# Triangular wave function: probabilities

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 1.4.

The square modulus of the wave function which is the solution to the Schrödinger equation is interpreted as a probability density. As an example consider the wave function given by

$\displaystyle \Psi\left(x,0\right)=\begin{cases} A\frac{x}{a} & 0\le x\le a\\ A\frac{b-x}{b-a} & a\le x\le b\\ 0 & \mbox{otherwise} \end{cases} \ \ \ \ \ (1)$

We can normalize ${\Psi}$ by requiring

$\displaystyle \int_{0}^{b}\left|\Psi\right|^{2}dx=1 \ \ \ \ \ (2)$

Plugging in the formula and doing the integral gives

 $\displaystyle \int_{0}^{b}\left|\Psi\right|^{2}dx$ $\displaystyle =$ $\displaystyle \left|A\right|^{2}\left[\int_{0}^{a}\frac{x^{2}}{a^{2}}dx+\int_{a}^{b}\left(\frac{b-x}{b-a}\right)^{2}dx\right]\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left|A\right|^{2}\frac{b}{3}\ \ \ \ \ (4)$ $\displaystyle A$ $\displaystyle =$ $\displaystyle \sqrt{\frac{3}{b}} \ \ \ \ \ (5)$

where we’ve taken the positive real root for ${A}$. Note that ${A}$ could also be multiplied by a phase factor ${e^{i\delta}}$ for any real ${\delta}$ without affecting normalization. This can be important in some applications where we need to add together wave functions.

Given this value for ${A}$, we can plot 1. Here, we’ve taken ${a=1}$ and ${b=3}$:

Since ${\Psi}$ has its maximum at ${x=a}$, that is where the particle is most likely to be found. The probability of the particle being found to the left of ${x=a}$ is

$\displaystyle P_{x

If ${b=a}$, then ${\Psi}$ drops to zero at ${x=a}$ so ${P_{x. If ${b=2a}$, then ${\Psi}$ is an isosceles triangle symmetric about ${x=a}$ so ${P_{x.

The expectation value of ${x}$ is

$\displaystyle \left\langle x\right\rangle =\int x\left|\Psi\right|^{2}dx=\frac{a}{2}+\frac{b}{4} \ \ \ \ \ (7)$

where we used Maple to simplify the integration. If ${b=2a}$, then ${\left\langle x\right\rangle =a}$ as expected.

## 8 thoughts on “Triangular wave function: probabilities”

1. Deniska

Oh the wave function given in the question is the solution of schrodinger equation ( which may or may not be the actual wave function of the particle) but after normalizing we get the wave function(actual/correct), is my reasoning correct?

1. gwrowe Post author

Essentially, yes. The normalization is required to make the wave function consistent with the probabilistic interpretation, that is, equation 2 says that the probability of finding the particle somewhere in space must add up to 1.

1. gwrowe Post author

If you plug equation 1 into the integral in 7, the gory details are:

 $\displaystyle \int x\left|\Psi\right|^{2}dx$ $\displaystyle =$ $\displaystyle \int_{0}^{a}x\left(\sqrt{\frac{3}{b}}\frac{x}{a}\right)^{2}dx+\int_{a}^{b}x\left(\sqrt{\frac{3}{b}}\frac{b-x}{b-a}\right)^{2}dx$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{3}{4}\,{\frac{{a}^{2}}{b}}+\frac{3}{4}\,{\frac{-{a}^{4}+{b}^{4}}{b\left(-b+a\right)^{2}}}-2\,{\frac{-{a}^{3}+{b}^{3}}{\left(-b+a\right)^{2}}}+\frac{3}{2}\,{\frac{b\left(-{a}^{2}+{b}^{2}\right)}{\left(-b+a\right)^{2}}}$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{3}{4}\,{\frac{{a}^{2}}{b}}+\frac{-\frac{3}{4}\,{a}^{4}+\frac{3}{4}\,{b}^{4}-2\,b\left(-{a}^{3}+{b}^{3}\right)+\frac{3}{2}\,{b}^{2}\left(-{a}^{2}+{b}^{2}\right)}{b\left(-b+a\right)^{2}}$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{3}{4}\,{\frac{{a}^{2}}{b}}+\frac{-\frac{3}{4}\,{a}^{4}+\frac{1}{4}\,{b}^{4}+2\,{a}^{3}b-\frac{3}{2}\,{a}^{2}{b}^{2}}{b\left(-b+a\right)^{2}}$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{3}{4}\,{\frac{{a}^{2}}{b}}-\frac{1}{4}\frac{\left(3\,a+b\right)\left(-b+a\right)^{3}}{b\left(-b+a\right)^{2}}$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{3}{4}\,{\frac{{a}^{2}}{b}}-\frac{1}{4}\,{\frac{3\,{a}^{2}-2\,ab-{b}^{2}}{b}}$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{a}{2}+\frac{b}{4}$

Now you can see why I use Maple to do the integrals and algebra(!)

1. gwrowe Post author

You just have to multiply the integrands out and then do the integrals. It’s a lot of work, but there’s nothing fancy about it.