Question about $\delta$-function in Fermi's Golden Rule I am aware that Fermi's Golden Rule states the rate $k_{if}$ of a process which moves a quantum system from an initial state $i$ to a final state $f$ can be expressed as $$\frac{2\pi}{\hbar}|V_{if}|^2\delta(E_i-E_f).$$ However, I am confused by the use of the Dirac Delta function here, since it implies that the rate should be $0$ for any two energy eigenstates that are not exactly equal. I know that this equation is an approximation - but even so, shouldn't there at least be a nonzero rate constant in cases where $E_f<E_i$ (so that the system is moving "downhill")?
 A: The delta function expresses energy conservation. It presupposes that all particles are already being included in the decay process. For example, in the decay rate for an excited state of a Hydrogen atom, the final should include the photon that is carrying away energy from the atom.
A: 
$$\frac{2\pi}{\hbar}|V_{if}|^2\delta(E_i-E_f).$$


... I am confused by the use of the Dirac Delta function here, since it implies that the rate should be 0 for any two energy eigenstates that are not exactly equal.

Yes, it is true that the total energy is conserved by the delta function. But, see below for a further explanation.
First, recall that a Dirac Delta function only really makes sense when it is integrated over. I.e., it is not really a "function," but rather a "functional."
This means, we should anticipate that we are going to be integrating the above expression against some measure that relates to a sum over final states.
For example, when a free electron scatters off an atomic electron, there are two particles in the initial state and two in the final state. Then Fermi's golden rule looks something like:
$$
\frac{2\pi}{\hbar}|\langle \vec k_f; n_f\ell_f m_f|\hat V|\vec k_i; n_i\ell_i m_i\rangle|^2 \delta(\omega + \epsilon_{n_i} - \epsilon_{n_f})\;,
$$
where $\omega = \frac{\hbar^2k_i^2}{2m_e} - \frac{\hbar^2 k_f^2}{2m_e}$ is the energy change of the probe, and where $\epsilon_{n_i} - \epsilon_{n_f}$ is the energy change of the atomic electron, and where I'm completely ignoring spin.
To calculate a rate, we typically sum over final states:
$$
R \sim \frac{2\pi}{\hbar}\sum_{n_f,\ell_f,m_f}\int\frac{d^3k_f}{(2\pi)^3}
|\langle \vec k_f; n_f\ell_f m_f|\hat V|\vec k_i; n_i\ell_i m_i\rangle|^2 \delta(\omega + \epsilon_{n_i} - \epsilon_{n_f})\;,
$$
where the $\sim$ symbol means I'm leaving out or forgetting about some constant factors (like all the $\hbar$s).
So, indeed, the delta function enforces total energy conservation in the scattering process. In our example it was the total initial energy of the probe electron and the atomic electron that must equal the total final energy of the probe electron and the atomic electron. But that energy conservation still allows for many different types of final states.
