This is a question I asked myself a couple of years back, and which a student recently reminded me of. My off-the-cuff answer is wrong, and whilst I can make some hand-waving responses I'd like a canonical one!

In the derivation of Fermi's Golden Rule (#2 of course), one first calculates the quantity $P(t)\equiv P_{a\rightarrow b}(t)$ to lowest order in $t$. This is the probability that, if the system was in initial state $a$, and a measurement is made after a time $t$, the system is found to be in state $b$. One finds that, to lowest order in the perturbation and for $\left< a \mid b \right > = 0$, $$P(t) \propto \left( \frac{\sin(\omega t/2)}{\omega/2} \right)^2, \qquad \hbar\omega = E_b - E_a $$

Then one says $P(t) = \text{const.} \times t \times f_\omega(t)$ where as $t$ increases $f$ becomes very sharply peaked around $\omega=0$, with peak of height $t$ and width $1/t$, and with total area below the curve fixed at $2\pi$. In other words, $f_t(\omega)$ looks like $2\pi \delta(\omega)$ for large $t$.

Now suppose we consider the total probability $Q(t)$ of jumping to any one of a family of interesting states, e.g. emitting photons of arbitrary momenta. Accordingly, let us assume a continuum of states with density in energy given by $\rho(\omega)$. Then one deduces that $Q(t) \sim \text{const.} \times t \rho$, and defines a "transition rate" by $Q(t)/t$ which we note is independent of time.

The issue I have with this is the following: $Q(t)/t$ has the very specific meaning of "The chance that a jump $a \to F$ (for a family $F$ of interesting states) occurs after making a measurement a time $t$ from the system being in state $a$, divided by the time we wait to make this measurement." It is not immediately clear to me why this is a quantity in a physical/experimental context which is deserving of the name "transition rate". In particular, note that

  • $t$ must be large enough that the $\delta$ function approximation is reasonable, so the small-$t$ regime of the formula is not trustworthy;
  • $t$ must be small enough that the perturbation expansion is reasonable (and also presumably so that the $\delta$ function approximation is not insanely sensitive to whether there is a genuine continuum of states or simply very finely spaced states) so the large-$t$ regime of the formula is not trustworthy.
  • Therefore the physical setup in which one measures $P(t)/t$ events per unit time must use properties as if some measurement/decoherence occurs made in some intermediate range of $t$. What is the microscopic detail of this physical setup, and why is this intermediate range interesting?
  • Edit: Also I would like to emphasize that the nature of $P(t),Q(t)$ is such that whenever one "makes a measurement", the "time since in initial state" is reset to 0. It seems that the "time between measurements" is in this intermediate range. (Of course, this isn't necessarily about measurements, but might be to do with decoherence times or similar too, I'm simply not sure.) People tell me that the Golden Rule is used in calculating lifetimes on occasion, so I would like to understand why this works!

Succinct question: In what sense is $Q(t)/t$ a transition rate?

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    $\begingroup$ I just wanted to signal this little and clear paper, about the Fermi golden's rule. I am afraid I don't understand clearly the question. Within the authorized range of time (see formulae $(20)$ and $(22)$), you could roughly write : $|\psi(t)\rangle \approx |\psi_i\rangle + \sum\limits_{f \neq i} \sqrt{R_{if}} \sqrt{t} |\psi_f\rangle$, where $R_{if}$ is the "transition rate". So, within the authorized range of time, the probability, at time $t$, to find the system in the state $|\psi_f\rangle$, is $R_{if} ~t$. $\endgroup$
    – Trimok
    Commented Dec 25, 2013 at 18:30
  • $\begingroup$ ....So you have not directly $R_{if}$. You must repeat the same experiment, with different allowed times , to be sure of the linear dependence, and get a precise value of $R_{if}$ $\endgroup$
    – Trimok
    Commented Dec 25, 2013 at 18:31
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    $\begingroup$ I do not think you can derive such approximate equation from the time-dependent theory. The Schr. equation is linear, so the first-approximation contributions to the function $\psi(t)$ from the eigenfunctions of $H_0$ is linear are $t$. $\endgroup$ Commented Dec 25, 2013 at 19:52
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    $\begingroup$ @Trimok I completely understand the derivation of the quantity, but do not understand what experiment this quantity predicts the outcome of, and why. $\endgroup$ Commented Dec 25, 2013 at 23:28
  • $\begingroup$ @Sharkos : If you repeat the same experiment, with the same initial state, but with a measurement done with different allowed times $t$, you are measuring the probability, supposed being $R_{if} ~t$, to find the system in a final state $f$, at time $t$. So, with several experiments, you are able to draw a line $P_{if}(t)=F(t)$, where the slope is precisely $R_{if}$. About the word "transition rate", this only means, that, inside the allowed range of time, the probability of finding the system in a final state $f$ at time $t$ is increasing linearily with $t$, and that's all. $\endgroup$
    – Trimok
    Commented Dec 26, 2013 at 11:09

5 Answers 5


In this answer, we shall not go into measurement-theoretical aspects of quantum mechanics. Here we shall just derive the range where (the derivation of) Fermi's golden rule holds and is trustworthy.

I) Let us for simplicity assume that all states are normalizable and live in a Hilbert space, so that we have an absolute notion of probability. (This effectively means that the system is put in a sufficiently large potential box to discretize the continuum spectrum.) First-order time-dependent perturbation theory

$$\begin{align} H~=~&H_0+V(t), \cr |\psi \rangle ~=~& \sum_n c_n \exp\left[\frac{E_nt}{\mathrm{i}\hbar}\right] |n \rangle,\cr \qquad c_n(t\!=\!0)~=~&\delta_n^i,\cr \mathrm{i}\hbar\frac{d|\psi \rangle}{dt}~=~&H|\psi \rangle,\cr H_0|n \rangle~=~&E_n|n \rangle, \end{align}\tag{1} $$

yields (under the assumption that the possible final states are sparsely populated in a probabilistic sense $|c_f| \ll 1$) that

$$\begin{align} \mathrm{i}\hbar~\dot{c}_f ~\stackrel{(1)+(3)+(4)}{=}&~\sum_n V_{fn}(t) e^{\mathrm{i}\omega_{fn}t}c_n\cr ~\stackrel{(1)}{\approx}~~~~~& V_{fi}(t)~ e^{\mathrm{i}\omega_{fi}t}. \end{align}\tag{2}$$

Here subscripts $i$ and $f$ refer to initial and final states for the non-perturbed system. Moreover,

$$ \hbar\omega_{fi}~:=~E_f-E_i\tag{3}$$

is the energy difference, and

$$ V_{fi}(t)~:=~\langle f | V(t)| i \rangle\tag{4}$$

is the pertinent matrix element of the interaction.

II) In the harmonic perturbation [1,2,3], the interaction potential reads

$$ V(t)~=~\sum_{\pm}W^{\pm} e^{\pm\mathrm{i}\Omega t},\tag{5}$$

where $\Omega$ is the angular frequency of absorption/stimulated emission. Integration of the eq. (2) wrt. time $t$ leads to

$$\begin{align} \mathrm{i}\hbar~ c_f~\stackrel{(2)+(5)}{\approx}& \sum_{\pm}W^{\pm}_{fi}\int_0^t \!dt^{\prime}~e^{\mathrm{i}(\omega_{fi}\pm\Omega)t^{\prime}}\cr ~=~~&\sum_{\pm}W^{\pm}_{fi}\frac{e^{\mathrm{i}(\omega_{fi}\pm\Omega)t}-1}{\mathrm{i}(\omega_{fi}\pm\Omega)}\cr ~\stackrel{(7)}{=}~~&\sum_{\pm}W^{\pm}_{fi}~\exp\left[\mathrm{i}\frac{\omega_{fi}\pm\Omega}{2}t\right]~t~ \sqrt{\eta((\omega_{fi}\pm\Omega)t)}.\end{align}\tag{6}$$

One may show that this favors transitions $\omega_{fi}\approx \mp\Omega $. (See eq. (11) below.)

III) In eq. (6) we have defined the function

$$ \eta(x)~:=~\left(\frac{\sin(x/2)}{x/2}\right)^2 ~\in~ [0,1]~\subseteq~\mathbb{R}.\tag{7} $$

While we are at it, let us also define its integral

$$ H(x)~:=~\int_{-x}^{x} \!dx^{\prime}~\eta(x^{\prime}), \tag{8}$$

cf. Fig. 1.

$\uparrow$ Fig. 1: The violet curve is the graph of the function $x\mapsto \frac{H(x)}{H(\infty)}$. It converges to 1 for $x\to \infty$.

Note that

$$\begin{align} H(\infty)~=~~~~&\lim_{n\to\infty} H(2\pi n)\cr \stackrel{\text{int. by parts}}{=}& \lim_{n\to\infty} 2 \int_{-2\pi n}^{2\pi n}\!dx~\frac{\sin x}{x}\cr =~~~~&2\pi,\end{align}\tag{9}$$

and that numerically

$$\begin{align} \frac{H(2\pi)}{H(\infty)}~\approx~& 90 \%, \cr \frac{H(4\pi)}{H(\infty)}~\approx~& 95 \%. \end{align}\tag{10}$$


$$ \lim_{t\to\infty} t~\eta(\omega t)~=~H(\infty)~\delta(\omega), \tag{11}$$

cf. e.g. this Phys.SE post.

IV) Here let us for simplicity assume a time-independent perturbation $\Omega=0$, and leave the case $\Omega>0$ as an exercise for the reader. (This means that $i$ and $f$ must have approximately the same energy.) Then the time integration of eq. (2) simplifies to

$$ \mathrm{i}\hbar ~c_f ~\stackrel{(2)}{\approx}~ V_{fi}~ \exp\left[\mathrm{i}\frac{\omega_{fi}}{2}t\right]~t~\sqrt{\eta(\omega_{fi}t)},\tag{12}$$

so that the probability reads

$$ P_f~:=~|c_f|^2~\stackrel{(12)}{\approx}~\left(\frac{|V_{fi}|t}{\hbar} \right)^2\eta(\omega_{fi}t).\tag{13}$$

First-order perturbation theory (12) is valid if $|c_f|\ll 1$, i.e. in the short time limit

$$ t ~\stackrel{(12)}{\ll}~ \frac{\hbar}{|V_{fi}|}. \tag{14}$$

V) Let $\rho_f(E_f)$ be the energy density of distinguished final states. (This might not include all possible final states. In particular, the distinguished final states do not include the initial state $i$, which is not sparsely populated: $c_i\approx 1$.) Define a function

$$ g(\omega_{fi})~:=~ |V_{fi}|^2\rho_f(E_f),\tag{15} $$

viewed as a function of the final state $f$ where the initial state $i$ is kept fixed. We are interested in a sufficiently small energy interval of final states

$$ F~:=~\left\{ E_f \left|~|E_f-E_i| ~\leq ~\hbar \Delta \omega\right.\right\} \tag{16}$$

(necessarily centred around the initial energy $E_i$, since we consider a time-independent perturbation), such that the function $g$ is constant on the whole interval $F$ to a good approximation

$$ g(\omega_{fi})~\approx~g(0).\tag{17}$$

The full probability becomes

$$\begin{align} P(F)~:=~~& \int_F \! dE_f ~\rho(E_f) ~P_f\cr \stackrel{(13)+(15)}{\approx}&~ \left(\frac{t}{\hbar} \right)^2\int_F \! dE_f~ g(\omega_{fi})~ \eta(\omega_{fi}t) \cr \stackrel{(17)}{\approx}~~& g(0)~ \left(\frac{t}{\hbar} \right)^2\int_F \! dE_f~ \eta(\omega_{fi}t)\cr \stackrel{(8)}{=}~~~& g(0)~\frac{t}{\hbar}~H(t\Delta \omega). \end{align}\tag{18}$$

The short time condition (14) for the $F$-interval (16) becomes

$$ t ~\stackrel{(14)}{\ll}~ \frac{\hbar}{\sup_{f\in F}|V_{fi}|}. \tag{19}$$

[In eq. (19) we have applied a slight abuse of notation where $F$ now also denotes the set of distinguished final states in the energy interval $F$.] Next Fermi's golden rule reads in our notation

$$ \fbox{$\frac{P(F)}{t}~\approx~ \frac{g(0)}{\hbar} H(\infty),$} \tag{20}$$

i.e. the probability $P(F)$ increases proportionally with $t$, so that

$$ \fbox{$\frac{dP(F)}{dt}~\approx~\frac{P(F)}{t}.$}\tag{21} $$

For eq. (18) to be a good approximation to eq. (20), we should choose the time

$$ t ~\stackrel{(10)}{\gtrsim}~ \frac{2\pi}{\Delta \omega}, \tag{22}$$

cf. Fig. 1. Altogether, the time $t$ should satisfy

$$ \frac{2\pi}{\Delta \omega} ~\stackrel{(22)}{\lesssim}~t~\stackrel{(19)}{\ll}~\frac{\hbar}{\sup_{f\in F}|V_{fi}|} . \tag{23}$$

This is only possible if the length of the $F$-interval (16) is much bigger than the corresponding matrix elements

$$ \sup_{f\in F}|V_{fi}|~\ll ~\hbar\Delta \omega. \tag{24}$$


  1. J.J. Sakurai, Modern Quantum Mechanics, 1994; Section 5.6.

  2. L.D. Landau & E.M. Lifshitz, QM, Vol. 3, 3rd ed, 1981; $\S$40-43.

  3. C. Cohen-Tannoudji, B. Diu & F. Laloe, QM, Vol. 2, 1978; Complement XIII.D.

  4. D. Tong, QFT lectures; section 3.6.1.

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    $\begingroup$ Sorry, so what is your answer to my question, "In what sense is $Q(t)/t$ a transition rate?" ? $\endgroup$ Commented Dec 30, 2013 at 23:43
  • $\begingroup$ @Sharkos Conditions (23) and (24) obtain in the overwhelming majority of particle decays and Q(t)/t is therefore constant, perhaps surprisingly! That's why the rule was dubbed "Golden"! $\endgroup$ Commented Jun 29, 2018 at 22:29
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    $\begingroup$ This answer does not address the main question posed by OP: Assuming we have already accepted that Fermi's golden rule holds, why can $Q(t)/t$ be interpreted as a transition rate? If one is thoughtful about what the variable t means, it should not have any interpretation as a transition rate $dP/dt$, because that would require the numerator to be the cumulative probability that the particle had been found at time up to t. Rather, by the postulates of QM, the numerator is the probability (density) that an energy measurement at time t yields the result $E_b$. Proof by confusion of variables. $\endgroup$ Commented Dec 14, 2018 at 14:37
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    $\begingroup$ That is explained in eq. (21). $\endgroup$
    – Qmechanic
    Commented Sep 17, 2019 at 21:03
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    $\begingroup$ This question is one I keep coming back to, and still do not know the answer. I do understand that in some regime, $P\propto t$. But I think OP's question is based on usage of the rate here for things like spontaneous emission, or the decay of a particle. Why should we consider the decay rate here to be given by the probability of transition for a fixed measurement at time t, divided by the time during which we wait before making that measurement? In fact, the postulates of QM do not seem to cover probabilities for anything but measurements at fixed, chosen times. $\endgroup$ Commented Aug 10, 2022 at 20:32

I think you're right being cautious calling the quantity $Q(t)/t$ a "transition rate" and it's a pity that this question has not received a complete answer yet.

The point is this: what the derivations of Fermi's Golden Rule (FGR) (e.g. the one given above in Qmechanic's answer) tells me is that there is a certain regime in which the expected value $p$ of the observable $P=\text{projection over final states}$ goes like $p=wt$.

Strictly speaking, this probability refers to the free, deterministic, evolution of the isolated system (the source of the interaction hamiltonian is understood as part of the system), in particular with no measurements being made upon it. From this point of view, I think that the term "population rate" would be more appropriate than "transition rate", since $w$ is effectively the rate at which the final states become populated.

Therefore, it is somewhat an extrapolation to say that $w$ is the rate at which some transition, like the decay of a neutron, actually occurs. For if we are able to count the number of decays in a certain sample of neutrons during the time $t$, that means that we are making measurements on the system, therefore breaking the unitary evolution of the ket which we calculate using FGR.

To theoretically justify the extrapolation, it is necessary to resort on some kind of model for the act of "measurement". Now, what a detector essentially does is to destroy coherence beetween the eigenstates of the measured observable. To be concrete, take the case of the neutron: the detector could be a screen that intercepts the proton produced in $n\to p e \nu$ and therefore destroys coherence beetween the states $\lvert 1\rangle =\lvert \text {"decayed"}\rangle$ and $\lvert 0\rangle =\lvert \text {"not decayed"}\rangle$ of the system. Now, if the decoherence time beetween $\lvert 0,1\rangle$ is of the order required by the application of FGR, and the act of decoherence is almost instantaneous (to be precise, if the decoherence factor has a graph which resembles Fermi's distribution at $T=0$, with Fermi's energy of the right order of magnitude), Trimok's analysis above shows that the density matrix obtained is approximately $\rho = e^ {-wt} \lvert 0 \rangle \langle 0 \rvert + (1-e^{-wt})\lvert 1 \rangle \langle 1 \rvert$ for a single neutron. This decoeherence interval could be the time of flight of the proton towards the detector, and the interaction of the proton with the detector can be assumed instantaneous. Therefore, in such an idealized experiment, the interpretation of $w$ as an actual transition rate can be justified.

In general, it seems to me that the interpretation of $w$ as the "number of counts per unit time" depends on the experimental arrangement and may possibly fail. An example is the so called "Zeno paradox", which consists in the "freezing" of the state of a system induced by repeated measurements. In fact it can be shown that for short times, the survival probability of a state is $1-\alpha t^2$, in contrast with FGR. In such cases the unitary evolution of the state simply takes too much to reach FGR regime, and all predictions based on the assumption of a transition rate $w$ would be simply wrong.

As a sidenote, I want to remark that the derivation of the quantum cross section which employs the concept of transition rate (see for example Mandl&Shaw) can be replaced by a much more physical one in which the cross section is calculated as (proability of scattering)/(surface density of incident particles) (see Taylor for a nonrelativistic derivation).

  • $\begingroup$ Thanks for your thoughts on this! I'm glad you think it's a sensible question, and also that you broadly agree with my thoughts on the matter. As I indicate in the question, I'm really interested in any justification for either why the set of additional assumptions on decoherence times which seem to be needed (a) work in specific situations of interest; or (b) generically work for some non-obvious reasons; or (c) can be circumvented. Do you have any thoughts on the matter? $\endgroup$ Commented Dec 27, 2015 at 11:37
  • $\begingroup$ You're welcome Sharkos. At the present I'm not able to give a satisfying justification. The main difficulty is that, while FGR describes an (approximately) unitary evolution of the state, to model measurement an additional non-unitary component must be included in the dynamics. My intuition tells that if the decoherence time happens to be in the $t\to \infty$ range required by FGR, then the unitary and non-unitary part of the evolution should decouple in some way. After a decoherence time $\tau _D$ the density matrix would be approximately diagonal with [...] $\endgroup$
    – pppqqq
    Commented Dec 28, 2015 at 12:18
  • $\begingroup$ [...] diagonal entries having the expected poissonian dependence $e^{-wt}$ or $1-e^{-wt}$. This is true in the opposite case, when the decoherence is much faster than the free evolution of the ket: for example, the density matrix of a free particle in a gas evolves following $\dot \rho = -i[\frac{\mathbf P ^2}{2M},\rho] +\cal D \rho $, where $\cal D \rho $ describes decohering effect of the scatterings. In this case the dynamics is totally decoupled and one can study the evolution of a gaussian packet and discover that this is decoupled too, that is [...] $\endgroup$
    – pppqqq
    Commented Dec 28, 2015 at 12:26
  • $\begingroup$ [...] The coherent spread of the packet and the damping of the diagonal terms take place on different time-scales. However while this looks like an indication, I suspect that a sound proof following this approach, if possible, would be complicated. $\endgroup$
    – pppqqq
    Commented Dec 28, 2015 at 12:35

As for the derivation of the Fermi golden rule, there is a crystal clear one by myself:



There is no hand-waving argument at all. It is completely rigorous in the mathematical sense. Of course, it is based on some assumptions on the continuum spectrum and the couplings.

In talking about Fermi golden rule, one must keep in mind that this is a first-order perturbation result. Up to this lowest approximation, for a generic case, we find that the population on the initial state decreases linearly in time. The transition rate is the slope, or the rate of decrease.

If you read my paper, you will see interesting, unexpected effects beyond this linear behaviour. Actually, what we found, under the assumptions above and still within the 1st order perturbation theory, is that the population on the initial state is a piecewise linear function of time!

  • $\begingroup$ I looked at paper arxiv.org/abs/1604.06916, but it seems to me that eq (11) only holds in the limit $t\rightarrow \infty$. I dont see how you can drop the limit for the remaining time variable. I.e. the equation should read $$\lim_{t\rightarrow\infty} P = \lim_{t\rightarrow \infty}\frac{2\pi t}{\hbar }\rho(E_b)|g(E_b)|^2$$. $\endgroup$
    – Hans Wurst
    Commented Sep 16, 2019 at 13:48
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    $\begingroup$ @HansWurst Exactly. This is another loose point in the usual argument for the golden rule. $\endgroup$ Commented Nov 24, 2019 at 2:55

I have a proposition. It seems that there is a contradiction between the domain of validity of the Fermi's golden rule, which is certainly only valid for times $t << \tau = \frac{1}{\Gamma}$, where $\Gamma$ is a decay rate, and $\tau$ a particle lifetime, where the state obeys a Schrodinger equation evolution, and the vision of a probabilistic exponential decay until the particle lifetime.

However, we may reconciliate the two, by making repeated measurements (and repeat a great number of experiments) at regular intervals $\Delta t = \frac{\tau}{M}$, with $M>>1$

This could be sketched like this: Initially, at time $t_0=0$, the state is pure, and in a basis $\psi_i, \psi_f$, the initial density matrix is $\rho(t_0) = Diag (N,0)$ . During a time $\Delta t$, there is an evolution following Schrodinger equation. At time $t_1 = (1)*\Delta t$, there are off-diagonal non zero terms for the density matrix. Now, we perform a measurement at time $t_1 = \Delta t$ which is going to project the system into a state $\psi_i$ of $\psi_f$. With a great number of experiments, this is equivalent to have, after measurement, a classical statistical density matrix $\rho_{a.m.}(t_1 = \frac{\tau}{M}) = Diag (N(1-\Gamma \Delta t), N \Gamma \Delta t) = Diag (N(1- \frac{\Gamma \tau}{M}), N \frac{\Gamma \tau}{M})$. Now the system evolves until the time $t_2 = 2 \frac{\tau}{M}$, with non-zero off-diagonal terms for the density matrix, where we perform an other measurement, so with a lot of experiments, this would be equivalent, after measurement, to a classical density matrix : $\rho_{a.m.}(t_2 = 2 \frac{\tau}{M}) = Diag (N(1- \frac{\Gamma \tau}{M})^2, N - N(1- \frac{\Gamma \tau}{M})^2)$.

At a time $t = M' \Delta t = M' \frac{\tau}{M}$, the density matrix would be : $\rho_{a.m.}(t_{M'} = t) = Diag (N(1- \frac{\Gamma \tau}{M})^{M'}, N - N(1- \frac{\Gamma \tau}{M})^{M'})$. For $M>>1$, this is equivalent to $\rho_{a.m.}(t) = Diag (N e^{-{\Gamma t}}, N - N e^{-{\Gamma t}})$

  • 1
    $\begingroup$ Thanks for the analysis. (I haven't looked through it in detail, but I imagine this is true.) However, as you observe, $\Delta t$ is required to be in the realm of applicability of the GR. The smallness of this quantity presumably corresponds to the system being entangled with its surroundings so that 'measurements' occur frequently. However, there is also a lower bound on how small it can be in the GR due to the density of states approximation. Is this an issue? Why, heuristically, would decoherence correspond to measurements on an acceptable timescale? $\endgroup$ Commented Dec 28, 2013 at 20:35

While reading Dicke&Wittke, pp. 274-275, it occured to me that a (maybe more) convincing derivation of the "transition rate formula", as an experimentalist would interpret such a thing, can be given as follows.

As usual, we assume that the system is in a pure state $\vert i \rangle$ at $t=0$, when the perturbation (which for simplicity I assume to be constant) is turned on. The system evolves unitarily until it interacts with some external environment; first order perturbation theory gives: $$\vert c_f (t)\vert ^2 = \dfrac{|V_{fi}|^2}{\hbar ^2}\dfrac{\sin ^2 (\frac {\omega_{fi}t}{2})}{(\frac {\omega_{fi}}{2})^2}.$$ We now go on assuming that the interaction with the environment occurs by scattering events, which are Poisson distributed: $P(t)=1-e^{- \gamma t}$, where $P(t)=\text{Probability of no collision beetween 0 and t}$. We also assume that each scattering event instantaneously destroys the phase relations in the unperturbed energy eigenbase, so that if before the scattering event the state was pure, immediately after the scattering the state is represented by a density matrix which is diagonal in the energy eigenbase. From now on, it's just elementary calculus plus a crucial observation. We use the integral:$$\intop \gamma e^{- \gamma t}\frac{1}{2}sin ^2{\frac {\alpha t}{2}}=\dfrac{1}{2}\dfrac{\alpha ^2}{\alpha^2+\gamma ^2},$$ to obtain the transition probability per collision to the state $f$: $$\intop _{0}^{\infty} \gamma e^{- \gamma t}|c_{f}(t)|^2\,\text d t=\dfrac{|V_{fi}|^2}{\hbar ^2} \dfrac{2}{\gamma ^2 +\omega _{fi} ^2}.$$ The crucial observation is that now we can surely talk of a transition probability in the experimentalist's sense, since after the collision the interference beetween the states "decayed" and "not decayed" is not observable anymore. We can therefore also talk of a definite transition rate to the state $f$, which is: $$w_{i\to f} = \dfrac{|V_{fi}|^2}{\hbar ^2} \dfrac{2\gamma}{\gamma ^2 + \omega _{fi} ^2}.$$ Now, if we assume a quasicontinuous energy spectrum, it's easy to obtain Fermi's golden rule, by integrating the above formula with respect to the density of final states $\rho _{f}(\omega_{fi})$, under the assumption that the function $|V_{fi}|^2 \rho _f(\omega _f)$ varies slowly in the range $\pm \gamma $.

Appendix: what is a small enough $\gamma$? Tipically, for a continuous spectrum, $A\equiv |V_{fi}|^2 \rho _f (\omega _fi)$ will be proportional to some power of $\omega _{f}=\frac{E_f}{\hbar}$, so that $$\frac{\text d A}{\text d \omega _{f}}\sim \frac {1}{\omega_{f}}A.$$ Therefore, "small enough" means $\gamma \ll \omega _{f}$ .


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