# Uncertainty principle: an example

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 1.17.

Here’s another example of calculating the uncertainty principle. We have a wave function defined as

$\displaystyle \Psi\left(x,0\right)=\begin{cases} A\left(a^{2}-x^{2}\right) & -a\le x\le a\\ 0 & \mbox{otherwise} \end{cases} \ \ \ \ \ (1)$

The constant ${A}$ is determined by normalization in the usual way:

 $\displaystyle \int_{-a}^{a}\left|\Psi\right|^{2}dx$ $\displaystyle =$ $\displaystyle 1\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle A^{2}\left.\left(\frac{x^{5}}{5}-\frac{2}{3}a^{2}x^{3}+a^{4}x\right)\right|_{-a}^{a}\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle A^{2}\frac{16a^{5}}{15}\ \ \ \ \ (4)$ $\displaystyle A$ $\displaystyle =$ $\displaystyle \frac{\sqrt{15}}{4a^{5/2}} \ \ \ \ \ (5)$

The expectation value of ${x}$ is ${\left\langle x\right\rangle =0}$ from the symmetry of the wave function. The expectation value of ${p}$ is

 $\displaystyle \left\langle p\right\rangle$ $\displaystyle =$ $\displaystyle -i\hbar\int_{-a}^{a}\Psi^*\frac{\partial}{\partial x}\Psi dx\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -i\hbar\int_{-a}^{a}\left(-2Ax\right)A\left(a^{2}-x^{2}\right)\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (8)$

[We can’t calculate ${\left\langle p\right\rangle =\frac{d}{dt}\left(m\left\langle x\right\rangle \right)}$ in this case, because we know the value of ${\left\langle x\right\rangle }$ only at one specific time (${t=0}$), so we don’t have enough information to calculate its derivative.]

The remaining statistics are (the integrals are all just integrals of polynomials, so nothing complicated):

 $\displaystyle \left\langle x^{2}\right\rangle$ $\displaystyle =$ $\displaystyle \int_{-a}^{a}x^{2}\left|\Psi\right|^{2}dx\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{15}{16a^{5}}\left.\left(\frac{x^{7}}{7}-\frac{2}{5}a^{2}x^{5}+\frac{1}{3}a^{4}x^{3}\right)\right|_{-a}^{a}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{a^{2}}{7}\ \ \ \ \ (11)$ $\displaystyle \left\langle p^{2}\right\rangle$ $\displaystyle =$ $\displaystyle -\hbar^{2}\int_{-a}^{a}\Psi^*\frac{\partial^{2}}{\partial x^{2}}\Psi dx\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{15\hbar^{2}}{8a^{5}}\left.\left(a^{2}-x^{2}\right)\right|_{-a}^{a}\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{5\hbar^{2}}{2a^{2}}\ \ \ \ \ (14)$ $\displaystyle \sigma_{x}$ $\displaystyle =$ $\displaystyle \sqrt{\left\langle x^{2}\right\rangle -\left\langle x\right\rangle ^{2}}\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{a}{\sqrt{7}}\ \ \ \ \ (16)$ $\displaystyle \sigma_{p}$ $\displaystyle =$ $\displaystyle \sqrt{\left\langle p^{2}\right\rangle -\left\langle p\right\rangle ^{2}}\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{5}{2}}\frac{\hbar}{a}\ \ \ \ \ (18)$ $\displaystyle \sigma_{x}\sigma_{p}$ $\displaystyle =$ $\displaystyle \sqrt{\frac{5}{14}}\hbar\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle \cong$ $\displaystyle 0.598\hbar>\frac{\hbar}{2} \ \ \ \ \ (20)$

Thus the uncertainty principle is satisfied in this case.

## 3 thoughts on “Uncertainty principle: an example”

1. Valco

Hi there….

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