A particle is in an infinite square well of width \(L\) and is in the \(n=3\) state. What is the probability that, when observed, the particle is found to be in the rightmost \(10.0 \%\) of the well?

Short Answer

Expert verified
Short Answer: To find the probability that a particle in the n=3 state is in the rightmost 10% of an infinite square well of width L, we need to calculate the square of the wavefunction in that region and integrate it over that range. The relevant region is from x = 0.9L to x = L. The probability is given by: \( P(x) = 0.1L + \frac{L}{6\pi}\sin(5.4\pi) \) This expression represents the probability that the particle will be found in the rightmost 10% of the well when observed.

Step by step solution

01

Wavefunction for an infinite square well

For an infinite square well, the wavefunction in terms of the width L and the quantum number n is given by: \( \psi_n(x) = \sqrt{\frac{2}{L}} \sin(\frac{n\pi x}{L}) \) Since the particle is in the n=3 state, the wavefunction is: \( \psi_3(x) = \sqrt{\frac{2}{L}} \sin(\frac{3\pi x}{L}) \)
02

Determine the relevant region of the well

We need to find the probability that the particle is in the rightmost 10% of the well. The width of this region (Δx) can be calculated as: \( \Delta x = 0.1 L \) So, the region we are interested in our integral is from x = 0.9L to x = L.
03

Calculate the probability

The probability that the particle is found in that region can be obtained by taking the square of the wavefunction and integrating it over that range. The probability P(x) is given by: \( P(x) = \int_{0.9L}^{L} {|\psi_3(x)|}^2 dx \) Substitute the wavefunction of n=3: \( P(x) = \int_{0.9L}^{L} (\frac{2}{L} \sin^2(\frac{3\pi x}{L})) dx \)
04

Evaluate the integral

To evaluate the integral, we can use a trigonometric identity: \( \sin^2(x) = \frac{1 - \cos(2x)}{2}\) Substitute this identity into the integral and simplify: \( P(x) = \int_{0.9L}^{L} (\frac{2}{L} (\frac{1 - \cos(2(\frac{3\pi x}{L}))}{2})) dx \) It becomes: \( P(x) = \int_{0.9L}^{L} (1 - \cos(\frac{6\pi x}{L})) dx \) Now, we can split this integral into two parts and evaluate them separately: \( P(x) = \int_{0.9L}^{L} 1 dx - \int_{0.9L}^{L} \cos(\frac{6\pi x}{L}) dx \) Evaluating these integrals, we get: \( P(x) = (L - 0.9L) - \frac{L}{6\pi}(\sin(\frac{6\pi L}{L}) - \sin(\frac{6\pi (0.9L)}{L})) \) Simplifying, we find the probability that the particle is in the rightmost 10% of the well: \( P(x) = 0.1L - \frac{L}{6\pi}(\sin(6\pi) - \sin(5.4\pi)) \) Since sin(6π) = 0, the probability is: \( P(x) = 0.1L + \frac{L}{6\pi}\sin(5.4\pi) \) This is the probability that the particle will be found in the rightmost 10% of the infinite square well when observed.

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Key Concepts

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

Quantum Mechanics
Quantum mechanics is a fundamental theory in physics that provides a description of the physical properties at microscopic scales, such as atoms and subatomic particles. Unlike classical mechanics, where objects follow deterministic paths, quantum mechanics is characterized by the probabilistic nature of the outcomes of measurements.

At the core of quantum mechanics is the concept that matter has properties of both particles and waves, known as wave-particle duality. This duality is captured mathematically by the wavefunction, which encapsulates all information about a quantum system. When we talk about a particle being in an infinite square well, we're examining how it behaves when confined within an infinitely deep potential well, with perfectly rigid walls.

In the exercise given, we're asked to calculate the likelihood of finding a quantum particle within a specific region. Why is it probabilistic? Because according to quantum mechanics, until a measurement is made, we can't say for certain where the particle is—instead, we can only talk about probabilities. Tools like wavefunctions and probability densities come into play when making such predictions.
Wavefunction
The wavefunction is a mathematical function that describes the quantum state of a system. For a particle in a one-dimensional infinite square well, the wavefunction, denoted by \( \psi_n(x) \) for the nth quantum state, is a sine wave confined within the well.

The wavefunction's properties are fascinating: it contains all the information about the quantum state of the particle and, when squared, gives us the probability density function for finding the particle at a certain position. This exercise features the \( n=3 \) state wavefunction, which has three half-wavelengths fitting within the width of the well.

What's crucial to grasp here is that the wavefunction itself is not the probability; instead, the probability of finding a particle at a position \( x \) is proportional to the square of the wavefunction's magnitude, \( |\psi_n(x)|^2 \) at that position.
Probability Density
When quantum mechanics tells us that we cannot know the position of a particle with certainty, we're directed to use the probability density to talk about where the particle might be found. Probability density for a quantum particle is derived from the square of the magnitude of its wavefunction.

In the solution provided, we calculate the integral of \( |\psi_3(x)|^2 \) across the rightmost 10% of the well to find the probability that the particle is found there. This integrates the probability density over that specific region—an integral that essentially adds up all the 'likelihoods' for each point within this segment of the well.

Thus, if you have a probability density function, you can integrate it over any region of interest to find the probability of locating the particle within that region, which is exactly what the step-by-step solution accomplishes.
Trigonometric Identities in Quantum Physics
Trigonometric identities are powerful tools in simplifying complex expressions, particularly in quantum physics where wavefunctions often involve sinusoidal functions. In our example, the wavefunction \( \sin(\frac{3\pi x}{L}) \) is squared, which might seem complicated to integrate.

However, by using the trigonometric identity \( \sin^2(x) = \frac{1 - \cos(2x)}{2} \) to simplify \( \sin^2(\frac{3\pi x}{L}) \) before integrating, it turns a tricky problem into a more manageable one.

Trigonometric identities in quantum physics are not just a mathematical convenience; they reflect the deep relationships between various properties of waves and their symmetries. They enable us to untangle the math behind physical phenomena, peeling back the layers to reveal the probabilities that govern the behavior of particles in quantum systems.

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Most popular questions from this chapter

Consider a water vapor molecule in a room \(4.00 \mathrm{~m} \times\) \(10.0 \mathrm{~m} \times 10.0 \mathrm{~m}\). a) What is the ground state energy of this molecule, treating it as a simple particle in a box? b) Compare this energy to the average thermal energy of such a molecule, taking the temperature to be \(300 . \mathrm{K}\). c) What can you conclude from the two numbers you just calculated?

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