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Chapter 35: Problem 12

What are the units of the wave function \(\psi(x)\) in a one-dimensional situation?

Short Answer

Expert verified
The units of the one-dimensional wave function \(\psi(x)\) are \(1/\sqrt{L}\), where \(L\) represents length.

Step by step solution

01

Understand the concept of wave function

In quantum mechanics, the wave function is a function that describes the state of a given quantum system. More specifically, the absolute square of the wave function, \(|\psi(x)|^2\), represents the probability density function which gives the likelihood of the system's quantum state at each point in space.
02

Understanding probability density

Probability density is typically measured in terms of units per length for a one-dimensional system. That is, if we are considering a region of length \(L\) in space, then the probability of finding the quantum state in that region is given by the integral over that region of the probability density, which has the dimension \(1/L\). This dimension must be equal to the square of the magnitude of \(\psi(x)\).
03

Find the units of the wave function

Because the square of the magnitude of the wave function, \(|\psi(x)|^2\), has the units of \(1/L\), the wave function itself, \(\psi(x)\), must have the units of \(1/\sqrt{L}\). This is so when the square of \(\psi\) is calculated, the correct units of \(1/L\) are achieved, satisfying the necessary condition for probability density function.

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Most popular questions from this chapter

Your roommate is taking Newtonian physics, while you've moved on to quantum mechanics. He claims that QM can't be right, because he didn't see any evidence of quantized energy levels in a mass-spring harmonic oscillator experiment. You reply by calculating the spacing between energy levels of this system, which consists of a \(1-\mathrm{g}\) mass on a spring with \(k=80 \mathrm{N} / \mathrm{m}\) What is that spacing, and how does this help your argument?

The wave functions of Problem \(58,\) as well as their derivatives, need to be continuous at \(x=L\) if these functions are to represent the quantum state of a particle in the finite square well. (a) Show that these conditions lead to two equations: $$ \begin{array}{c} A \sin (\sqrt{\epsilon})=B e^{-\sqrt{\mu-\epsilon}} \\ \sqrt{\epsilon} A \cos (\sqrt{\epsilon})=-\sqrt{\mu-\epsilon} B e^{-\sqrt{\mu-\epsilon}} \end{array} $$ (b) then show that these lead to the single equation $$ \tan (\sqrt{\epsilon})=-\sqrt{\frac{\epsilon}{\mu-\epsilon}} $$

Suppose \(\psi_{1}\) and \(\psi_{2}\) are solutions of the Schr枚dinger equation for the same energy \(E .\) Show that the linear combination \(a \psi_{1}+b \psi_{2}\) is also a solution, where \(a\) and \(b\) are arbitrary constants.

The ground-state wave function for a quantum harmonic oscillator has a single central peak. Why is this at odds with classical physics?

In terms of de Broglie's matter-wave hypothesis, how does making the sides of a box different lengths remove the degeneracy associated with a particle confined to that box?
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