# Infinite square well – particle in left half

Required math: calculus

Required physics: Schrödinger equation

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 2.8.

As another example of an explicit case of a particle in the infinite square well, we can consider a particle that starts off in a state where it is equally likely to be found anywhere in the left half of the well. This means that the wave function is:

$\displaystyle \Psi(x,0)=\begin{cases} A & 0

and zero everywhere else.

Normalizing, we require

$\displaystyle \int_{0}^{a/2}A^{2}dx=1 \ \ \ \ \ (2)$

so

$\displaystyle A=\sqrt{\frac{2}{a}} \ \ \ \ \ (3)$

To find the probability that the particle is in the ground state (with energy ${\pi^{2}\hbar^{2}/2ma^{2}}$), we need to find the coefficient ${c_{1}}$ in the expansion of ${\Psi(x,0)}$ in terms of the orthonormal function set. Thus:

 $\displaystyle c_{1}$ $\displaystyle =$ $\displaystyle \int_{0}^{a}\Psi(x,0)\psi_{1}(x)dx\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2}{a}\int_{0}^{a/2}\sin(\pi x/a)dx\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2}{\pi} \ \ \ \ \ (6)$

The probability of this energy is then ${c_{1}^{2}=4/\pi^{2}\approx0.405285}$.

## 5 thoughts on “Infinite square well – particle in left half”

1. growescience

The wave function is discontinuous at ${x=0}$ and at ${x=\frac{a}{2}}$, so technically it doesn’t satisfy Born’s conditions, but we can take it as a limiting case of a wave function that declines rapidly to zero at ${x=0}$ and at ${x=\frac{a}{2}}$.

1. andromeda

Why does the likelihood of the particle being equal on either side indicate an initial wave function of x? Wouldn’t that make the particle more likely to be on the right side, since there will be more area under the function there?

1. gwrowe Post author

The wave function in equation 1 isn’t equal to ${x}$. It’s a constant ${A}$ in the left half of the well and zero in the right half.