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Chapter 9: Problem 17

A particle of mass \(m\) is initially in the ground state of the (onedimensional) infinite square well. At time \(t=0\) a "brick" is dropped into the well, so that the potential becomes $$ V(x)=\left\\{\begin{array}{ll} V_{0}, & \text { if } 0 \leq x \leq a / 2 \\ 0, & \text { if } a / 2

Short Answer

Expert verified
The transition probability to state with energy \(E_2\) is \(P_{1 \to 2} \approx \left(\frac{8 V_0}{3\pi^2 E_1}\right)^2 \sin^2\left(\frac{3\pi^2 T \hbar}{a^2 m}\right).\)

Step by step solution

01

Determine Initial State

Initially, the particle is in the ground state of the infinite square well, with the wave function \( \psi_1(x) = \sqrt{\frac{2}{a}} \sin\left( \frac{\pi x}{a} \right) \), and energy \( E_1 = \frac{\hbar^2 \pi^2}{2ma^2} \). Since \( V_0 \ll E_1 \), we can consider \( V(x) \) as a small perturbation.
02

Express the New Perturbation Term

The perturbation introduced by the 'brick' is given by \( V(x) = V_0 \), only affecting the region \( 0 \leq x \leq \frac{a}{2} \). The perturbation potential is \( V(x) = V_0\theta(\frac{a}{2} - x) \), where \( \theta \) is the Heaviside step function.
03

Calculate the Matrix Element

Use the matrix element for the perturbation: \[ \langle \psi_2 | H' | \psi_1 \rangle = \int_{0}^{a} \psi_2^*(x) V(x) \psi_1(x) \, dx, \] where \( \psi_2(x) = \sqrt{\frac{2}{a}} \sin\left( \frac{2\pi x}{a} \right) \) is the wave function for the first excited state.
04

Integrate to Find Matrix Element

The integral becomes: \[ \langle \psi_2 | H' | \psi_1 \rangle = \frac{2V_0}{a} \int_{0}^{a/2} \sin\left(\frac{2\pi x}{a}\right) \sin\left(\frac{\pi x}{a}\right) \, dx. \] Solve this integral by using trigonometric identities and standard integrals.
05

Solve the Integral

Using \( \sin A \sin B = \frac{1}{2} \left[ \cos(A-B) - \cos(A+B) \right] \), evaluate the integral: \[ \langle \psi_2 | H' | \psi_1 \rangle = \frac{V_0}{a} \int_{0}^{a/2} \left[ \frac{1}{2}(\cos\left(\frac{\pi x}{a}\right) - \cos\left(\frac{3\pi x}{a}\right)) \right] dx. \] Compute this integral to find the matrix element.
06

Compute Transition Probability

The probability of transition, according to first-order perturbation theory, is given by: \[ P_{1 \to 2} = \left| \frac{\langle \psi_2 | H' | \psi_1 \rangle}{E_2 - E_1} \right|^2 \sin^2\left( \frac{(E_2 - E_1)T}{2\hbar} \right), \] where \( E_2 = \frac{4\hbar^2\pi^2}{2ma^2} \). Substitute the values obtained from previous steps to find the transition probability.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Infinite Square Well
The infinite square well is a foundational concept in quantum mechanics. It describes a particle confined to a box with infinitely high walls, meaning the particle cannot escape. The box has a perfect 'trap' which imposes boundary conditions on the particle's wave function. Inside the well, the potential energy is zero, while it's infinite outside.
  • Wave function inside: Nodes at walls and known forms like sine waves.
  • Energy levels: Quantized and discrete, given by the formula \( E_n = \frac{n^2\pi^2\hbar^2}{2ma^2} \).
  • Ground state: The lowest energy state, given by \( n=1 \).
The infinite square well problem is solved using these quantized solutions, exemplifying the core principles of quantization in quantum mechanics.
Perturbation Theory
Perturbation theory helps us understand how small changes in a system鈥攍ike introducing a weak potential鈥攁ffect its behavior without solving complex equations from scratch. In this problem, when a 'brick' potential is introduced, it is considered a small alteration to the infinite square well:
  • Small perturbation: The change is minor compared to the ground state's energy \( E_1 \).
  • Time-dependent perturbations: Allows us to estimate effects like transitions between energy states over time.
  • Matrix elements: Calculated to find the expected influence of the perturbation between states.
Using perturbation theory, we assess how the energy and states are affected by this small popping in of potential without excessively complex recalculations.
Transition Probability
Transition probability is crucial in determining how likely it is for a particle to move from one state to another due to perturbations. For a particle initially in a ground state \( E_1 \) transitioning to an excited state \( E_2 \), it's defined as:\[P_{1 \to 2} = \left| \frac{\langle \psi_2 | H' | \psi_1 \rangle}{E_2 - E_1} \right|^2 \sin^2\left( \frac{(E_2 - E_1)T}{2\hbar} \right)\]
  • Probability rises when \( T \) (interaction time) and matrix element \( \langle \psi_2 | H' | \psi_1 \rangle \) are significant.
  • Energy difference: Transition depends on the gap between \( E_2 \) and \( E_1 \).
This concept is valuable for predicting system behaviors when small perturbations alter their quantum states.
Wave Function
Wave functions are fundamental in quantum mechanics to understand a particle's state and properties. In the infinite square well, wave functions are quantized and specific sine functions:
  • Form: For the particle in different states, given by \( \psi_n(x) = \sqrt{\frac{2}{a}} \sin\left(\frac{n\pi x}{a}\right) \).
  • Normalization: Ensures that the probability of finding the particle in the well sums to one.
  • State Representation: Designates probabilities and expected behaviors of a particle within a defined system.
Especially under perturbation theory, wave functions determine how the system's states and energies evolve in response to external influences.

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Most popular questions from this chapter

The half-life \(\left(t_{1 / 2}\right)\) of an excited state is the time it would take for half the atoms in a large sample to make a transition. Find the relation between \(t_{1 / 2}\) and \(\tau\) (the "lifetime" of the state).

As a mechanism for downward transitions, spontaneous emission competes with thermally stimulated emission (stimulated emission for which Planck radiation is the source). Show that at room temperature \((T=300 \mathrm{~K})\) thermal stimulation dominates for frequencies well below \(5 \times 10^{12} \mathrm{~Hz}\), whereas spontaneous emission dominates for frequencies well above \(5 \times 10^{12} \mathrm{~Hz}\). Which mechanism dominates for visible light?

A hydrogen atom is placed in a (time-dependent) electric field \(\mathbf{E}=\) \(E(t) \hat{k} .\) Calculate all four matrix elements \(H_{i j}^{\prime}\) of the perturbation \(H^{\prime}=-e E z\) between the ground state \((n=1)\) and the (quadruply degenerate) first excited states \((n=2)\). Also show that \(H_{i i}^{\prime}=0\) for all five states. Note: There is only one integral to be done here, if you exploit oddness with respect to \(z\). As a result, only one of the \(n=2\) states is "accessible" from the ground state by a perturbation of this form, and therefore the system functions as a two-level configuration- assuming transitions to higher excited states can be ignored.

We have encountered stimulated emission, (stimulated) absorption, and spontaneous emission ... how come there's no such thing as spontaneous \(a b\). sorption?
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