I was reading second chapter of Introductory Nuclear Physics by Kenneth S.Krane, and in that chapter he was giving about the logic of why there must be a factor of $\rho(E_{f})$ in the decay probability term, i.e. $$\lambda=\frac{2\pi}{\hbar}|V_{fi}|^2\rho(E_{f}).$$

He says,

"The quantity $\rho(E_{f})$ is known as the density of final states. It is the number of states per unit energy interval at $E$, and it must be included for the following reason: if the final state $E_{f}$, is a single isolated state, then the decay probability will be much smaller than it would be in the case that there are many, many states in a narrow band near $E_{f}$. If there is a large density of states near $E_{f}$, there are more possible final states that can be reached by the transition and thus a larger transition probability".

Now I personally did not get the logic as to why we must incorporate this factor, and don't the same thing apply for the ground state too also.

  • 2
    $\begingroup$ That argument is just sooooo bad. Instead, I would separate between transitions from discrete to discrete, discrete to continuous, continuous to discrete, and continuous to continuous, and the transition probabilities are different for all 4 cases. Those that involve the continuous spectrum have the density of states appear in the transition probability. $\endgroup$ Jun 27, 2023 at 13:26
  • $\begingroup$ There is a discussion in this answer and also here. A higher level approach is here. $\endgroup$
    – Roger V.
    Jul 4, 2023 at 8:39

4 Answers 4


Intuitively, the probability for going from a state $\vert \phi_i\rangle$ to a manifold of states $\{\vert \phi_f\rangle\}$ at a particular energy $E_f$ should depend on how many states are in the available energy region.

Imagine the following scenario within Fermi's golden rule drawn in the cartoon below. It shows a schematic of a single hydrogen atom which is adsorbed on a solid surface. We shall ask ourselves the question "at which rate does the electron in the hydrogen atom decay into the surfaces' states, ending up with a particular final energy $E_f$?".

Let's describe the hydrogen by a single level at energy $\varepsilon_i$, which corresponds to the initial state $\vert \phi_i\rangle$ (black bar on the left side of the plot). The solid shall be characterized by its surface density of states, which is the number of states per volume of the solid "seen" from outside the solid (not in the bulk). This is shown on the right side by the blue semi-circles. Here I have chosen an example in which the density of states is a function of energy and has a gap (meaning, in a certain energy range there are no states, which would correspond to a band insulator). The coupling $W_{if}$ is depicted by the lines connecting the single level with the density of states of the surface. These are also in general energy-dependent (which I tried to draw using a color coding from dark black (very strong $W_{if}$) to gray (very weak $W_{if}$)). The reason for the energy dependence of $W_{if}$ is that some surface states have a larger overlap with the hydrogen state than others (think, for example, about delocalized bloch states vs. localized surface states far away from the hydrogen atom).

Now in this picture, I think, it is pretty clear that if we ask ourselves "at which rate does the electron go from $\varepsilon_i$ in the hydrogen atom to $\varepsilon_i$ in the solid?", the answer is "zero", because there are no states the electron can go to at that particular final energy (due to the gap).

If we would ask ourselves "which states in the solid get populated the fastest?", the answer within Fermi's golden rule would be "those states with eigenenergies in the energy window $(E,E+\Delta E)$, for which the product of the transition probability and the density of states is maximal." This is simply FGR written out in words, but I think that the picture shown below makes it pretty clear, as $W_{if}$ act as "wires" connecting $\vert \phi_i\rangle$ with the manifold of states $\{\vert \phi_f\rangle\}$ (and their density given by $\rho(E_f)$).

enter image description here

  • $\begingroup$ My doubt is why we consider final states of final energy to be reaches only and not initial state. Also why does it only appears for time dependent potentials only s.t these potentials are also weak and for short time only these potential effect takes place and not for time independent potentials. The second doubt is when there is a transition from initial to final states, and say initial state is some excited state and my final state is ground state, then does that means ground state have bunch of many continuum states. $\endgroup$ Jul 4, 2023 at 12:14
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    $\begingroup$ 1) One considers the initial state. Its just conceptually simpler to understand the content of FGR using a single initial state. In principle, the system can be prepared in multiple states in a superposition. The probabilities would then simply sum up over all initial states. 2) In the example above the potential is not time-dependent. You still do time-dependent perturb. theory, but in the interaction-picture (then the potential is time-dep.). 3) The potential has to be weak, since perturb. theory is then most likely expected to work. 4) I'm not sure if I understand properly, but I'd say no. $\endgroup$ Jul 4, 2023 at 13:21
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    $\begingroup$ Edit/comment to point 2): The interaction does not have to be necessarily time-dependent. It can be also constant. Then the perturbation theory is even simpler since the elements $\langle \psi_f\vert V\vert \psi_i\rangle$ (with interaction potential $V$) can be taken outside of the time-integral. $\endgroup$ Jul 4, 2023 at 14:59
  • $\begingroup$ So a potential which is time independent, does it not make sense then to just simply use time independent perturbation theory, then why we account it in time dependent perturbation problem? $\endgroup$ Jul 8, 2023 at 8:38
  • 1
    $\begingroup$ A rate tells you how much a thing changes in time. For this the time-dependence of the wavefunction is needed. $\endgroup$ Jul 8, 2023 at 15:46



Now I personally did not get the logic as to why we must incorporate this factor, and don't the same thing apply for the ground state too also.

Fermi's golden rule provides an expression for the decay rate. But maybe you also want to know where that comes from. I will use units where $\hbar=1$.

I'll start from first order time-dependent perturbation theory, and I'll elide a few of the steps for time.

In the interaction picture, to lowest order in the interaction, the state is: $$ |\Psi\rangle = |\psi_i\rangle - i\lim_{\epsilon\to 0}\int_{-\infty}^t dt' e^{\epsilon t'}V_I(t')|\psi_i\rangle\;, $$ where $\psi_i$ is the initial state of the non-interacting hamiltonian and $V_I(t)$ is the interaction in the interaction picture. And where I have inserted a $e^{\epsilon t}$ by hand, where $\epsilon$ is infinitesimal. The $e^{\epsilon t}$ indicates that the interaction is "turned on" in the distant past (since $e^{\epsilon t}\to 0$ as $t\to-\infty$).

Using the above expression, the transition rate for $\psi_i\to\phi_f$ is found to be: $$ R_{if} = \lim_{\epsilon\to 0}|\langle\psi_f|V|\psi_i\rangle|^2\frac{2\epsilon e^{2\epsilon t}}{|E_i - E_f + i\epsilon|^2}\;,\tag{1} $$ where $E_i$ and $E_f$ are the unperturbed energies of the initial and final states, respectively.

Notice however that Eq. (1) is zero unless $E_i = E_f$. Indeed, we can use the definition of the delta function to see that $$ R_{if} = 2\pi |\langle\psi_f|V|\psi_i\rangle|^2\delta(E_i - E_f)\;.\tag{2} $$

Eq. (2) gives the expression for the transition rate from a fixed initial state to a fixed final state. However, in most experiments we have no way to control which final state we decay into. Therefore, in most experiments we really want to know the transition rate to any final state: $$ R_{i\to{\text{any}}}=\sum_f 2\pi |\langle\psi_f|V|\psi_i\rangle|^2\delta(E_i - E_f) $$

Now, we use the definition of the density of (final) states $\rho_f(E)$ in energy to write $$ R_{i\to{\text{any}}}=\int dE' \rho(E') 2\pi |\langle\psi_f|V|\psi_i\rangle|^2\delta(E_i - E') $$ $$ =2\pi\rho(E_i)|\langle\psi_f|V|\psi_i\rangle|^2\;, $$ where the matrix elements has been assumed to only depend on the energy and where both the initial and final states in the matrix element are evaluated at the same energy ($E_f=E_i$).

Note also, that in cases where the final state of the interacting system includes more than one particle (say the decaying particle along with a secondary particle such as a photon that carries off energy) then the density of states for the final states includes the final states of both the decaying particle and the final states of the secondary particle. And the energy must be apportioned appropriately to each final state particle.

For example, if the final state also contains a photon, then we would write the total final state as $|\psi_f\rangle = |\phi_f\rangle\otimes|\vec k_f\epsilon_f\rangle$, where the $\phi_f$ now refers to the electronic (or nuclear or whatever) state and the rest refers to the photon.

In such a case, the interaction term may be simply factorable, for example, in the case of an electron-photon interaction at lowest order: $$ V \sim \hat{\vec p} \cdot \vec A \sim \sum_{\vec q}\vec \epsilon\cdot\hat{\vec p} \hat a_{\vec q}^\dagger+\text{h.c.}\;, $$ where $\hat p$ operates on electronic state and the $\hat a$ operates on photon states.

In such a case, we can write: $$ R_{i\to{\text{any}}}\sim\int d\omega \rho_\gamma(\omega)\int dE' \rho_{e}(E') 2\pi \epsilon_i\epsilon_j^* \langle\phi_f|\hat p_i|\phi_i\rangle \langle\phi_f|\hat p_j|\phi_i\rangle^*\delta(E_i - \omega - E')\;, $$

Or, assuming we are insensitive to detecting the outgoing photon polarization, we write the spectrum as a function of photon frequency as: $$ \frac{dR}{d\omega}\sim 2\pi\rho_e(E_i-\omega) |\langle\phi_f| p_i|\phi_i\rangle|^2\;, $$ where we see that the electronic (or nuclear or whatever) density of states is evaluated at the "final energy" $E_i-\omega$ due to the photon carrying away the rest of the energy.


Densities of state always show up in contexts where we have to sum some function of state over the different states, and the states form a continuum. Often, the states are wave-like, i.e., defined by a wave vector $\vec{k}$.$^1$ Thus, we have to perform a sum $$ S = \sum_{\vec{k}}f(\vec{k})\,. $$ In the context of the form of Fermi's golden rule that the OP is considering, this sum is over the set of possible final states in which the system can end up. The main idea is that there is a rate of transition into each of these final states, and so the total rate is just the sum of all of these rates. For the case of sinusoidal perturbations of the form $2\hat{V}\cos(\omega t)$, if the rate of transition from state $i$ to state $\vec{k}$ is $$ \Gamma_{i\to\vec{k}} = \frac{2\pi}{\hbar}|\langle \vec{k} | \hat{V} | i\rangle|^2\delta\left(|E_i-E_{\vec{k}}|-\hbar\omega\right)\,, $$ then we are evaluating the sum $$ \Gamma = \sum_{\vec{k}}\Gamma_{i\to\vec{k}} = \sum_{\vec{k}}\frac{2\pi}{\hbar}|\langle \vec{k} | \hat{V} | i\rangle|^2\delta\left(|E_i-E_{\vec{k}}|-\hbar\omega\right)\,. $$ Since the states $\vec{k}$ form a continuum of final states, this sum is actually the integral $$ \sum_\vec{k} \to \frac{V}{(2\pi)^3}\int d^3 k\,, $$ where $V$ is the quantization volume.$^2$ Finally, we perform a change of variables by using the dispersion relation $E = E(k)$ that relates the energy of the final state to the wave vector magnitude $k$.$^3$ The result is that $$ \sum_\vec{k} \to \frac{V}{(2\pi)^3}\int d^3 k \to \int dE\, \left(\frac{V}{(2\pi)^3}k^2\frac{dk}{dE} d\Omega\right)\,, $$ where $d\Omega=\sin\theta\,d\theta\,d\varphi$ in $k$-space. In the case where the situation is spherically symmetric, we just replace $d\Omega$ with $4\pi$, and then the quantity in parentheses is called the density of (final) states $\rho(E)$.

Finally, then, for the total transition rate out of state $i$, we have \begin{align} \Gamma &= \sum_{\vec{k}}\frac{2\pi}{\hbar}|\langle \vec{k} | \hat{V} | i\rangle|^2\delta\left(|E_i-E_{\vec{k}}|-\hbar\omega\right) \\&\to \int dE\,\rho(E)\,\frac{2\pi}{\hbar}|\langle \vec{k} | \hat{V} | i\rangle|^2\delta\left(|E_i-E|-\hbar\omega\right)\,. \end{align} For the case of spontaneous emission, for instance, $\omega=0$ and so this is \begin{align} \Gamma &= \int dE\,\rho(E)\,\frac{2\pi}{\hbar}|\langle \vec{k} | \hat{V} | i\rangle|^2\delta\left(E_i-E\right) = \frac{2\pi}{\hbar}|\langle \vec{k} | \hat{V} | i\rangle|^2\rho(E_f=E_i)\,. \end{align}

$^1$ There can also be a polarization $\vec{\epsilon}$, but since quantum particles don't have a polarization, we'll ignore this.

$^2$ To see how this comes about requires thinking about how, for instance, plane-wave states are quantized in a box, and then counting the number of states with values of $|vec{k}|$ less than some $k$. This is covered in any standard solid state textbook, for instance. Luckily, the quantization volume typically cancels out in the calculation anyway, so it is physically irrelevant. Alternatively, one can choose an energy normalization for the final states that essentially does the same thing.

$^3$ The energy can sometimes depend on the direction of $\vec{k}$, but it's not hard to generalize to these cases, and the extra complication doesn't provide any extra illumination.


I think for these type of more theoretical questions it is more useful to use a grounded example.

Transitions and energy levels for a four-level Nd:YAG laser. Energy scale is in 10 3 <span class=$cm^−1$ " />

In the picture above you can find the transitions and energy levels for a four-level Nd:YAG laser. Source here.

In case you are not too familiar with laser theory: For lasers you typically pump light at 808 nm (this can be another laser or a lamp, for example), and the material then is excited and radiatively emits (e.g. creates a laser light) at a longer wavelength (1064nm in this case). The underlying mechanism is that you excite electrons from the ground state of the active material (Nd:YAG in this case) and due to how when working with lasers you generally prefer shorter-lived radiative emission states (here ^4 F_{3/2}), you essentially pump to a much higher level than you intend to emit, where the excited electrons can stay longer and interact with the active medium, and then be pumped out by the shorter-lived radiative emission state.

If the tl;dr is not satisfactory, you can also take a brief look at this article in RP Photonics with a more detailed explanation about 4-level lasers.

In lasers the energy states are often depicted not as levels but as Stark level manifolds because some levels are so close to each other that only when there is an actual level separation due to selection rules, then you consider the manifolds to be energetically separated.

Referring to your question you can see here a close relation to the density of states for such complex atomic configurations: The fact that you would, for example, use $^4 F_{5/2} - ^2 H_{9/2}$ as your final state $E_f$ then you can immediately see why a density of states would prove useful, as there is not a single amount of excited electrons that would decay and emit radiatively (the 1064 nm), but rather the contributions from neighboring states that also contribute to the transition to the ground state.

The reason why the ground state does not apply in this case is because $|V_ii| =0$ since electrons are already in the relaxed state, therefore there are no significant contributions for any form of radiation from neighboring states (except for the case of ground state fluctuations but I think that is a bit out of the scope of the original question).

I know that introducing a different topic is dangerous to explain specific phenomena but I think lasers are an application where it is immediately obvious when certain atomic physics situations are introduced.


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