91Ó°ÊÓ

A particle in the infinite square well has as its initial wave function an even mixture of the first two stationary states: $$\Psi(x, 0)=A\left[\psi_{1}(x)+\psi_{2}(x)\right].$$ (a) Normalize \(\Psi(x, 0)\). (That is, find \(A\). This is very easy, if you exploit the orthonormality of \(\psi_{1}\) and \(\psi_{2} .\) Recall that, having normalized \(\Psi\) at \(t=0\) you can rest assured that it stays normalized- if you doubt this, check it explicitly after doing part (b).) (b) Find \(\Psi(x, t)\) and \(|\Psi(x, t)|^{2} .\) Express the latter as a sinusoidal function of time, as in Example \(2.1 .\) To simplify the result, let \(\omega \equiv \pi^{2} \hbar / 2 m a^{2}.\) (c) Compute \(\langle x\rangle .\) Notice that it oscillates in time. What is the angular frequency of the oscillation? What is the amplitude of the oscillation? (If your amplitude is greater than \(a / 2,\) go directly to jail.) (d) Compute \(\langle p\rangle .\) (As Peter Lorre would say, "Do it ze kveek vay, Johnny!") (e) If you measured the energy of this particle, what values might you get, and what is the probability of getting each of them? Find the expectation value of \(H .\) How does it compare with \(E_{1}\) and \(E_{2}\) ?

Step by step solution

01

Normalize the Wave Function

To normalize the wave function \(\Psi(x,0) = A[\psi_1(x) + \psi_2(x)]\), we need to set the integral of the probability density over the entire space equal to 1:\[\int_{0}^{a} |\Psi(x, 0)|^2 \ dx = 1.\] Substituting \(\Psi(x, 0)\), we have:\[A^2 \int_{0}^{a} (\psi_1(x) + \psi_2(x))^2 \ dx = 1.\] Using orthonormality, \(\int \psi_1(x)^2 dx = \int \psi_2(x)^2 dx = 1\) and \(\int \psi_1(x) \psi_2(x) dx = 0\), the equation simplifies to \[2A^2 = 1\]. Solving for \(A\), we get \(A = \frac{1}{\sqrt{2}}\).

02

Find Time-Dependent Wave Function

The time evolution of the wave function is given by \(\Psi(x, t) = A[\psi_1(x)e^{-iE_1t/\hbar} + \psi_2(x)e^{-iE_2t/\hbar}]\). Here, \(E_n = \frac{n^2\pi^2\hbar^2}{2ma^2}\) for an infinite square well. Therefore, \(E_1 = \frac{\pi^2\hbar^2}{2ma^2}\) and \(E_2 = \frac{4\pi^2\hbar^2}{2ma^2}\). Using \(\omega = \frac{\pi^2\hbar}{2ma^2}\), we have \(E_1 = \omega\) and \(E_2 = 4\omega\). Thus, \[\Psi(x, t) = \frac{1}{\sqrt{2}}[\psi_1(x)e^{-i\omega t} + \psi_2(x)e^{-i4\omega t}].\]

03

Calculate Probability Density

The probability density \(|\Psi(x, t)|^2\) is given by \(\Psi(x, t)\cdot \Psi^*(x, t)\). Using the expansion from Step 2, we find\[|\Psi(x, t)|^2 = \frac{1}{2}[\psi_1(x)^2 + \psi_2(x)^2 + 2\psi_1(x)\psi_2(x)\cos(3\omega t)].\] Thus, the probability density is a sinusoidal function in time.

04

Compute Expectation Value of Position \(\langle x \rangle\)

The expectation value \(\langle x \rangle\) is given by the integral \(\langle x \rangle = \int_{0}^{a} x |\Psi(x,t)|^2 dx\). Calculating it over time, we find it oscillates as \[\langle x \rangle = \frac{a}{2} + A\cos(3\omega t),\] where \(A\) is the oscillation amplitude. The angular frequency of oscillation is \(3\omega\), and for \(\langle x \rangle\), the amplitude should not exceed \(a/2\).

05

Compute Expectation Value of Momentum \(\langle p \rangle\)

The expectation value \(\langle p \rangle\) for a state with wave function \(\Psi(x,t)\) is found using \(\langle p \rangle = \int_{0}^{a} \Psi^*(x,t) \left(-i\hbar \frac{\partial}{\partial x}\right) \Psi(x,t) dx\). For this mixture of energy eigenstates, \(\langle p \rangle = 0\), since \(\Psi\) is symmetric and the momentum operator \(-i\hbar \frac{\partial}{\partial x}\) is anti-symmetric, leading to cancellation.

06

Find Energy Measurement Outcomes and Probabilities

If the energy is measured, it could be \(E_1\) with probability \(|\langle \psi_1 | \Psi \rangle|^2 = \frac{1}{2}\), or \(E_2\) with the same probability \(= \frac{1}{2}\). The expectation value of energy \(\langle H \rangle = \frac{E_1 + E_2}{2} = \frac{5\omega}{2}\), which is the average of the two energies weighted by their probabilities, lying between \(E_1\) and \(E_2\).

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Wave Function Normalization

In quantum mechanics, the wave function \(\Psi(x)\) describes the state of a particle and its behavior. To gain physical meaning, it must be normalized. This means that the total probability of finding the particle in all space equals 1. The normalization condition can be expressed as \(\int_{-\infty}^{\infty} |\Psi(x)|^2 \, dx = 1\).
For a particle in an infinite square well, where the potential is zero inside the well and infinite outside, the wave function must be zero outside the well. So, the normalization needs to be computed over the interval \([0, a]\), where \(a\) is the width of the well.
In the given problem, the wave function is an even mixture of the first two stationary states: \(\Psi(x,0) = A[\psi_1(x) + \psi_2(x)]\). By using the orthonormality of stationary states, you can find the normalization constant \(A\). For our example, this simplifies to \(|A| = \frac{1}{\sqrt{2}}\). It ensures the wave function's squared integral over the interval \([0, a]\) equals 1. Normalizing is crucial because it allows predictions about probabilities and observables to be made accurately.

Time-Dependent Schrödinger Equation

The time-dependent Schrödinger equation governs how the wave function \(\Psi(x,t)\) evolves over time. It's crucial when dealing with quantum systems, especially in an infinite square well, where the solutions are energy eigenstates.
For a particle in an infinite square well, the energy eigenstates are known as stationary states \(\psi_n(x)\), and the general wave function is expressed as a superposition of these states.
The time evolution of these states includes an exponential factor, \(e^{-iE_nt/\hbar}\), where \(E_n\) is the energy of the \(n\)th state. In the problem, the time-dependent wave function is \(\Psi(x, t) = \frac{1}{\sqrt{2}}[\psi_1(x)e^{-i\omega t} + \psi_2(x)e^{-i4\omega t}]\), with \(\omega = \frac{\pi^2\hbar}{2ma^2}\).
This expression captures the oscillatory nature of quantum states in a potential well, allowing us to see how probability distributions shift over time.

Expectation Value

In quantum mechanics, the expectation value provides insight into physical quantities such as position, momentum, or energy. It's like a weighted average of all possible outcomes, reflecting what can be expected if many identical measurements are made.
To find the expectation value of position \(\langle x \rangle\), you compute the integral \(\langle x \rangle = \int_{0}^{a} x |\Psi(x,t)|^2 \, dx\). For our scenario, it represents how the average position of a particle varies over time inside an infinite square well.
The result \(\langle x \rangle = \frac{a}{2} + A \cos(3\omega t)\) shows oscillation around the center of the well \(\frac{a}{2}\) with angular frequency \(3\omega\). The amplitude depends on the overlap of wave functions. This oscillatory behavior is insightful for understanding the dynamic nature of quantum systems.

Infinite Square Well

The infinite square well is a simple yet fundamental concept in quantum mechanics. It describes a particle confined to a one-dimensional box with impenetrable walls, making it a perfect potential well. The potential energy is zero inside the well and infinite outside.
This setup results in quantized energy levels that only allow specific states, expressed as \(E_n = \frac{n^2\pi^2\hbar^2}{2ma^2}\), where \(n\) is an integer representing the quantum number. Since the potential is infinite at the boundaries, the wave function must be zero outside and at the boundary of the well.
Because of its simplicity, the infinite square well is a classic model for understanding more complex quantum systems. It demonstrates the concept of quantization and boundary conditions effectively and is solvable in both time-independent and time-dependent scenarios.

Energy Measurement Probabilities

Measuring the energy of a particle in a quantum state involves probabilities, as different energies can be measured depending on the state's composition.
In our exercise dealing with the infinite square well, the wave function is a superposition of the first two eigenstates \(\psi_1(x)\) and \(\psi_2(x)\), each associated with energies \(E_1\) and \(E_2\), respectively. When measuring energy, you are likely to get either \(E_1\) or \(E_2\) with certain probabilities.
The probability of measuring a specific energy \(E_n\) is given by \(|\langle \psi_n | \Psi \rangle|^2\). For our problem, each energy has a probability of \(\frac{1}{2}\). The expectation value of the energy, a weighted average of possible outcomes, is found to be \(\langle H \rangle = \frac{E_1 + E_2}{2}\). It shows how the average energy value relates to the possible measurements, offering insight into the energy characteristics of superposed states.