91影视

From the masses of the weak bosons given in Table 12.1, show that range is of weak part of electroweak force should be about ${\mathbf{10}}^{\mathbf{鈭�}\mathbf{3}}\mathbf{\text{鈥塮尘}}$.

Someone proposes the existence of a new force whose range is10^-20 m. We found in Chapter 2 that accelerator turn kinetic energy into mass. About how much energy do you estimate an accelerator would need to create the mediating particle for such a force.

As in electron-positron annihilation, when a proton and antiproton annihilate. Twophoton can be produced. Ofwhat wavelength are of these photons?

To show that the Klein-Gordon equation has valid solutions for negative values of $\mathit{E}$, verify that equation (12-4) is satisfied by a wave function of the form .$\mathit{蠄}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{t}\mathbf{)}\mathbf{=}{\mathbf{Ae}}^{\mathbf{卤}\mathbf{ipx}\mathbf{/}\mathbf{鈩�}\mathbf{卤}\mathbf{iEt}\mathbf{/}\mathbf{鈩�}}$

To show,

(a) ${\mathit{唯}}_{\mathbf{1}}\mathbf{(}\mathit{x}\mathbf{,}\mathit{t}\mathbf{)}\mathbf{=}\mathit{A}{\mathit{e}}^{\mathbf{i}\mathbf{k}\mathbf{x}\mathbf{鈭�}\mathbf{i}\mathbf{蝇}\mathbf{t}}$is the solution of both Klein-Gordon and the Schrodinger equations.

(b) ${\mathit{唯}}_{\mathbf{2}}\mathbf{(}\mathit{x}\mathbf{,}\mathit{t}\mathbf{)}\mathbf{=}\mathit{A}{\mathit{e}}^{\mathbf{i}\mathbf{k}\mathbf{x}}\mathit{c}\mathit{o}\mathit{s}\mathit{蝇}\mathit{t}$is the solution of both Klein-Gordon but not the Schrodinger equations.

(c) The${\mathit{唯}}_{\mathbf{2}}$ is a combination of positive and negative energy solutions of the Klein-Gordon equation.

(d) To compare the time dependence of${\mathit{唯}}^{\mathbf{2}}$ for ${\mathit{唯}}_{\mathbf{1}}$and ${\mathit{唯}}_{\mathbf{2}}$.

For solutions of Klein-Gordon equation, the quantity,

$i{蠄}^{鈭�}\frac{鈭�}{鈭�t}蠄鈭�i蠄\frac{鈭�}{鈭�t}{蠄}^{鈭�}$is interpreted as charge density. Show that for a positive-energy plane-wavesolution. It is a real constant, and for negative-energy solution.It is a negative of that constant.

In non-relavistic quantum mechanics, governed by the Schrodinger equation, the probability of finding a particle does not change with time.

(a)

Prove it, Begin with the time derivative of the total probability

$\frac{d}{dt}鈭�{唯}^{*}(x,t){唯}^{\text{'}}(x,t)dx=鈭�\left(唯(x,t)\frac{鈭�}{鈭�t}{唯}^{*}(x,t)+{唯}^{*}(x,t)\frac{鈭�}{鈭�t}唯(x,t)\right)dx$

Then use the Schrodinger equation to eliminate the partial time derivatives, integrate by parts, and show that the result is zero. Assume that the particle is well localised, so that $蠄\text{鈥塧苍诲鈥�}\frac{鈭�蠄}{鈭�x}$are 0 when evaluated at .$卤\mathrm{鈭�}$

(b) Does this procedure lead to the same conclusion if Wave function obeyKlein-Gordon rather than Shrodinger equation? Why and why not?

The electron mentioned in Section 12.3 for deep inelastic scattering experiments is$20\text{GeV}$ , and the momentum is given as .$20\text{鈥塆别痴}/\text{c}$ Why so simple a conversion?

Approximately what energy would electrons need to have to be useful as probes that could reveal features as small as ${\mathbf{10}}^{\mathbf{-}\mathbf{18}}\mathbf{}\mathbf{m}$,the approximate range of the weak force?

Suppose a force between two particles decreases distance according to $F=k/r^b$. What is the limit on b if the energy required to separate the particlesInfinitely far is not to be infinite?