Fermi's golden rule

 In quantum physics, Fermi's golden rule is a formula that describes the transition rate (the probability of a transition per unit time) from one energy eigenstate of a quantum system to a group of energy eigenstates in a continuum, as a result of a weak perturbation. This transition rate is effectively independent of time (so long as the strength of the perturbation is independent of time) and is proportional to the strength of the coupling between the initial and final states of the system (described by the square of the matrix element of the perturbation) as well as the density of states. It is also applicable when the final state is discrete, i.e. it is not part of a continuum, if there is some decoherence in the process, like relaxation or collision of the atoms, or like noise in the perturbation, in which case the density of states is replaced by the reciprocal of the decoherence bandwidth.

GeneralEdit

Although named after Enrico Fermi, most of the work leading to the "golden rule" is due to Paul Dirac, who formulated 20 years earlier 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", however, Fermi's "golden rule No. 1" is of a similar form and considers the probability of indirect transitions per unit time.^[4]

The rate and its derivationEdit

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 (in bra–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 ${\displaystyle |f\rangle }$ 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]

Derivation in time-dependent perturbation theory

Main article: Perturbation theory (quantum mechanics) § Time-dependent perturbation theory Statement of the problemEdit The golden rule is a straightforward consequence of the Schrödinger equation, solved to lowest order in the perturbation H' of the Hamiltonian. The total Hamiltonian is the sum of an “original” Hamiltonian H0 and a perturbation: ${\displaystyle H=H_{0}+H'}$. In the interaction picture, we can expand an arbitrary quantum state’s time evolution in terms of energy eigenstates of the unperturbed system ${\displaystyle |n\rangle }$, with ${\displaystyle H_{0}|n\rangle =E_{n}|n\rangle }$. Discrete spectrum of final statesEdit We first consider the case where the final states are discrete. The expansion of a state in the perturbed system at a time t is ${\displaystyle |\psi (t)\rangle =\sum _{n}a_{n}(t)e^{-iE_{n}t/\hbar }|n\rangle }$. The coefficients an(t) are yet unknown functions of time yielding the probability amplitudes in the Dirac picture. This state obeys the time-dependent Schrödinger equation: ${\displaystyle H|\psi (t)\rangle =i\hbar {\frac {\partial }{\partial t}}|\psi (t)\rangle .}$ Expanding the Hamiltonian and the state, we see that, to first order, ${\displaystyle \left(H_{0}+H'-\mathrm {i} \hbar {\frac {\partial }{\partial t}}\right)\sum _{n}a_{n}(t)|n\rangle e^{-\mathrm {i} tE_{n}/\hbar }=0,}$ where En and |n⟩ are the stationary eigenvalues and eigenfunctions of H0. This equation can be rewritten as a system of differential equations specifying the time evolution of the coefficients ${\displaystyle a_{n}(t)}$: ${\displaystyle \mathrm {i} \hbar {\frac {da_{k}(t)}{dt}}=\sum _{n}\langle k|H'|n\rangle a_{n}(t)e^{\mathrm {i} t(E_{k}-E_{n})/\hbar }.}$ This equation is exact, but normally cannot be solved in practice. For a weak constant perturbation H' that turns on at t = 0, we can use perturbation theory. Namely, if ${\displaystyle H'=0}$, it is evident that ${\displaystyle a_{n}(t)=\delta _{n,i}}$, which simply says that the system stays in the initial state ${\displaystyle i}$. For states ${\displaystyle k\neq i}$, ${\displaystyle a_{k}(t)}$ becomes non-zero due to ${\displaystyle H'\neq 0}$, and these are assumed to be small due to the weak perturbation. Hence, one can plug in the zeroth-order form ${\displaystyle a_{n}(t)=\delta _{n,i}}$ into the above equation to get the first correction for the amplitudes ${\displaystyle a_{k}(t)}$: ${\displaystyle \mathrm {i} \hbar {\frac {da_{k}(t)}{dt}}=\langle k|H'|i\rangle e^{\mathrm {i} t(E_{k}-E_{i})/\hbar },}$ whose integral can be expressed via the identity ${\displaystyle e^{ix}-1=2e^{ix/2}\sin x/2}$ as ${\displaystyle \mathrm {i} \hbar a_{k}(t)=2\langle k|H'|i\rangle e^{\mathrm {i} \omega t/2}{\frac {\sin \omega t/2}{\omega }}}$ with ${\displaystyle \omega \equiv (E_{k}-E_{i})/\hbar }$, for a state with ai(0) = 1, ak(0) = 0, transitioning to a state with ak(t) (again, ${\displaystyle k\neq i}$). This is the same as the generic result for the time evolution of any two-state system in a basis where the Hamiltonian is not diagonal. The transition rate is then ${\displaystyle \Gamma _{i\to k}={\frac {d}{dt}}|a_{k}(t)|^{2}={\frac {2|\langle k|H'|i\rangle |^{2}}{\hbar ^{2}}}{\frac {\sin \omega t}{\omega }},}$ a sinc function peaking sharply for small ω. At ${\displaystyle \omega =0}$, ${\displaystyle \sin(\omega t)/\omega =t}$, so the transition rate varies linearly with t for an isolated state ${\displaystyle |k\rangle }$! Continuous spectrum of final statesEdit By dramatic contrast, for states of energy E embedded in a continuum, they must be all accounted for collectively. For a density of states per unit energy interval ρ(E), they must be integrated over their energies, and whence the corresponding ω values, ${\displaystyle \Gamma _{i\to f}={\frac {2}{\hbar }}\int _{-\infty }^{\infty }d\omega \,\rho (\omega )|\langle f|H'|i\rangle |^{2}{\frac {\sin \omega t}{\omega }}.}$ For large t, the sinc function is sharply peaked at ω ≈ 0, so the density of states can be taken out of the integral. We also assume that the transition element can be approximated as a constant. The rate is then ${\displaystyle \Gamma _{i\to f}={\frac {2\rho |\langle f|H'|i\rangle |^{2}}{\hbar }}\int _{-\infty }^{\infty }d\omega \,{\frac {\sin \omega t}{\omega }}.}$ A change of variables shows that the integral is independent of t, the definite integral being π. The time dependence has vanished, and the constant decay rate of the golden rule follows.[7] As a constant, it underlies the exponential particle decay laws of radioactivity. (For excessively long times, however, the secular growth of the ak(t) terms invalidates lowest-order perturbation theory, which requires ak ≪ ai.)

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.^[8]

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 ${\displaystyle \langle f|f\rangle =\int d^{3}r|f(\mathbf {r} )|^{2}}$ 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 ${\displaystyle \epsilon }$ labelled ${\displaystyle |\epsilon \rangle }$, by writing ${\displaystyle \langle \epsilon |\epsilon '\rangle =\delta (\epsilon -\epsilon ')}$ where ${\displaystyle \delta }$ is the Dirac delta function, and effectively a factor of the square-root of the density of states is included into ${\displaystyle |\epsilon _{i}\rangle }$ .^[9] In this case, the continuum wave function has dimensions of ${\displaystyle 1/\surd }$[energy], and the Golden Rule is now

${\displaystyle \Gamma _{i\to \epsilon _{i}}={\frac {2\pi }{\hbar }}|\langle \epsilon _{i}|H'|i\rangle |^{2}.}$

where ${\displaystyle \epsilon _{i}}$ 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 .^[10]

Normalized Derivation in time-dependent perturbation theory

Main article: Perturbation theory (quantum mechanics) § Time-dependent perturbation theory The following paraphrases the treatment of Cohen-Tannoudji.[9] 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 picture is ${\displaystyle |\psi (t)\rangle =a_{i}e^{-\mathrm {i} \omega _{i}t}|i\rangle +\int _{C}d\epsilon a_{\epsilon }e^{-\mathrm {i} \omega t}|\epsilon \rangle .}$ where ${\displaystyle \omega _{i}=\epsilon _{i}/\hbar ,\omega =\epsilon /\hbar }$ and ${\displaystyle \epsilon _{i},\epsilon }$ are the energies of states ${\displaystyle |i\rangle ,|\epsilon \rangle }$. The integral is over the continuum ${\displaystyle \epsilon \in C}$, i.e. ${\displaystyle |\epsilon \rangle }$ 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 ${\displaystyle {\frac {da_{i}(t)}{dt}}=-\mathrm {i} \int _{C}d\epsilon \Omega _{i\epsilon }e^{-\mathrm {i} (\omega -\omega _{i})t}a_{\epsilon }(t),}$ where ${\displaystyle \Omega _{i\epsilon }=\langle i|H'|\epsilon \rangle /\hbar }$, and premultiplying by ${\displaystyle \langle \epsilon '|}$ produces ${\displaystyle {\frac {da_{\epsilon }(t)}{dt}}=-\mathrm {i} \Omega _{\epsilon i}e^{\mathrm {i} (\omega -\omega _{i})t}a_{i}(t).}$ We made use of the normalisation ${\displaystyle \langle \epsilon '|\epsilon \rangle =\delta (\epsilon '-\epsilon )}$. Integrating the latter and substituting into the former, ${\displaystyle {\frac {da_{i}(t)}{dt}}=-\int _{C}d\epsilon \Omega _{i\epsilon }\Omega _{\epsilon i}\int _{0}^{t}dt'e^{-\mathrm {i} (\omega -\omega _{i})(t-t')}a_{i}(t').}$ 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 ${\displaystyle a_{i}}$≈1 used above, and allows the perturbation to be strong) ${\displaystyle {\frac {da_{i}(t)}{dt}}=-\int _{C}d\epsilon |\Omega _{i\epsilon }|^{2}a_{i}(t)\int _{0}^{t}dTe^{-\mathrm {i} \Delta T}.}$ where ${\displaystyle T=t-t'}$ and ${\displaystyle \Delta =\omega -\omega _{i}}$. Integrating over ${\displaystyle T}$, ${\displaystyle {\frac {da_{i}(t)}{dt}}=2\pi \hbar \int _{C}d\epsilon |\Omega _{i\epsilon }|^{2}a_{i}(t){\frac {e^{-\mathrm {i} \Delta t/2}\sin(\Delta t/2)}{\pi \hbar \Delta }},}$ The fraction on the right is a nascent Dirac delta function, meaning it tends to ${\displaystyle \delta (\epsilon -\epsilon _{i})}$ as ${\displaystyle t\to \infty }$ (ignoring its imaginary part which leads to an unimportant energy shift, while the real part produces decay [9]). Finally ${\displaystyle {\frac {da_{i}(t)}{dt}}=2\pi \hbar |\Omega _{i\epsilon _{i}}|^{2}a_{i}(t)}$ which has solutions ${\displaystyle a_{i}(t)=\exp(-\Gamma _{i\to \epsilon _{i}}t/2)}$, i.e. the decay of population in the initial discrete state is ${\displaystyle P_{i}(t)=|a_{i}(t)|^{2}=\exp(-\Gamma _{i\to \epsilon _{i}}t)}$ where ${\displaystyle \Gamma _{i\to \epsilon _{i}}=2\pi \hbar |\Omega _{i\epsilon _{i}}|^{2}={\frac {2\pi }{\hbar }}|\langle i|H'|\epsilon \rangle |^{2}}$

ApplicationsEdit

SemiconductorsEdit

The Fermi 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.^[11] 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 the EM wave is given by ${\displaystyle {\textbf {A}}=A_{0}{\vec {\epsilon }}e^{i({\textbf {q}}\cdot {\textbf {r}}-\omega t)}+C}$ where the resulting electric field is

${\displaystyle {\textbf {E}}=-\partial {\textbf {A}}/\partial t=i\omega A_{0}{\vec {\epsilon }}e^{i({\textbf {q}}\cdot {\textbf {r}}-\omega t)}}$

For a charged particle in the valence band, the Hamiltonian is

${\displaystyle H={\frac {({\textbf {p}}-Q{\textbf {A}})^{2}}{2m_{0}}}+V({\textbf {r}})}$

where ${\displaystyle V({\textbf {r}})}$ is the potential of the crystal. If our particle is an electron (${\displaystyle Q=-e}$) and 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 the EM wave.

From here on we have transition probability based on time-dependent perturbation theory that

${\displaystyle \Gamma _{if}={\frac {2\pi }{\hbar }}|\langle f|H'|i\rangle |^{2}\delta (E_{f}-E_{i}\pm \hbar \omega )}$

${\displaystyle H'\approx {\frac {eA_{0}}{m_{0}}}{\vec {\epsilon }}\cdot {\vec {p}}}$

where ${\displaystyle {\vec {\epsilon }}}$ is the light polarization vector. From perturbation it is evident that the heart of the calculation lies in the matrix elements shown in the braket.

For the initial and final states in valence and conduction bands respectively, we have ${\displaystyle |i\rangle =\Psi _{v,{\textbf {k}}_{i},s_{i}}({\textbf {r}})}$ and ${\displaystyle |f\rangle =\Psi _{c,{\textbf {k}}_{f},s_{f}}({\textbf {r}})}$, 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 wavefunctions as Bloch waves so

${\displaystyle \Psi _{v,{\textbf {k}}_{i}}({\textbf {r}})={\frac {1}{\sqrt {N\Omega _{0}}}}u_{n_{v},{\textbf {k}}_{i}}({\textbf {r}})e^{i{\textbf {k}}_{i}\cdot {\textbf {r}}}}$

${\displaystyle \Psi _{c,{\textbf {k}}_{f}}({\textbf {r}})={\frac {1}{\sqrt {N\Omega _{0}}}}u_{n_{c},{\textbf {k}}_{f}}({\textbf {r}})e^{i{\textbf {k}}_{f}\cdot {\textbf {r}}}}$

where ${\displaystyle N}$ is the number of unit cells with volume ${\displaystyle \Omega _{0}}$. Using these wavefunctions and with some more mathematics, and focusing on emission (photoluminescence) rather than absorption, we are led to the transition rate

${\displaystyle \Gamma _{cv}={\frac {2\pi }{\hbar }}\left({\frac {eA_{0}}{m_{0}}}\right)^{2}|{\vec {\epsilon }}\cdot {\boldsymbol {\mu }}_{cv}({\textbf {k}})|^{2}\delta (E_{c}-E_{v}-\hbar \omega )}$

where ${\displaystyle {\boldsymbol {\mu }}_{cv}}$ is the transition dipole moment matrix element 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 initial and final states (i.e. an integral of the Brillouin zone in the k-space), and take into account spin degeneracy, which through some mathematics results in

${\displaystyle \Gamma (\omega )={\frac {4\pi }{\hbar }}\left({\frac {eA_{0}}{m_{0}}}\right)^{2}|{\vec {\epsilon }}\cdot {\boldsymbol {\mu }}_{cv}|^{2}\rho _{cv}(\omega )}$

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.

Finally we note that in a general way we can express the Fermi golden rule for semiconductors as^[12]

${\displaystyle \Gamma _{vc}={\frac {2\pi }{\hbar }}\int _{\text{BZ}}{\frac {d{\textbf {k}}}{4\pi ^{3}}}|H_{vc}'|^{2}\delta (E_{c}({\textbf {k}})-E_{v}({\textbf {k}})-\hbar \omega )}$

Scanning tunneling microscopyEdit

Main article: Scanning tunneling microscope § Principle of operation

In a scanning tunneling microscope, the Fermi 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.

Quantum opticsEdit

When considering energy level transitions between two discrete states, Fermi's golden rule is written as

${\displaystyle \Gamma _{i\to f}={\frac {2\pi }{\hbar }}|\langle f|H'|i\rangle |^{2}g(\hbar \omega ),}$

where ${\displaystyle g(\hbar \omega )}$ is the density of photon states at a given energy, ${\displaystyle \hbar \omega }$ is the photon energy, and ${\displaystyle \omega }$ is the angular frequency. This alternative expression relies on the fact that there is a continuum of final (photon) states, i.e. the range of allowed photon energies is continuous.^[13]

Drexhage experimentEdit

Both the radiation pattern and the total emitted power (which is proportional to the decay rate) of a dipole depend on its distance from a mirror.

Fermi's golden rule predicts that the probability that an excited state will decay depends on the density of states. This can be seen experimentally by measuring the decay rate of a dipole near a mirror: as the presence of the mirror creates regions of higher and lower density of states, the measured decay rate depends on the distance between the mirror and the dipole.^[14][15]

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