91影视

(a)

The original formula of the electric field is given by:

$\mathbf{E}={\mathrm{E}}_0\widehat{n}\cos (\mathbf{k}鈰�\mathbf{r}鈭�\mathrm{蝇迟})$ 鈥︹��. (1)

Here, $\mathrm{k}$is the direction of propagation,$\hat{\mathrm{n}}$is the direction of polarization and$\mathrm{蝇}=\mathrm{kc}$is the angular frequency. Assume that the wavelength is fairly large, then$\mathrm{kr}=2\mathrm{蟺谤}/\mathrm{位}$is fairly small, and thus we can use the first order approximations, to get:

$\mathbf{E}={\mathrm{E}}_0\widehat{n}\left[\cos \right(\mathbf{k}鈰�\mathbf{r}\left)\cos \right(\mathrm{蝇迟})+\sin (\mathbf{k}鈰�\mathbf{r}\left)\sin \right(\mathrm{蝇迟}\left)\right]$ 鈥︹�︹�� (2)

$E鈮�E_0\widehat{n}\left[\cos \right(蝇t)+\mathbf{k}鈰�\mathbf{r}\sin (蝇t\left)\right]$ 鈥︹�︹�� (3)

let used $\sin \left(\mathrm{x}\right)鈮�\mathrm{x}$for small $\mathrm{x}$and also, we use the subtract identity of the sine, that is $\sin (\mathrm{a}鈭�\mathrm{b})=\sin \left(\mathrm{a}\right)\cos \left(\mathrm{b}\right)鈭�\cos \left(\mathrm{a}\right)\sin \left(\mathrm{b}\right)$

To find the perturbation$H^{\text{'}}$to the Hamiltonian, we need to look up on the work done to move a charge of $q$from the origin (the centre of the atom) to the field point $r$that is:

$\mathrm{W}=鈭�\mathrm{q}{鈭�}_0^b\mathbf{E}鈰�\mathrm{d}{\mathbf{r}}^{\text{'}}$

The energy due to the second term of (2) is therefore:

${\mathrm{H}}^{\text{'}}=鈭�{\mathrm{qE}}_0\sin \left(\mathrm{蝇迟}\right){鈭�}_0^r(\mathbf{k}鈰�{\mathbf{r}}^{\text{'}})\widehat{n}鈰�\mathrm{d}{\mathbf{r}}^{\text{'}}$

Here, ${\mathbf{r}}^{\text{'}}$ is constant, and hence the angles between $\mathrm{k}$and ${\mathbf{r}}^{\text{'}}$,and between $\widehat{n}$and ${\mathbf{r}}^{\text{'}}$say ${胃}_k$and ${胃}_n$so we can write:

role="math" localid="1659008855557" $\begin{array}{c}\mathbf{k}鈰�{\mathbf{r}}^{\text{'}}=\left(\mathrm{kcos}\left({\mathrm{胃}}_{\mathrm{k}}\right)\right){\mathrm{r}}^{\text{'}}\\ \widehat{n}鈰�\mathrm{d}{\mathbf{r}}^{\text{'}}=\left(\cos \left({\mathrm{胃}}_{\mathrm{n}}\right)\right){\mathrm{dr}}^{\text{'}}.\end{array}$

Substitute into (3), so we get:

$\begin{array}{c}{\mathrm{H}}^{\text{'}}=鈭�{\mathrm{qE}}_0\mathrm{ksin}\left(\mathrm{蝇迟}\right)\cos \left({\mathrm{胃}}_{\mathrm{k}}\right)\cos \left({\mathrm{胃}}_{\mathrm{n}}\right){鈭�}_0^{\mathrm{r}}{\mathrm{r}}^{\text{'}}{\mathrm{dr}}^{\text{'}}\\ =鈭�\frac{1}{2}{\mathrm{qE}}_0\mathrm{ksin}\left(\mathrm{蝇迟}\right)\cos \left({\mathrm{胃}}_{\mathrm{k}}\right)\cos \left({\mathrm{胃}}_{\mathrm{n}}\right){\mathrm{r}}^2\\ =鈭�\frac{1}{2}{\mathrm{qE}}_0\sin \left(\mathrm{蝇迟}\right)\left[\mathrm{krcos}\left({\mathrm{胃}}_{\mathrm{k}}\right)\right]\left[\mathrm{rcos}\left({\mathrm{胃}}_{\mathrm{n}}\right)\right]\\ =鈭�\frac{1}{2}{\mathrm{qE}}_0\sin \left(\mathrm{蝇迟}\right)(\mathbf{k}鈰�\mathbf{r})(\widehat{n}鈰�\mathbf{r})\\ =鈭�\frac{1}{2}{\mathrm{qE}}_0\frac{\mathrm{蝇}}{\mathrm{c}}\sin \left(\mathrm{蝇迟}\right)(\widehat{k}鈰�\mathbf{r})(\widehat{n}鈰�\mathbf{r})\end{array}$

The matrix element of the perturbation to the Hamiltonian is therefore:

${\mathrm{H}}_{\mathrm{ba}}^{\text{'}}=鈭�\frac{1}{2}{\mathrm{qE}}_0\frac{\mathrm{蝇}}{\mathrm{c}}鉄�\mathrm{b}\left|\right(\widehat{k}鈰�\mathbf{r}\left)\right(\widehat{n}鈰�\mathbf{r}\left)\right|\mathrm{a}鉄�\sin \left(\mathrm{蝇迟}\right)$

role="math" localid="1659008966623" ${\mathrm{R}}_{\mathrm{a}鈫�\mathrm{b}}={\left(\frac{\mathrm{蝇q}}{2\mathrm{鈩�}\mathrm{c}}\right)}^2\frac{\mathrm{蟺}}{{\mathrm{系}}_0}\mathrm{蚁}\left({\mathrm{蝇}}_0\right)\left|鉄�\mathrm{b}\left|\right(\widehat{k}鈰�\mathbf{r}\left)\right(\widehat{n}鈰�\mathbf{r}\left)\right|\mathrm{a}鉄�\right|{}^2$

The spontaneous emission rate$A,$ is given by:

$\mathrm{A}=\frac{\mathrm{鈩�}{\mathrm{蝇}}^3}{{\mathrm{蟺}}^2{\mathrm{c}}^3}{\mathrm{R}}_{\mathrm{a}鈫�\mathrm{b}}$

Thus,

$\mathrm{A}=\frac{{\mathrm{q}}^2{\mathrm{蝇}}^5}{4\mathrm{蟺}\mathrm{鈩�}{\mathrm{系}}_0{\mathrm{c}}^5}\left|鉄�\mathrm{a}\left|\right(\widehat{k}鈰�\mathbf{r}\left)\right(\widehat{n}鈰�\mathbf{r}\left)\right|\mathrm{b}鉄�\right|{}^2$ 鈥︹�︹��. (4)

(b)

Now consider first a one-dimensional harmonic oscillator, to find the forbidden transitions we need to work out the average, it is easier with the spherical coordinates, so we use:

$\begin{array}{c}\widehat{k}=\widehat{z}\widehat{n}\\ =\cos \left(蠒\right)\widehat{x}+\sin \left(蠒\right)\widehat{y}\mathbf{r}\\ =x\sin \left(胃\right)\widehat{y}+x\cos \left(胃\right)\widehat{z}\end{array}$

So, we can write:

$(\widehat{k}鈰�\mathbf{r})(\widehat{n}鈰�\mathbf{r})={\mathrm{x}}^2(\widehat{k}鈰�\widehat{r})(\widehat{n}鈰�\widehat{r})$

Take out the $\mathrm{x}$and replace the vector $r$ with its unit vector $\hat{r}$. because the eigenvectors of the one-dimensional oscillator depend only on $x$ so we can pull out $(\widehat{k}鈰�\widehat{r})(\widehat{n}鈰�\widehat{r})$from the average in equation (4), and leave $x^2$inside.

Now we can average$\left|\right(\widehat{k}鈰�\widehat{r}\left)\right(\widehat{n}鈰�\widehat{r}){|}^2$ over all values of$胃$and $蠒$as:

$\begin{array}{c}\left|\right(\widehat{k}鈰�\widehat{r}\left)\right(\widehat{n}鈰�\widehat{r}){|}_{ave}^2=\frac{1}{4蟺}{鈭�}_0^{蟺}{鈭�}_0^{2蟺}{\left(\cos 胃\right)}^2{\left(\sin 胃\sin 蠒\right)}^2\sin 胃d蠒d胃\\ =\frac{1}{4蟺}{鈭�}_0^{蟺}{\sin }^3胃(1鈭�{\sin }^2胃)d胃{鈭�}_0^{2蟺}{\sin }^2蠒d蠒\\ =\frac{1}{15}\end{array}$

Now we need to do $鉄�\mathrm{a}\left|{\mathrm{x}}^2\right|\mathrm{b}鉄�$with the raising and lowering operators,

${\mathrm{x}}^2=\frac{\mathrm{鈩�}}{2{\mathrm{m蝇}}_{\mathrm{o}}}({\mathrm{a}}_{+}^2+{\mathrm{a}}_{+}{\mathrm{a}}_{鈭�}+{\mathrm{a}}_{鈭�}{\mathrm{a}}_{+}+{\mathrm{a}}_{鈭�}^2)$

So we want the elements of the matrix with $a$and $b$states such that the energy of state is less than the only term that can do this is the last one, $\frac{\mathrm{鈩�}}{2m{蝇}_o}{a}_{鈭�}^2$which converts the state with energy of $\mathrm{鈩�}{蝇}_o\left(n+\frac{1}{2}\right)$to a state with energy of $\mathrm{鈩�}{蝇}_o\left(n鈭�2+\frac{1}{2}\right)$since $a_{鈭�}{蠄}_n=\sqrt{n}{蠄}_{n鈭�1}$then we must chose $鉄�\mathrm{a}|=鉄�\mathrm{n}鈭�2||$and, $鈭�b鉄�=|n鉄�$that is:

$鉄�\mathrm{n}鈭�2\left|{\mathrm{x}}^2\right|\mathrm{n}鉄�=\frac{\mathrm{鈩�}}{2{\mathrm{m蝇}}_{\mathrm{o}}}\sqrt{\mathrm{n}(\mathrm{n}鈭�1)}$

The energy of the emitted photon is the energy difference between the two states, that is:

$\begin{array}{c}\mathrm{鈩�}\mathrm{蝇}=\mathrm{鈩�}{\mathrm{蝇}}_{\mathrm{o}}\left(\mathrm{n}+\frac{1}{2}\right)鈭�\mathrm{鈩�}{\mathrm{蝇}}_{\mathrm{o}}\left(\mathrm{n}鈭�2+\frac{1}{2}\right)\\ =2\mathrm{鈩�}{\mathrm{蝇}}_{\mathrm{o}}\\ ={\mathrm{蝇}}_{\mathrm{o}}\\ =\frac{\mathrm{蝇}}{2}\end{array}$

Now plug all the results in equation (4), note that we will use the answer given by Griffith, which is the same as (4) without the factor $1/4$so we get:

$\begin{array}{c}{A}_f=\frac{q^2{蝇}^5}{蟺\mathrm{鈩�}{系}_0{c}^5}\left(\frac{1}{15}\right){\left(\frac{\mathrm{鈩�}}{m蝇}\sqrt{n(n鈭�1)}\right)}^2\\ =\frac{\mathrm{鈩�}{q}^2{蝇}^{3}n(n鈭�1)}{15蟺{系}_0{m}^2{c}^5}\\ =\frac{8\mathrm{鈩�}{q}^2{蝇}_o^{3}n(n鈭�1)}{15蟺{系}_0{m}^2{c}^5}\end{array}$

The spontaneous emission rate for a one-dimensional oscillator is given by equation as (if we ignore $\mathbf{k}鈰�\mathbf{r}$):

${\mathrm{A}}_{\mathrm{a}}=\frac{{\mathrm{nq}}^2{\mathrm{蝇}}_{\mathrm{o}}^2}{6{\mathrm{蟺系}}_0{\mathrm{mc}}^3}$

Thus, the ratio of$A_f$ to$A_a$ is:

$\frac{{\mathrm{A}}_{\mathrm{f}}}{{\mathrm{A}}_{\mathrm{a}}}=\frac{16\mathrm{鈩�}{\mathrm{蝇}}_{\mathrm{o}}(\mathrm{n}鈭�1)}{5{\mathrm{mc}}^2}$

For a nonrelativistic system,$\mathrm{鈩�}蝇鈮�m{c}^2$ therefore, the transitions involving the first-order$\mathbf{k}鈰�\mathbf{r}$ are forbidden.

(c)

Now for $\mathrm{l}=0$ in both states (the initial and the final), then the wave function is independent of angle (the spherical harmonics for $\mathrm{l}=0$is constant), thus the angular integral yields,

${鈭�}_0^{\mathrm{蟺}}{鈭�}_0^{2\mathrm{蟺}}(\widehat{k}鈰�\widehat{r})(\widehat{n}鈰�\widehat{r})\sin \left(\mathrm{胃}\right)\mathrm{诲蠒诲胃}=\widehat{k}鈰�\widehat{n}$

But$\widehat{k}鈰�\widehat{n}=0$

So, the rate is zero for both the allowed and the forbidden transitions.