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Chapter 40: Problem 15

Consider a particle moving in one dimension, which we shall call the \(x\)-axis. (a) What does it mean for the wave function of this particle to be \(normalized\)? (b) Is the wave function \(\psi(x) = e^{ax}\) , where a is a positive real number, normalized? Could this be a valid wave function? (c) If the particle described by the wave function \(\psi(x) = Ae^{-bx}\), where \(A\) and \(b\) are positive real numbers, is confined to the range \(x \geq 0\), determine A (including its units) so that the wave function is normalized.

Short Answer

Expert verified
(a) Normalization means the integral of the wave function's squared modulus over all space equals 1. (b) \(e^{ax}\) is not normalized and not a valid wave function. (c) For \(Ae^{-bx}\), \(A = \sqrt{2b}\) with units \(m^{-1/2}\).

Step by step solution

01

Define Normalization

For a wave function \( \psi(x) \) to be normalized, the integral of the square of its absolute value over all space must be equal to 1:\[\int_{-\infty}^{\infty} |\psi(x)|^2 \, dx = 1.\] It ensures the total probability of finding the particle somewhere on the x-axis is 1.
02

Check Normalization of \( e^{ax} \)

Consider the wave function \( \psi(x) = e^{ax} \). Compute its normalization:\[\int_{-\infty}^{\infty} |e^{ax}|^2 \, dx = \int_{-\infty}^{\infty} e^{2ax} \, dx.\] Since \(a > 0\), \( e^{2ax} \) grows indefinitely for \(x \to \infty\), thus the integral does not converge. Hence, \( e^{ax} \) cannot be normalized.
03

Evaluate Validity of \( e^{ax} \)

A valid wave function must be normalizable and typically tends to zero as \(x\) approaches \(\pm \infty\). Since \( \psi(x) = e^{ax} \) doesn't satisfy these conditions, it can't be a valid wave function in this context.
04

Normalize \( Ae^{-bx} \) for \(x \geq 0\)

Given \( \psi(x) = Ae^{-bx} \) for \(x \geq 0\), its normalization condition is:\[ \int_0^{\infty} |Ae^{-bx}|^2 \, dx = 1. \]Computing this gives:\[ \int_0^{\infty} A^2 e^{-2bx} \, dx = 1. \]
05

Solve the Integral

Solve the integral \( \int_0^{\infty} e^{-2bx} \, dx = \frac{1}{2b} \). Thus, the normalization condition becomes:\[ A^2 \cdot \frac{1}{2b} = 1. \]Solving for \(A\) gives:\[ A^2 = 2b \quad \Rightarrow \quad A = \sqrt{2b}. \]
06

Determine Units of \(A\)

Since \([\psi]\) must be \(m^{-1/2}\) for the wave function probability density \(|\psi(x)|^2\) to have units of \(length^{-1}\) (to ensure dimensionless integration), and \(b\) itself is in \(length^{-1}\), \(A\) should also have units \(m^{-1/2}\) to balance units.

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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 branch of physics. It describes the peculiar behaviors of particles at atomic and subatomic levels. In this realm, particles like electrons behave both as particles and waves, leading to fascinating consequences. One key concept of quantum mechanics is the wave function: a mathematical description of the quantum state of a system. It contains all the information about a particle's position and momentum.
  • Wave functions are typically denoted by the Greek letter \(\psi\).
  • The square of the wave function鈥檚 absolute value, \(|\psi(x)|^2\), gives the probability density of finding a particle at a particular position \(x\).
This probability view is revolutionary, as it shifts the perspective from determinism to probabilities. Instead of knowing exactly where a particle is, we calculate where it might be. This inherently probabilistic nature of quantum mechanics is one of its most iconic features.
Particle in One Dimension
Imagine a particle moving along a straight line, such as the x-axis. In quantum mechanics, this is known as a one-dimensional system. Studying a particle in one dimension simplifies its description and helps in understanding broader quantum mechanics concepts.
In this scenario, the wave function \(\psi(x)\) encapsulates all potential information about the particle's position.
  • For a meaningful interpretation of quantum mechanics, the wave function should be normalizable.
  • Normalizable means that the total probability of finding the particle somewhere along the x-axis is 1, which requires a proper wave function behavior as \(x\) moves towards infinity.
Simplifying to one dimension doesn't sacrifice the complexity but provides more tractable equations. This is crucial for focusing on key quantum behaviors and mathematical techniques.
Normalization Condition
The concept of wave function normalization is essential in quantum mechanics. For a wave function \(\psi(x)\) to be physically meaningful, it must satisfy the normalization condition: \[ \int_{-\infty}^{\infty} |\psi(x)|^2 \, dx = 1.\]This integral equation ensures that the total probability of locating the particle somewhere on the x-axis sums up to one.
In practical terms, this means:
  • Normalizable wave functions should approach zero as \(x\) tends to \(\pm\infty\).
  • For a specific given range, like \(x \geq 0\), integration limits adjust accordingly to find the normalization constant.
For instance, the wave function \(\psi(x) = Ae^{-bx}\) for \(x \geq 0\) is normalized by adjusting the constant \(A\). To find \(A\), we solve its integral:\[ \int_0^{\infty} |Ae^{-bx}|^2 \, dx = 1.\]Solving this gives \(A = \sqrt{2b}\), allowing for a completely normalized wave function. Understanding this process is pivotal in ensuring the use of correct wave functions and accurate quantum predictions.

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Most popular questions from this chapter

Consider a beam of free particles that move with velocity \(v = p/m\) in the \(x\)-direction and are incident on a potentialenergy step \(U(x)\) = 0, for \(x <\) 0, and \(U(x) = U_0 < E\), for \(x >\) 0. The wave function for \(x <\) 0 is \(\psi(x) = Ae^{ik_1x} + Be^{-ik_1x}\), representing incident and reflected particles, and for \(x >\) 0 is \(\psi(x) = Ce^{ik_2x}\), representing transmitted particles. Use the conditions that both \(\psi\) and its first derivative must be continuous at \(x\) = 0 to find the constants \(B\) and \(C\) in terms of \(k_1\), \(k_2\), and \(A\).

(a) Show by direct substitution in the Schr\(\ddot{o}\)dinger equation for the one-dimensional harmonic oscillator that the wave function \(\psi_1(x) = A_1xe^{-a^2x^2/2}\), where \(\alpha^2 = m\omega/\hslash\), is a solution with energy corresponding to \(n\) = 1 in Eq. (40.46). (b) Find the normalization constant A1. (c) Show that the probability density has a minimum at \(x\) = 0 and maxima at \(x = \pm1/\alpha\), corresponding to the classical turning points for the ground state \(n\) = 0.

An electron is moving as a free particle in the -\(x\)-direction with momentum that has magnitude 4.50 \(\times\) 10\(^{-24}\) kg \(\bullet\) m/s. What is the one- dimensional time-dependent wave function of the electron?

A free particle moving in one dimension has wave function $$\Psi(x, t) = A[e^{i(kx-vt)} - e^{i(2kx-4vt)}]$$ where \(k\) and \(\omega\) are positive real constants. (a) At \(t\) = 0 what are the two smallest positive values of \(x\) for which the probability function \(\mid \Psi(x, t) \mid ^2\) is a maximum? (b) Repeat part (a) for time \(t = 2\pi/\omega\). (c) Calculate \(v_{av}\) as the distance the maxima have moved divided by the elapsed time. Compare your result to the expression \(v_{av} = (\omega_2 - \omega_1)/(k_2 - k_1)\) from Example 40.1.

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