What Is the Normalization Constant and Lowest Energy of an Electron in a 1D Box?

  • Thread starter Thread starter Domnu
  • Start date Start date
  • Tags Tags
    Box Electron
Click For Summary
SUMMARY

The normalization constant for an electron in a one-dimensional box with walls at x=(0,a) is determined to be A = √(1/a). The lowest energy of the electron in this quantum state is non-zero due to confinement, as dictated by the Heisenberg uncertainty principle. The wavefunction provided, which is a step function, does not correspond to the typical sine functions of energy eigenstates, necessitating a Fourier series representation to express measurable energies. This highlights the unique characteristics of the quantum state in question.

PREREQUISITES
  • Quantum mechanics fundamentals
  • Understanding of wavefunctions and normalization
  • Familiarity with the Heisenberg uncertainty principle
  • Knowledge of Fourier series and energy eigenfunctions
NEXT STEPS
  • Study the derivation of wavefunctions for a particle in a one-dimensional infinite potential well
  • Learn about the implications of the Heisenberg uncertainty principle in quantum mechanics
  • Explore Fourier series and their application in quantum state analysis
  • Investigate the properties of delta functions and their role in quantum mechanics
USEFUL FOR

Students and professionals in physics, particularly those focusing on quantum mechanics, wavefunction analysis, and energy quantization in confined systems.

Domnu
Messages
176
Reaction score
0
Problem
An electron in a one-dimensional box with walls at [tex]x=(0,a)[/tex] is in the quantum state

[tex]\psi(x) = A, 0<x<a/2[/tex]
[tex]\psi(x) = -A, a/2 < x < a[/tex]

Obtain an expression for the normalization constant, [tex]A[/tex]. What is the lowest energy of the electron that will be measured in this state?

Solution

Well, we know that

[tex]\int_{0}^A |\psi(x)|^2 dx = 1 \iff A^2 \cdot a = 1 \iff A = \sqrt{\frac{1}{a}}[/tex]

so this is our normalization constant, [tex]A[/tex]. Now, to find the lowest energy of the electrong that will be measured in this state, we have:

[tex]-\frac{\hbar^2}{2m} \frac{\partial^2 \psi}{\partial x^2} = E \psi[/tex]

but the left hand side evaluates to 0 since [tex]\psi[/tex] is independent of [tex]x[/tex]. So this is the lowest (and only observed) energy in this state... is this right?
 
Physics news on Phys.org
Well, [tex]\psi[/tex] is independent of [tex]x[/tex] everywhere except around [tex]x = a / 2[/tex]. Around there things could get a little weird. (In fact, as this problem is stated, [tex]\psi[/tex] changes infinitely quickly from one side of [tex]a / 2[/tex] to the other, so around there things could get very weird.)

Even without doing any calculations, though, you should know that since the electron is stuck in the box, it must have nonzero energy. If nothing else, the box gives it [tex]\Delta x < \infty[/tex], so it must have [tex]\Delta p > 0[/tex] by Heisenberg.

If you've already studied the infinite potential well (one-dimensional box), you might also know that the energy eigenfunctions (or stationary states) are all [tex]sin[/tex] functions (since one of the walls is at [tex]x = 0[/tex]). Now, the given wavefunction [tex]\psi[/tex] is not such a [tex]sin[/tex] function, so to get the measurable energies we have to rewrite it as a sum of the energy eigenfunctions -- that is, as a Fourier series of [tex]sin[/tex] functions.

Hopefully that's enough to get you started.
 
Well, its a step function so if you compute the derivative you get a delta function...and then a derivative of a delta function and so on.

By the way, if you solve for the wavefunction of a particle in a 1D box, you get sin or cosine functions. Any state the particle is in can be written as a linear combination of these eigenfunctions. This is a rather strange state...and interesting too :smile:
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 24 ·
Replies
24
Views
3K
  • · Replies 8 ·
Replies
8
Views
3K
Replies
29
Views
3K
  • · Replies 20 ·
Replies
20
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
16
Views
3K
  • · Replies 30 ·
2
Replies
30
Views
5K
Replies
9
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K