Fermi's Golden Rule - Discrete state coupled to a continuum

Question

I'm studying a discrete state coupled to a continuum of states. The Hamiltonian of this interaction, which i called Friedrichs Hamiltonian, is given by: $$H_{F}=E_{e}|e\rangle\langle e|+\int U|\chi_{U}\rangle\langle\chi_{U}|dU+\int\left(\xi_{U}|\chi_{U}\rangle\langle e|+\xi_{U}^{*}|e\rangle\langle\chi_{U}|\right)dU.$$ $E_{e}$ is the energy of the discrete state, $|e\rangle$, $U$ is the energy of the continuum of states, $|\chi_{U}\rangle$ and $\xi_{U}$ is the coupling coefficient of the discrete state to the continuum.

After some calculations, i got that the amplitude of probability is given by: $$\langle e\vert e^{-iHt}\vert e\rangle=\int\vert A(E)\vert^{2}e^{-iEt}dE,$$ so, the time evolution of the probability is: $$P=\vert\langle e\vert e^{-iHt}\vert e\rangle\vert^{2}=\int\vert A(E)\vert^{2}e^{-iEt}dE\int\vert A(E^{\prime})\vert^{2}e^{iE^{\prime}t}dE^{\prime}.$$ Where $E$ and $E'$ are energies of the continuum. $\vert A(E)\vert$ is the coeficient of the wave function associated with the discrete state. The value of $\vert A(E)\vert^{2}$ was also calculated and given by: $$\Rightarrow\vert A(E)\vert^{2}=\frac{\left|\xi_{E}\right|^{2}}{\left[E-E_{e}-F(E)\right]^{2}+\left|\xi_{E}\right|^{4}\pi^{2}}.$$ So, from this, i can suppose that the bandwith of the continuous spectrum is given by $\Gamma=2\pi\left|\xi_{E}\right|^{2}$ and the decayment time by $\tau=\Gamma^{-1}=\frac{1}{2\pi\left|\xi_{E}\right|^{2}}$.

I'm trying to calculate the transition rate from the discrete state to the continuum, Fermi's Golden Rule.

Can someone help me calculating for this case, please? I'm having some difficulties.

Answer 1

Lowest-order Fermi's golden rule for a single state is \begin{equation} \Gamma_f = \frac{2\pi}{\hbar} |\langle f|V|i\rangle|^2 \delta(E_i-E_f) \end{equation} and taking $|i\rangle = |e\rangle$, $|\chi_U\rangle = |f\rangle$, the matrix element of your $V$ is \begin{equation} \langle \chi_U| \int dU' \left (\xi_{U'}|\chi_{U'}\rangle \langle e| +\xi_{U'}^* |e\rangle\langle \chi_{U'}|\right )|e\rangle = \xi_U \,. \end{equation} Since you don't care which continuum state you transition to, the total transition rate, assuming the normalization \begin{equation} 1 = |e\rangle\langle e| + \int dU |\chi_U\rangle \langle \chi_U| \,, \end{equation} gives \begin{equation} \Gamma = \int dU \Gamma_U = \int dU \frac{2\pi}{\hbar} |\xi_U|^2 \delta(U-E) = \frac{2\pi}{\hbar} |\xi_E|^2 \,. \end{equation} Taking units with $\hbar = 1$ gives the same result you found.