Question

Classically, if a particle is not observed, the probability per unit length of finding it in a box is a constant \(1 / L\) along the entire length of the box. With this, show that the classical expectation value of the position is \(\frac{1}{2} L\), that the expectation value of the square of the position is \(\frac{1}{3} L^{2}\). and that the uncertainty in position is \(L / \sqrt{12}\).

Short answer

The expectation value of the position is \(\frac{1}{2} L\), the expectation value of the square of the position is \(\frac{1}{3} L^{2}\), and the uncertainty in position is \(\frac{L}{\sqrt{12}}\).

Step-by-step solution

Expectation Value of Position

The expectation value of a variable is its average value in terms of probability. The expectation value of the position \(x\) is given by \(\langle x \rangle = \int x p(x) dx\), where \(p(x) = 1/L\) is the constant probability, and \(x\) ranges from \(0\) to \(L\). So, \(\langle x \rangle = \int_{0}^{L} x \cdot \frac{1}{L} dx = \frac{1}{L}[\frac{1}{2}x^2]_{0}^{L} = \frac{1}{2}L\).

Expectation Value of Square of Position

The expectation value of the square of the position is given by \(\langle x^2 \rangle = \int x^2 p(x) dx\), where \(p(x) = 1/L\), and \(x\) ranges from \(0\) to \(L\). So, \(\langle x^2 \rangle = \int_{0}^{L} x^2 \cdot \frac{1}{L} dx = \frac{1}{L}[\frac{1}{3}x^3]_{0}^{L} = \frac{1}{3}L^2\).

Uncertainty in Position

The uncertainty in a variable is given by the square root of the difference between the expectation value of the square of the variable and the square of the expectation value of the variable. So, the uncertainty in position is given by \(\sqrt{\langle x^2 \rangle - \langle x \rangle^2} = \sqrt{\frac{1}{3}L^2 - (\frac{1}{2}L)^2} = \sqrt{\frac{1}{12}}L = \frac{L}{\sqrt{12}}.\)

Key concepts

Expectation Value of Position

The concept of an expectation value is central to understanding probability distributions in classical mechanics. It represents the average or most probable value of a given quantity based on its probability distribution. In the context of a particle's position in a box, assuming a uniform distribution, the expectation value indicates where we are most likely to find the particle on average. To calculate the expectation value of position, \, for a particle inside a box with length L, where the probability is evenly distributed, we integrate the position x multiplied by the probability density function p(x).Let's start off with the basic equation for the expectation value of position: \[\langle x \rangle = \int x p(x) dx\].In our scenario, the probability density, p(x), is constant, signifying that the particle has an equal chance of being anywhere within the box, leading to \[p(x) = \frac{1}{L}\].Integrating from 0 to L yields: \[\langle x \rangle = \int_{0}^{L} x \cdot \frac{1}{L} dx = \frac{1}{L}\left[\frac{1}{2}x^2\right]_{0}^{L} = \frac{1}{2}L\].As a result, the expectation value of position \ in a uniformly distributed system is exactly at the midpoint of the box, \ = \frac{1}{2}L, which intuitively makes sense as the average of positions distributed continuously and evenly from 0 to L.

Uncertainty in Position

Uncertainty is a fundamental concept in physics, reflecting the limitations in our knowledge about a system's state. It is even more prominent in quantum mechanics but holds relevance in classical contexts as well. The uncertainty in position, denoted as \<\Delta x\>, quantifies how spread out the possible values of a particle's position are from the expectation value. It indicates the precision of a particle’s location within the system.The formula to calculate the uncertainty in position for such a system is as follows: \[\Delta x = \sqrt{\langle x^2 \rangle - \langle x \rangle^2}\].Using the results from the previous steps where \<\langle x^2 \rangle = \frac{1}{3}L^2\> and \<\langle x \rangle = \frac{1}{2}L\>, the uncertainty in position is computed by taking the square root of the difference between these two expectation values:\[\Delta x = \sqrt{\frac{1}{3}L^2 - \left(\frac{1}{2}L\right)^2} = L\sqrt{\frac{1}{12}} = \frac{L}{\sqrt{12}}\].This expresses a measure of the dispersion of a particle's position in the box, highlighting that despite a uniform distribution, there is still a quantifiable spread in its possible positions.

Probability Distributions in Physics

In classical mechanics and other areas of physics, probability distributions describe the spread of different possible outcomes of a system or an experiment. Rather than providing a single predicted result, they give a range of potential outcomes along with the probabilities of each outcome occurring. Probability distributions give valuable insights into phenomena where determinism falls short, accommodating for variations and uncertainties inherent in real-life measurements.The uniform probability distribution encountered in the box problem is just one example of how these distributions are represented mathematically. The uniform distribution assigns equal probability to all outcomes in a specified range, indicative of a situation with maximal uncertainty since no position within the box is favored over another. However, different situations might call for other types of distributions, such as Gaussian distributions, which represent many natural phenomena and processes.Understanding probability distributions is crucial when predicting outcomes. The ability to calculate expectation values and uncertainties is an essential skill for scientists and engineers as it quantifies both the average outcome and the reliability of this prediction. Familiarizing oneself with various probability distributions can improve one's ability to model and understand complex systems across different domains of physics and beyond.