Question

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?

Short answer

The uncertainty principle is satisfied.

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

The required values are:

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

Step-by-step solution

Step 1: Define the Schrodinger equation

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.

Determine the uncertainty principle

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.$

Determine the expectation position and momentum

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:

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

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

Determine the variance of position and momentum

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:

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