What Are the Eigenfunctions for the 1D Infinite Square Well?

Click For Summary
SUMMARY

The eigenfunctions for the 1D infinite square well with boundaries at -L/2 and +L/2 are definitively given by the equations $$\psi_1(x) = \sqrt{\frac{2}{L}}\cos{\frac{\pi x}{L}}$$ for the ground state and $$\psi_2(x) = \sqrt{\frac{2}{L}}\sin{\frac{2\pi x}{L}}$$ for the first excited state. These solutions arise from solving the time-independent Schrödinger equation $$\frac{-\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\psi(x) = E\psi(x)$$ under the specified boundary conditions. Each eigenfunction corresponds to a quantized energy level, with the number of nodes in the wave function reflecting the state of excitation.

PREREQUISITES
  • Understanding of quantum mechanics principles, particularly the Schrödinger equation.
  • Familiarity with boundary conditions in quantum systems.
  • Knowledge of eigenfunctions and eigenvalues in the context of differential equations.
  • Basic grasp of trigonometric functions and their properties.
NEXT STEPS
  • Study the derivation of eigenfunctions for the 1D infinite square well in greater detail.
  • Explore the implications of boundary conditions on quantum states.
  • Learn about the quantization of energy levels in quantum mechanics.
  • Investigate the role of nodes in wave functions and their physical significance.
USEFUL FOR

Students and educators in quantum mechanics, physicists studying wave functions, and anyone interested in the mathematical foundations of quantum systems.

andre220
Messages
75
Reaction score
1

Homework Statement



Find the ground and first excited state eigenfunctions of for the 1D infinite square well with boundaries -L/2 and +L/2

Homework Equations


$$\frac{-\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\psi(x) = E\psi(x)$$

The Attempt at a Solution


Okay so I know how to solve it and get that $$\psi_1(x) = \sqrt{\frac{2}{L}}\cos{\frac{\pi x}{L}}$$. Next, I also know that $$\psi_2(x) = \sqrt{\frac{2}{L}}\sin{\frac{2\pi x}{L}}$$ one could reason this by arguing that each eigenfunction should have ##n-1## nodes. However, what is a more mathematical reasoning to ##\psi## for a given excited state. I am sure it is quite simple, I just can't seem to see it.
 
Physics news on Phys.org
I'm not sure exactly what your question is. If you solve the Schrödinger equation with the appropriate boundary conditions — that is, solve the math problem — those are two of the solutions you get.
 

Similar threads

Replies
16
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
7
Views
2K
Replies
7
Views
3K
Replies
3
Views
2K
Replies
4
Views
5K
  • · Replies 39 ·
2
Replies
39
Views
13K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K