# 物理代写|量子力学代写quantum mechanics代考|PHYS3040

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

Now, as previously, we are in a position to iterate these equations and obtain a power series in $H^{\prime}$. Since the r.h.s. of Eqs. (5.12) is already linear in $H^{\prime}$, we can just make use of our previous coefficients $c_{n_1^{\prime}, n_2^{\prime}}^0(t)=c_{n_1^{\prime}, n_2^{\prime}}^0$ on the r.h.s.! This gives
$$i \hbar \frac{d c_{n_1, n_2}(t)}{d t}=\sum_{n_1^{\prime}, n_2^{\prime}}\left\langle n_1, n_2\left|H^{\prime}\right| n_1^{\prime}, n_2^{\prime}\right\rangle c_{n_1^{\prime}, n_2^{\prime}}^0 e^{i\left(E_{n_1}+E_{n_2}-E_{n_1^{\prime}}-E_{n_2^{\prime}}\right) / \hbar}+\cdots$$
Suppose it is the state $\psi_{n_1^0}\left(x_1\right) \psi_{n_2^0}\left(x_2\right)$ that is occupied at the initial time $t=0$, so that
$$c_{n_1^{\prime}, n_2^{\prime}}^0=\delta_{n_1^{\prime}, n_1^0} \delta_{n_2^{\prime}, n_2^0} \quad \text {; given initial state }$$
Then, at a later time, the amplitude for finding the system in a different two-particle state satisfies
$$\begin{array}{r} i \hbar \frac{d c_{n_1, n_2}(t)}{d t}=\left\langle n_1, n_2\left|H^{\prime}\right| n_1^0, n_2^0\right\rangle e^{i\left(E_{n_1}+E_{n_2}-E_{n_1^0}-E_{n_2^0}\right) t / \hbar} \ ;\left(n_1, n_2\right) \neq\left(n_1^0, n_2^0\right) \end{array}$$
Integration of this relation between the initial time $t=0$, and the total elapsed time $t=T$, gives
$$c_{n_1, n_2}(T)=-\frac{1}{\hbar}\left\langle n_1, n_2\left|H^{\prime}\right| n_1^0, n_2^0\right\rangle \frac{1}{\omega}\left(e^{i \omega T}-1\right)$$
where the initial and final energies of the pair, and energy differences, are defined by
\begin{aligned} E_0 & \equiv E_{n_1^0}+E_{n_2^0} \ E & \equiv E_{n_1}+E_{n_2} \ \hbar \omega & \equiv E-E_0 \end{aligned}

## 物理代写|量子力学代写quantum mechanics代考|Density of Final States

Suppose we are doing a scattering experiment in our simple model. We can prepare the target in a given state with energy $E_{n_2^0}$, and we can prepare an incident beam with a well-defined energy $E_{n_1^0}=\hbar^2 k_0^2 / 2 m_1$, where $k_0=$ $2 \pi n_1^0 / L_1$. We certainly can achieve the energy resolution to determine that the target ends up in another state with discrete energy $E_{n_2}$; however, with the scattered particle, the situation is more complicated. Let us, for simplicity, call the size of the big region in which the first particle moves $L_1 \equiv L$. The final particle energy is $E_{n_1}=\hbar^2 k^2 / 2 m_1$ with $k=2 \pi n_1 / L$, and as $L$ becomes very large, these energies are very closely spaced. Thus no matter how small our resolution $d k$ is on the final particle, many final states will lie within this resolution! For large $L$, the number of these states $d n_f$ is
$$d n_f=\frac{L}{2 \pi} d k \quad ; L \rightarrow \infty$$
Thus all of these states will get into our final detector, and the transition rate that we actually measure is of necessity
$$R_{f i} d n_f=R_{f i}\left(\frac{L}{2 \pi} d k\right) \quad ; \text { measured rate }$$
Equation (5.28) then reads
$$R_{f i} d n_f=\frac{2 \pi}{\hbar}\left|\left\langle n_1, n_2\left|H^{\prime}\right| n_1^0, n_2^0\right\rangle\right|^2 \delta\left(E-E_0\right)\left(\frac{L}{2 \pi} d k\right)$$
Multiply and divide this expression by $d E$. It is then possible to immediately do the integral over $E$ using Eq. (5.27), where we have summed over all of the energy-conserving events that get into our detector.

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

$$i \hbar \frac{d c_{n_1, n_2}(t)}{d t}=\sum_{n_1^{\prime}, n_2^{\prime}}\left\langle n_1, n_2\left|H^{\prime}\right| n_1^{\prime}, n_2^{\prime}\right\rangle c_{n_1^{\prime}, n_2^{\prime}}^0 e^{i\left(E_{n_1}+E_{n_2}-E_{n_1^{\prime}}-E_{n_2^{\prime}}\right) / \hbar}+\cdots$$

$$c_{n_1^{\prime}, n_2^{\prime}}^0=\delta_{n_1^{\prime}, n_1^0} \delta_{n_2^{\prime}, n_2^0} \quad ; \text { given initial state }$$

$$i \hbar \frac{d c_{n_1, n_2}(t)}{d t}=\left\langle n_1, n_2\left|H^{\prime}\right| n_1^0, n_2^0\right\rangle e^{i\left(E_{n_1}+E_{n_2}-E_{n_1^0}-E_{n_2^0}\right) t / \hbar} ;\left(n_1, n_2\right) \neq\left(n_1^0, n_2^0\right)$$

$$c_{n_1, n_2}(T)=-\frac{1}{\hbar}\left\langle n_1, n_2\left|H^{\prime}\right| n_1^0, n_2^0\right\rangle \frac{1}{\omega}\left(e^{i \omega T}-1\right)$$

$$E_0 \equiv E_{n_1^0}+E_{n_2^0} E \equiv E_{n_1}+E_{n_2} \hbar \omega \equiv E-E_0$$

## 物理代写|量子力学代写quantum mechanics代考|Density of Final States

$$d n_f=\frac{L}{2 \pi} d k \quad ; L \rightarrow \infty$$

$$R_{f i} d n_f=R_{f i}\left(\frac{L}{2 \pi} d k\right) \quad ; \text { measured rate }$$

$$R_{f i} d n_f=\frac{2 \pi}{\hbar}\left|\left\langle n_1, n_2\left|H^{\prime}\right| n_1^0, n_2^0\right\rangle\right|^2 \delta\left(E-E_0\right)\left(\frac{L}{2 \pi} d k\right)$$

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