# 物理代写|粒子物理代写Particle Physics代考|FY3403 The transition probability

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## 物理代写|粒子物理代写Particle Physics代考|The transition probability

Let us assume that at $t=t_{i}$ the system is in an eigenstate $|i\rangle$ of $H_{0}$. Under the influence of the perturbation, at time $t=t_{f}$ the system will be in some state $|f\rangle$, which, under our assumptions, we can write as a linear superposition of the eigenstates of $H_{0}$. We would like to compute the probability to find the system in a particular eigenstate of $H_{0}$. It will be given by the square of the amplitude (2.41):
$$\mathcal{P}{f i}=\left|\left\langle f\left|U\left(t{f}, t_{i}\right)\right| i\right\rangle\right|^{2}$$
We wish to compute this matrix element in perturbation theory using the expansion (2.49). In all interesting cases, the initial and the final states are different and orthogonal to each other, so the zeroth order term, the unit operator, drops out. At first order we obtain
$$\mathcal{P}{f i}=\left|\int{0}^{t} \mathrm{e}^{\mathrm{i}\left(E_{i}-E_{f}\right) t_{1}}\left\langle f\left|H_{P}\right| i\right\rangle \mathrm{d} t_{1}\right|^{2}$$
where we have set $t_{i}=0$ and $t_{f}=t$.
In our problem of spontaneous emission, as well as in most physically interesting problems, the perturbation Hamiltonian is time-independent. In this case the integration over $t_{1}$ is trivial and gives
$$\mathcal{P}{f i}=\left|\left\langle f\left|H{P}\right| i\right\rangle\right|^{2} f(\omega, t) t^{2}$$
where $\omega=E_{i}-E_{f}$ and
$$f(\omega, t)=\left(\frac{\sin \gamma}{\gamma}\right)^{2}$$
with $\gamma=\omega t / 2$. This is shown graphically in Figure 2.1. The maximum around $\omega=$ 0 gets more and more pronounced as $t$ increases. In fact, we can show that, as a distribution, $f$ satisfies
$$\lim _{t \rightarrow \infty} t f(\omega, t)=2 \pi \delta(\omega)$$

## 物理代写|粒子物理代写Particle Physics代考|Application to the problem of spontaneous emission

After all this general discussion, let us return to our problem of spontaneous emission. We start by splitting the Hamiltonian (2.40) into an unperturbed part $H_{0}$ and a perturbation: $H=H_{0}+H_{P} . H_{0}$ must be exactly solvable; in other words we must be able to find its eigenvalues and eigenvectors exactly, so we choose
$$H_{0}=H_{a}+H_{r}, \quad H_{P}=H_{I}$$
in the notation of the equations (2.40) and (2.39). We denote the eigenvectors of $H_{0}$ by $\left|\mathcal{E}{m},{n(\boldsymbol{k}, \lambda)}\right\rangle$ in an obvious notation in which $\mathcal{E}{m}$ is the energy of the $m$-th level of the atom and $n(\boldsymbol{k}, \lambda)$ the number of photons with momentum $\boldsymbol{k}$ and polarisation $\lambda$. In other words, the space of states of $H_{0}$ is the direct product of the Hilbert space of the atomic eigenfunctions and the Fock space of the photons.
$$\begin{gathered} H_{a}\left|\mathcal{E}{m},{n(\boldsymbol{k}, \lambda)}\right\rangle=\mathcal{E}{m}\left|\mathcal{E}{m},{n(\boldsymbol{k}, \lambda)}\right\rangle \ H{r}\left|\mathcal{E}{m},{n(\boldsymbol{k}, \lambda)}\right\rangle=\sum{\lambda, \boldsymbol{k}} n(\boldsymbol{k}, \lambda)|\boldsymbol{k}|\left|\mathcal{E}_{m},{n(\boldsymbol{k}, \lambda)}\right\rangle \end{gathered}$$

## 物理代写|粒子物理代写PARTICLE PHYSICS代考|THE TRANSITION PROBABILITY

$$\mathcal{P} f i=\left|\left\langle f\left|U\left(t f, t_{i}\right)\right| i\right\rangle\right|^{2}$$

$$\mathcal{P f i}=\left|\int 0^{t} \mathrm{e}^{\mathrm{i}\left(E_{i}-E_{f}\right) t_{1}}\left\langle f\left|H_{P}\right| i\right\rangle \mathrm{d} t_{1}\right|^{2}$$

$$\mathcal{P} f i=|\langle f|H P| i\rangle|^{2} f(\omega, t) t^{2}$$

$$f(\omega, t)=\left(\frac{\sin \gamma}{\gamma}\right)^{2}$$

$$\lim {t \rightarrow \infty} t f(\omega, t)=2 \pi \delta(\omega)$$

## 物理代写|粒子物理代写PARTICLE PHYSICS代 考|APPLICATION TO THE PROBLEM OF SPONTANEOUS EMISSION

$$H_{0}=H_{a}+H_{r}, \quad H_{P}=H_{I}$$

$$H_{a}|\mathcal{E} m, n(\boldsymbol{k}, \lambda)\rangle=\mathcal{E} m|\mathcal{E} m, n(\boldsymbol{k}, \lambda)\rangle \quad H r|\mathcal{E} m, n(\boldsymbol{k}, \lambda)\rangle=\sum \lambda, \boldsymbol{k} n(\boldsymbol{k}, \lambda)|\boldsymbol{k}|\left|\mathcal{E}_{m}, n(\boldsymbol{k}, \lambda)\right\rangle$$

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