Question

In Equation 9.31 assumed that the atom is so small (in comparison to the wavelength of light) that spatial variations in the field can be ignored. The true electric field would be $\mathbf{E}\mathbf{(}\mathbf{r}\mathbf{,}\mathbf{t}\mathbf{)}\mathbf{=}{\mathbf{E}}_{\mathbf{0}}\mathbf{cos}\mathbf{(}\mathbf{k}\mathbf{\cdot }\mathbf{r}\mathbf{-}\mathbf{\omega t}\mathbf{)}\mathbf{.}$

If the atom is centered at the origin, then$\mathbf{k}\mathbf{\cdot }\mathbf{r}\mathbf{\ll }\mathbf{1}$ over the relevant volume,$\mathbf{\left|}\mathbf{k}\mathbf{\right|}\mathbf{=}\mathbf{2}\mathbf{\pi }\mathbf{/}\mathbf{\lambda }$ so$\mathbf{k}\mathbf{\cdot }\mathbf{r}\mathbf{~}\mathbf{r}\mathbf{/}\mathbf{\lambda }\mathbf{\ll }\mathbf{1}\mathbf{)}$ and that's why we could afford to drop this term. Suppose we keep the first-order correction:

$\mathbf{E}\mathbf{(}\mathbf{r}\mathbf{,}\mathit{t}\mathbf{)}\mathbf{=}{\mathbf{E}}_{\mathbf{0}}\mathbf{\left[}\mathbf{cos}\mathbf{\right(}\mathit{\omega }\mathit{t}\mathbf{)}\mathbf{+}\mathbf{(}\mathbf{k}\mathbf{\cdot }\mathbf{r}\mathbf{\left)}\mathbf{sin}\mathbf{\right(}\mathit{\omega }\mathit{t}\mathbf{\left)}\mathbf{\right]}\mathbf{.}$

The first term gives rise to the allowed (electric dipole) transitions we considered in the text; the second leads to so-called forbidden (magnetic dipole and electric quadrupole) transitions (higher powers of $\mathit{k}\mathbf{.}\mathit{r}$lead to even more "forbidden" transitions, associated with higher multipole moments).

(a) Obtain the spontaneous emission rate for forbidden transitions (don't bother to average over polarization and propagation directions, though this should really be done to complete the calculation). Answer:role="math" localid="1659008133999" ${\mathit{R}}_{\mathbf{b}\mathbf{\to }\mathbf{a}}\mathbf{=}\frac{{\mathbf{q}}^{\mathbf{2}}{\mathbf{\omega }}^{\mathbf{5}}}{\mathbf{\pi }{\mathbf{ϵ}}_{\mathbf{0}}\mathbf{\hslash }{\mathbf{c}}^{\mathbf{5}}}\mathbf{|}\mathbf{⟨}\mathit{a}\mathbf{|}\mathbf{(}\widehat{\mathbf{n}}\mathbf{\cdot }\mathbf{r}\mathbf{)}\mathbf{(}\widehat{\mathbf{k}}\mathbf{\cdot }\mathbf{r}\mathbf{)}\mathbf{|}\mathit{b}\mathbf{⟩}{\mathbf{|}}^{\mathbf{2}}\mathbf{.}$

(b) Show that for a one-dimensional oscillator the forbidden transitions go from level$\mathit{n}$ to levelrole="math" localid="1659008239387" $\mathit{n}\mathbf{-}\mathbf{2}$ and the transition rate (suitably averaged over $\widehat{\mathbf{n}}\mathit{a}\mathit{n}\mathit{d}\widehat{\mathbf{k}}$) is$\mathbf{R}\mathbf{=}\frac{\mathbf{\hslash }{\mathbf{q}}^{\mathbf{2}}{\mathbf{\omega }}^{\mathbf{3}}\mathbf{n}\mathbf{(}\mathbf{n}\mathbf{-}\mathbf{1}\mathbf{)}}{\mathbf{15}{\mathbf{\pi ϵ}}_{\mathbf{0}}{\mathbf{m}}^{\mathbf{2}}{\mathbf{c}}^{\mathbf{5}}}\mathbf{.}$

(Note: Here$\mathbf{\omega }$ is the frequency of the photon, not the oscillator.) Find the ratio of the "forbidden" rate to the "allowed" rate, and comment on the terminology.

(c) Show that the$\mathbf{2}\mathit{S}\mathbf{\to }\mathbf{1}\mathit{S}$ transition in hydrogen is not possible even by a "forbidden" transition. (As it turns out, this is true for all the higher multipoles as well; the dominant decay is in fact by two-photon emission, and the lifetime it is about a tenth of a second

Short answer

(a)$A=\frac{q^2{\omega }^5}{4\pi \hslash {ϵ}_0{c}^5}\left|⟨a\left|\right(\widehat{k}\cdot \mathbf{r}\left)\right(\widehat{n}\cdot \mathbf{r}\left)\right|b⟩\right|{}^2$

(b)$\frac{{\mathrm{A}}_{\mathrm{f}}}{{\mathrm{A}}_{\mathrm{a}}}=\frac{16\mathrm{\hslash }{\mathrm{\omega }}_{\mathrm{o}}(\mathrm{n}-1)}{5{\mathrm{mc}}^2}$

(c) The rate is zero for both the allowed and the forbidden transitions.

Step-by-step solution

The Electric Field

(a)

The original formula of the electric field is given by:

$\mathbf{E}={\mathrm{E}}_0\widehat{n}\cos (\mathbf{k}\cdot \mathbf{r}-\mathrm{\omega t})$ ……. (1)

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

$\mathbf{E}={\mathrm{E}}_0\widehat{n}\left[\cos \right(\mathbf{k}\cdot \mathbf{r}\left)\cos \right(\mathrm{\omega t})+\sin (\mathbf{k}\cdot \mathbf{r}\left)\sin \right(\mathrm{\omega t}\left)\right]$ ……… (2)

$E\approx E_0\widehat{n}\left[\cos \right(\omega t)+\mathbf{k}\cdot \mathbf{r}\sin (\omega t\left)\right]$ ……… (3)

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

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:

$\mathrm{W}=-\mathrm{q}{\int }_0^b\mathbf{E}\cdot \mathrm{d}{\mathbf{r}}^{\text{'}}$

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

${\mathrm{H}}^{\text{'}}=-{\mathrm{qE}}_0\sin \left(\mathrm{\omega t}\right){\int }_0^r(\mathbf{k}\cdot {\mathbf{r}}^{\text{'}})\widehat{n}\cdot \mathrm{d}{\mathbf{r}}^{\text{'}}$

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 ${\theta }_k$and ${\theta }_n$so we can write:

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

Substitute into (3), so we get:

$\begin{array}{c}{\mathrm{H}}^{\text{'}}=-{\mathrm{qE}}_0\mathrm{ksin}\left(\mathrm{\omega t}\right)\cos \left({\mathrm{\theta }}_{\mathrm{k}}\right)\cos \left({\mathrm{\theta }}_{\mathrm{n}}\right){\int }_0^{\mathrm{r}}{\mathrm{r}}^{\text{'}}{\mathrm{dr}}^{\text{'}}\\ =-\frac{1}{2}{\mathrm{qE}}_0\mathrm{ksin}\left(\mathrm{\omega t}\right)\cos \left({\mathrm{\theta }}_{\mathrm{k}}\right)\cos \left({\mathrm{\theta }}_{\mathrm{n}}\right){\mathrm{r}}^2\\ =-\frac{1}{2}{\mathrm{qE}}_0\sin \left(\mathrm{\omega t}\right)\left[\mathrm{krcos}\left({\mathrm{\theta }}_{\mathrm{k}}\right)\right]\left[\mathrm{rcos}\left({\mathrm{\theta }}_{\mathrm{n}}\right)\right]\\ =-\frac{1}{2}{\mathrm{qE}}_0\sin \left(\mathrm{\omega t}\right)(\mathbf{k}\cdot \mathbf{r})(\widehat{n}\cdot \mathbf{r})\\ =-\frac{1}{2}{\mathrm{qE}}_0\frac{\mathrm{\omega }}{\mathrm{c}}\sin \left(\mathrm{\omega t}\right)(\widehat{k}\cdot \mathbf{r})(\widehat{n}\cdot \mathbf{r})\end{array}$

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

${\mathrm{H}}_{\mathrm{ba}}^{\text{'}}=-\frac{1}{2}{\mathrm{qE}}_0\frac{\mathrm{\omega }}{\mathrm{c}}⟨\mathrm{b}\left|\right(\widehat{k}\cdot \mathbf{r}\left)\right(\widehat{n}\cdot \mathbf{r}\left)\right|\mathrm{a}⟩\sin \left(\mathrm{\omega t}\right)$

role="math" localid="1659008966623" ${\mathrm{R}}_{\mathrm{a}\to \mathrm{b}}={\left(\frac{\mathrm{\omega q}}{2\mathrm{\hslash }\mathrm{c}}\right)}^2\frac{\mathrm{\pi }}{{\mathrm{ϵ}}_0}\mathrm{\rho }\left({\mathrm{\omega }}_0\right)\left|⟨\mathrm{b}\left|\right(\widehat{k}\cdot \mathbf{r}\left)\right(\widehat{n}\cdot \mathbf{r}\left)\right|\mathrm{a}⟩\right|{}^2$

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

$\mathrm{A}=\frac{\mathrm{\hslash }{\mathrm{\omega }}^3}{{\mathrm{\pi }}^2{\mathrm{c}}^3}{\mathrm{R}}_{\mathrm{a}\to \mathrm{b}}$

Thus,

$\mathrm{A}=\frac{{\mathrm{q}}^2{\mathrm{\omega }}^5}{4\mathrm{\pi }\mathrm{\hslash }{\mathrm{ϵ}}_0{\mathrm{c}}^5}\left|⟨\mathrm{a}\left|\right(\widehat{k}\cdot \mathbf{r}\left)\right(\widehat{n}\cdot \mathbf{r}\left)\right|\mathrm{b}⟩\right|{}^2$ ………. (4)

One dimensional Harmonic oscillator

(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:

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

So, we can write:

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

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 $(\widehat{k}\cdot \widehat{r})(\widehat{n}\cdot \widehat{r})$from the average in equation (4), and leave $x^2$inside.

Now we can average$\left|\right(\widehat{k}\cdot \widehat{r}\left)\right(\widehat{n}\cdot \widehat{r}){|}^2$ over all values of$\theta$and $\varphi$as:

$\begin{array}{c}\left|\right(\widehat{k}\cdot \widehat{r}\left)\right(\widehat{n}\cdot \widehat{r}){|}_{ave}^2=\frac{1}{4\pi }{\int }_0^{\pi }{\int }_0^{2\pi }{\left(\cos \theta \right)}^2{\left(\sin \theta \sin \varphi \right)}^2\sin \theta d\varphi d\theta \\ =\frac{1}{4\pi }{\int }_0^{\pi }{\sin }^3\theta (1-{\sin }^2\theta )d\theta {\int }_0^{2\pi }{\sin }^2\varphi d\varphi \\ =\frac{1}{15}\end{array}$

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

${\mathrm{x}}^2=\frac{\mathrm{\hslash }}{2{\mathrm{m\omega }}_{\mathrm{o}}}({\mathrm{a}}_{+}^2+{\mathrm{a}}_{+}{\mathrm{a}}_{-}+{\mathrm{a}}_{-}{\mathrm{a}}_{+}+{\mathrm{a}}_{-}^2)$

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, $\frac{\hslash }{2m{\omega }_o}{a}_{-}^2$which converts the state with energy of $\hslash {\omega }_o\left(n+\frac{1}{2}\right)$to a state with energy of $\hslash {\omega }_o\left(n-2+\frac{1}{2}\right)$since $a_{-}{\psi }_n=\sqrt{n}{\psi }_{n-1}$then we must chose $⟨\mathrm{a}|=⟨\mathrm{n}-2||$and, $\mid b⟩=|n⟩$that is:

$⟨\mathrm{n}-2\left|{\mathrm{x}}^2\right|\mathrm{n}⟩=\frac{\mathrm{\hslash }}{2{\mathrm{m\omega }}_{\mathrm{o}}}\sqrt{\mathrm{n}(\mathrm{n}-1)}$

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

$\begin{array}{c}\mathrm{\hslash }\mathrm{\omega }=\mathrm{\hslash }{\mathrm{\omega }}_{\mathrm{o}}\left(\mathrm{n}+\frac{1}{2}\right)-\mathrm{\hslash }{\mathrm{\omega }}_{\mathrm{o}}\left(\mathrm{n}-2+\frac{1}{2}\right)\\ =2\mathrm{\hslash }{\mathrm{\omega }}_{\mathrm{o}}\\ ={\mathrm{\omega }}_{\mathrm{o}}\\ =\frac{\mathrm{\omega }}{2}\end{array}$

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:

$\begin{array}{c}{A}_f=\frac{q^2{\omega }^5}{\pi \hslash {ϵ}_0{c}^5}\left(\frac{1}{15}\right){\left(\frac{\hslash }{m\omega }\sqrt{n(n-1)}\right)}^2\\ =\frac{\hslash q^2{\omega }^{3}n(n-1)}{15\pi {ϵ}_0{m}^2{c}^5}\\ =\frac{8\hslash q^2{\omega }_o^{3}n(n-1)}{15\pi {ϵ}_0{m}^2{c}^5}\end{array}$

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

${\mathrm{A}}_{\mathrm{a}}=\frac{{\mathrm{nq}}^2{\mathrm{\omega }}_{\mathrm{o}}^2}{6{\mathrm{\pi ϵ}}_0{\mathrm{mc}}^3}$

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

$\frac{{\mathrm{A}}_{\mathrm{f}}}{{\mathrm{A}}_{\mathrm{a}}}=\frac{16\mathrm{\hslash }{\mathrm{\omega }}_{\mathrm{o}}(\mathrm{n}-1)}{5{\mathrm{mc}}^2}$

For a nonrelativistic system,$\hslash \omega \ll m{c}^2$ therefore, the transitions involving the first-order$\mathbf{k}\cdot \mathbf{r}$ are forbidden.

The wave function is independent

(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,

${\int }_0^{\mathrm{\pi }}{\int }_0^{2\mathrm{\pi }}(\widehat{k}\cdot \widehat{r})(\widehat{n}\cdot \widehat{r})\sin \left(\mathrm{\theta }\right)\mathrm{d\varphi d\theta }=\widehat{k}\cdot \widehat{n}$

But$\widehat{k}\cdot \widehat{n}=0$

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