### 物理代写|量子力学代写quantum mechanics代考|PHYSICS 3544

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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|de Broglie Relation

In attempting to write a wave relation for a non-relativistic particle of mass $m$, de Broglie appealed to the analogous photon relations from the above. He associated a wavelength with the momentum according to
$$p=m v=\hbar k=\frac{h}{\lambda} \quad ; \text { de Broglie wavelength }$$
As one immediate consequence, if one fits an integral number $n$ of wavelengths around a circle of radius $a$, then
$$2 \pi a=n \lambda=\frac{n h}{m v}$$
The angular momentum $|\vec{L}|$ of a particle moving around in the circle is then
$$|\vec{L}|=m v a=n \hbar \quad ; \text { angular momentum }$$
As we have seen, this is precisely the quantization condition that leads to the Bohr theory of the one-electron atom!

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

With the de Broglie relation, and the angular frequency $\omega(k)$ given by
$$\varepsilon=\hbar \omega(k)=\frac{p^{2}}{2 m}=\frac{(\hbar k)^{2}}{2 m}$$

the wave in Eq. (1.1) now takes the dispersive form
$$\Psi(x, t)=e^{i[k x-\omega(k) t]}=e^{i\left[k x-\left(h k^{2} / 2 m\right) t\right]}$$
Appropriate linear combinations of these waves can again describe a localized wave packet. ${ }^{2}$
Let us ask what wave equation this $\Psi(x, t)$ satisfies. Evidently
$$i \hbar \frac{\partial \Psi(x, t)}{\partial t}=-\frac{\hbar^{2}}{2 m} \frac{\partial^{2} \Psi(x, t)}{\partial x^{2}}$$
The momentum of the particle is $p=\hbar k$. This quantity is obtained from the wave in Eq. (2.5) by taking a partial derivative with respect to $x$. Let us therefore define the momentum $p$ to be the differential operator
$$p \equiv \frac{\hbar}{i} \frac{\partial}{\partial x} \quad ; \text { momentum }$$
and write the hamiltonian $H(p)$ for a free particle as
$$H=\frac{p^{2}}{2 m}=-\frac{\hbar^{2}}{2 m} \frac{\partial^{2}}{\partial x^{2}} \quad ; \text { hamiltonian }$$
Then this wave $\Psi(x, t)$ for a free particle satisfies the Schrödinger equation
$$i \hbar \frac{\partial \Psi(x, t)}{\partial t}=H \Psi(x, t) \quad ; \text { Schrödinger equation }$$

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

The complex Schrödinger wave function $\Psi(x, t)$, for which we write the underlying differential equation in quantum mechanics, is not a physical observable. As you might imagine, this leads to substantial complications. On the other hand, we do know intuitively that the wave function should be large where the particle is, and small where it is not. Born suggested that we interpret the square of the modulus of $\Psi(x, t)$ as the probability density of finding the particle at the position $x$ at the time $t$
$$\rho(x, t) \equiv|\Psi(x, t)|^{2}=\Psi^{}(x, t) \Psi(x, t) \quad ; \text { probability density (2.12) }$$ Here $\Psi^{}(x, t)$ its the cumplex conjugute of $\Psi(x, t)$.
We should at least find a continuity equation for the probability density $\rho(x, t)$, and the consequent conservation of probability, in the theory. Let us try to establish that. Consider
$$\frac{\partial \rho}{\partial t}=\frac{\partial \Psi^{}(x, t)}{\partial t} \Psi(x, t)+\Psi^{}(x, t) \frac{\partial \Psi(x, t)}{\partial t}$$
The complex conjugate of the Schrödinger equation gives
$$-i \hbar \frac{\partial \Psi^{}(x, t)}{\partial t}=[H \Psi(x, t)]^{}$$
Hence
$$\frac{\partial \rho}{\partial t}=\frac{1}{i \hbar}\left{\Psi^{}(x, t)[H \Psi(x, t)]-[H \Psi(x, t)]^{} \Psi(x, t)\right}$$

## 物理代写|量子力学代写quantum mechanics代考|de Broglie Relation

p=米在=⁇ķ=Hλ; 德布罗意波长

2圆周率一个=nλ=nH米在

|大号→|=米在一个=n⁇; 角动量

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

e=⁇ω(ķ)=p22米=(⁇ķ)22米

Ψ(X,吨)=和一世[ķX−ω(ķ)吨]=和一世[ķX−(Hķ2/2米)吨]

p≡⁇一世∂∂X; 势头

H=p22米=−⁇22米∂2∂X2; 汉密尔顿

ö一世⁇∂Ψ(X,吨)∂吨=HΨ(X,吨); 薛定谔方程

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

ρ(X,吨)≡|Ψ(X,吨)|2=Ψ(X,吨)Ψ(X,吨); 概率密度 (2.12) 这里Ψ(X,吨)它的复杂共轭Ψ(X,吨).

∂ρ∂吨=∂Ψ(X,吨)∂吨Ψ(X,吨)+Ψ(X,吨)∂Ψ(X,吨)∂吨

−一世⁇∂Ψ(X,吨)∂吨=[HΨ(X,吨)]

\frac{\partial \rho}{\partial t}=\frac{1}{i \hbar}\left{\Psi^{}(x, t)[H \Psi(x, t)]-[H \Psi(x, t)]^{} \Psi(x, t)\right}

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## MATLAB代写

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