(x), \psi^{\prime}(x) \text { continuous } \quad ; \text { boundary conditions }$$This sounds so obvious, but as we shall now see, this has essential, and quite unexpected, consequences.

Consider a non-relativistic particle moving in one dimension against a barrier of height $V_{0}$ extending for all $x>0$. Suppose its energy is less than the barrier height. Then classically it can never get into the barrier, since its kinetic energy is a positive definite quantity
$$\begin{array}{rlr} E & =\frac{m}{2} \dot{x}^{2}+V_{0} \quad ; x>0 \ E-V_{0} & =\frac{m}{2} \dot{x}^{2} \geq 0 \end{array}$$
Let us now ask what happens in quantum mechanics with the above boundary conditions. Consider a stationary state with an energy $E<V_{0}$ below the barrier. To the left of the barrier, we have both an incident and reflected wave (see Fig. 3.4)
$$\psi(x)=e^{i k x}+a e^{-i k x} \quad ; x<0$$

## 物理代写|量子力学代写quantum mechanics代考|Schr¨odinger Equation

p22米→p22米+在(X)

H(p,X)=p22米+在(X); 汉密尔顿

p=⁇一世∂∂X; 势头

∫dXψ(X)在ψ(X)=∫dX[在ψ(X)]ψ(X); 厄米特人

Ψ(X,吨)=ψ(X)和−一世和吨/⁇

∫dXψ∗(X)Hψ(X)∫dX|ψ(X)|2=和

## 物理代写|量子力学代写quantum mechanics代考|Particle in a Box

ψ(0)=ψ(大号)=0; 盒子里的颗粒

ψn(X)=2大号罪⁡ķnX ķn=n圆周率大号;n=1,2,3,⋯

[ψ米∗(X)ψn(X)]0大号=0

## 物理代写|量子力学代写quantum mechanics代考|Boundary Conditions

ψ(X),ψ′(X) 连续的 ; 边界条件 这听起来很明显，但正如我们现在将看到的，这会产生根本性的、非常出乎意料的后果。

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