91Ó°ÊÓ

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:

${\mathrm{ψ}}_{\mathrm{n}}\left(\mathrm{x}\right)=\sqrt{\frac{2}{\mathrm{a}}}\sin \left(\frac{\mathrm{²ÔÏ€}}{\mathrm{a}}\mathrm{x}\right)$

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

$\left(x\right)=\frac{2}{a}{∫}_0^a×\sin \left(\frac{n\mathrm{π}}{a}x\right)dx$

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:

$$\begin{gathered} ⟨x^2âŸ\copyright =\frac{2}{a}{\left(\frac{a}{n\mathrm{Ï€}}\right)}^3{∫}_0^{n\mathrm{Ï€}}{y}^2\sin ydy \\ ⟨x^2âŸ\copyright =a^2\left(\frac{1}{3}-\frac{1}{2n^2{\mathrm{Ï€}}^2}\right) \end{gathered}$$

The expectation value for the momentum operator is:

$$\begin{gathered} \left(p\right)=-i\mathrm{Ä\S }\frac{2}{a}{∫}_0^a\sin \left(\frac{n\mathrm{Ï€}}{a}x\right)\cos \left(\frac{n\mathrm{Ï€}}{a}x\right)dx \\ \left(p\right)=\frac{-i\mathrm{Ä\S }2}{a}\frac{a}{n\mathrm{Ï€}}{∫}_0^{n\mathrm{Ï€}}\sin ysocydy \\ \left(p\right)=0 \end{gathered}$$

$$\begin{gathered} \left(p^2\right)=-\mathrm{Ä\S }\frac{2}{a}{\left(\frac{n\mathrm{Ï€}}{a}\right)}^2{∫}_0^a{\sin }^2\left(\frac{n\mathrm{Ï€}}{a}x\right)dx \\ \left(p^2\right)={\left(\frac{\mathrm{Ä\S }\mathrm{Ï€²Ô}}{a}\right)}^2 \\ \left(p^2\right)={\left(\frac{\mathrm{Ä\S }\mathrm{Ï€²Ô}}{a}\right)}^2 \end{gathered}$$

Find the variance for position and momentum:

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

Substitute the values, and we get,

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

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

Substitute the values, and we get,

${\mathrm{σ}}_p=\frac{\mathrm{Ä\S }n\mathrm{Ï€}}{a}$

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" $$\begin{gathered} {\mathrm{σ}}_x{\mathrm{σ}}_p=\frac{\mathrm{Ä\S }}{2}\sqrt{\frac{{\mathrm{Ï€}}^2}{3}}-2 \\ =1.136\frac{\mathrm{Ä\S }}{2} \end{gathered}$$

And

$1.136\frac{\mathrm{Ä\S }}{2}>\frac{\mathrm{Ä\S }}{2}$

The uncertainty principle is satisfied.

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

The required values are:

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