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Consider a particle in a two-dimensional potential $$ V_{0}= \begin{cases}0 & \text { for } 0 \leq x \leq L, 0 \leq y \leq L \\\ \infty & \text { otherwise }\end{cases} $$ Write the energy eigenfunctions for the ground and first excited states. We now add a time-independent perturbation of the form $$ V_{1}= \begin{cases}\lambda x y & \text { for } 0 \leq x \leq L, 0 \leq y \leq L \\ 0 & \text { otherwise }\end{cases} $$ Obtain the zeroth-order energy eigenfunctions and the first-order energy shifts for the ground and first excited states.

Short Answer

Expert verified
The zeroth-order eigenfunctions remain unchanged, and first-order energy shifts for both ground and first excited states are zero.

Step by step solution

01

Identify the system and formulate the boundary conditions

The problem presents a two-dimensional infinite potential well (or box). The particle is free within the region defined by \(0 \leq x \leq L\) and \(0 \leq y \leq L\). Outside this region, the potential is infinite, effectively confining the particle. The Schrödinger equation in this region is given by \(-\frac{\hbar^2}{2m}abla^2 \psi(x,y) = E \psi(x,y)\) with boundary conditions \(\psi(0, y) = \psi(L, y) = \psi(x, 0) = \psi(x, L) = 0\).
02

Write the energy eigenfunctions for the unperturbed system

The eigenfunctions \(\psi(x, y)\) for a 2D box are given by the product of sine functions, \(\psi_{n_x, n_y}(x, y) = \sqrt{\frac{4}{L^2}} \sin\left(\frac{n_x \pi x}{L}\right) \sin\left(\frac{n_y \pi y}{L}\right)\), where \(n_x, n_y = 1, 2, 3, \ldots\). The respective energy eigenvalues are \(E_{n_x, n_y} = \frac{\hbar^2 \pi^2}{2mL^2} (n_x^2 + n_y^2)\).
03

Find the ground and first excited states

The ground state corresponds to \(n_x = 1\) and \(n_y = 1\), giving \(\psi_{1,1}(x, y) = \sqrt{\frac{4}{L^2}} \sin\left(\frac{\pi x}{L}\right) \sin\left(\frac{\pi y}{L}\right)\) with energy \(E_{1,1} = \frac{\hbar^2 \pi^2}{mL^2}\). The first excited state can be degenerate, with either \(n_x = 2, n_y = 1\) or \(n_x = 1, n_y = 2\). Thus, \(\psi_{2,1}(x, y) = \sqrt{\frac{4}{L^2}} \sin\left(\frac{2 \pi x}{L}\right) \sin\left(\frac{\pi y}{L}\right)\) and \(\psi_{1,2}(x, y) = \sqrt{\frac{4}{L^2}} \sin\left(\frac{\pi x}{L}\right) \sin\left(\frac{2\pi y}{L}\right)\) with energies \(E_{2,1} = E_{1,2} = \frac{5\hbar^2 \pi^2}{2mL^2}\).
04

Introduce the perturbation and find first-order energy shifts

The time-independent perturbation is given by \(V_1 = \lambda xy\). The first-order energy shift for a state \(\psi_{n_x, n_y}\) is calculated as \(E^{(1)}_{n_x, n_y} = \langle \psi_{n_x, n_y} | V_1 | \psi_{n_x, n_y} \rangle = \int_0^L \int_0^L \psi_{n_x, n_y}^*(x, y) \lambda xy \psi_{n_x, n_y}(x, y) \, dx \, dy\). For the ground state \(n_x = 1\) and \(n_y = 1\), this integral evaluates to zero because the integral of odd functions over symmetric limits results in zero. Thus, \(E^{(1)}_{1,1} = 0\). For the first excited state, they are cross terms, so it will also be zero due to symmetry and properties of sine functions, \(E^{(1)}_{2,1} = E^{(1)}_{1,2} = 0\).
05

Conclusion on energy shifts and state functions

The zeroth-order eigenfunctions are the same as found in the unperturbed system: \(\psi_{1,1}\) for the ground state and \(\psi_{2,1}, \psi_{1,2}\) for the first excited state. Due to the zero result of the first-order eigenvalue corrections from this specific perturbation, the energy levels remain unshifted in first-order perturbation theory.

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Key Concepts

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

Two-Dimensional Potential Well
A two-dimensional potential well is a concept in quantum mechanics that represents a confined region where a particle can exist. Within this well, the particle is free to move, but at the boundaries, the potential suddenly becomes infinitely large, essentially trapping the particle inside. Imagine it like a square box with infinitely high walls. The important part about this potential well is that it defines a region between specific limits: \(0 \leq x \leq L\) and \(0 \leq y \leq L\). Outside these boundaries, the particle cannot exist because the potential is infinite. This conceptual box forces the particle to behave in certain ways, affecting how we describe it mathematically, especially in terms of its wave functions and energy states.
Energy Eigenfunctions
In a two-dimensional potential well, the energy eigenfunctions describe the state of a particle within this confined space. These eigenfunctions are solutions to the Schrödinger equation, and they tell us how a particle behaves.For a two-dimensional box, the eigenfunctions take the form of sine waves that show how the particle's probability density is distributed. They are given by:
  • \(\psi_{n_x, n_y}(x, y) = \sqrt{\frac{4}{L^2}} \sin\left(\frac{n_x \pi x}{L}\right) \sin\left(\frac{n_y \pi y}{L}\right)\)
  • Where \(n_x\) and \(n_y\) are integers called quantum numbers.
The energy levels of the particle are associated with these functions, and higher quantum numbers correspond to higher energy states. The ground state, normally the lowest energy state, has \(n_x = 1\) and \(n_y = 1\), while first excited states have slightly higher quantum numbers. The energy levels arise from the kinetic energy of a particle trapped in this kind of potential.
Perturbation Theory
Perturbation theory is a method used in quantum mechanics to handle small changes in a system. Imagine setting up our well-defined system and then adding a small bump (or perturbation) to it. In our context of a two-dimensional potential well, perturbation theory helps us understand how a small additional potential, like \(V_{1} = \lambda xy\), affects the energy levels. The perturbation is usually small, enabling us to treat it as a correction rather than creating a completely new problem. The primary goal here is to find the change, known as the first-order energy shift, which we calculate using:
  • \(E^{(1)}_{n_x, n_y} = \langle \psi_{n_x, n_y} | V_1 | \psi_{n_x, n_y} \rangle \)
This involves integrating over the region of interest, showing how much this perturbation affects previously known states. If the energy shift is zero, the perturbation doesn't affect the energy levels of those states.
Infinite Potential Well
The infinite potential well is a fundamental concept in quantum mechanics. It describes a system where a particle is trapped in a region with zero potential energy, while outside this region, the potential energy is infinite. This creates an effective boundary where the wave function of a particle must go to zero, because it cannot exist where the potential is infinite. It's a perfect example of ideal confinement. This kind of well is incredibly useful for understanding quantum behaviors, as it allows us to start with a simple model before dealing with more complex potentials. In calculations, this helps us derive wave functions and understand particle arrangements and behaviors in quantum systems.
Schrödinger Equation
The Schrödinger equation is the king of equations when it comes to quantum mechanics. It’s a differential equation that describes how the quantum state of a system changes over time. In a two-dimensional potential well, we use the time-independent Schrödinger equation to solve for the energy eigenfunctions, which are key to understanding particle behavior. Written as:
  • \(-\frac{\hbar^2}{2m} abla^2 \psi(x,y) = E \psi(x,y)\)
Where \(\hbar\) is the reduced Planck's constant and \(m\) is the particle's mass.The equation is solved by applying boundary conditions relevant to the potential well—meaning we ensure the wave function is zero at the boundaries—helping us find valid solutions that describe real-world phenomena. Understanding this equation is fundamental for knowing how particles exist and evolve in quantum mechanical systems.

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Most popular questions from this chapter

In nondegenerate time-independent perturbation theory, what is the probability of finding in a perturbed energy eigenstate \((|k\rangle)\) the corresponding unperturbed eigenstate \(\left(\left|k^{(0)}\right\rangle\right)\) ? Solve this up to terms of order \(\lambda^{2}\).

A diatomic molecule can be modeled as a rigid rotor with moment of inertia \(I\) and an electric dipole moment \(d\) along the axis of the rotor. The rotor is constrained to rotate in a plane, and a weak uniform electric field \(\mathscr{E}\) lies in the plane. Write the classical Hamiltonian for the rotor, and find the unperturbed energy levels by quantizing the angular- momentum operator. Then treat the electric field as a perturbation, and find the first nonvanishing corrections to the energy levels.

Compute the Stark effect for the \(2 s_{1 / 2}\) and \(2 p_{1 / 2}\) levels of hydrogen for a field \(\mathscr{E}\) sufficiently weak so that \(e \mathscr{E} a_{0}\) is small compared to the fine structure, but take the Lamb \(\operatorname{shift} \delta(\delta=1057 \mathrm{MHz})\) into account (that is, ignore \(2 p_{3 / 2}\) in this calculation). Show that for \(e \mathscr{E} a_{0} \ll \delta\), the energy shifts are quadratic in \(\mathscr{E}\), whereas for \(e \mathscr{E} a_{0} \gg \delta\) they are linear in \(\mathscr{E}\). Briefly discuss the consequences (if any) of time reversal for this problem. This problem is from Gottfried (1966), Problem 7-3.

Consider a composite system made up of two spin \(\frac{1}{2}\) objects. For \(t<0\), the Hamiltonian does not depend on spin and can be taken to be zero by suitably adjusting the energy scale. For \(t>0\), the Hamiltonian is given by $$ H=\left(\frac{4 \Delta}{\hbar^{2}}\right) \mathbf{S}_{1} \cdot \mathbf{S}_{2} $$ Suppose the system is in \(|+-\rangle\) for \(t \leq 0\). Find, as a function of time, the probability for being found in each of the following states \(|++\rangle,|+-\rangle,|-+\rangle\), and \(|--\rangle\). a. By solving the problem exactly. b. By solving the problem assuming the validity of first-order time-dependent perturbation theory with \(H\) as a perturbation switched on at \(t=0\). Under what condition does (b) give the correct results?

A one-electron atom whose ground state is nondegenerate is placed in a uniform electric field in the \(z\)-direction. Obtain an approximate expression for the induced electric dipole moment of the ground state by considering the expectation value of ez with respect to the perturbed state vector computed to first order. Show that the same expression can also be obtained from the energy shift \(\Delta=-\alpha|\mathbf{E}|^{2} / 2\) of the ground state computed to second order. (Note: \(\alpha\) stands for the polarizability.) Ignore spin.

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