Problem 42 - University Physics with Modern Physics

Question

CALC A particle is described by the normalized wave function \(\psi(x, y, z)=A x e^{-\alpha x^{2}} e^{-\beta y^{2}} e^{-\gamma y^{2}},\) 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 \(\left|\psi\left(x_{0}, y_{0}, z_{0}\right)\right|^{2} d x d y d z z\) (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 \(x_{0} ?\)

Step-by-Step Solution

Answer

(a) Maximum at \(x_0 = \pm \frac{1}{\sqrt{2\alpha}}\). (b) Zero at \(x_0 = 0\).

Step 1: Interpret the Probability Density

The probability of finding a particle in a volume element is given by the square of the wave function's absolute value multiplied by that volume element, i.e., \[P(x, y, z) = \left| \psi(x, y, z) \right|^2 dx \, dy \, dz.\]In this exercise, the wave function \(\psi(x, y, z) = A x e^{-\alpha x^2} e^{-\beta y^2} e^{-\gamma z^2}\). Hence, \[\left|\psi(x, y, z)\right|^2 = |A|^2 x^2 e^{-2\alpha x^2} e^{-2\beta y^2} e^{-2\gamma z^2}.\]

Step 2: Find the Position for Maximum Probability

To find the value of \(x_0\) where the probability is maximized, consider only the \(x\)-dependent part of the probability density:\[P(x) = |A|^2 x^2 e^{-2\alpha x^2}.\]Maximize this by taking its derivative with respect to \(x\), setting the derivative to zero, and solving for \(x\):\[\frac{d}{dx}(x^2 e^{-2\alpha x^2}) = 0.\]This involves using the product rule and solving for \(x\) gives\[x = 0.\]But for a meaningful maximum away from zero, solve \[2x - 4\alpha x^3 = 0 \Rightarrow x(2 - 4\alpha x^2) = 0.\]Thus, the non-trivial solution is\[x_0 = \pm \frac{1}{\sqrt{2\alpha}}.\]

Step 3: Determine Values with Zero Probability

The probability density is zero when \(|\psi(x, y, z)|^2\) is zero. In our specific function, this happens when \(x^2 e^{-2\alpha x^2} = 0\), but exponential terms never equal zero. The only way \(x^2 = 0\) is when \(x = 0\), but given the product \(x e^{-\alpha x^2},\) then \(\psi(x, y, z)\) is zero at \[x_0 = 0.\]

Key Concepts

Wave Function

In quantum mechanics, the wave function is a crucial concept that describes the quantum state of a particle or system of particles. It is typically represented by the Greek letter \( \psi \), and in mathematical terms, it is a complex function that contains all the information about a particle's position and momentum.
For the given exercise, the wave function \( \psi(x, y, z) = A x e^{-\alpha x^2} e^{-\beta y^2} e^{-\gamma z^2} \) includes arbitrary constants \( A, \alpha, \beta, \) and \( \gamma \) to tailor the function's behavior under specific conditions.
The wave function is vital because when you take its absolute square, you get the probability density function, which reveals where a particle is likely to be found.

Probability Density

Probability density is derived from the wave function and describes the likelihood of finding a particle in a given space. Mathematically, it is represented as the absolute square of the wave function: \( P(x, y, z) = |\psi(x, y, z)|^2 \).
This exercise has the probability density given by \( |A|^2 x^2 e^{-2\alpha x^2} e^{-2\beta y^2} e^{-2\gamma z^2} \). Essentially this function provides the probability per unit volume, \( dx \, dy \, dz \), at any point \((x_0, y_0, z_0)\).
Importantly, due to the squared term, the probability density is always non-negative, meaning there are no negative probabilities in quantum mechanics.

Normalization

Normalization ensures that the total probability of finding the particle somewhere in space is equal to 1. For wave functions, this is done using an integral over all space: \( \int |\psi(x, y, z)|^2 \, dx \, dy \, dz = 1 \). This requirement assures that probabilities are meaningful and adhere to the mathematical constraints of probability theory.
When a wave function is specified to be normalized, like in the exercise, it implies the norm of the function—here represented by the constant \( A \)—is adjusted to satisfy the normalization condition.
Without normalization, predictions made using the wave function could extend to greater probabilities, sidestepping realistic outcomes.

Particle in a Potential

The concept of a particle in a potential is a cornerstone of quantum mechanics, reflecting how particles behave when subjected to forces or fields. Unlike classical physics, quantum particles are described statistically, allowing them to exist in several states until measured.
The wave function of a particle in a potential generally contains terms influenced by external forces, indicated by variables like \( \alpha, \beta, \) and \( \gamma \). These decide the particle's spread and behavior within the field or potential landscape.
Understanding the interplay between a particle's wave function and the potential field is key to predicting behaviors such as tunneling, energy levels, and state distributions, providing profound insights into nature's microscopic intricacies.