# Delta function well: statistics

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 1.5.

The delta function well gives rise to a wave function that decays exponentially either side of the delta function:

$\displaystyle \Psi\left(x,t\right)=Ae^{-\lambda\left|x\right|}e^{-i\omega t} \ \ \ \ \ (1)$

We can normalize ${\Psi}$ in the usual way:

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

By symmetry, ${\left\langle x\right\rangle =0}$ and

$\displaystyle \left\langle x^{2}\right\rangle =2\lambda\int_{0}^{\infty}x^{2}e^{-2\lambda x}dx=\frac{1}{2\lambda^{2}} \ \ \ \ \ (6)$

Therefore

$\displaystyle \sigma=\sqrt{\left\langle x^{2}\right\rangle -\left\langle x\right\rangle ^{2}}=\frac{1}{\sqrt{2}\lambda} \ \ \ \ \ (7)$

A plot of ${\left|\Psi\right|^{2}}$ is shown, with vertical yellow lines indicating ${x=\pm\frac{1}{\sqrt{2}\lambda}}$, for the case ${\lambda=2}$:

The probability that the particle lies outside ${x=\pm\frac{1}{\sqrt{2}\lambda}}$ is

$\displaystyle P_{\left|x\right|>\sigma}=2\lambda\int_{1/\sqrt{2}\lambda}^{\infty}e^{-2\lambda x}dx=\frac{1}{e^{\sqrt{2}}}=0.2431 \ \ \ \ \ (8)$

In this case, the probability of ${x}$ being greater than one standard deviation is a constant, independent of ${\lambda}$.

# 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.