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Question: From equation (6.33), we conclude that the group velocity of a matter wave group equals the velocity V₀ of the massive particle with which it is associated. However both the dispersion relation used to show that ${\mathit{v}}_{\mathbf{g}\mathbf{r}\mathbf{o}\mathbf{u}\mathbf{p}}\mathbf{=}\mathit{h}{\mathit{k}}_{\mathbf{0}}\mathbf{/}\mathit{m}$ and the formula used to relatethis to the particle velocity are relativistically incorrect. It might be argued that we proved what we wished to prove by making an even number of mistakes.

a. Using the relativistically correct dispersion relation given in Exercise 4.1show that the group velocity of a wave pulse is actually given by

${\mathit{v}}_{\mathbf{g}\mathbf{r}\mathbf{o}\mathbf{u}\mathbf{p}}\mathbf{=}\frac{\mathbf{h}{\mathbf{k}}_{\mathbf{0}}{\mathbf{c}}^{\mathbf{2}}}{\sqrt{{\left(h{k}_0\right)}^{\mathbf{2}}{\mathbf{c}}^{\mathbf{2}}\mathbf{+}{\mathbf{m}}^{\mathbf{2}}{\mathbf{c}}^{\mathbf{4}}}}$

b. The fundamental relationship$\mathit{ÒÏ}\mathbf{=}\mathit{h}\mathit{k}$ is universally connect. So is indeed the particle momentum $\mathit{ÒÏ}$. (it is not well defined. but this is its approximate or central value) Making this substitution in the expression for${\mathit{v}}_{\mathbf{g}\mathbf{r}\mathbf{o}\mathbf{u}\mathbf{p}}$from part (a). then using the relativistically correct, relationship between momentum$\mathit{ÒÏ}$ and panicle velocity v,show that the group velocity again is equal to the panicle velocity.

Step by step solution

01

A concept:

Write the expression for angular frequency.

$\mathrm{Ó\neg }=\sqrt{k^2{c}^2+\frac{m^2{c}^4{k}^2}{h^2}}$ ….. (1)

The relativistic correct dispersion relation connects is the angular frequency, is the wave number, is the speed of light, is the mass of particles and is the Plank constant.

02

(a) Determine Vgroup : 

The group velocity is derived from angular velocity$\mathrm{Ó\neg }$ and the wave number k .and this is evaluated by the median wave number of group k₀ .

$$\begin{gathered} v_{group}={\left.\frac{d\mathrm{Ó\neg }}{dk}\right|}_{k_0} \\ =\frac{1}{2}{\left(k^2{c}^2+\frac{m^2{c}^4}{h^2}\right)}^{-1/2}\left(2k{c}^2\right) \\ ==\frac{h{k}_0{c}^2}{\sqrt{{\left(h{k}_{0}c\right)}^2+m^2{c}^4}} \end{gathered}$$

Thus, the group velocity of group pulse is $v_{group}=\frac{h{k}_0{c}^2}{\sqrt{{\left(h{k}_{0}c\right)}^2+m^2{c}^4}}$.

03

 Step 3: (b) Determine the group velocity again is equal to the panicle velocity:

Replace Planks constant and the median wave number with momentum .

$$\begin{gathered} v_{group}=\frac{h{k}_0{c}^2}{\sqrt{{\left(h{k}_{0}c\right)}^2+m^2{c}^4}} \\ =\frac{p{c}^2}{\sqrt{{\left(pc\right)}^2+m^2{c}^4}} \end{gathered}$$

Here, k is the wave number, c is the speed of light, m is the mass of particles and is the Plank constant.

$p=\frac{m{v}_{part}}{\sqrt{1-\frac{{V_{part}}^2}{c^2}}}$

Substitute the expression for momentum into equation for group velocity.

$$\begin{gathered} v_{group}=\frac{m{v}_{part}{c}^2}{\sqrt{1-\frac{{V_{part}}^2}{c^2}}\sqrt{\frac{m{v}_{part}}{\sqrt{1-\frac{{V_{part}}^2}{c^2}}}+m^2{c}^4}} \\ =\frac{v_{part}{c}^2}{\sqrt{{\left(v_{part}c\right)}^2+\left(1-\frac{{V_{part}}^2}{c^2}\right)}} \\ =v_{part} \end{gathered}$$

Hence, the group velocity is equal to the partial velocity.

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