Chapter 35: Problem 52

A particle is in the $n$ th quantum state of an infinite square well. (a) Show that the probability of finding it in the left-hand quarter of the well is $$ P=\frac{1}{4}-\frac{\sin (n \pi / 2)}{2 n \pi} $$ (b) Show that for odd $n$, the probability approaches the classical value $\frac{1}{4}$ as $n \rightarrow \infty$

Short Answer

Expert verified

The probability of finding the particle in the left quarter is $ P=\frac{1}{4}-\frac{\sin (n \pi / 2)}{2 n \pi} $. If $ n $ is an odd number and tends to infinity, this probability approaches $\frac{1}{4}$.

Step by step solution

Initialize Variables

Initialize quantum state $n$,and the width of the box $L$. We will also use $x$ coordinate to work through probability density function.

Calculate Wave Function

The wave function of a particle in an infinite square well, normalized, is given by: $\psi (x) = \sqrt{\frac{2}{L}} \sin\left(\frac{n\pi x}{L}\right)$ for $0 < x < L$. The wave function is a solution to the Schrödinger equation.

Calculate Probability Density

The probability density is now calculated as the absolute square of the wave function: $P(x) = |\psi(x)|^2 = \frac{2}{L} \sin^2\left(\frac{n\pi x}{L}\right)$

Integrate over Desired Area

To find the probability of finding the particle in the left-hand quarter of the well, integrate the probability density from $0$ to $L/4$: $P = \int_{0}^{L/4} P(x) dx = \int_{0}^{L/4} \left(\frac{2}{L} \sin^2\left(\frac{n\pi x}{L}\right)\right) dx$ After calculating the integral, we get $ P=\frac{1}{4}-\frac{\sin(n \pi / 2)}{2n\pi} $

Evaluate Limt as $n \rightarrow \infty$

We are asked to show that for odd $n$, the probability approaches the classical value $\frac{1}{4}$ as $n \rightarrow \infty$. If $n$ is odd, we can write it as $n=2m+1$ where $m$ is an integer. The term $\sin((2m+1)\pi/2)$ becomes $-1^\frac{m+1}{2}$. And as $m \rightarrow \infty$ the term $\frac{\sin(n \pi / 2)}{2n\pi}$ is basically a small oscillating value that averages to zero in the limit. So in the limit $n \rightarrow \infty$, $ P=\frac{1}{4}-\frac{\sin(n \pi / 2)}{2n\pi} =\frac{1}{4} $

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Short Answer

Step by step solution

Initialize Variables

Calculate Wave Function

Calculate Probability Density

Integrate over Desired Area

Evaluate Limt as \(n \rightarrow \infty\)

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