D.39 Selection rules

This note derives the selection rules for electric dipole transitions between two hydrogen states $\psi_{\rm {L}}$ and $\psi_{\rm {H}}$. Some selection rules for forbidden transitions are also derived. The derivations for forbidden transitions use some more advanced results from later chapters. It may be noted that in any case, the Hamiltonian assumes that the velocity of the electrons is small compared to the speed of light.

According to chapter 4.3, the hydrogen states take the form $\psi_{\rm {L}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\psi_{n_{\rm {L}}l_{\rm {L}}m_{\rm {L}}}{\updownarrow}$ and $\psi_{\rm {H}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\psi_{n_{\rm {H}}l_{\rm {H}}m_{\rm {H}}}{\updownarrow}$. Here 1 $\raisebox{-.3pt}{$\leqslant$}$ $n$, 0 $\raisebox{-.3pt}{$\leqslant$}$ $l$ $\raisebox{-.3pt}{$\leqslant$}$ $n$ and $\vert m\vert$ $\raisebox{-.3pt}{$\leqslant$}$ $l$ are integer quantum numbers. The final ${\updownarrow}$ represents the electron spin state, up or down.

As noted in the text, allowed electric dipole transitions must respond to at least one component of a constant ambient electric field. That means that they must have a nonzero value for at least one electrical dipole moment,

\begin{displaymath}
\langle \psi_{\rm {L}}\vert r_i\vert\psi_{\rm {H}}\rangle \ne 0
\end{displaymath}

where $r_i$ can be one of $r_1$ $\vphantom0\raisebox{1.5pt}{$=$}$ $x$, $r_2$ $\vphantom0\raisebox{1.5pt}{$=$}$ $y$, or $r_3$ $\vphantom0\raisebox{1.5pt}{$=$}$ $z$ for the three different components of the electric field.

The trick in identifying when these inner products are zero is based on taking inner products with cleverly chosen commutators. Since the hydrogen states are eigenfunctions of $\L _z$, the following commutator is useful

\begin{displaymath}
\langle \psi_{\rm {L}}\vert [r_i,\L _z]\vert\psi_{\rm {H}}...
...{\rm {L}}\vert r_i\L _z - \L _zr_i \vert\psi_{\rm {H}}\rangle
\end{displaymath}

For the $r_i\L _z$ term in the right hand side, the operator $\L _z$ acts on $\psi_{\rm {H}}$ and produces a factor $m_{\rm {H}}\hbar$, while for the $\L _zr_i$ term, $\L _z$ can be taken to the other side of the inner product and then acts on $\psi_{\rm {L}}$, producing a factor $m_{\rm {L}}\hbar$. So:
\begin{displaymath}
\langle \psi_{\rm {L}}\vert [r_i,\L _z]\vert\psi_{\rm {H}}...
...
\langle\psi_{\rm {L}}\vert r_i\vert\psi_{\rm {H}}\rangle %
\end{displaymath} (D.24)

The final inner product is the dipole moment of interest. Therefore, if a suitable expression for the commutator in the left hand side can be found, it will fix the dipole moment.

In particular, according to chapter 4.5.4 $[z,\L _z]$ is zero. That means according to equation (D.24) above that the dipole moment $\langle\psi_{\rm {L}}\vert z\vert\psi_{\rm {H}}\rangle$ in the right hand side will have to be zero too, unless $m_{\rm {H}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $m_{\rm {L}}$. So the first conclusion is that the $z$-​component of the electric field does not do anything unless $m_{\rm {H}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $m_{\rm {L}}$. One down, two to go.

For the $x$ and $y$ components, from chapter 4.5.4

\begin{displaymath}[x,\L _z]= -{\rm i}\hbar y \qquad [y,\L _z] = {\rm i}\hbar x
\end{displaymath}

Plugging that into (D.24) produces

\begin{displaymath}
- {\rm i}\hbar \langle\psi_{\rm {L}}\vert y\vert\psi_{\rm ...
...)\hbar \langle\psi_{\rm {L}}\vert y\vert\psi_{\rm {H}}\rangle
\end{displaymath}

From these equations it is seen that the $y$ dipole moment is zero if the $x$ one is, and vice-versa. Further, plugging the $y$ dipole moment from the first equation into the second produces

\begin{displaymath}
{\rm i}\hbar \langle\psi_{\rm {L}}\vert x\vert\psi_{\rm {H...
...bar}
\langle\psi_{\rm {L}}\vert x\vert\psi_{\rm {H}}\rangle
\end{displaymath}

and if the $x$ dipole moment is nonzero, that requires that $(m_{\rm {H}}-m_{\rm {L}})^2$ is one, so $m_{\rm {H}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $m_{\rm {L}}\pm1$. It follows that dipole transitions can only occur if $m_{\rm {H}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $m_{\rm {L}}$, through the $z$ component of the electric field, or if $m_{\rm {H}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $m_{\rm {L}}\pm1$, through the $x$ and $y$ components.

To derive selection rules involving the azimuthal quantum numbers $l_{\rm {H}}$ and $l_{\rm {L}}$, the obvious approach would be to try the commutator $[r_i,\L ^2]$ since $\L ^2$ produces $l(l+1)\hbar^2$. However, according to chapter 4.5.4, (4.68), this commutator will bring in the ${\skew 2\widehat{\skew{-1}\vec r}}$ $\times$ ${\skew 4\widehat{\vec L}}$ operator, which cannot be handled. The commutator that works is the second of (4.73):

\begin{displaymath}[[r_i,\L ^2],\L ^2] = 2\hbar^2(r_i\L ^2+\L ^2r_i)
\end{displaymath}

where by the definition of the commutator

\begin{displaymath}[[r_i,\L ^2],\L ^2] = (r_i\L ^2-\L ^2r_i)\L ^2 - \L ^2(r_i\L ^2-\L ^2r_i)
= r_i\L ^2\L ^2-2\L ^2r_i\L ^2+\L ^2\L ^2r_i
\end{displaymath}

Evaluating $\langle\psi_{\rm {L}}\vert[[r_i,\L ^2],\L ^2]\vert\psi_{\rm {H}}\rangle$ according to each of the two equations above and equating the results gives

\begin{displaymath}
2\hbar^2[l_{\rm {H}}(l_{\rm {H}}+1)+l_{\rm {L}}(l_{\rm {L}...
...^2
\langle\psi_{\rm {L}}\vert r_i\vert\psi_{\rm {H}}\rangle
\end{displaymath}

For $\langle\psi_{\rm {L}}\vert r_i\vert\psi_{\rm {H}}\rangle$ to be nonzero, the numerical factors in the left and right hand sides must be equal,

\begin{displaymath}
2 [l_{\rm {H}}(l_{\rm {H}}+1)+l_{\rm {L}}(l_{\rm {L}}+1)]
= [l_{\rm {H}}(l_{\rm {H}}+1)-l_{\rm {L}}(l_{\rm {L}}+1)]^2
\end{displaymath}

The right hand side is obviously zero for $l_{\rm {H}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $l_{\rm {L}}$, so $l_{\rm {H}}-l_{\rm {L}}$ can be factored out of it as

\begin{displaymath}[l_{\rm {H}}(l_{\rm {H}}+1)-l_{\rm {L}}(l_{\rm {L}}+1)]^2
= (l_{\rm {H}}-l_{\rm {L}})^2(l_{\rm {H}}+l_{\rm {L}}+1)^2
\end{displaymath}

and the left hand side can be written in terms of these same factors as

\begin{displaymath}
2 [l_{\rm {H}}(l_{\rm {H}}+1)+l_{\rm {L}}(l_{\rm {L}}+1)]
...
..._{\rm {H}}-l_{\rm {L}})^2 + (l_{\rm {H}}+l_{\rm {L}}+1)^2 - 1
\end{displaymath}

Equating the two results and simplifying gives

\begin{displaymath}[(l_{\rm {H}}-l_{\rm {L}})^2-1][(l_{\rm {H}}+l_{\rm {L}}+1)^2-1] = 0
\end{displaymath}

The second factor is only zero if $l_{\rm {H}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $l_{\rm {L}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0, but then $\langle\psi_{\rm {L}}\vert r_i\vert\psi_{\rm {H}}\rangle$ is still zero because both states are spherically symmetric. It follows that the first factor will have to be zero for dipole transitions to be possible, and that means that $l_{\rm {H}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $l_{\rm {L}}\pm1$.

The spin is not affected by the perturbation Hamiltonian, so the dipole moment inner products are still zero unless the spin magnetic quantum numbers $m_s$ are the same, both spin-up or both spin-down. Indeed, if the electron spin is not affected by the electric field to the approximations made, then obviously it cannot change. That completes the selection rules as given in chapter 7.4.4 for electric dipole transitions.

Now consider the effect of the magnetic field on transitions. For such transitions to be possible, the matrix element formed with the magnetic field must be nonzero. Like the electric field, the magnetic field can be approximated as spatially constant and quasi-steady. The perturbation Hamiltonian of a constant magnetic field is according to chapter 13.4

\begin{displaymath}
H_1 = \frac{e}{2m_{\rm e}} \skew2\vec{\cal B}\cdot \left({\skew 4\widehat{\vec L}}+ 2 {\skew 6\widehat{\vec S}}\right)
\end{displaymath}

Note that now electron spin must be included in the discussion.

According to this perturbation Hamiltonian, the perturbation coefficient $H_{\rm {HL}}$ for the $z$-​component of the magnetic field is proportional to

\begin{displaymath}
\langle \psi_{\rm {L}}\vert\L _z+2{\widehat S}_z\vert\psi_{\rm {H}}\rangle
\end{displaymath}

and that is zero because $\psi_{\rm {H}}{\updownarrow}$ is an eigenfunction of both operators and orthogonal to $\psi_{\rm {L}}{\updownarrow}$. So the $z$-​component of the magnetic field does not produce transitions to different states.

However, the $x$-​component (and similarly the $y$-​component) produces a perturbation coefficient proportional to

\begin{displaymath}
\langle \psi_{\rm {L}}\vert\L _x\vert\psi_{\rm {H}}\rangle...
...e \psi_{\rm {L}}\vert{\widehat S}_x\vert\psi_{\rm {H}}\rangle
\end{displaymath}

According to chapter 12.11, the effect of $\L _x$ on a state with magnetic quantum number $m_{\rm {H}}$ is to turn it into a linear combination of two similar states with magnetic quantum numbers $m_{\rm {H}}+1$ and $m_{\rm {H}}-1$. Therefore, for the first inner product above to be nonzero, $m_{\rm {L}}$ will have to be either $m_{\rm {H}}+1$ or $m_{\rm {H}}-1$. Also the orbital azimuthal momentum numbers $l$ will need to be the same, and so will the spin magnetic quantum numbers $m_s$. And the principal quantum numbers $n$, for that matter; otherwise the radial parts of the wave fuctions are orthogonal.

The magnetic field simply wants to rotate the orbital angular momentum vector in the hydrogen atom. That does not change the energy, in the absence of an average ambient magnetic field. For the second inner product, the spin magnetic quantum numbers have to be different by one unit, while the orbital magnetic quantum numbers must now be equal. So, all together

\begin{displaymath}
l_{\rm {H}} = l_{\rm {L}}
\qquad m_{\rm {H}} = m_{\rm {L...
...\rm {H}}} = m_{s,{\rm {L}}} \mbox{ or } m_{s,{\rm {L}}} \pm 1
\end{displaymath}

and either the orbital or the spin magnetic quantum numbers must be unequal. That are the selection rules as given in chapter 7.4.4 for magnetic dipole transitions. Since the energy does not change in these transitions, Fermi’s golden rule would have the decay rate zero. Fermi’s analysis is not exact, but such transitions should be very rare.

The logical way to proceed to electric quadrupole transitions would be to expand the electric field in a Taylor series in terms of $y$:

\begin{displaymath}
\skew3\vec{\cal E}= {\hat k}{\cal E}_{\rm {f}} \cos\Big(\o...
...\frac{\omega}{c} {\cal E}_{\rm {f}} \sin(\omega t - \alpha) y
\end{displaymath}

The first term is the constant electric field of the electric dipole approximation, and the second would then give the electric quadrupole approximation. However, an electric field in which ${\cal E}_z$ is a multiple of $y$ is not conservative, so the electrostatic potential does no longer exist.

It is necessary to retreat to the so-called vector potential $\skew3\vec A$. It is then simplest to chose this potential to get rid of the electrostatic potential altogether. In that case the typical electromagnetic wave is described by the vector potential

\begin{displaymath}
\skew3\vec A
= - {\hat k}\frac{1}{\omega} {\cal E}_{\rm ...
...ial t}
\quad \skew2\vec{\cal B}= \nabla \times \skew3\vec A
\end{displaymath}

In terms of the vector potential, the perturbation Hamiltonian is, chapter 13.1 and 13.4, and assuming a weak field,

\begin{displaymath}
H_1 = \frac{e}{2m_{\rm e}} (\skew3\vec A\cdot{\skew 4\wide...
...e}{m_{\rm e}}{\skew 6\widehat{\vec S}}\cdot\skew2\vec{\cal B}
\end{displaymath}

Ignoring the spatial variation of $\skew3\vec A$, this expression produces an Hamiltonian coefficient

\begin{displaymath}
H_{\rm {HL}} = - \frac{e}{m_{\rm e}\omega} {\cal E}_{\rm {...
...e \psi_{\rm {L}}\vert{\widehat p}_z\vert\psi_{\rm {H}}\rangle
\end{displaymath}

That should be same as for the electric dipole approximation, since the field is now completely described by $\skew3\vec A$, but it is not quite. The earlier derivation assumed that the electric field is quasi-steady. However, ${\widehat p}_z$ is equal to the commutator ${{\rm i}}m_{\rm e}[H_0,z]$$\raisebox{.5pt}{$/$}$$\hbar$ where $H_0$ is the unperturbed hydrogen atom Hamiltonian. If that is plugged in and expanded, it is found that the expressions are equivalent, provided that the perturbation frequency is close to the frequency of the photon released in the transition, and that that frequency is sufficiently rapid that the phase shift from sine to cosine can be ignored. Those are in fact the normal conditions.

Now consider the second term in the Taylor series of $\skew3\vec A$ with respect to $y$. It produces a perturbation Hamiltonian

\begin{displaymath}
\frac{e}{m_{\rm e}}\frac{1}{c}{\cal E}_{\rm {f}}\cos(\omega t-\alpha) y {\widehat p}_z
\end{displaymath}

The factor $y{\widehat p}_z$ can be trivially rewritten to give

\begin{displaymath}
\frac{e}{2m_{\rm e}}\frac{1}{c}{\cal E}_{\rm {f}}\cos(\ome...
...rm {f}}\cos(\omega t-\alpha)(y{\widehat p}_z+z{\widehat p}_y)
\end{displaymath}

The first term has already been accounted for in the magnetic dipole transitions discussed above, because the factor within parentheses is $\L _x$. The second term is the electric quadrupole Hamiltonian for the considered wave.

As second terms in the Taylor series, both Hamiltonians will be much smaller than the electric dipole one. The factor that they are smaller can be estimated from comparing the first and second term in the Taylor series. Note that $c$$\raisebox{.5pt}{$/$}$$\omega$ is proportional to the wave length $\lambda$ of the electromagnetic wave. Also, the additional position coordinate in the operator scales with the atom size, call it $R$. So the factor that the magnetic dipole and electric quadrupole matrix elements are smaller than the electric dipole one is $R$$\raisebox{.5pt}{$/$}$$\lambda$. Since transition probabilities are proportional to the square of the corresponding matrix element, it follows that, all else being the same, magnetic dipole and electric quadrupole transitions are slower than electric dipole ones by a factor $(R/\lambda)^2$. (But note the earlier remark on the problem for the hydrogen atom that the energy does not change in magnetic dipole transitions.)

The selection rules for the electric quadrupole Hamiltonian can be narrowed down with a bit of simple reasoning. First, since the hydrogen eigenfunctions are complete, applying any operator on an eigenfunction will always produce a linear combination of eigenfunctions. Now reconsider the derivation of the electric dipole selection rules above from that point of view. It is then seen that $z$ only produces eigenfunctions with the same values of $m$ and the values of $l$ exactly one unit different. The operators $x$ and $y$ change both $m$ and $l$ by exactly one unit. And the components of linear momentum do the same as the corresponding components of position, since ${\widehat p}_i$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${{\rm i}}m_{\rm e}[H_0,r_i]$$\raisebox{.5pt}{$/$}$$\hbar$ and $H_0$ does not change the eigenfunctions, just their coefficients. Therefore $y{\widehat p}_z+z{\widehat p}_y$ produces only eigenfunctions with azimuthal quantum number $l$ either equal to $l_{\rm {H}}$ or to $l_{\rm {H}}\pm2$, depending on whether the two unit changes reinforce or cancel each other. Furthermore, it produces only eigenfunctions with $m$ equal to $m_{\rm {H}}\pm1$. However, $x{\widehat p}_y+y{\widehat p}_x$, corresponding to a wave along another axis, will produce values of $m$ equal to $m_{\rm {H}}$ or to $m_{\rm {H}}\pm2$. Therefore the selection rules become:

\begin{displaymath}
l_{\rm {H}} = l_{\rm {L}} \mbox{ or } l_{\rm {L}} \pm 2
...
... m_{\rm {L}} \pm 2
\qquad m_{s,{\rm {H}}} = m_{s,{\rm {L}}}
\end{displaymath}

That are the selection rules as given in chapter 7.4.4 for electric quadrupole transitions. These arguments apply equally well to the magnetic dipole transition, but there the possibilities are narrowed down much further because the angular momentum operators only produce a couple of eigenfunctions. It may be noted that in addition, electric quadrupole transitions from $l_{\rm {H}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 to $l_{\rm {L}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 are not possible because of spherical symmetry.