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Chapter 2: Problem 39

Solve the time-independent Schr枚dinger equation for an infinite square well with a delta-function barrier at the center: $$ V(x)=\left\\{\begin{array}{ll} \alpha \delta(x), & \text { for }(-a

Short Answer

Expert verified
The delta function modifies even solutions but not odd. Energy changes for even states; odd states remain unchanged. In the limit \(\alpha \rightarrow 0\), energies match the simple well; for \(\alpha \rightarrow \infty\), the well splits into two.

Step by step solution

01

Understand the Problem Setup

The problem describes a particle in an infinite potential well with an additional delta-function barrier at the center. We need to solve the time-independent Schr枚dinger equation for this system, treating the even and odd solutions separately, and find the allowed energy levels.
02

Write Down the Schr枚dinger Equation

The time-independent Schr枚dinger equation is given by \( -\frac{\hbar^2}{2m}\frac{d^2 \psi(x)}{dx^2} + V(x)\psi(x) = E\psi(x) \), where \( V(x) = \alpha \delta(x) \) within \( -a < x < a \) and \( V(x) = \infty \) outside this range.
03

Consider Boundary Conditions

Since \( V(x) = \infty \) outside \( -a < x < a \), the wave function \( \psi(x) \) must be zero for \( |x| \geq a \). At \( x = 0 \), we need to consider the effect of the delta function, which causes a discontinuity in the derivative of \( \psi(x) \).
04

Solve for Even Solutions

For the even solutions, assume \( \psi(x) = A \cos(kx) \) for \( -a < x < a \). The boundary condition at the center imposed by the delta function is \( \frac{d \psi}{dx}\Big|_{0^+} - \frac{d \psi}{dx}\Big|_{0^-} = \frac{2m\alpha}{\hbar^2} \psi(0) \). This leads to \( k \tan(ka) = \frac{m\alpha}{\hbar^2 k} \).
05

Solve for Odd Solutions

For the odd solutions, assume \( \psi(x) = B \sin(kx) \) for \( -a < x < a \). The odd solutions automatically satisfy \( \psi(0) = 0 \), so there is no contribution from the delta function. This simplifies to \( k \cot(ka) = 0 \), leading to \( k = \frac{n\pi}{a} \) for \( n = 1, 2, 3, \ldots \).
06

Determine Energy Levels

For the even solutions, the equation \( k \tan(ka) = \frac{m\alpha}{\hbar^2 k} \) must be solved graphically or numerically, while for the odd solutions, the energy levels are simply \( E_n = \frac{n^2\pi^2\hbar^2}{2ma^2} \). In the absence of the delta function, all solutions are of the form \( E_n = \frac{n^2\pi^2\hbar^2}{2ma^2} \) regardless of parity.
07

Analyze Limiting Cases

For \( \alpha \rightarrow 0 \), the effect of the delta function vanishes, and the solutions reduce to those of a simple infinite square well. For \( \alpha \rightarrow \infty \), the delta function imposes an infinite barrier at \( x = 0 \), splitting the potential into two separate wells \((0, a)\) and \((-a, 0)\), with each having its own set of quantized energy levels.

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Key Concepts

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

Schr枚dinger Equation
The Schr枚dinger Equation is a fundamental equation in quantum mechanics that describes how the quantum state of a physical system changes over time. In the context of our problem, we deal with the time-independent version of the equation, which is used to determine the stationary states of a system. These are states in which the probability distribution associated with the wave function doesn鈥檛 change over time. The time-independent Schr枚dinger equation is expressed as:
  • \[ -\frac{\hbar^2}{2m}\frac{d^2 \psi(x)}{dx^2} + V(x)\psi(x) = E\psi(x) \]
Here, \(\psi(x)\) represents the wave function of the particle, \(V(x)\) is the potential energy, \(E\) is the energy of the particle, \(\hbar\) is the reduced Planck鈥檚 constant, and \(m\) is the particle's mass. Solving this equation provides us with the wave functions (\(\psi\)) and associated energies (\(E\)) for the quantum states of a system enclosed by specific boundaries, such as in an infinite square well or those modified by additional potential barriers like a delta function. These solutions are critical for understanding the behavior of quantum particles.
Infinite Square Well
The Infinite Square Well is a classic problem in quantum mechanics that helps illustrate fundamental concepts. Imagine a particle confined to a 'well' with infinitely high walls. This means the particle has zero potential energy inside the well, and cannot exist outside of it. In mathematical terms, this potential is defined as:
  • \[ V(x) = \begin{cases} 0, & \text{if } -a < x < a \ \infty, & \text{otherwise} \end{cases} \]
The walls at \(x = -a\) and \(x = +a\) confine the particle, allowing it to exist only within this interval. At the boundary, the wave function (\(\psi(x)\)) must drop to zero, representing that the particle has no probability of being found there. In our exercise, this well is modified by introducing a delta function barrier at the center. This setup leads us to solve for even and odd wave functions separately, with each type subject to unique boundary conditions due to the symmetric nature and additional delta function potential at \(x = 0\).
Delta Function Potential
A Delta Function Potential introduces a specific type of discontinuity to the system's potential energy profile. It is represented by a spike of infinite height and zero width, yet is localized to a single point with finite area when integrated. Mathematically, it is expressed using the Dirac delta function \(\delta(x)\), as seen in this exercise:
  • \[ V(x) = \alpha \delta(x) \text{ for } (-a < x < a) \]
The parameter \(\alpha\) determines the strength or 'height' of this potential spike. This potential impacts the system by introducing a discontinuity in the derivative of the wave function at the point where the delta function is located. As it affects only one point, it doesn't directly alter the form of the wave function, but it imposes conditions on its derivatives. The influence of the delta function becomes evident in its limiting cases: as \(\alpha\) approaches zero or infinity, altering how the particle is confined within the well.
Wave Function Solutions
Wave Function Solutions describe the allowed states of a quantum particle within a potential well. For the infinite square well with a delta function at the center, these solutions are split into even and odd functions due to the symmetry of the problem:
  • **Even Solutions:** These can be expressed using cosine functions: \(\psi(x) = A \cos(kx)\). The boundary condition introduced by the delta function requires a special condition for these states, leading to the equation \(k \tan(ka) = \frac{m\alpha}{\hbar^2 k}\) to determine the allowed wave frequencies \(k\) and subsequently the energies.
  • **Odd Solutions:** Expressed as sine functions: \(\psi(x) = B \sin(kx)\), these automatically satisfy the condition \(\psi(0) = 0\), simplifying the problem to standard conditions of the square well without modification: \(k = \frac{n\pi}{a}\).
Each solution, even or odd, corresponds to a specific energy level given by \(E = \frac{\hbar^2 k^2}{2m}\). Understanding how these solutions differ with and without the delta function helps visualize the impact of potential barriers and symmetries in quantum systems.

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Most popular questions from this chapter

Consider the double delta-function potential $$ V(x)=-\alpha[\delta(x+a)+\delta(x-a)] \text { . } $$ where \(\alpha\) and \(a\) are positive constants. (a) Sketch this potential. (b) How many bound states does it possess? Find the allowed energies, for \(\alpha=\) \(\hbar^{2} / m a\) and for \(\alpha=\hbar^{2} / 4 m a\), and sketch the wave functions.

Find the allowed energies of the half-harmonic oscillator $$ V(x)=\left\\{\begin{array}{ll} (1 / 2) m \omega^{2} x^{2}, & \text { for }(x>0) \\ \infty, & \text { for }(x<0) \end{array}\right. $$ (This represents, for example, a spring that can be stretched, but not compressed.) Hint: This requires some careful thought, but very little actual computation.

In this problem we explore some of the more useful theorems (stated without proof) involving Hermite polynomials. (a) The Rodrigues formula states that $$ H_{n}(\xi)=(-1)^{n} e^{\xi^{2}}\left(\frac{d}{d \xi}\right)^{n} e^{-\xi^{2}} $$ Use it to derive \(H_{3}\) and \(H_{4}\).

A free particle has the initial wave function $$ \Psi(x, 0)=A e^{-a x^{2}} $$ where \(A\) and \(a\) are constants \((a\) is real and positive). (a) Normalize \(\Psi(x, 0)\). (b) Find \(\Psi(x, t)\). Hint: Integrals of the form $$ \int_{-\infty}^{+\infty} e^{-\left(a x^{2}+b x\right)} d x $$ can be handled by "completing the square." Let \(y \equiv \sqrt{a}[x+(b / 2 a)]\), and note that \(\left(a x^{2}+b x\right)=y^{2}-\left(b^{2} / 4 a\right)\). Answer: $$ \Psi(x, t)=\left(\frac{2 a}{\pi}\right)^{1 / 4} \frac{e^{\left.-a x^{2} / 11+(2 i \hbar a t / m)\right]}}{\sqrt{1+(2 i \hbar a t / m)}} $$ (c) Find \(|\Psi(x, t)|^{2}\). Express your answer in terms of the quantity \(w \equiv\) for some very large \(t\). Qualitatively, what happens to \(|\Psi|^{2}\) as time goes on? (d) Find \(\langle x\rangle,\langle p\rangle,\left(x^{2}\right\rangle,\left\langle p^{2}\right\rangle, \sigma_{x}\), and \(\sigma_{p} .\) Partial answer: \(\left\langle p^{2}\right\rangle=a \hbar^{2}\), but it may take some algebra to reduce it to this simple form. (e) Does the uncertainty principle hold? At what time \(t\) does the system come closest to the uncertainty limit?

This problem is designed to guide you through a "proof" of Plancherel's theorem, by starting with the theory of ordinary Fourier series on a finite interval, and allowing that interval to expand to infinity. (a) Dirichlet's theorem says that "any" function \(f(x)\) on the interval \([-a,+a]\) can be expanded as a Fourier series: $$ f(x)=\sum_{n=0}^{\infty}\left[a_{n} \sin (n \pi x / a)+b_{n} \cos (n \pi x / a)\right] . $$ Show that this can be written equivalently as $$ f(x)=\sum_{n=-\infty}^{\infty} c_{n} e^{i n x x / a} $$ What is \(c_{n}\), in terms of \(a_{n}\) and \(b_{n}\) ? (b) Show (by appropriate modification of Fourier's trick) that $$ c_{n}=\frac{1}{2 a} \int_{-a}^{+a} f(x) e^{-i n \pi x / a} d x $$ (c) Eliminate \(n\) and \(c_{n}\) in favor of the new variables \(k=(n \pi / a)\) and \(F(k)=\) \(\sqrt{2 / \pi} a c_{n} .\) Show that (a) and (b) now become $$ f(x)=\frac{1}{\sqrt{2 \pi}} \sum_{n=-\infty}^{\infty} F(k) e^{i k x} \Delta k ; \quad F(k)=\frac{1}{\sqrt{2 \pi}} \int_{-a}^{+a} f(x) e^{-i k x} d x $$ where \(\Delta k\) is the increment in \(k\) from one \(n\) to the next. (d) Take the limit \(a \rightarrow \infty\) to obtain Plancherel's theorem. Note: In view of their quite different origins, it is surprising (and delightful) that the two formulas [one for \(F(k)\) in terms of \(f(x)\), the other for \(f(x)\) in terms of \(F(k)]\) have such a similar structure in the limit \(a \rightarrow \infty\).
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