# 物理代写|量子力学代写quantum mechanics代考|PHYSICS3544

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

Lèt us rèturn to our threee-dimensional problem of the scattering of a particle from a potential $V(r)$, and we calculate to lowest order in $V(r)$. This is now a one-body problem. We work in a large box of volume $L^3$ and apply p.b.c. The initial and final particle wave functions and energies are
\begin{aligned} \psi_i(\vec{x}) &=\frac{1}{\sqrt{L^3}} e^{i \vec{k} \cdot \vec{x}} & ; E=\frac{(\hbar k)^2}{2 m} \ \psi_f(\vec{x}) &=\frac{1}{\sqrt{L^3}} e^{i \vec{k}^{\prime} \cdot \vec{x}} & ; E^{\prime} &=\frac{\left(\hbar k^{\prime}\right)^2}{2 m} \end{aligned}
The initial probability flux is
$$I_{\mathrm{inc}}=\hat{k} \cdot \vec{S}(\vec{x})=\frac{1}{L^3} \frac{\hbar k}{m}$$
The transition rate multiplied by the number of final states is
$$R_{f i} d n_f=\frac{2 \pi}{\hbar}|\langle f|V| i\rangle|^2 \delta\left(E^{\prime}-E\right)\left[\frac{L^3}{(2 \pi)^3} d^3 k^{\prime}\right]$$
Here the matrix element of the potential is given by
$$\langle f|V| i\rangle=\frac{1}{L^3} \int d^3 x e^{i \vec{q} \cdot \vec{x}} V(r) \quad ; \vec{q} \equiv \vec{k}-\vec{k}^{\prime}$$
Multiply and divide the transition rate by $d E^{\prime}$, do the integral over the Dirac delta function, and invoke the resulting energy conservation to obtain
$$R_{f i} d n_f=\frac{2 \pi}{\hbar}|\langle f|V| i\rangle|^2\left[\frac{L^3}{(2 \pi)^3} k^2\left(\frac{d k}{d E}\right) d \Omega\right]$$
where $d \Omega$ is the solid angle into which the particle is scattered. Now use
$$\frac{d E}{d k}=\frac{\hbar^2 k}{m}$$

## 物理代写|量子力学代写quantum mechanics代考|Two-State Mixing

So far in looking at transition rates we have worked to leading order in $H^{\prime}$. We now simplify the problem enough so that we can treat $H^{\prime}$ exactly. We still seek separated solutions to the Schrödinger equation as in Eqs. (2.18)(2.22), so that we have
\begin{aligned} &\Psi(x, t)=\psi(x) e^{-i E t / \hbar} \ &H \psi(x)=E \psi(x) \quad ; H=H_0+H^{\prime} \end{aligned}
The eigenfunction $\psi(x)$ can be expanded in the complete set of solutions to the unperturbed problem
\begin{aligned} \psi(x) &=\sum_n a_n \psi_n(x) \ H_0 \psi_n(x) &=E_n^0 \psi_n(x) \end{aligned}
Substitution into the eigenvalue equation, and the use of the orthonormality of the eigenfunctions $\psi_n(x)$, gives
$$\sum_{n^{\prime}}\left[\left(E_n^0-E\right) \delta_{n, n^{\prime}}+\left\langle n\left|H^{\prime}\right| n^{\prime}\right\rangle\right] a_{n^{\prime}}=0$$

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

$$\psi_i(\vec{x})=\frac{1}{\sqrt{L^3}} e^{i \vec{k} \cdot \vec{x}} \quad ; E=\frac{(\hbar k)^2}{2 m} \psi_f(\vec{x})=\frac{1}{\sqrt{L^3}} e^{i \vec{k} \cdot \vec{x}} \quad ; E^{\prime}=\frac{\left(\hbar k^{\prime}\right)^2}{2 m}$$

$$I_{\mathrm{inc}}=\hat{k} \cdot \vec{S}(\vec{x})=\frac{1}{L^3} \frac{\hbar k}{m}$$

$$R_{f i} d n_f=\frac{2 \pi}{\hbar}|\langle f|V| i\rangle|^2 \delta\left(E^{\prime}-E\right)\left[\frac{L^3}{(2 \pi)^3} d^3 k^{\prime}\right]$$

$$\langle f|V| i\rangle=\frac{1}{L^3} \int d^3 x e^{i \vec{q} \cdot \vec{x}} V(r) \quad ; \vec{q} \equiv \vec{k}-\vec{k}^{\prime}$$

$$R_{f i} d n_f=\frac{2 \pi}{\hbar}|\langle f|V| i\rangle|^2\left[\frac{L^3}{(2 \pi)^3} k^2\left(\frac{d k}{d E}\right) d \Omega\right]$$

$$\frac{d E}{d k}=\frac{\hbar^2 k}{m}$$

## 物理代写|量子力学代写quantum mechanics代考|Two-State Mixing

$$\Psi(x, t)=\psi(x) e^{-i E t / \hbar} \quad H \psi(x)=E \psi(x) \quad ; H=H_0+H^{\prime}$$

$$\psi(x)=\sum_n a_n \psi_n(x) H_0 \psi_n(x) \quad=E_n^0 \psi_n(x)$$

$$\sum_{n^{\prime}}\left[\left(E_n^0-E\right) \delta_{n, n^{\prime}}+\left\langle n\left|H^{\prime}\right| n^{\prime}\right\rangle\right] a_{n^{\prime}}=0$$

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