Question

From equations (6-23) and (6-29) obtain the dispersion coefficient for matter waves (in vacuum), then show that probability density (6-35) follows from (6-28)

Short answer

The dispersion coefficient of matter waves in vacuum is $\frac{h}{m}$and it is verified that probability density is followed from equation (6-28)

Step-by-step solution

Definition of dispersion relation

The relation between angular frequency and the wave number of matter wave is termed as the Dispersion relation

Determination of the value of coefficient of dispersion

Write the expression of matter wave dispersion relation from (6–23).

$\mathrm{\omega }\left(\mathrm{k}\right)=\frac{{\mathrm{hk}}^2}{2\mathrm{m}}$

Here, $\omega$is angular frequency of matter wave, k is wave number and m is mass of corresponding particle.

Write the expression of coefficient of dispersion from (6–29).

$\mathrm{D}=\frac{{\mathrm{d}}^2\mathrm{\omega }\left(\mathrm{k}\right)}{{\mathrm{dk}}^2}$

Substitute $\frac{{\mathrm{hk}}^2}{2\mathrm{m}}$for $\omega \left(k\right)$in the above expression.

$\begin{array}{rcl}D& =& \frac{d^2\left(\frac{h{k}^2}{2m}\right)}{d{k}^2}\\ & =& \frac{h}{m}\end{array}$

Verification of the given condition

Write the expression of probability density from (6–28).

$\left|\mathrm{\Psi }\right(\mathrm{x},\mathrm{t}){|}^2=\frac{{\mathrm{c}}^2}{{\left(1+\frac{{\mathrm{D}}^2{\mathrm{t}}^2}{4{\mathrm{\epsilon }}^4}\right)}^{\frac{1}{2}}}\exp \left(\frac{-{(\mathrm{x}-\mathrm{st})}^2}{2{\mathrm{\epsilon }}^2\left(1+\frac{{\mathrm{D}}^2{\mathrm{t}}^2}{4{\mathrm{\epsilon }}^4}\right)}\right)$

Here, C is amplitude of wave function and t is any arbitrary time.

Substitute$\frac{h}{m}$for D in above expression.

$\begin{array}{rcl}\left|\mathrm{\Psi }\right(\mathrm{x},\mathrm{t}){|}^2& =& \frac{{\mathrm{c}}^2}{{\left(1+\frac{{\left(\frac{\mathrm{h}}{\mathrm{m}}\right)}^2{\mathrm{t}}^2}{4{\mathrm{\epsilon }}^4}\right)}^{\frac{1}{2}}}\exp \left(\frac{-{(\mathrm{x}-\mathrm{st})}^2}{2{\mathrm{\epsilon }}^2\left(1+\frac{{\left(\frac{\mathrm{h}}{\mathrm{m}}\right)}^2{\mathrm{t}}^2}{4{\mathrm{\epsilon }}^4}\right)}\right)\\ & =& \frac{{\mathrm{C}}^2}{{\left(1+\frac{{\mathrm{h}}^2{\mathrm{t}}^2}{4{\mathrm{m}}^2{\mathrm{\epsilon }}^4}\right)}^{1/2}}\exp \left(\frac{-{(\mathrm{x}-\mathrm{st})}^2}{2{\mathrm{\epsilon }}^2\left(1+\frac{{\mathrm{h}}^2{\mathrm{t}}^2}{4{\mathrm{m}}^2{\mathrm{\epsilon }}^4}\right)}\right)\end{array}$

It can be observed that the above expression resembles equation (6-35) therefore it is verified that it follows from equation (6-28).

Therefore, the dispersion coefficient of matter waves in vacuum is $\frac{h}{m}$and it is verified that probability density is followed from equation (6–28).