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A finite potential energy function \(U(x)\) allows \(\psi(x)\). the solution of the time-independent Schrödinger equation. to penetrate the classically forbidden region. Without assuming any particular function for \(U(x)\). show that W. \(x\) ) must have an inflection point at any value of \(x\) where it enters a classically forbidden region.

Short Answer

Expert verified
Every point where \(\psi(x)\) penetrates a classically forbidden region is an inflection point, where the second derivative of \(\psi(x)\) equals zero and the third derivative does not equal zero.

Step by step solution

01

- Define concept of Inflection Point

In mathematics, an inflection point or point of inflection is a point on a curve at which the curve changes concavity — that is, changes between being concave up to concave down, or vice versa.
02

- Compose Schrödinger Equation

First, we write the time-independent Schrödinger equation, which is \(\frac{-\hbar^2}{2m}\frac{d^2\psi}{dx^2} +U\psi=E\psi\), where \(-\hbar^2/(2m)d^2\psi/dx^2\) represents the kinetic energy and \(U\psi\) represents the potential energy. Here \(E\) is the total energy.
03

- Define Classically Forbidden Region

A classically forbidden region is where the potential energy \(U\) is greater than the total energy \(E\). Therefore, in a classically forbidden region, the Schrödinger equation becomes \(\frac{-\hbar^2}{2m}\frac{d^2\psi}{dx^2} = (U - E)\psi\). As \(U > E\), we can write it as \(\frac{-\hbar^2}{2m}\frac{d^2\psi}{dx^2} = F\psi\) for some positive constant \(F\).
04

- Solve the Equation

We differentiate the equation \(\frac{-\hbar^2}{2m}\frac{d^2\psi}{dx^2} = F\psi\) with respect to \(x\) and get \(\frac{-\hbar^2}{2m}\frac{d^3\psi}{dx^3} = F\frac{d\psi}{dx}\). Now, for an inflection point, the second derivative of \(\psi\) with respect to \(x\), i.e., \(\frac{d^2\psi}{dx^2}\) must be equal to zero (from definition of inflection point). At this point, the third derivative is given by \(\frac{d^3\psi}{dx^3} = 2mF/\hbar^2\frac{d\psi}{dx}\), which is non-zero if the first derivative, \( \frac{d\psi}{dx}\) is non-zero. So, we conclude that the wave function must have an inflection point at any value of x where it enters a classically forbidden region.

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Key Concepts

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

Schrödinger equation
The Schrödinger equation is a fundamental equation in quantum mechanics. It describes how the quantum state of a physical system changes over time. In our specific case, we look at the **time-independent Schrödinger equation**. This form is used when the system's properties do not change with time.

The equation is given by:
  • \[ \frac{-\hbar^2}{2m}\frac{d^2\psi}{dx^2} + U\psi = E\psi \]
Here:
  • \(\hbar\) is the reduced Planck’s constant.
  • \(m\) is the mass of the particle.
  • \(\psi(x)\) is the wave function of the particle.
  • \(U(x)\) represents the potential energy.
  • \(E\) denotes the total energy of the system.
The equation balances two sides: kinetic energy on the left, and on the right, it's the difference between the total and potential energy.

By solving this equation, we can find the wave function \(\psi(x)\), which gives valuable information about the quantum system, such as the likelihood of finding a particle at a certain position.
potential energy
Potential energy in quantum mechanics deals with the energy stored within a system due to its position, condition, or configuration. In the Schrödinger equation, it's represented by the function \(U(x)\). This function can vary with respect to position \(x\).

Unlike classical mechanics, which defines potential energy using well-known forces like gravity, in quantum mechanics, the potential energy landscape \(U(x)\) can determine how a particle behaves. It can include barriers, wells, or other forms that influence the wave function of the particle.

Potential energy creates regions of space where particles might find themselves unable to enter classically. These regions demand special consideration because they can affect the movement and likelihood of finding particles, notably seen through phenomena like quantum tunneling.
wave function
The wave function, denoted as \(\psi(x)\), is a core concept in quantum mechanics. It provides a probability amplitude for a particle's position in space. This function allows us to predict how a particle behaves at different locations along the path.

The wave function is often a complex number. Its magnitude squared, \(|\psi(x)|^2\), gives us the probability density of finding a particle at a point \(x\). The wave function must be **continuous and normalized**, meaning that the integral of the probability density over all space equals one.
The wave function changes in response to the Schrödinger equation. Its shape and nature depend on the potential energy \(U(x)\), defining how particles interact with their surroundings. Therefore, understanding this function helps illuminate broader quantum behaviors, including energy levels and probabilities.
quantum tunneling
Quantum tunneling is a fascinating phenomenon predicted by quantum mechanics, where particles can traverse potential energy barriers even when they seem to lack sufficient energy based on classical physics. It highlights the unique nature of quantum particles, which behave differently from classical objects.

When a particle encounters a barrier where its potential energy \(U\) exceeds its total energy \(E\), it enters what we call a "classically forbidden region". In classical physics, a particle would simply bounce back unless it has more energy than the barrier. However, due to the wave-like properties of particles, there is a non-zero probability that the particle will "tunnel" through the barrier and appear on the other side.

Quantum tunneling has important implications and applications in fields such as nuclear physics, semiconductor physics, and even biological processes. For instance, it is a key principle underlying the operation of quantum mechanically enabled technologies like tunnel diodes and some types of transistors.

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Most popular questions from this chapter

Consider a particle of mass \(m\) and energy \(E\) in a region where the potential energy is a constant \(U_{0}\). greaterthan E. and the region extends to \(x=+\infty\). (a) Guess a physically acceptable solution of the Schrödinger equation in this region and demonstrate that it is a solution. (b) The region noted in part (a) extends from \(x=+1 \mathrm{~nm}\) to \(+\infty\). To the left of \(x=1 \mathrm{~nm}\), the particle's wave function is \(D \cos \left(10^{9} \mathrm{~m}^{-1} x\right)\). Is \(U(x)\) also greater than \(E\) here? (c) The particle's mass \(m\) is \(10^{-30} \mathrm{~kg}\). By how much (in \(\mathrm{eV}\) ) does \(U_{0}\), the potential energy prevailing from \(x=1 \mathrm{~nm}\) to \(+\infty\), exceed the particle's energy?

Quantum-mechanical stationary states are of the general form \(\Psi(x, t)=\psi(x) e^{-i \omega t},\) For the basic plane wave (Chapter 4 ), this is \(\Psi(x, t)=A e^{i k x} e^{-i \omega t}=\) \(A e^{i(k x-\omega t)},\) and for a particle in a box, it is \(\Psi(x, r)=\) A \(\sin (k x) e^{-i \omega t}\). Although both are sinusoidal, we claim that the plane wave alone is the prototype function whose momentum is pure-a well-defined value in one direction. Reinforcing the claim is the fact that the plane wave alone lacks features that we expect to see only when, effectively, waves are moving in both directions. What features are these, and, considering the probability densities, are they indeed present for a particle in a box and absent for a plane wave?

Exercises \(94-97\) refer to a bound particle of mass \(m\) described by the wave function $$ \psi(x)=A x e^{-x^{2} / 2 b^{2}} $$ Sketch \(\psi(x)\). Would you expect this wave function to be the ground state? Why or why not?

Consider the differential equation \(d^{2} f(x) / d x^{2}=b f(x)\). (a) Suppose that \(f_{1}(x)\) and \(f_{2}(x)\) are solutions. That is, $$ \frac{d^{2} f_{1}(x)}{d x^{2}}=b f_{1}(x) \text { and } \frac{d^{2} f_{2}(x)}{d x^{2}}=b f_{2}(x) $$ Show that the equation also holds when the linear combination \(A_{1} f_{1}(x)+A_{2} f_{2}(x)\) is inserted. (b) Suppose that \(f_{3}(x)\) and \(f_{4}(x)\) are solutions of \(d^{2} f(x) / d x^{2}=b f^{2}(x)\). Is \(A_{3} f_{3}(x)+A_{4} f_{4}(x)\) a solution? Justify your answer.

Exercises \(78-88\) refer to a particle of mass \(m\) described by the wave function $$ \psi(x)=\left\\{\begin{array}{ll} 2 \sqrt{a^{3}} x e^{-a x} & x>0 \\ 0 & x<0 \end{array}\right. $$ Determine the expectation value of the momentum of the particle. Explain.

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