Question

Consider a wave function given by \(\psi(x) = A \space sin \space kx\), where \(k = 2\pi/\lambda\) and \(A\) is a real constant. (a) For what values of \(x\) is there the highest probability of finding the particle described by this wave function? Explain. (b) For which values of \(x\) is the probability \(zero\)? Explain.

Short answer

Maximum probability occurs at \( x = (2m+1)\frac{\lambda}{4} \). Zero probability occurs at \( x = n\frac{\lambda}{2} \).

Step-by-step solution

Understanding Maximum Probability

The probability density of finding a particle at a position \( x \) is represented by the square of the absolute value of the wave function, \( |\psi(x)|^2 \). Here, \( \psi(x) = A \sin kx \), so the probability density becomes \( |A \sin kx|^2 = A^2 \sin^2 kx \). The values of \( \sin^2 kx \) range from 0 to 1. The maximum value of \( \sin^2 kx \) is 1.

Finding Maximum Probability Points

To achieve maximum probability (1), \( \sin kx \) must be \( \pm 1 \). This occurs for angles \( kx = n\pi/2 + \pi/2 \), where \( n \) is an odd integer (1, 3, 5,...). Solving these conditions for \( x \), we get \( x = (2m+1)\frac{\lambda}{4} \), where \( m \) is an integer.

Understanding Zero Probability

Zero probability occurs when \( |\psi(x)|^2 = 0 \). This happens when \( \sin kx = 0 \).

Finding Zero Probability Points

The values for which \( kx = n\pi \), where \( n \) is an integer (0, 1, 2,...), will make \( \sin kx = 0 \). Solving for \( x \), we find \( x = n\frac{\lambda}{2} \).

Key concepts

Wave Function

The wave function, denoted as \( \psi(x) \), is one of the cornerstones in quantum mechanics. It serves as a mathematical representation of a quantum system and describes the quantum state of a particle. In this specific exercise, the wave function is given by \( \psi(x) = A \sin kx \). Let's break it down to understand its components and significance:

• \( A \) is a constant that normalizes the wave function. This ensures that the total probability of finding the particle across the entire space equals 1.
• \( k = \frac{2\pi}{\lambda} \) is the wavenumber, connected to the wavelength \( \lambda \) of the wave. It defines how many wave peaks fit into a unit length.
The wave function is essential because it gives us all the necessary information to determine where a particle might be located and how to calculate probabilities for different positions. However, it's important to note that the wave function itself does not provide probability directly—it needs to be squared to get those probabilities.

Probability Density

Probability density is crucial in quantum mechanics for predicting where a particle may be found. We derive it from the wave function by taking its absolute square. This transformation results in \( |\psi(x)|^2 = A^2 \sin^2 kx \), for this particular exercise.

What do the components mean for probability density?

• \( A^2 \) scales the probability density, ensuring the total probability sums to 1 when integrated over all space.
• \( \sin^2 kx \) ranges from 0 to 1, affecting where these probabilities are higher or lower.

Here's why this matters:

• When \( \sin^2 kx \) is 1, the probability density is at its maximum, indicating the highest probability of finding the particle at that position. This happens when \( \sin kx \) equals \( \pm 1 \).


• Conversely, when \( \sin^2 kx \) is 0, there is zero probability of finding the particle, occurring when \( \sin kx \) equals 0.

The picture painted by probability density gives a probabilistic "map" of where the particle is most likely to exist, based on the wave function.

Particle Position

The particle's position in this context is analyzed through the values of \( x \) that maximize or minimize the probability density. In this exercise, particle positions of maximum likelihood are derived from the condition \( \sin kx = \pm 1 \).

To find those positions, we solve for:

• \( kx = (2m+1)\frac{\pi}{2} \), with \( m \) being an integer. This yields positions \( x = (2m+1)\frac{\lambda}{4} \).

These solutions correspond to the peaks of the sinusoidal function \( \sin kx \), where the probability density is highest.

On the flip side, zero probability positions occur when \( \sin kx = 0 \). Solving this condition gives:

• \( kx = n\pi \), where \( n \) is an integer, resulting in positions \( x = n\frac{\lambda}{2} \).

These positions occur at the nodes of the wave function, meaning there is virtually no chance of finding the particle there. Understanding these maxima and minima is fundamental to predicting where a particle might be observed experimentally.

Sinusoidal Wave

A sinusoidal wave, as seen in the wave function \( \psi(x) = A \sin kx \), oscillates periodically, akin to a classic sine wave. Let's explore its characteristics:

• Its periodic and repetitive nature translates to regular patterns of high and low probability in quantum systems.
• The sinusoidal pattern comes from the angle \( kx \), where \( k = \frac{2\pi}{\lambda} \), imparting the wave a cyclic nature dependent on \( \lambda \).

The peaks (where \( \sin kx = \pm 1 \)) and nodes (where \( \sin kx = 0 \)) of the sinusoidal wave are key to understanding the particle's probabilistic behavior.

• Peak points correspond to maximum probability density and likely positions for finding the particle. They represent the amplitude heights in the wave cycle.
• Nodes, on the other hand, indicate points of zero probability density, creating a clear delineation between regions of potential particle presence.

In summary, the sinusoidal wave pattern of a wave function provides a clear and intuitive way to visualize how probabilities change with particle position. It beautifully illustrates the wave-particle duality intrinsic to quantum mechanics.