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A particle is described by the normalized wave function \(\psi(x, y, z)=A x e^{-\alpha x^{2}} e^{-\beta \beta} e^{-\gamma x^{2}},\) where \(A, \alpha, \beta,\) and \(\gamma\) are all real, positive constants. The probability that the particle will be found in the infinitesimal volume \(d x d y d z\) centered at the point \(\left(x_{0}, y_{0}, z_{0}\right)\) is \(\left|\psi\left(x_{0}, y_{0}, z_{0}\right)\right|^{2} d x d y d z\) (a) At what value of \(x_{0}\) is the particle most likely to be found? (b) Are there values of \(x_{0}\) for which the probability of the particle being found is zero? If so, at what \(x_{0} ?\)

Short Answer

Expert verified
(a) Most likely at \( x = \pm \sqrt{\frac{1}{2\alpha}} \). (b) Zero probability at \( x = 0 \).

Step by step solution

01

Understand the Probability Density

The probability density of the particle being located at a specific point \( (x_0, y_0, z_0) \) is given by the square of the wave function, \( \left| \psi(x_0, y_0, z_0) \right|^2 \). We need to find the values of \( x_0 \) where the probability density is maximum and zero.
02

Express the Probability Density Function

The given wave function is \( \psi(x, y, z) = A x e^{-\alpha x^2} e^{-\beta y} e^{-\gamma z^2} \). The probability density is \( \left| \psi(x, y, z) \right|^2 = A^2 x^2 e^{-2\alpha x^2} e^{-2\beta y} e^{-2\gamma z^2} \). Note that the terms involving \( y \) and \( z \) do not affect \( x \).
03

Analyze the Function for Maximum Probability in x

The expression for \( x \) is \( x^2 e^{-2\alpha x^2} \). To find the maximum, find the derivative \( \frac{d}{dx}(x^2 e^{-2\alpha x^2}) \), set it to zero, and solve for \( x \).
04

Differentiate and Set to Zero

Differentiate \( f(x) = x^2 e^{-2\alpha x^2} \) using the product rule: \( f'(x) = 2x e^{-2\alpha x^2} - 4\alpha x^3 e^{-2\alpha x^2} \), and set \( f'(x) = 0 \): \( e^{-2\alpha x^2}(2x - 4\alpha x^3) = 0 \).
05

Solve Derivative Equation

Solving \( 2x - 4\alpha x^3 = 0 \) gives: \( x(2 - 4\alpha x^2) = 0 \). This yields solutions \( x = 0 \) and \( x = \pm \sqrt{\frac{1}{2\alpha}} \). Check for maxima using second derivative test or consider the physical context for valid \( x \).
06

Check Values for Zero Probability

A zero probability occurs when \( \psi(x, y, z) = 0 \). Since \( A, \alpha, \beta, \gamma \) are positive, the \( e \)-terms never become zero. The term \( x \) causes the wave function to be zero at \( x = 0 \).
07

Conclusion

The particle is most likely found at \( x = \pm \sqrt{\frac{1}{2\alpha}} \). The probability is zero at \( x = 0 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Wave Function
The wave function is a fundamental concept in quantum mechanics that encapsulates everything we can know about a quantum system at a given time. In our exercise, the wave function is represented as \( \psi(x, y, z) = A x e^{-\alpha x^2} e^{-\beta y} e^{-\gamma z^2} \). This expression involves mathematical elements such as exponential functions and constants \( A, \alpha, \beta, \) and \( \gamma \), which describe how the particle behaves in space.

Understanding the wave function is crucial because it provides the basis to compute other physical quantities, such as the probability density. Importantly, note that this wave function is normalized, meaning it has been adjusted to ensure the total probability of finding the particle somewhere in space is 1. By analyzing the structure of the wave function, particularly the exponential terms, we can infer about the particle's behavior and its most probable locations at any given time in space.
Probability Density
Probability density tells us where a particle, described by the wave function, is likely to be found. Mathematically, the probability density is given by the square of the absolute value of the wave function: \( \left| \psi(x, y, z) \right|^2 \). In our exercise, after squaring the wave function, we focus on the expression: \( A^2 x^2 e^{-2\alpha x^2} e^{-2\beta y} e^{-2\gamma z^2} \).

This formula shows that probability density changes with \( x, y, \) and \( z \). However, for our main focus on finding where the particle is most likely in the x-direction, the main concern is \( x^2 e^{-2\alpha x^2} \). The exponential terms, which decay rapidly, help determine where the probability density is significantly high or low. This is crucial for understanding particle localization and finding the probability maxima or zero points.
Particle Localization
Particle localization refers to the determination of where a particle is most likely to be found in space, which is derived from the probability density. In our problem, we identify the potential locations for the particle by examining the function related to x, \( x^2 e^{-2\alpha x^2} \).

To find where the probability density is maximized, we need to find critical points of this expression through differentiation. By setting the derivative to zero, \( f'(x) = 2x e^{-2\alpha x^2} - 4\alpha x^3 e^{-2\alpha x^2} = 0 \), we solve for \( x \), resulting in the points \( x = 0 \) and \( x = \pm \sqrt{\frac{1}{2\alpha}} \). Here, the latter represents where the particle is most likely to be found, and these points emphasize regions of high probability density in x-direction.

In practical terms, adjustments like these require critical thinking about the problem, including the use of the second derivative test or simply considering the physical context to confirm the maxima and minima.
Normalization Constants
Normalization constants play a pivotal role in quantum mechanics, especially when dealing with wave functions. The constant \( A \) in the wave function \( \psi(x, y, z) = A x e^{-\alpha x^2} e^{-\beta y} e^{-\gamma z^2} \) ensures that the wave function is normalized.

Normalization means that the total integral of the probability density over all space equals 1, reflecting that the particle must be somewhere in space with complete certainty. This is calculated by integrating the probability density over all possible values of \( (x, y, z) \) and setting that equal to 1. So, the normalization constant \( A \) is determined as part of this process to scale the wave function appropriately.

This step is crucial in solving problems within quantum mechanics as it ensures the results are physically meaningful. Normalization doesn't affect where the particle is most likely to be found or the shape of the wave function but guarantees that probabilities are calculated correctly.

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Most popular questions from this chapter

Why Don't We Diffract? (a) Calculate the de Broglie wavelength of a typical person walking through a doorway. Make reasonable approximations for the necessary quantitics. (b) Will the person in part (a) exhibit wave-like behavior when walking through the "single slit" of a doorway? Why?

Atomic Spectra Uncertainties. A certain atom has an energy level \(2.58 \mathrm{cV}\) above the ground level. Once excited to this Ievel, the atom remains in this level for \(1.04 \times 10^{-7} \mathrm{~s}\) (on average) before emitting a photon and returning to the ground level. (a) What is the energy of the photon (in electron volts)? What is its wavelength (in nanomcters)? (b) What is the smallest possible uncertainty in energy of the photon? Give your answer in electron volts. (c) Show that \(|\Delta E| E|=| \Delta \lambda / \lambda \mid\) if \(|\Delta \lambda / \lambda|<1 .\) Use this to calculate the magnitude of the smallest possible uncertainty in the wavelength of the photon. Give your answer in nanometers.

(a) An electron moves with a speed of \(4.70 \times 10^{6} \mathrm{m} / \mathrm{s}\) . What is its de Broglie wavelength? (b) A proton moves with the same speed. Determine its de Broglie wavelength.

Particle \(A\) is described by the wave function \(\psi(x, y, z)\) . Particle \(B\) is described by the wave function \(\psi(x, y, z) e^{i \phi}\) , where \(\phi\) is a real constant. How does the probability of finding particle \(A\) within a volume \(d V\) around a certain point in space compare with the probability of finding particle \(B\) within this same volume?

Imagine another universe in which the value of Planck's constant is \(0.0663 \mathrm{J} \cdot \mathrm{s},\) but in which the physical laws and all other physical constants are the same as in our universe. In this universe, two physics students are playing catch. They are 12 \(\mathrm{m}\) apart, and one throws a \(0.25-\mathrm{kg}\) ball directly toward the other with a speed of 6.0 \(\mathrm{m} / \mathrm{s}\) (a) What is the uncertainty in the ball's horizontal momentum, in a direction perpendicular to that in which it is being thrown, if the student throwing the ball knows that it is located within a cube with volume 125 \(\mathrm{cm}^{3}\) at the time she throws it? (b) By what horizontal distance could the ball miss the second student?

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