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Chapter 5: Problem 96

Exercises \(94-97\) refer to a bound particle of mass \(m\) described by the wave function $$ \psi(x)=A x e^{-x^{2} / 2 b^{2}} $$ Given that the particle's total energy \(E\) is 0 , show that the potential energy is \(U(x)=\frac{h^{2}}{2 m b^{4}} x^{2}-\frac{3 h^{2}}{2 m b^{2}}\)

Short Answer

Expert verified
The potential energy of the particle is given by \(U(x)=\frac{h^{2}}{2 m b^{4}} x^{2}-\frac{3 h^{2}}{2 m b^{2}}\)

Step by step solution

01

Compute the first derivative of the wave function

The wave function \(\psi(x)\) is given as \(A x e^{-x^{2} / 2 b^{2}}\). We need to compute its first derivative. Using the product and chain rule, it is:\(\psi'(x) = Ae^{-x^2/2b^2} - Ax^2/b^2 e^{-x^2/2b^2}\)
02

Compute the second derivative of the wave function

We compute the second derivative of the wave function from the result in Step 1. This also involves application of product and chain rule:\(\psi''(x) = -Ax^2/b^2 e^{-x^2/{2b^2}} - A/b^2 e^{-x^2/{2b^2}} + 2Ax^3/{b^4} e^{-x^2/{2b^2}}\)
03

Calculate the kinetic energy

From the Schr枚dinger equation, the kinetic energy is given by \(-\frac{h^2}{2m}\psi''(x)\). Substituting \(\psi''(x)\) with the result from Step 2 gives:\(K(x) = -\frac{h^2}{2m}[-Ax^2/b^2 e^{-x^2/{2b^2}}- A/b^2 e^{-x^2/{2b^2}}+ 2Ax^3/b^4 e^{-x^2/{2b^2}}]\) = \[\frac{h^2x^2}{2mb^2} - \frac{h^2}{2mb^2} + \frac{h^2x^3}{mb^4}\] \(\psi(x)\)
04

Find the expression for the potential energy

Knowing that the total energy is zero, we can express the potential energy as \(U=E - K\). Hence \(U(x) = - K(x)\). Substitute \(K(x)\) from Step 3 to get:\(U(x) = -\[\frac{h^2x^2}{2mb^2} - \frac{h^2}{2mb^2} + \frac{h^2x^3}{mb^4}\] \(\psi(x)\) = \frac{h^2x^2}{2mb^2} - \frac{3h^2}{2mb^2}\)

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Key Concepts

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

Wave Function
In the realm of quantum mechanics, the wave function is a fundamental concept. It represents the quantum state of a particle in space and time. The wave function, usually denoted by the Greek letter \( \psi(x) \), is a crucial part of describing the behavior of particles at a subatomic level.

  • It's a complex-valued function of position \( x \) and possibly time.
  • The square of the wave function's absolute value, \(|\psi(x)|^2\), gives us the probability density of finding a particle at a particular position \( x \).
In the given exercise, the wave function is provided as \( \psi(x)=A x e^{-x^2 / 2b^2} \). This indicates a Gaussian-type function, which illustrates how the particle's probability amplitude distributes around a center point.

Understanding this is essential because at the quantum level, particles don't possess definite positions. Instead, they are defined by probabilities, which the wave function helps calculate.
Potential Energy
Potential energy in quantum mechanics is a measure of the potential energy landscape that affects a particle's dynamics. It is integral in understanding how particles behave under different forces and configurations.

  • In classical physics, potential energy is often associated with forces like gravity or electromagnetism.
  • In quantum mechanics, it forms a core component of the total energy of a system, often represented as \( U(x) \).
The exercise derives the potential energy using the expression \( U(x) = \frac{h^2}{2mb^4}x^2 - \frac{3h^2}{2mb^2} \). This formula shows a potential that changes with \( x \), indicating variable influence on particle behavior as its position changes.

The expression reflects the system's potential energy where the total energy is set to zero, aligning with the principle that waves can oscillate around potential energy wells. Grasping how potential energy shapes the wave function is key to mastering quantum mechanics.
Schr枚dinger Equation
The Schr枚dinger Equation is a cornerstone of quantum mechanics. It allows us to calculate how quantum systems evolve over time and describes how a particle's wave function changes.

  • The equation is given as \( -\frac{h^2}{2m} \frac{d^2\psi}{dx^2} + U(x)\psi = E\psi \).
  • This equation combines both kinetic and potential energy terms into one framework.
For the given exercise, we use the time-independent Schr枚dinger Equation. Here, the total energy \( E \) is a constant, set to zero. It is used to equate potential and kinetic energy in the expression \( U = E - K \), as done in the step-by-step solution.

Understanding and applying the Schr枚dinger Equation lets us predict how quantum states progress, making it pivotal for anyone delving into quantum behavior. It integrates the wave function, potential energy, and total energy, highlighting their interconnections.

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Most popular questions from this chapter

A student of classical physics says, "A charged particle. like an electron orbiting in u simple atom. shouldn't have only certain stable energies: in fact, it should lose energy by electromagnetic radiation until the atom collapses." Answer these two complaints qualitatively. appealing to as few fundumentul claims of quantum mechanics as possible

In Section \(5.3\), we leamed that to be normaliable, a wave function (1) must not itself diverge and \((2)\) must fall to 0 faster than \(|\mathrm{r}|^{-1 / 2}\) as \(x\) gets lar ge. Nevertheless, we find two functions that slightly violate these requirements very useful. Consider the quantum mechanical plane wave Aelue-aul and the weird function \(\psi_{x_{1}}(x)\) pictured in Figure \(5.19\). which we here call by its preper name, the Direc delta function. (a) Which of the iwo normalizability requirements is violated by the plane wave, and which by the Dirac delta function? (b) Nomalization of the plane Wave could be accomplished if it were simply truncated, restricted to the rejion \(-b

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 Schr枚dinger 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?

Exercises \(90-92\) refer to a particle described by the wave function $$ \psi(x)=\sqrt{\frac{2}{\pi}} a^{3 / 2} \frac{1}{x^{2}+a^{2}} $$ Calculate the uncertainty in the particle's position.

A comet in an extremely elliptical orbit about a star has, of course, a maximum orbit radius. By comparison, its minimum orbit radius may be nearly \(0 .\) Make plots of the potential energy and a plausible total energy \(E\) versus radius on the same set of axes. Identify the classical turning points on your plot.
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