Fermi's Golden Rule for Continuum Transitions

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The freely available resources this lesson will direct you to.

Hello! Welcome back to our study of quantum dynamics.

In our last lesson, we developed the framework of first-order time-dependent perturbation theory. We found that for a constant perturbation, the probability of transitioning to a non-degenerate state oscillates, while the probability of transitioning to a degenerate state grows quadratically with time, as $t^{2}$ . This result, while correct, doesn't seem to describe many common physical processes, like the decay of an excited atomic state, where we observe a constant rate of transition, implying a probability that grows linearly with time.

Today, we will bridge that gap. We'll see how a constant transition rate naturally emerges when we consider transitions not to a single, isolated final state, but to a dense continuum of final states. This will lead us to one of the most useful formulas in quantum mechanics: Fermi's Golden Rule. This rule is indispensable for calculating scattering rates in solids, decay rates in nuclear and particle physics, and rates of light absorption and emission in atoms—a cornerstone of your interest in condensed matter physics.

1. From a Single State to a Continuum

Let's begin where we left off. For a constant perturbation $V$ turned on at $t = 0$ , the probability of transition from state $∣ i ⟩$ to state $∣ f ⟩$ is:
$P_{i \to f}^{(1)} (t) = \frac{∣ ⟨ f ∣ V ∣ i ⟩ ∣ ^{2}}{ℏ ^{2}} \frac{4 s i n ^{2} ( ω _{f i} t /2 )}{( ω _{f i} /2 ) ^{2}}$
where $ω_{f i} = (E_{f} - E_{i}) /ℏ$ .

The key insight is that in many physical situations—an atom being ionized, a neutron undergoing beta decay, or an electron scattering off an impurity in a crystal—the final state is not one discrete level. Instead, there's a continuous band of available final states. For instance, an ejected electron can have any kinetic energy within a certain range.

To find the total probability of making a transition, we must sum the probabilities of transitioning to all possible final states. For a continuum, this sum becomes an integral. We introduce a crucial quantity, the density of states, $ρ (E)$ , which is defined as the number of available states per unit energy around energy $E$ . The number of states in a small energy interval $d E_{f}$ is $d N = ρ (E_{f}) d E_{f}$ .

The total transition probability $P_{total} (t)$ is then:
$P_{total} (t) = \sum_{f} P_{i \to f}^{(1)} (t) \to \int P_{i \to f}^{(1)} (t) ρ (E_{f}) d E_{f}$

Substituting our expression for $P_{i \to f}^{(1)} (t)$ gives:
$P_{total} (t) = \int \frac{∣ ⟨ f ∣ V ∣ i ⟩ ∣ ^{2}}{ℏ ^{2}} \frac{4 s i n ^{2} (( E _{f} - E _{i} ) t /2ℏ )}{(( E _{f} - E _{i} ) /2ℏ ) ^{2}} ρ (E_{f}) d E_{f}$

This integral looks daunting, but as you'll see, it simplifies beautifully under a few well-justified approximations.

2. Deriving the Golden Rule

The following video provides a clear and visual derivation of Fermi's Golden Rule, starting from the integral we just set up. It masterfully explains the central argument involving the properties of the $\frac{s i n ^{2} ( x )}{x ^{2}}$ function.

L10.3 Integrating over the continuum to find Fermi's Golden Rule

This MIT OpenCourseWare video, 'L10.3 Integrating over the continuum to find Fermi's Golden Rule', walks through the complete derivation. It shows how summing over a continuum of states transforms the quadratic time dependence we found earlier into a linear one, yielding a constant transition rate.

Please watch the entire video. Pay close attention to these key steps: The approximations made: why can we treat the matrix element and the density of states as constants and pull them out of the integral? (02:00) The behavior of the integrand: how the central peak of the sin²(x)/x² function dominates the integral and becomes narrower for longer times, enforcing energy conservation. (04:40) The final change of variables that reveals the linear dependence on time, t, which allows us to define a rate. (11:25) The discussion of the conditions for validity at the end of the video. (17:07)

Let's summarize the elegant argument from the video.

The Main Contribution: The function $\frac{s i n ^{2} ( α x )}{x ^{2}}$ has a very sharp peak at $x = 0$ , and its width is proportional to $1/ α$ . In our integral, $x = E_{f} - E_{i}$ and $α = t /2ℏ$ . This means for a sufficiently long time $t$ , the integrand is sharply peaked around $E_{f} = E_{i}$ . The transition overwhelmingly favors final states that conserve energy.
The Key Approximation: Because this peak is so narrow, we can assume that the matrix element $⟨ f ∣ V ∣ i ⟩$ and the density of states $ρ (E_{f})$ are approximately constant over the narrow range of energies that contribute to the integral. We can therefore evaluate them at the peak, $E_{f} = E_{i}$ , and pull them outside the integral.
Evaluating the Integral: After pulling out the slowly-varying terms, the core integral becomes $\int_{- \infty}^{\infty} \frac{s i n ^{2} ( x t /2ℏ )}{(( x ) /2ℏ ) ^{2}} d x$ , where $x = E_{f} - E_{i}$ . A change of variables $u = x t /2ℏ$ shows that the integral is proportional to $t$ . The standard result is $\int_{- \infty}^{\infty} \frac{s i n ^{2} u}{u ^{2}} d u = π$ , which leads to the total integral being equal to $\frac{2 π ℏ}{t} \cdot (\frac{t}{2ℏ})^{2} \cdot t = \frac{π t}{2ℏ}$ . No, let's re-do that based on the video/notes... The integral from the notes I(t0) is ∫ [sin²(ω_fi t₀/2) / ω_fi²] dE_f. With x = E_f - E_i, ω_fi = x/ħ. The integral is ∫ [sin²(xt/2ħ) / (x²/ħ²)] dx. Let u = xt/2ħ, so x = 2ħu/t and dx = (2ħ/t)du. The integral becomes ∫ [sin²(u) / (4ħ²u²/t²)] * (2ħ/t) du = ∫ sin²(u)/u² * (t/2ħ) du. The integral of sin²(u)/u² is π. So the result is (πt/2ħ).
Putting it all together, the total probability becomes:
$P_{total} (t) = ∣ ⟨ f ∣ V ∣ i ⟩ ∣^{2} ρ (E_{i}) \frac{1}{ℏ ^{2}} (\frac{π ℏ t}{2} \times 4)$
No, let me retrace the derivation from the resources.
The total probability is $P (t) = \int ∣ c_{f}^{(1)} (t) ∣^{2} ρ (E_{f}) d E_{f} = \int \frac{4∣ V _{f i} ∣ ^{2}}{ℏ ^{2}} \frac{s i n ^{2} (( E _{f} - E _{i} ) t /2ℏ )}{(( E _{f} - E _{i} ) /ℏ ) ^{2}} ρ (E_{f}) d E_{f}$ .
This is incorrect. The denominator in the probability formula is $(E_{f} - E_{i})^{2}$ , not $ω_{f i}^{2}$ . So it's:
$P_{total} (t) = \int \frac{4∣ V _{f i} ∣ ^{2} s i n ^{2} (( E _{f} - E _{i} ) t /2ℏ )}{( E _{f} - E _{i} ) ^{2}} ρ (E_{f}) d E_{f}$
Pulling constants out: $∣ V_{f i} ∣^{2} ρ (E_{i}) \int \dots$ . Let $x = E_{f} - E_{i}$ . The integral is $\int \frac{4 s i n ^{2} ( x t /2ℏ )}{x ^{2}} d x$ . Let $u = x t /2ℏ$ . Then $x = 2ℏ u / t$ and $d x = (2ℏ/ t) d u$ . The integral becomes $\int \frac{4 s i n ^{2} ( u )}{( 2ℏ u / t ) ^{2}} \frac{2ℏ}{t} d u = \int \frac{4 s i n ^{2} ( u )}{4 ℏ ^{2} u ^{2} / t ^{2}} \frac{2ℏ}{t} d u = \frac{2 π t}{ℏ}$ .
So, $P_{total} (t) = ∣ V_{f i} ∣^{2} ρ (E_{i}) \frac{2 π t}{ℏ}$ . Something is still off with a factor.
Let's check Zwiebach's notes (resource LINK, section 4.3.1).
He has $P (t) = \frac{4∣ V _{f i} ∣ ^{2}}{ℏ ^{2}} ρ (E_{i}) I (t)$ , with $I (t) = \int \frac{s i n ^{2} ( ω _{f i} t /2 )}{ω _{f i}^{2}} d E_{f}$ . He changes variables $u = ω_{f i} t /2$ , noting $d E_{f} = ℏ d ω_{f i}$ , so $I (t) = ℏ \int \frac{s i n ^{2} ( u t /2 )}{...}$ no...
His change of variables is $u = ω_{f i} t_{0} /2$ . Then $d E_{f} = ℏ d ω_{f i} = \frac{2ℏ}{t _{0}} d u$ . The integral $I (t_{0})$ becomes $ℏ \int \frac{s i n ^{2} u}{( \frac{2 u}{t _{0}} ) ^{2}} \frac{2ℏ}{t _{0}} d u$ . Something is not right in my scratchpad.
Let's follow the notes carefully. $I (t_{0}) = ℏ \int \frac{s i n ^{2} ( ω _{f i} t _{0} /2 )}{ω _{f i}^{2}} d ω_{f i}$ . Change variable to $u = ω_{f i} t_{0} /2$ . $d ω_{f i} = 2 d u / t_{0}$ . $I (t_{0}) = ℏ \int \frac{s i n ^{2} u}{( 2 u / t _{0} ) ^{2}} \frac{2}{t _{0}} d u = ℏ \frac{t _{0}^{2}}{4} \frac{2}{t _{0}} \int \frac{s i n ^{2} u}{u ^{2}} d u = \frac{ℏ t _{0}}{2} π$ .
So $P_{total} (t) = \frac{4∣ V _{f i} ∣ ^{2}}{ℏ ^{2}} ρ (E_{i}) \frac{π ℏ t}{2} = \frac{2 π}{ℏ} ∣ V_{f i} ∣^{2} ρ (E_{i}) t$ . This is correct.
The Result: The total probability grows linearly with time!
$P_{total} (t) = (\frac{2 π}{ℏ} ∣ ⟨ f ∣ V ∣ i ⟩ ∣^{2} ρ (E_{i})) t$
This allows us to define a constant transition rate, $w_{i \to f} = d P_{total} / d t$ .

This gives us the celebrated Fermi's Golden Rule:

$w_{i \to f} = \frac{2 π}{ℏ} ∣ ⟨ f ∣ V ∣ i ⟩ ∣^{2} ρ (E_{f}) with E_{f} = E_{i}$

This formula is profoundly important. It states that the transition rate is proportional to the square of the matrix element connecting the initial and final states, multiplied by the density of available final states that conserve energy.

Test your understanding!

The derivation of Fermi's Golden Rule relied on two seemingly contradictory assumptions about the time $t$ :

$t$ must be "large enough" so that the central peak of the $\frac{s i n ^{2} ( ω _{f i} t /2 )}{ω _{f i}^{2}}$ function is narrow enough to treat $ρ (E_{f})$ and $V_{f i}$ as constant.
$t$ must be "small enough" so that the total probability $w_{i \to f} \cdot t$ remains much less than 1, keeping us within the valid regime of first-order perturbation theory.

How can both of these conditions be satisfied simultaneously?

Show answer

This is a crucial point about the validity of the Golden Rule. The apparent contradiction is resolved if the perturbation $V$ is sufficiently weak.

Condition 1 requires the energy width of the central peak, $Δ E \sim ℏ/ t$ , to be much smaller than the energy scale over which $ρ (E)$ or $V_{f i}$ vary. Let's call this energy scale $E_{char}$ . So we need $ℏ/ t ≪ E_{char}$ , or $t ≫ ℏ/ E_{char}$ .

Condition 2 requires $w \cdot t ≪ 1$ . Substituting the Golden Rule for $w$ , this is $\frac{2 π}{ℏ} ∣ V_{f i} ∣^{2} ρ (E_{i}) t ≪ 1$ .

For both to hold, we need to be able to find a time $t$ such that $ℏ/ E_{char} ≪ t ≪ \frac{ℏ}{2 π ∣ V _{f i} ∣ ^{2} ρ ( E _{i} )}$ . A window for such a $t$ exists if $ℏ/ E_{char} ≪ \frac{ℏ}{2 π ∣ V _{f i} ∣ ^{2} ρ ( E _{i} )}$ , which simplifies to $∣ V_{f i} ∣^{2} ≪ \frac{E _{char}}{2 π ρ ( E _{i} )}$ . Since $E_{char}$ and $ρ (E_{i})$ are properties of the unperturbed system, this condition can always be met by making the perturbation strength $∣ V_{f i} ∣$ small enough. In essence, Fermi's Golden Rule is an approximation valid for weak perturbations that cause slow transitions.

3. Application: Scattering in a Solid

To make this more concrete, let's see how the ingredients of the Golden Rule are calculated in a scenario relevant to your interest in condensed matter physics. We will consider an electron in a semiconductor scattering off a point-like impurity. The following reading provides a different, more formal derivation of the Golden Rule that is common in advanced texts, and then applies it to this exact problem.

This approach uses a mathematical trick of "slowly turning on" the perturbation with a factor $e^{η t}$ and taking the limit $η \to 0^{+}$ at the end. This automatically produces a Dirac delta function $δ (E_{f} - E_{i})$ , which is a formal statement of energy conservation.

Chapter 24 - Fermi's Golden Rule

The document 'Chapter 24 - Fermi's Golden Rule' from Cornell University provides an alternative derivation and a fantastic condensed matter application. This will give you a glimpse of how these tools are used in practice.

Please read sections 24.2 ('Fermi's Golden Rule') and 24.4 ('Transitions to a continuum of states'). In section 24.2, focus on how the η → 0⁺ limit leads to a Dirac delta function, providing a more formal statement of energy conservation. In section 24.4, follow the example of an electron scattering from a delta-function potential W(r) = V₀δ(r). Pay close attention to: How the matrix element ⟨k'|W|k⟩ is calculated. How the sum over final states Σ_k' is converted into an integral over the density of states D(E_k). How the macroscopic volume V in the matrix element and the density of states cancels out to give a physically meaningful scattering rate.

The example in the reading is particularly illustrative. The total scattering rate of an electron with initial momentum $k$ is found by summing the rates to all possible final states $k^{'}$ :
$\frac{1}{τ ( k )} = \sum_{k^{'}} w_{k \to k^{'}} = \sum_{k^{'}} \frac{2 π}{ℏ} ∣ ⟨ k^{'} ∣ W ∣ k ⟩ ∣^{2} δ (E_{k^{'}} - E_{k})$
The sum over the delta function, $\sum_{k^{'}} δ (E_{k^{'}} - E_{k})$ , is precisely the definition of the density of states at energy $E_{k}$ . This elegantly shows how the two formalisms connect. For a free electron in 3D, $ρ (E) \propto E$ , so the scattering rate will depend on the energy of the incident electron. This kind of calculation is fundamental to understanding electrical resistivity in metals and semiconductors.

4. Extension to Harmonic Perturbations

We derived the Golden Rule for a constant perturbation $V$ . A similar analysis can be done for a harmonically oscillating perturbation, like that from an electromagnetic wave. This is the subject of our next lesson, but I'll give you the result here as it follows the same logic. For a perturbation $δ H (t) = 2 H^{'} cos (ω t)$ , the derivation yields two distinct processes:

Absorption: The system absorbs energy $ℏ ω$ from the field.
$w_{i \to f}^{(abs)} = \frac{2 π}{ℏ} ∣ ⟨ f ∣ H^{'} ∣ i ⟩ ∣^{2} ρ (E_{f}), with E_{f} = E_{i} + ℏ ω$
Stimulated Emission: The system is stimulated by the field to release energy $ℏ ω$ .
$w_{i \to f}^{(stim)} = \frac{2 π}{ℏ} ∣ ⟨ f ∣ H^{'} ∣ i ⟩ ∣^{2} ρ (E_{f}), with E_{f} = E_{i} - ℏ ω$

Notice that the structure is identical to the rule we derived, but energy conservation is now satisfied by accounting for the energy quantum $ℏ ω$ from the oscillating field.

Conclusion

Today we derived and applied one of quantum mechanics' most powerful calculational tools. By considering transitions into a continuum, we saw how a constant transition rate arises naturally from first-order perturbation theory.

Key Takeaways:

When a quantum system can transition to a continuum of final states, the total transition probability grows linearly with time, defining a constant transition rate.
This rate is given by Fermi's Golden Rule: $w_{i \to f} = \frac{2 π}{ℏ} ∣ ⟨ f ∣ V ∣ i ⟩ ∣^{2} ρ (E_{f})$ .
The rule's three essential components are:
1. The matrix element $∣ ⟨ f ∣ V ∣ i ⟩ ∣^{2}$ , which quantifies how strongly the perturbation couples the states.
2. The density of states $ρ (E_{f})$ , which counts how many final states are available to transition into.
3. An implicit energy conservation condition ( $E_{f} = E_{i}$ ), enforced by the derivation.
The rule is valid for weak perturbations and for times that are long enough for a rate to be established but short enough that the initial state is not significantly depleted.

Preview of the next lesson:
We have now established the rule for calculating transition rates. In the next lesson, we will apply it to one of the most important physical phenomena: the interaction of atoms with light. We will use the Golden Rule for harmonic perturbations to analyze the processes of absorption and emission of radiation, and derive the so-called dipole selection rules that govern which atomic transitions are allowed.