The Sudden Approximation

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Hello! Welcome to our lesson on approximation methods for quantum dynamics.

In the last two lessons, we've explored two distinct limits of time-dependent Hamiltonians. We started with rapidly oscillating perturbations, leading to Fermi's Golden Rule for transition rates. Then, we examined the opposite extreme of very slow changes, where the adiabatic theorem ensures a system can adapt without making transitions.

Today, we complete this picture by investigating the third case: what happens when a Hamiltonian changes abruptly and dramatically over a very short period? This is the domain of the sudden approximation. Our goal is to understand and apply this approximation to calculate the state of a system after such a rapid change. This method is surprisingly simple yet powerful, with direct applications ranging from nuclear decay to the "quantum quenches" used to study non-equilibrium physics in condensed matter systems.

1. The Sudden Approximation: Concept and Condition

Imagine a quantum system is happily sitting in an eigenstate $∣ ψ_{i} ⟩$ of a time-independent Hamiltonian $H_{0}$ . Then, over a very short time interval $τ$ centered around $t = 0$ , the Hamiltonian changes to a new form, $H$ , and remains so for $t > 0$ .

What does it mean for the change to be "sudden"? Intuitively, it means the change happens so fast that the system has no time to react. The timescale of the change, $τ$ , must be much shorter than any natural timescale of the initial system. These internal timescales are determined by the energy differences between states, $T_{in t er na l} \sim ℏ/Δ E$ .

Therefore, the condition for the sudden approximation is:
$τ ≪ \frac{ℏ}{Δ E _{ik}} for all states k \neq = i$
where $Δ E_{ik} = ∣ E_{i} - E_{k} ∣$ are the energy gaps of the initial Hamiltonian $H_{0}$ .

If this condition holds, the state of the system at time $t = 0^{+}$ (just after the change) is essentially the same as it was at $t = 0^{-}$ (just before the change). Why? Let's consider the time evolution operator $U (t_{f}, t_{i})$ . For a very small interval $τ$ , we have:
$∣Ψ (τ /2)⟩ = U (τ /2, - τ /2) ∣Ψ (- τ /2)⟩ = exp (- \frac{i}{ℏ} \int_{- τ /2}^{τ /2} H (t) d t) ∣Ψ (- τ /2)⟩$
If $τ$ is extremely small, the integral in the exponent is also very small, and the exponential is approximately the identity operator. Thus, $∣Ψ (τ /2)⟩ \approx ∣Ψ (- τ /2)⟩$ .

The Sudden Approximation Rule:
If a system is in state $∣ ψ_{i} ⟩$ before a sudden change in its Hamiltonian, its state immediately after the change is still $∣ ψ_{i} ⟩$ .

The crucial point is that while the state vector has not changed, the Hamiltonian has. This means $∣ ψ_{i} ⟩$ is no longer an eigenstate of the new Hamiltonian $H$ . To find out what happens for $t > 0$ , we must express the initial state as a superposition of the new eigenstates, $∣ ϕ_{f} ⟩$ , which satisfy $H ∣ ϕ_{f} ⟩ = E_{f}^{'} ∣ ϕ_{f} ⟩$ :
$∣ ψ_{i} ⟩ = \sum_{f} c_{f} ∣ ϕ_{f} ⟩ where c_{f} = ⟨ ϕ_{f} ∣ ψ_{i} ⟩$
The probability of finding the system in a specific new eigenstate $∣ ϕ_{f} ⟩$ is then simply:
$P_{i \to f} = ∣ c_{f} ∣^{2} = ∣ ⟨ ϕ_{f} ∣ ψ_{i} ⟩ ∣^{2}$
This is the central formula of the sudden approximation. It states that the transition probability is just the squared overlap between the initial state and the final target eigenstate.

2. Application 1: Beta Decay of Tritium

A classic physical example of a sudden change occurs in nuclear beta decay. Your background in physics makes this a natural starting point.

6. Approximation Methods

Let's examine how the sudden approximation applies to the beta decay of a tritium atom. This section from David Tong's lecture notes provides a clear, physically motivated example.

Please read section 6.3.1, 'The Sudden Approximation', and the example of Tritium (pages 165-166). Pay close attention to two things: The justification for using the sudden approximation by comparing the timescales. The calculation of the probability that the electron remains in the ground state after the decay.

Let's break down this example.

Beta Decay of Tritium (3H) to Helium-3 (3He) — CAPTION: This diagram illustrates the beta decay of a tritium nucleus ($^{3}\text{H}$). One of its two neutrons transforms into a proton, changing the nucleus into helium-3 ($^{3}\text{He}$) and emitting an electron and an antineutrino. For the orbital electron of the original atom, this is an almost instantaneous change in the nuclear charge from Z=1 to Z=2.

The Setup: A tritium atom ( $^{3} H$ ) consists of a nucleus with charge $Z = 1$ and a single electron. We assume this electron is initially in the ground state. The nucleus then undergoes beta decay, becoming a helium-3 nucleus ( $^{3} He$ ) with charge $Z = 2$ . From the electron's point of view, the Coulomb potential it experiences changes abruptly.
Validity Check: As calculated in the notes, the emitted beta particle (electron) is highly energetic and leaves the atomic radius in about $τ \approx 1 0^{- 19}$ s. The characteristic timescale of the initial hydrogen-like atom is set by the energy gap between the ground and first excited states, which is $T \sim ℏ/Δ E \approx 6.5 \times 1 0^{- 17}$ s. Since $τ ≪ T$ , the sudden approximation is well-justified.
Calculation:
- Initial State $∣ ψ_{i} ⟩$ : The ground state wavefunction of a hydrogen-like atom with $Z = 1$ . This is $∣ ψ_{1 s}^{Z = 1} ⟩$ . Its spatial form is $ψ_{i} (r) = \frac{1}{π a _{0}^{3}} e^{- r / a_{0}}$ .
- Final Eigenstate $∣ ϕ_{f} ⟩$ : We want the probability of the electron ending up in the ground state of the new atom, which is a helium ion ( $^{3} He^{+}$ ) with $Z = 2$ . This is $∣ ϕ_{1 s}^{Z = 2} ⟩$ . Its spatial form is $ϕ_{f} (r) = \frac{1}{π ( a _{0} /2 ) ^{3}} e^{- 2 r / a_{0}}$ .
- Probability: The probability is the squared overlap:
  $P = ∣ ⟨ ϕ_{f} ∣ ψ_{i} ⟩ ∣^{2} = \int ϕ_{f}^{*} (r) ψ_{i} (r) d^{3} x^{2}$
  As the calculation in the notes shows, this integral evaluates to $P = (88 /27)^{2} = 512/729 \approx 0.702$ . So, there is a ~70% chance that the electron remains in the 1s ground state, even though the nucleus has changed. The other ~30% of the time, it will be found in an excited state of the helium ion (or be ejected entirely).

3. Application 2: The Expanding Infinite Square Well

This is a canonical textbook problem that cleanly illustrates the mechanics of the calculation without the physical complexities of atomic wavefunctions.

Phys 506 lecture 32: Fermi's golden rule and the sudden ...

Now, let's work through another standard example: a particle in an infinite square well whose walls are moved suddenly. These notes from Georgetown University provide a clear, step-by-step derivation.

Please read section 3 'The sudden approximation' and section 4 'Example of the sudden approximation' (pages 3-4). Follow the calculation for a particle initially in the ground state of a box of length L that suddenly expands to 2L. Note how the probability is calculated for each new eigenstate.

The key steps are:

Initial State $∣ ψ_{i} ⟩$ : The particle is in the ground state ( $n = 1$ ) of a box of length $L$ .
$ψ_{i} (x) = \frac{2}{L} sin (\frac{π x}{L}) for 0 < x < L$
Final Eigenstates $∣ ϕ_{n^{'}} ⟩$ : The Hamiltonian changes to that of a box of length $2 L$ . The new eigenstates are:
$ϕ_{n^{'}} (x) = \frac{2}{2 L} sin (\frac{n ^{'} π x}{2 L}) = \frac{1}{L} sin (\frac{n ^{'} π x}{2 L}) for 0 < x < 2 L$
Probability Calculation: We need to compute $P_{1 \to n^{'}} = ∣ ⟨ ϕ_{n^{'}} ∣ ψ_{i} ⟩ ∣^{2}$ . The overlap integral is:
$⟨ ϕ_{n^{'}} ∣ ψ_{i} ⟩ = \int_{0}^{L} (\frac{1}{L} sin (\frac{n ^{'} π x}{2 L}))^{*} (\frac{2}{L} sin (\frac{π x}{L})) d x$
Note that the integral only goes up to $L$ , because the initial wavefunction is zero for $x > L$ . As the resource shows, this calculation yields non-zero probabilities only for odd $n^{'}$ (and a special case for $n^{'} = 2$ ). This is a consequence of the symmetry of the wavefunctions.

Test your understanding!

In the expanding box example, the initial state is the ground state $ψ_{1} (x)$ , which is symmetric about the center of the original box ( $x = L /2$ ). The new eigenfunctions $ϕ_{n^{'}} (x)$ have definite parity (symmetry) about the center of the new box ( $x = L$ ).

Why is the transition probability to the new ground state ( $n^{'} = 1$ ) non-zero, but the transition probability to the first excited state ( $n^{'} = 2$ ) is exactly zero? (Hint: Consider the symmetry of the wavefunctions involved in the overlap integral.)

Show answer

The initial state $ψ_{1} (x)$ is a single "hump" between 0 and $L$ .
The new ground state $ϕ_{1} (x)$ is a single "hump" between 0 and $2 L$ . Its wavefunction is positive over the entire range of integration [0, L]. The overlap integral $\int_{0}^{L} ϕ_{1} (x) ψ_{1} (x) d x$ will be non-zero because we are integrating the product of two positive functions.

The new first excited state $ϕ_{2} (x)$ has a node at $x = L$ . It is positive for $0 < x < L$ and negative for $L < x < 2 L$ . However, when we compute the overlap, the integral $\int_{0}^{L} ϕ_{2} (x) ψ_{1} (x) d x$ is non-zero (as shown in the resource, $P_{1 \to 2} = 1/2$ ).

Wait, the resource calculation gives $P = 1/2$ for n=2. Let me re-read it. Ah, the result is n=even, n!=2 -> 0. And for n=2 -> 1/2. For n=odd -> 32/.... My hint was slightly off. Let's correct the explanation.

The calculation shows the probability for $n^{'} = 2$ is $1/2$ . The probability is zero for all other even $n^{'}$ . Why is this? The overlap integral for even $n^{'}$ involves integrating $sin (n^{'} π x /2 L) sin (π x / L)$ . Let $n^{'} = 2 k$ . The term becomes $sin (k π x / L) sin (π x / L)$ . For $k = 1$ (i.e., $n^{'} = 2$ ), the integral is $\int_{0}^{L} sin^{2} (π x / L) d x = L /2$ , leading to the non-zero probability. For other even $n^{'}$ ( $k > 1$ ), the functions $sin (k π x / L)$ and $sin (π x / L)$ are orthogonal over the interval $[0, L]$ , so their integral is zero.

The probability is non-zero for odd $n^{'}$ because the symmetry doesn't lead to a perfect cancellation in the integral.

4. Application 3: Quantum Quench of a Harmonic Oscillator

Your interest lies in condensed matter physics, where the concept of a quantum quench is a vital tool for studying systems out of equilibrium. A quench is simply a rapid change in a system's parameters, making it a direct application of the sudden approximation.

6. Approximation Methods

Let's explore a quantum quench in a harmonic oscillator, where the spring constant is changed suddenly. This is a foundational model for more complex quench dynamics studied in many-body systems.

Please read section 6.3.2, 'An Example: Quantum Quench of a Harmonic Oscillator' (pages 166-167). Focus on how the final state is constructed. Note that the initial ground state is not an eigenstate of the new Hamiltonian and becomes a superposition of the new even-numbered energy states.

In this example, we start with a harmonic oscillator of frequency $ω_{0}$ in its ground state $∣ 0_{0} ⟩$ . We then suddenly change the frequency to $ω$ .

The initial ground state $∣ 0_{0} ⟩$ is defined by $a_{0} ∣ 0_{0} ⟩ = 0$ , where $a_{0}$ is the annihilation operator for the initial Hamiltonian.
After the quench, the Hamiltonian is governed by the new operators $a$ and $a^{†}$ .
As the notes show, the old annihilation operator $a_{0}$ can be written as a linear combination of the new operators $a$ and $a^{†}$ . Consequently, the old ground state $∣ 0_{0} ⟩$ is no longer annihilated by the new operator $a$ .
When expanded in the basis of the new eigenstates $∣ n ⟩$ , the state $∣ 0_{0} ⟩$ becomes a superposition of all the even parity states: $∣0 ⟩, ∣2 ⟩, ∣4 ⟩, \dots$ . This is because the ground state has even parity, and changing the frequency symmetrically doesn't break this parity.

This simple example demonstrates how a quench can inject energy into a system and populate a whole spectrum of excited states, kicking the system into a non-trivial dynamical evolution. This is a fundamental technique used in experiments with ultracold atoms to explore quantum dynamics and thermalization.

Conclusion

Today, we've explored the sudden approximation, the last of our three key time-dependent approximation methods. We've seen that its core idea is remarkably simple: if a change is fast enough, the system's state vector has no time to change.

Key Takeaways:

The Sudden Approximation is valid when the Hamiltonian changes on a timescale $τ$ much shorter than the system's characteristic internal timescales, $τ ≪ ℏ/Δ E$ .
The central rule is that the state vector is continuous across the rapid change: $∣Ψ (t_{a f t er})⟩ = ∣Ψ (t_{b e f or e})⟩$ .
Transition probabilities to the new eigenstates $∣ ϕ_{f} ⟩$ are calculated by projecting the initial state $∣ ψ_{i} ⟩$ onto the new basis: $P_{i \to f} = ∣ ⟨ ϕ_{f} ∣ ψ_{i} ⟩ ∣^{2}$ .
We saw applications in nuclear physics (Tritium decay), classic quantum mechanics (expanding box), and modern condensed matter concepts (quantum quench).

In our previous lesson on adiabatic evolution, we saw that very slow changes keep a system in its instantaneous eigenstate. Today, we saw that very fast changes keep a system in its instantaneous state, which is then no longer an eigenstate. The Landau-Zener formula you studied provides a perfect bridge: in the limit of very fast sweeping through a crossing ( $∣ α ∣ \to \infty$ ), the diabatic transition probability $P_{L Z} \to 1$ , meaning the system behaves as if the levels crossed without interacting—it stays in a state analogous to its initial one, exactly as the sudden approximation predicts.

Preview of the next lesson:
We have now completed our survey of the main approximation methods for time-dependent problems. We will now shift our focus back to time-independent problems, but for potentials that are too complex for exact solution. In the next lesson, we will introduce the WKB (Wentzel-Kramers-Brillouin) approximation, a powerful semi-classical method for finding approximate energy eigenvalues and wavefunctions for systems with slowly varying potentials.