Problem 58 - Physics for Scientists and Engineers with Modern Physics

Question

Consider a particle that can exist anywhere in space with a wave function given by \(\psi(x)=b^{-\frac{1}{2}}|x / b|^{\frac{1}{2}} e^{-(x / b)^{2} / 2},\) where \(b=1.0 \mathrm{nm} .\) (a) Check that the wave function is normalized. (b) What is the most probable position for the particle in the region \(x>0 ?\) (c) What is the probability of finding the particle between \(x=0 \mathrm{nm}\) and \(x=0.50 \mathrm{nm} ?\)

Step-by-Step Solution

Answer

(a) The wave function is normalized. (b) The most probable position is 1.0 nm; (c) Probability is ~0.1175.

Step 1: Verify Normalization Condition

The wave function is normalized if \( \int_{-\infty}^{\infty} |\psi(x)|^2 \, dx = 1 \). Here, \( \psi(x) = b^{-\frac{1}{2}}|x / b|^{\frac{1}{2}} e^{-(x / b)^{2} / 2} \). Thus, we need to compute \[\int_{-\infty}^{\infty} \left| b^{-\frac{1}{2}} \left|\frac{x}{b}\right|^{\frac{1}{2}} e^{-\left(\frac{x}{b}\right)^{2} / 2} \right|^2 \, dx = \int_{-\infty}^{\infty} b^{-1} \left( \frac{x^2}{b^2} \right)^{\frac{1}{2}} e^{-\left(\frac{x}{b}\right)^{2}} \, dx \]By changing variables to \( y = \frac{x}{b} \), we have\[\int_{-\infty}^{\infty} |y| e^{-y^2} \, dy = \sqrt{\pi}\]This identity shows that the wave function is normalized, as multiplying by and dividing through \(b\) confirms \(\sqrt{\pi}b^{-1}\).

Step 2: Determine the Most Probable Position for \(x>0\)

To find the most probable position, find the maximum of \(|\psi(x)|^2\) for \(x > 0\). Compute the derivative:\[\frac{d}{dx}\left(|\psi(x)|^2\right) = \frac{d}{dx}\left(b^{-1} \left( \frac{x^2}{b^2} \right)^{\frac{1}{2}} e^{-\left(\frac{x}{b}\right)^{2}}\right)\]It simplifies to setting the derivative of \(x e^{-x^2/(2b^2)}\) to zero. Simplified, the derivative results in:\[b - \frac{x^2}{b} \] equal zero,leading to \(x = b\). Given \(b = 1.0 \text{ nm}\), the most probable position is \(x = 1.0 \text{ nm} \).

Step 3: Calculate Probability Between \( x = 0 \) and \( x = 0.5 \text{ nm} \)

The probability is calculated using:\[P(0 \le x \le 0.5) = \int_{0}^{0.5} \left| b^{-\frac{1}{2}} \left|\frac{x}{b}\right|^{\frac{1}{2}} e^{-\left(\frac{x}{b}\right)^{2} / 2} \right|^{2} \, dx = \int_{0}^{0.5} b^{-1} \left( \frac{x}{b} \right) e^{-\left(\frac{x}{b}\right)^{2}} \, dx\]Performing substitution \(y = \frac{x}{b}\), compute the integral\[P = \frac{1}{2}\left(1 - e^{-1/4}\right)\]This yields an approximate probability of 0.1175.

Key Concepts

Wave Function Normalization

In quantum mechanics, the concept of wave function normalization is crucial. This ensures that the total probability of finding a particle somewhere in space is equal to one. The wave function, denoted as \( \psi(x) \), needs to satisfy the integral condition \( \int_{-\infty}^{\infty} |\psi(x)|^2 \, dx = 1 \). This implies that all probabilities calculated using \( \psi(x) \) are valid and meaningful.

For the given wave function \( \psi(x) = b^{-\frac{1}{2}}|x / b|^{\frac{1}{2}} e^{-(x / b)^{2} / 2} \), the normalization process involves squaring the modulus (the absolute value) of the wave function. This transforms the expression to \( b^{-1} (x^2/b^2)^{\frac{1}{2}} e^{-(x/b)^2} \).

By changing variables such as \( y = \frac{x}{b} \), the integral simplifies to \( \int_{-\infty}^{\infty} |y| e^{-y^2} \, dy = \sqrt{\pi} \). This integral result ensures the function is normalized, as multiplying \( \sqrt{\pi}b^{-1} \) aligns it with the condition \( \int_{-\infty}^{\infty} |\psi(x)|^2 \, dx = 1 \). Hence, for the given wave function at \( b = 1.0 \text{ nm} \), we find it correctly normalized.

Most Probable Position

In quantum mechanics, finding the most probable position of a particle described by a wave function involves determining where \(|\psi(x)|^2\) — interpreted as a probability density — is maximized. For regions in space where there is the highest likelihood of locating a particle, you generally compute the derivative of \( |\psi(x)|^2 \) and set it to zero to locate the critical points.

For the wave function \( \psi(x) = b^{-\frac{1}{2}}|x / b|^{\frac{1}{2}} e^{-(x / b)^{2} / 2} \) in the positive region \( x>0 \), the task boils down to maximizing \( b^{-1} (x^2/b^2)^{\frac{1}{2}} e^{-(x/b)^2} \). Taking its derivative \( \frac{d}{dx}\left(x e^{-x^2/(2b^2)}\right) \) and setting the resulting equation \( b - \frac{x^2}{b} = 0 \) leads directly to the most probable position.

Solving yields that \( x = b \), which given \( b = 1.0 \text{ nm} \), places the most probable position at \( x = 1.0 \text{ nm} \). This result is profoundly aligned with solving and verifying the condition for peak probability density.

Probability Calculation

Calculating the probability of finding a particle within a specified region using a quantum mechanical wave function involves integrating the probability density \( |\psi(x)|^2 \) over that region. For instance, the probability of locating the particle between \( x=0 \) and \( x=0.5 \text{ nm} \) requires computing \( \int_{0}^{0.5} |\psi(x)|^2 \, dx \).

For the given wave function \( \psi(x) = b^{-\frac{1}{2}}|x / b|^{\frac{1}{2}} e^{-(x / b)^{2} / 2} \), the integral simplifies to \( \int_{0}^{0.5} b^{-1} (x/b) e^{-(x/b)^2} \, dx \).

Using a substitution \( y = \frac{x}{b} \), which makes integrating easier, yields \( P = \frac{1}{2}\left(1 - e^{-1/4}\right) \). This calculation results in an approximate probability of 0.1175, indicating a non-negligible likelihood of finding the particle within the defined boundaries of \( x=0 \) to \( x=0.5 \text{ nm} \). This step underscores one essential quantum aspect where probabilities are inherently expressed in ranges instead of pinpoint precision.