Calculate $\left(x\right)\mathbf{,}\left(x^2\right)\mathbf{,}\left(p\right)\mathbf{,}\left(p^2\right)\mathbf{,}{\mathit{\sigma }}_{\mathbf{x}}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}{\mathit{\sigma }}_{\mathbf{p}}$,for the nth stationary state of the infinite square well. Check that the uncertainty principle is satisfied. Which state comes closest to the uncertainty limit?

A differential equation that describes matter in quantum mechanics in terms of the wave-like properties of particles in a field. Its answer is related to a particle's probability density in space and time.

The stationary state for the infinite potential well is:

${\mathrm{\psi }}_{\mathrm{n}}\left(\mathrm{x}\right)=\sqrt{\frac{2}{\mathrm{a}}}\sin \left(\frac{\mathrm{n\pi }}{\mathrm{a}}\mathrm{x}\right)$

Calculate all the expectation values and the variance in that values as we did in chapter one. The expectation value of the position is:

$\left(x\right)=\frac{2}{a}{\int }_0^a\times \sin \left(\frac{n\mathrm{\pi }}{a}x\right)dx$

Using integration by parts, so we get:

$\left(x\right)=\raisebox{1ex}{$a$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$

The expectation for the position squared is:

$$\begin{gathered} ⟨x^2⟩=\frac{2}{a}{\left(\frac{a}{n\pi }\right)}^3{\int }_0^{n\mathrm{\pi }}{y}^2\sin ydy \\ ⟨x^2⟩=a^2\left(\frac{1}{3}-\frac{1}{2n^2{\pi }^2}\right) \end{gathered}$$

The expectation value for the momentum operator is:

$$\begin{gathered} \left(p\right)=-iħ\frac{2}{a}{\int }_0^a\sin \left(\frac{n\mathrm{\pi }}{a}x\right)\cos \left(\frac{n\mathrm{\pi }}{a}x\right)dx \\ \left(p\right)=\frac{-iħ2}{a}\frac{a}{n\mathrm{\pi }}{\int }_0^{n\mathrm{\pi }}\sin ysocydy \\ \left(p\right)=0 \end{gathered}$$

$$\begin{gathered} \left(p^2\right)=-ħ\frac{2}{a}{\left(\frac{n\mathrm{\pi }}{a}\right)}^2{\int }_0^a{\sin }^2\left(\frac{n\mathrm{\pi }}{a}x\right)dx \\ \left(p^2\right)={\left(\frac{ħ\mathrm{\pi n}}{a}\right)}^2 \\ \left(p^2\right)={\left(\frac{ħ\mathrm{\pi n}}{a}\right)}^2 \end{gathered}$$

Find the variance for position and momentum:

${\sigma }_x=\sqrt{\left(x^2\right)-\left(x^2\right)}$

Substitute the values, and we get,

${\sigma }_x=\frac{a}{2}\sqrt{\frac{1}{3}-\frac{2}{{\left(n\mathrm{\pi }\right)}^2}}$

${\sigma }_p=\sqrt{\left(p^2\right)-\left(p^2\right)}$

Substitute the values, and we get,

${\sigma }_p=\frac{ħn\mathrm{\pi }}{a}$

Finally, the closet state to the uncertainty limit is the state with the lowest possible energy (n = 1), where we can prove this by:

role="math" localid="1658122114260" $$\begin{gathered} {\sigma }_x{\sigma }_p=\frac{ħ}{2}\sqrt{\frac{{\mathrm{\pi }}^2}{3}}-2 \\ =1.136\frac{ħ}{2} \end{gathered}$$

And

$1.136\frac{ħ}{2}>\frac{ħ}{2}$

The uncertainty principle is satisfied.

For $1.136\frac{ħ}{2}>\frac{ħ}{2}$n = 1 is the state that comes closest to the uncertainty limit.

The required values are:

$$\begin{gathered} \left(x\right)=\raisebox{1ex}{$a$}\!\left/ \!\raisebox{-1ex}{$2$}\right. \\ \left(x^2\right)=a^2\left(\frac{1}{3}-\frac{1}{2n^2{\mathrm{\pi }}^2}\right) \\ \left(p\right)=0 \\ \left(p^2\right)={\left(\frac{ħn\mathrm{\pi }}{a}\right)}^2 \\ {\sigma }_x=\frac{a}{2}\sqrt{\frac{1}{3}-\frac{2}{{\left(n\mathrm{\pi }\right)}^2}} \\ {\sigma }_p=\frac{ħn\mathrm{\pi }}{a} \end{gathered}$$