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….

    first of all, thanks for this amazing work. It is very helpful in my study.
    About this exercise, equation 9 is wrong, a small copy-paste mistake, I think.
    x^2 is missing at the integral.

    Thanks

    Reply

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