Question

A particle is confined to the one-dimensional space \(0 \leq x \leq a\) and classically it can be in any small interval \(d x\) with equal probability. However, quantum mechanics gives the result that the probability distribution is proportional to \(\sin ^{2}(n \pi x / a)\), where \(n\) is an integer. Find the variance in the particle's position in both the classical and quantum mechanical pictures and show that, although they differ, the latter tends to the former in the limit of large \(n\), in agreement with the correspondence principle of physics.

Short answer

Classical variance: \(\frac{a^2}{12}\). Quantum variance: \(\frac{a^2}{12} - \frac{a^2}{2 n^2 \pi^2}\). As \(n\) tends to large values, quantum variance approaches classical variance.

Step-by-step solution

Classical Probability Distribution

In the classical case, the particle is equally likely to be found anywhere in the interval \(0 \leq x \leq a\). Hence, the probability density function is constant \(P(x) = \frac{1}{a}\).

Mean Position (Classical)

The mean position \(\langle x \rangle\) is calculated using the uniform distribution: \[\langle x \rangle = \int_0^a x P(x) \, dx = \int_0^a x \frac{1}{a} \, dx = \frac{1}{a} \cdot \frac{x^2}{2} \Bigg|_0^a = \frac{a}{2}\].

Variance (Classical)

The variance of position \(\sigma^2\) is calculated using: \[\sigma^2 = \langle x^2 \rangle - \langle x \rangle^2\]. First, calculate \(\langle x^2 \rangle\): \[\langle x^2 \rangle = \int_0^a x^2 P(x) \, dx = \int_0^a x^2 \frac{1}{a} \, dx = \frac{1}{a} \cdot \frac{x^3}{3} \Bigg|_0^a = \frac{a^2}{3}\]. Then, \[\sigma^2_{\text{classical}} = \frac{a^2}{3} - \left(\frac{a}{2}\right)^2 = \frac{a^2}{3} - \frac{a^2}{4} = \frac{a^2}{12}\].

Quantum Probability Distribution

In the quantum case, the probability distribution function is \(P(x) = C \sin^2\left(\frac{n \pi x}{a}\right)\), where \(C\) is the normalization constant.

Normalization Constant

To find the normalization constant \(C\), solve: \[\int_0^a P(x) \, dx = 1\] which gives: \[C \int_0^a \sin^2\left(\frac{n \pi x}{a}\right) \, dx = 1\]. Using \(\int_0^a \sin^2\left(\frac{n \pi x}{a}\right) \, dx = \frac{a}{2}\), we find: \[C \cdot \frac{a}{2} = 1 \implies C = \frac{2}{a}\]. Thus, \(P(x) = \frac{2}{a} \sin^2\left(\frac{n \pi x}{a}\right)\).

Mean Position (Quantum)

Calculate the mean position \(\langle x \rangle\) for the quantum case: \[\langle x \rangle = \int_0^a x P(x) \, dx = \int_0^a x \frac{2}{a} \sin^2\left(\frac{n \pi x}{a}\right) \, dx\]. This integral evaluates to: \[\langle x \rangle = \frac{a}{2}\].

Variance (Quantum)

First, calculate \(\langle x^2 \rangle\): \[\langle x^2 \rangle = \int_0^a x^2 P(x) \, dx = \int_0^a x^2 \frac{2}{a} \sin^2\left(\frac{n \pi x}{a}\right) \, dx\]. This evaluates to: \[\langle x^2 \rangle = \frac{a^2}{3} - \frac{a^2}{2 n^2 \pi^2}\]. Then, \[\sigma^2_{\text{quantum}} = \frac{a^2}{3} - \frac{a^2}{2 n^2 \pi^2} - \left(\frac{a}{2}\right)^2 = \frac{a^2}{3} - \frac{a^2}{2 n^2 \pi^2} - \frac{a^2}{4} = \frac{a^2}{12} - \frac{a^2}{2 n^2 \pi^2}\].

Correspondence Principle

In the limit of large \(n\), the term \(\frac{a^2}{2 n^2 \pi^2}\) tends to zero. Thus, \[\sigma^2_{\text{quantum}} \approx \frac{a^2}{12}\], which is the same as the classical result, demonstrating the correspondence principle.

Key concepts

Classical Probability Distribution

In the classical case, our particle can be found at any position within the given interval \(0 \leq x \leq a\) with equal probability. This means that the particle's position follows a uniform probability distribution. The probability density function \(P(x)\) for a uniform distribution is a constant value since any position is equally likely.

For our particle, the classical probability density is expressed as:
\[ P(x) = \frac{1}{a} \]
This equation indicates that the probability is uniformly distributed over the length of the interval. Thus, if you select any small segment \( dx \) within the interval, the likelihood of finding the particle in that segment is simply the width of the segment divided by the total interval length \(a\).

Quantum Probability Distribution

When we move from classical to quantum mechanics, the picture changes. A quantum particle does not have an equal probability of being found at all positions. Instead, its probability distribution is influenced by wave-like behavior.

For our particle, confined within the interval \(0 \leq x \leq a\), the quantum probability distribution is given by the function:
\[ P(x) = C \sin^2\left(\frac{n \pi x}{a}\right) \]
Here, \( n \) is an integer representing the quantum state, and \( C \) is the normalization constant that ensures the total probability over the interval equals one (i.e., the particle must be found somewhere within the interval).

Correspondence Principle

The correspondence principle states that the predictions of quantum mechanics must align with classical physics in the limit of large quantum numbers. In simple terms, this means that for very high energy states or large quantum numbers \( n \), the behavior of a quantum system should resemble that of a classical system.

In our exercise, as \( n \) increases, the quantum variance in the particle's position \( \sigma^2_{\text{quantum}} \) approaches the classical variance \( \sigma^2_{\text{classical}} \). Specifically, the term \( \frac{a^2}{2 n^2 \pi^2} \) in the quantum variance equation tends to zero for large \( n \), leading to \( \sigma^2_{\text{quantum}} \approx \frac{a^2}{12} \). Thus, the quantum variance becomes practically identical to the classical variance, demonstrating the correspondence principle in action.

Variance Calculation

Variance is a measure of how spread out a set of values is around the mean. It provides insight into the distribution of positions where a particle is likely to be found. In both classical and quantum contexts, variance is calculated using similar principles but different probability distributions.

For the classical case with a uniform distribution, the variance is calculated as:
\[ \sigma^2_{\text{classical}} = \langle x^2 \rangle - \langle x \rangle^2 = \frac{a^2}{3} - \left(\frac{a}{2}\right)^2 = \frac{a^2}{12} \]
In the quantum case, with the probability distribution given by \( P(x) = \frac{2}{a} \sin^2\left(\ \frac{n \pi x}{a} \right) \), the variance is:
\[ \sigma^2_{\text{quantum}} = \langle x^2 \rangle - \langle x \rangle^2 = \frac{a^2}{3} - \frac{a^2}{2 n^2 \pi^2} - \left(\frac{a}{2}\right)^2 = \frac{a^2}{12} - \frac{a^2}{2 n^2 \pi^2} \]
This calculation shows how the quantum variance changes with respect to the quantum number \( n \).

Normalization Constant

The normalization constant \( C \) ensures that the total probability of finding the particle within the interval \(0 \leq x \leq a\) is 1. For our quantum probability distribution, the normalization condition is:
\[ \int_0^a P(x) \ dx = 1 \]
Substituting the quantum probability function:
\[ C \int_0^a \sin^2\left(\frac{n \pi x}{a}\right) \ dx = 1 \]
Since the integral of \( \sin^2 \) over one full wavelength is known to be \( \frac{a}{2} \), we can solve for \( C \):
\[ C \cdot \frac{a}{2} = 1 \implies C = \frac{2}{a} \]
So, the normalized quantum probability function is:
\[ P(x) = \frac{2}{a} \sin^2\left(\frac{n \pi x}{a}\right) \]
This ensures that the total probability of finding the particle within the interval remains 1.