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Fermi's golden rule

Although the rule is named after Enrico Fermi, most of the work leading to it is due to Paul Dirac, who twenty years earlier had formulated a virtually identical equation, including the three components of a constant, the matrix element of the perturbation and an energy difference.^[1][2] It was given this name because, on account of its importance, Fermi called it "golden rule No. 2".^[3]

Most uses of the term Fermi's golden rule are referring to "golden rule No. 2", but Fermi's "golden rule No. 1" is of a similar form and considers the probability of indirect transitions per unit time.^[4]

Fermi's golden rule describes a system that begins in an eigenstate ${\displaystyle |i\rangle }$ of an unperturbed Hamiltonian H₀ and considers the effect of a perturbing Hamiltonian H' applied to the system. If H' is time-independent, the system goes only into those states in the continuum that have the same energy as the initial state. If H' is oscillating sinusoidally as a function of time (i.e. it is a harmonic perturbation) with an angular frequency ω, the transition is into states with energies that differ by ħω from the energy of the initial state.

In both cases, the transition probability per unit of time from the initial state ${\displaystyle |i\rangle }$ to a set of final states ${\displaystyle |f\rangle }$ is essentially constant. It is given, to first-order approximation, by $${\displaystyle \Gamma _{i\to f}={\frac {2\pi }{\hbar }}\left|\langle f|H'|i\rangle \right|^{2}\rho (E_{f}),}$$ where ${\displaystyle \langle f|H'|i\rangle }$ is the matrix element (inbra–ket notation) of the perturbation H' between the final and initial states, and ${\displaystyle \rho (E_{f})}$ is the density of states (number of continuum states divided by ${\displaystyle dE}$ in the infinitesimally small energy interval ${\displaystyle E}$to${\displaystyle E+dE}$) at the energy ${\displaystyle E_{f}}$ of the final states. This transition probability is also called "decay probability" and is related to the inverse of the mean lifetime. Thus, the probability of finding the system in state ${\displaystyle |i\rangle }$ is proportional to ${\displaystyle e^{-\Gamma _{i\to f}t}}$.

The standard way to derive the equation is to start with time-dependent perturbation theory and to take the limit for absorption under the assumption that the time of the measurement is much larger than the time needed for the transition.^[5][6]

Only the magnitude of the matrix element ${\displaystyle \langle f|H'|i\rangle }$ enters the Fermi's golden rule. The phase of this matrix element, however, contains separate information about the transition process. It appears in expressions that complement the golden rule in the semiclassical Boltzmann equation approach to electron transport.^[9]

While the Golden rule is commonly stated and derived in the terms above, the final state (continuum) wave function is often rather vaguely described, and not normalized correctly (and the normalisation is used in the derivation). The problem is that in order to produce a continuum there can be no spatial confinement (which would necessarily discretise the spectrum), and therefore the continuum wave functions must have infinite extent, and in turn this means that the normalisation ${\textstyle \langle f|f\rangle =\int d^{3}\mathbf {r} \left|f(\mathbf {r} )\right|^{2}}$ is infinite, not unity. If the interactions depend on the energy of the continuum state, but not any other quantum numbers, it is usual to normalise continuum wave-functions with energy ${\displaystyle \varepsilon }$ labelled ${\displaystyle |\varepsilon \rangle }$, by writing ${\displaystyle \langle \varepsilon |\varepsilon '\rangle =\delta (\varepsilon -\varepsilon ')}$ where ${\displaystyle \delta }$ is the Dirac delta function, and effectively a factor of the square-root of the density of states is included into ${\displaystyle |\varepsilon _{i}\rangle }$.^[10] In this case, the continuum wave function has dimensions of ${\textstyle 1/{\sqrt {\text{[energy]}}}}$, and the Golden Rule is now $${\displaystyle \Gamma _{i\to \varepsilon _{i}}={\frac {2\pi }{\hbar }}|\langle \varepsilon _{i}|H'|i\rangle |^{2}.}$$ where ${\displaystyle \varepsilon _{i}}$ refers to the continuum state with the same energy as the discrete state ${\displaystyle i}$. For example, correctly normalized continuum wave functions for the case of a free electron in the vicinity of a hydrogen atom are available in Bethe and Salpeter.^[11]

The following paraphrases the treatment of Cohen-Tannoudji.^[10] As before, the total Hamiltonian is the sum of an “original” Hamiltonian H₀ and a perturbation: ${\displaystyle H=H_{0}+H'}$. We can still expand an arbitrary quantum state’s time evolution in terms of energy eigenstates of the unperturbed system, but these now consist of discrete states and continuum states. We assume that the interactions depend on the energy of the continuum state, but not any other quantum numbers. The expansion in the relevant states in the Dirac pictureis$${\displaystyle |\psi ($$t)\rangle =a_{i}e^{-\mathrm {i} \omega _{i}t}|i\rangle +\int _{C}d\varepsilon a_{\varepsilon }e^{-\mathrm {i} \omega t}|\varepsilon \rangle ,} where ${\displaystyle \omega _{i}=\varepsilon _{i}/\hbar }$, ${\displaystyle \omega =\varepsilon /\hbar }$ and ${\displaystyle \varepsilon _{i},\varepsilon }$ are the energies of states ${\displaystyle |i\rangle ,|\varepsilon \rangle }$, respectively. The integral is over the continuum ${\displaystyle \varepsilon \in C}$, i.e. ${\displaystyle |\varepsilon \rangle }$ is in the continuum.

Substituting into the time-dependent Schrödinger equation $${\displaystyle H|\psi ($$t)\rangle =\mathrm {i} \hbar {\frac {\partial }{\partial t}}|\psi (t)\rangle } and premultiplying by ${\displaystyle \langle i|}$ produces $${\displaystyle {\frac {da_{i}($$t)}{dt}}=-\mathrm {i} \int _{C}d\varepsilon \Omega _{i\varepsilon }e^{-\mathrm {i} (\omega -\omega _{i})t}a_{\varepsilon }(t),} where ${\displaystyle \Omega _{i\varepsilon }=\langle i|H'|\varepsilon \rangle /\hbar }$, and premultiplying by ${\displaystyle \langle \varepsilon '|}$ produces $${\displaystyle {\frac {da_{\varepsilon }($$t)}{dt}}=-\mathrm {i} \Omega _{\varepsilon i}e^{\mathrm {i} (\omega -\omega _{i})t}a_{i}(t).} We made use of the normalisation ${\displaystyle \langle \varepsilon '|\varepsilon \rangle =\delta (\varepsilon '-\varepsilon )}$. Integrating the latter and substituting into the former, $${\displaystyle {\frac {da_{i}($$t)}{dt}}=-\int _{C}d\varepsilon \Omega _{i\varepsilon }\Omega _{\varepsilon i}\int _{0}^{t}dt'e^{-\mathrm {i} (\omega -\omega _{i})(t-t')}a_{i}(t').} It can be seen here that ${\displaystyle da_{i}/dt}$ at time ${\displaystyle t}$ depends on ${\displaystyle a_{i}}$ at all earlier times ${\displaystyle t'}$, i.e. it is non-Markovian. We make the Markov approximation, i.e. that it only depends on ${\displaystyle a_{i}}$ at time ${\displaystyle t}$ (which is less restrictive than the approximation that ${\displaystyle a_{i}\approx 1}$ used above, and allows the perturbation to be strong) $${\displaystyle {\frac {da_{i}($$t)}{dt}}=\int _{C}d\varepsilon |\Omega _{i\varepsilon }|^{2}a_{i}(t)\int _{0}^{t}dTe^{-\mathrm {i} \Delta T},} where ${\displaystyle T=t-t'}$ and ${\displaystyle \Delta =\omega -\omega _{i}}$. Integrating over ${\displaystyle T}$, $${\displaystyle {\frac {da_{i}($$t)}{dt}}=-2\pi \hbar \int _{C}d\varepsilon |\Omega _{i\varepsilon }|^{2}a_{i}(t){\frac {e^{-\mathrm {i} \Delta t/2}\sin(\Delta t/2)}{\pi \hbar \Delta }},} The fraction on the right is a nascent Dirac delta function, meaning it tends to ${\displaystyle \delta (\varepsilon -\varepsilon _{i})}$as${\displaystyle t\to \infty }$ (ignoring its imaginary part which leads to an unimportant energy shift, while the real part produces decay ^[10]). Finally $${\displaystyle {\frac {da_{i}($$t)}{dt}}=-2\pi \hbar |\Omega _{i\varepsilon _{i}}|^{2}a_{i}(t),} which can have solutions: ${\displaystyle a_{i}($t)=\exp(-\Gamma _{i\to \varepsilon _{i}}t/2)} , i.e., the decay of population in the initial discrete state is ${\displaystyle P_{i}($t)=|a_{i}(t)|^{2}=\exp(-\Gamma _{i\to \varepsilon _{i}}t)} where $${\displaystyle \Gamma _{i\to \varepsilon _{i}}=2\pi \hbar |\Omega _{i\varepsilon _{i}}|^{2}={\frac {2\pi }{\hbar }}|\langle i|H'|\varepsilon \rangle |^{2}.}$$

The Fermi's golden rule can be used for calculating the transition probability rate for an electron that is excited by a photon from the valence band to the conduction band in a direct band-gap semiconductor, and also for when the electron recombines with the hole and emits a photon.^[12] Consider a photon of frequency ${\displaystyle \omega }$ and wavevector ${\displaystyle {\textbf {q}}}$, where the light dispersion relation is ${\displaystyle \omega =(c/n)\left|{\textbf {q}}\right|}$ and ${\displaystyle n}$ is the index of refraction.

Using the Coulomb gauge where ${\displaystyle \nabla \cdot {\textbf {A}}=0}$ and ${\displaystyle V=0}$, the vector potential of light is given by ${\displaystyle {\textbf {A}}=A_{0}{\boldsymbol {\varepsilon }}e^{\mathrm {i} ({\textbf {q}}\cdot {\textbf {r}}-\omega t)}+C}$ where the resulting electric field is $${\displaystyle {\textbf {E}}=-{\frac {\partial {\textbf {A}}}{\partial t}}=\mathrm {i} \omega A_{0}{\boldsymbol {\varepsilon }}e^{\mathrm {i} .({\textbf {q}}\cdot {\textbf {r}}-\omega t)}.}$$

For an electron in the valence band, the Hamiltonian is $${\displaystyle H={\frac {({\textbf {p}}+e{\textbf {A}})^{2}}{2m_{0}}}+V({\textbf {r}}),}$$ where ${\displaystyle V({\textbf {r}})}$ is the potential of the crystal, ${\displaystyle e}$ and ${\displaystyle m_{0}}$ are the charge and mass of an electron, and ${\displaystyle {\textbf {p}}}$ is the momentum operator. Here we consider process involving one photon and first order in ${\displaystyle {\textbf {A}}}$. The resulting Hamiltonian is $${\displaystyle H=H_{0}+H'=\left[{\frac {{\textbf {p}}^{2}}{2m_{0}}}+V({\textbf {r}})\right]+\left[{\frac {e}{2m_{0}}}({\textbf {p}}\cdot {\textbf {A}}+{\textbf {A}}\cdot {\textbf {p}})\right],}$$ where ${\displaystyle H'}$ is the perturbation of light.

From here on we consider vertical optical dipole transition, and thus have transition probability based on time-dependent perturbation theory that $${\displaystyle \Gamma _{if}={\frac {2\pi }{\hbar }}\left|\langle f|H'|i\rangle \right|^{2}\delta (E_{f}-E_{i}\pm \hbar \omega ),}$$ with $${\displaystyle H'\approx {\frac {eA_{0}}{m_{0}}}{\boldsymbol {\varepsilon }}\cdot \mathbf {p} ,}$$ where ${\displaystyle {\boldsymbol {\varepsilon }}}$ is the light polarization vector. ${\displaystyle |i\rangle }$ and ${\displaystyle |f\rangle }$ are the Bloch wavefunction of the initial and final states. Here the transition probability needs to satisfy the energy conservation given by ${\displaystyle \delta (E_{f}-E_{i}\pm \hbar \omega )}$. From perturbation it is evident that the heart of the calculation lies in the matrix elements shown in the bracket.

For the initial and final states in valence and conduction bands, we have ${\displaystyle |i\rangle =\Psi _{v,{\textbf {k}}_{i},s_{i}}({\textbf {r}})}$ and ${\displaystyle |f\rangle =\Psi _{c,{\textbf {k}}_{f},s_{f}}({\textbf {r}})}$, respectively and if the ${\displaystyle H'}$ operator does not act on the spin, the electron stays in the same spin state and hence we can write the Bloch wavefunction of the initial and final states as $${\displaystyle \Psi _{v,{\textbf {k}}_{i}}({\textbf {r}})={\frac {1}{\sqrt {N\Omega _{0}}}}u_{n_{v},{\textbf {k}}_{i}}({\textbf {r}})e^{i{\textbf {k}}_{i}\cdot {\textbf {r}}},}$$ $${\displaystyle \Psi _{c,{\textbf {k}}_{f}}({\textbf {r}})={\frac {1}{\sqrt {N\Omega _{0}}}}u_{n_{c},{\textbf {k}}_{f}}({\textbf {r}})e^{i{\textbf {k}}_{f}\cdot {\textbf {r}}},}$$ where ${\displaystyle N}$ is the number of unit cells with volume ${\displaystyle \Omega _{0}}$. Calculating using these wavefunctions, and focusing on emission (photoluminescence) rather than absorption, we are led to the transition rate $${\displaystyle \Gamma _{cv}={\frac {2\pi }{\hbar }}\left({\frac {eA_{0}}{m_{0}}}\right)^{2}|{\boldsymbol {\varepsilon }}\cdot {\boldsymbol {\mu }}_{cv}({\textbf {k}})|^{2}\delta (E_{c}-E_{v}-\hbar \omega ),}$$ where ${\displaystyle {\boldsymbol {\mu }}_{cv}}$ defined as the optical transition dipole moment is qualitatively the expectation value ${\displaystyle \langle c|({\text{charge}})\times ({\text{distance}})|v\rangle }$ and in this situation takes the form $${\displaystyle {\boldsymbol {\mu }}_{cv}=-{\frac {i\hbar }{\Omega _{0}}}\int _{\Omega _{0}}d{\textbf {r}}'u_{n_{c},{\textbf {k}}}^{*}({\textbf {r}}')\nabla u_{n_{v},{\textbf {k}}}({\textbf {r}}').}$$

Finally, we want to know the total transition rate ${\displaystyle \Gamma (\omega )}$. Hence we need to sum over all possible initial and final states that can satisfy the energy conservation (i.e. an integral of the Brillouin zone in the k-space), and take into account spin degeneracy, which after calculation results in $${\displaystyle \Gamma (\omega )={\frac {4\pi }{\hbar }}\left({\frac {eA_{0}}{m_{0}}}\right)^{2}|{\boldsymbol {\varepsilon }}\cdot {\boldsymbol {\mu }}_{cv}|^{2}\rho _{cv}(\omega )}$$ where ${\displaystyle \rho _{cv}(\omega )}$ is the joint valence-conduction density of states (i.e. the density of pair of states; one occupied valence state, one empty conduction state). In 3D, this is $${\displaystyle \rho _{cv}(\omega )=2\pi \left({\frac {2m^{*}}{\hbar ^{2}}}\right)^{3/2}{\sqrt {\hbar \omega -E_{g}}},}$$ but the joint DOS is different for 2D, 1D, and 0D.

We note that in a general way we can express the Fermi's golden rule for semiconductorsas^[13] $${\displaystyle \Gamma _{vc}={\frac {2\pi }{\hbar }}\int _{\text{BZ}}{\frac {d{\textbf {k}}}{4\pi ^{3}}}|H_{vc}'|^{2}\delta (E_{c}({\textbf {k}})-E_{v}({\textbf {k}})-\hbar \omega ).}$$

In the same manner, the stationary photocurrent proportional to intensity of light is $${\displaystyle {\textbf {J}}=-{\frac {2\pi e\tau }{\hbar }}\sum _{i,f}\int _{\text{BZ}}{\frac {d{\textbf {k}}}{(2\pi )^{D}}}|{\textbf {v}}_{i}-{\textbf {v}}_{f}|(f_{i}({\textbf {k}})-f_{f}({\textbf {k}}))|H_{if}'|^{2}\delta (E_{f}({\textbf {k}})-E_{i}({\textbf {k}})-\hbar \omega ),}$$ where ${\displaystyle \tau }$ is the relaxation time, ${\displaystyle {\textbf {v}}_{i}-{\textbf {v}}_{f}}$ and ${\displaystyle f_{i}({\textbf {k}})-f_{f}({\textbf {k}})}$ are the difference of the group velocity and Fermi-Dirac distribution between possible the initial and final states. Here ${\displaystyle |H_{if}'|^{2}}$ defines the optical transition dipole. Due to the commutation relation between position ${\displaystyle {\textbf {r}}}$ and the Hamiltonian, we can also rewrite the transition dipole and photocurrent in terms of position operator matrix using ${\displaystyle \langle i|{\textbf {p}}|f\rangle =-im_{0}\omega \langle i|{\textbf {r}}|f\rangle }$.

In a scanning tunneling microscope, the Fermi's golden rule is used in deriving the tunneling current. It takes the form $${\displaystyle w={\frac {2\pi }{\hbar }}|M|^{2}\delta (E_{\psi }-E_{\chi }),}$$ where ${\displaystyle M}$ is the tunneling matrix element.