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A particle of mass \(m\) in a one-dimensional box has the following wave function in the region \(x=0\) to \(x=L :\) $$\Psi(x, t)=\frac{1}{\sqrt{2}} \psi_{1}(x) e^{-i E_{1} t / \hbar}+\frac{1}{\sqrt{2}} \psi_{3}(x) e^{-i E_{3} t / \hbar}$$ Here \(\psi_{1}(x)\) and \(\psi_{3}(x)\) are the normalized stationary-state wave functions for the \(n=1\) and \(n=3\) levels, and \(E_{1}\) and \(E_{3}\) are the energies of these levels. The wave function is zero for \(x<0\) and for \(x>L .\) (a) Find the value of the probability distribution function at \(x=L / 2\) as a function of time. (b) Find the angular frequency at which the probability distribution function oscillates.

Short Answer

Expert verified
(a) At \( x = L/2 \), \( |\Psi(L/2, t)|^2 = 0 \) at all times. (b) The frequency is \( (E_3 - E_1)/\hbar \).

Step by step solution

01

Write the Total Wave Function

The total wave function \( \Psi(x, t) \) is a superposition of two stationary states given by:\[ \Psi(x, t) = \frac{1}{\sqrt{2}} \psi_{1}(x) e^{-i E_{1} t / \hbar} + \frac{1}{\sqrt{2}} \psi_{3}(x) e^{-i E_{3} t / \hbar} \]
02

Find the Probability Distribution Function

The probability distribution function is given by the magnitude square of the wave function:\[ \left| \Psi(x, t) \right|^2 = \Psi(x, t) \cdot \Psi^*(x, t) \]Substitute the expression for \( \Psi(x, t) \):\[ \left| \Psi(x, t) \right|^2 = \left( \frac{1}{\sqrt{2}} \psi_1(x) e^{-i E_1 t / \hbar} + \frac{1}{\sqrt{2}} \psi_3(x) e^{-i E_3 t / \hbar} \right) \times \left( \frac{1}{\sqrt{2}} \psi_1(x) e^{i E_1 t / \hbar} + \frac{1}{\sqrt{2}} \psi_3(x) e^{i E_3 t / \hbar} \right) \]
03

Expand and Simplify

Expand the expression:\[ \left| \Psi(x, t) \right|^2 = \frac{1}{2} \left| \psi_1(x) \right|^2 + \frac{1}{2} \left| \psi_3(x) \right|^2 + \frac{1}{2} \left( \psi_1(x) \psi_3^*(x) e^{-i(E_1 - E_3)t / \hbar} + \psi_3(x) \psi_1^*(x) e^{i(E_1 - E_3)t / \hbar} \right) \]
04

Recognize Interference Terms

The cross terms represent the interference between the two states:\[ \psi_1(x) \psi_3^*(x) e^{-i (E_1 - E_3) t / \hbar} + \psi_3(x) \psi_1^*(x) e^{i (E_1 - E_3) t / \hbar} \]These terms lead to oscillations in the probability distribution with time.
05

Evaluate at x = L/2

At \( x = L/2 \), substitute \( \psi_1(L/2) = \sqrt{\frac{2}{L}} \sin\left(\frac{\pi \cdot 1 \cdot L/2}{L}\right) \) and \( \psi_3(L/2) = \sqrt{\frac{2}{L}} \sin\left(\frac{\pi \cdot 3 \cdot L/2}{L}\right) \).Since both these results equal zero due to their sine component, the cross terms become zero, and the magnitude squared becomes just the sum of the probabilities of the individual states.
06

Identify Angular Frequency

The probability distribution oscillates with an angular frequency given by the energy difference between the two states:\[ \omega = \frac{E_3 - E_1}{\hbar} \]

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Key Concepts

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

Wave Function
In quantum mechanics, the wave function is a crucial concept that describes the quantum state of a particle. It contains all information about a system's state. For a particle of mass \(m\) in a one-dimensional box, the wave function \(\Psi(x, t)\) is given by a superposition of individual stationary-state wave functions. Here, it combines \(\psi_1(x)\) and \(\psi_3(x)\).

The wave function is generally complex, containing both real and imaginary parts. This is represented mathematically as an exponential function involving imaginary numbers, which is often expressed with Euler's formula. Each stationary state term includes an energy component \(e^{-i E t / \hbar}\), which causes time dependency.
  • The wave function \(\Psi(x, t)\) is normalized, meaning that over its entire domain its total probability is equal to 1.
  • Due to the linearity of quantum mechanics, the wave function can be expressed as a linear combination of basis functions (stationary states in this case).
  • The superposition principle allows particles to exist in all possible states at once until measured.
Probability Distribution
The wave function itself is not enough to describe physical properties. Instead, the probability distribution function \(|\Psi(x, t)|^2\) gives us meaningful information, representing the probability density of finding a particle at a position \(x\) at time \(t\).

To find this, we take the square of the absolute value of the wave function. This calculation involves multiplying the wave function by its complex conjugate, which cancels out the imaginary components, leaving a real-valued result.
  • This probability is a crucial outcome because it tells us how likely a particle is to be found at a particular location.
  • At \(x = L/2\), due to the specific forms of \(\psi_1(x)\) and \(\psi_3(x)\), the computed distribution reveals the contribution of each state in the superposition.
  • The probability distribution may oscillate as time changes due to interference between the different wave functions involved.
Stationary States
Stationary states are particular solutions to the Schrödinger equation for a system where the potential energy is not a function of time. These states have wave functions, such as \(\psi_1(x)\) and \(\psi_3(x)\), which are eigenfunctions with corresponding energy eigenvalues \(E_1\) and \(E_3\).

In a boxed system, these stationary-state wave functions are typically sinusoidal, reflecting the boundary conditions and quantized energy levels. Since they are stationary, the probability distribution associated with each does not change over time.
  • Each stationary state has a specific, quantized energy level, which only changes if the parameters of the system are altered (e.g., if the dimensions of the 'box' change).
  • The superposition of stationary states, as seen in \(\Psi(x, t)\), results in time-dependent behavior of the overall wave function, even though individual stationary states are time-independent.
  • Understanding these states helps in analyzing more complex quantum systems.
Angular Frequency
In the context of oscillating wave functions, the angular frequency \(\omega\) indicates how rapidly the probability distribution oscillates with time. For our problem, it reflects the energy difference between two states.This angular frequency is derived from the energy difference \(E_3 - E_1\) divided by Planck's constant \(\hbar\). It's the bridge between energy difference and the time evolution of the system:
  • It provides the rate at which the probability distribution interference pattern cycles through phases within time.
  • In quantum mechanics, these frequencies arise naturally whenever you have a superposition of different energy states.
  • The concept is analogous to oscillations seen in classical physics, offering insight into how quantum systems evolve temporally.
Understanding angular frequency allows us to model and predict the behavior of systems with fundamental time-dependent characteristics due to their quantum state superpositions.

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Most popular questions from this chapter

(a) An electron with initial kinetic energy 32 eV encounters a square barrier with height 41 \(\mathrm{eV}\) and width 0.25 \(\mathrm{nm}\) . What is the probability that the electron will tunnel through the barrier? (b) A proton with the same kinetic energy encounters the same barrier. What is the probability that the proton will tunnel through the barrier?

The penetration distance \(\eta\) in a finite potential well is the distance at which the wave function has decreased to 1\(/ e\) of the wave function at the classical turning point: $$\psi(x=L+\eta)=\frac{1}{e} \psi(L)$$ The penetration distance can be shown to be $$\eta=\frac{\hbar}{\sqrt{2 m\left(U_{0}-E\right)}}$$ The probability of finding the particle beyond the penetration distance is nearly zero. (a) Find \(\eta\) for an electron having a kinetic energy of 13 eV in a potential well with \(U_{0}=20 \mathrm{eV} .\) (b) Find \(\eta\) for a 20.0 -MeV proton trapped in a 30.0 -Me \(\mathrm{V}\) -deep potential well.

The ground-state energy of a harmonic oscillator is 5.60 \(\mathrm{eV} .\) If the oscillator undergoes a transition from its \(n=3\) to \(n=2\) level by emitting a photon, what is the wavelength of the photon?

Compute \(|\Psi|^{2}\) for \(\Psi=\psi \sin \omega t,\) where \(\psi\) is time independent and \(\omega\) is a real constant. Is this a wave function for a stationary state? Why or why not?

(a) The wave nature of particles results in the quantum-mechanical situation that a particle confined in a box can assume only wavelengths that result in standing waves in the box, with nodes at the box walls. Use this to show that an electron confined in a one-dimensional box of length \(L\) will have energy levels given by $$E_{n}=\frac{n^{2} h^{2}}{8 m L^{2}}$$ (Hint: Recall that the relationship between the de Broglie wave-length and the speed of a nonrelativistic particle is \(m v=h / \lambda\) . The energy of the particle is \(\frac{1}{2} m v^{2} . )\) (b) If a hydrogen atom is modeled as a one- dimensional box with length equal to the Bohr radius, what is the energy (in electron volts) of the lowest energy level of the electron?

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