Chapter 40: Problem 5
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
Step by step solution
Understanding Maximum Probability
Finding Maximum Probability Points
Understanding Zero Probability
Finding Zero Probability Points
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Wave Function
- \( A \) 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
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.
- 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.
Particle Position
To find those positions, we solve for:
- \( kx = (2m+1)\frac{\pi}{2} \), with \( m \) being an integer. This yields positions \( x = (2m+1)\frac{\lambda}{4} \).
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 \( x = n\frac{\lambda}{2} \).
Sinusoidal Wave
- 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 \( \lambda \).
- 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.