Wouthuysen Field effect

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Reference Materials

• Page 32 of Furlanetto et. al, 2009
• Detailed quantum mechanics lecture notes from UCB Physics 221A (bottom of page)
• A Primer on Quantum Numbers and Spectroscopic Notation (Walter, Stony Brook)
• Orbitals (Guetta, Columbia)

Need to Review

• 21cm Transition
• Einstein Coefficients
• Classical Bohr Atom
• Atomic and Molecular Quantum Numbers
• Collisional Excitations
• Black-Body Radiation

The Wouthuysen-Field Effect

The Wouthuysen-Field effect is a coupling of the 21cm hyperfine transition to Ly-$\alpha$ radiation.

This is important for possible high-redshift observations of the 21cm line, during the epoch of reionization. The 21cm hyperfine transition is forbidden by normal dipole selection rules, but transitions from one hyperfine state, up to the $n=2$ state, and back down to the other hyperfine state are not forbidden. So, if there is sufficient Ly-$\alpha$ radiation to cause this intermediate transition, it will dominate over the direct (forbidden) hyperfine transition. Cosmologically speaking, this happens towards the end of the “Dark Ages” as reionization begins, and the WF effect remains the dominant effect until reionization is complete.

Fine structure of hydrogen

Energy levels in hydrogen atoms are split due to spin orbit interaction (fine splitting), and the smaller effect of electron-proton spin interaction (hyperfine splitting). Anti-aligned spins lead to lower energy levels. The splittings of the lowest energy levels are:

Energy level diagram with spectroscopic notation

Here we have used spectroscopic notation $n\ _{F}L_{J}$, where $n$ is the principal quantum number, $L=0(S),1(P)$ are the electron orbital angular momentum. $S_{e}$ ($S_{p}$) shall denote the electron (proton) spin (not to be confused with the $S$ orbital). Then $J=|L+S_{e}|$ is the electron total angular momentum, and $F=|L+S_{e}+S_{p}|$ is the hydrogen total angular momentum. Note these are vector sums.

The splitting between the two $1\ S$ levels is of particular interest. It has energy difference $\nu =1.42GHz$, corresponding to wavelength $\lambda =21cm$. We also define $T_{*}=E_{21cm}/k_{B}=0.068K$. In the $21cm$ regime, the Rayleigh-Jeans limit holds and we can define the “brightness temperature”

$T_{b}\approx I_{\nu }c^{2}/2k_{B}\nu ^{2}.\,\!$

$T_{b}$ only serves to redefine $I_{\nu }$, and should be distinguished from the ambient temperature of the CMB $T_{\gamma }=2.74(1+z){\mbox{ K}}$.

Cosmological context

After recombination $(z\sim 1100)$, the photons in the cosmic fluid no longer has free charges to interact with and thus free stream through the “dark ages” until the Epoch of Reionization, believed to be around $z\sim 6-9$. Studying the 21cm absorption of neutral hydrogen can thus potentially probe the universe during the dark ages and epoch of reionization. The frequency $\nu$ of 21cm undergoes cosmological redshift and hence the actual observed brightness temperature is

$T_{b}=T_{b}^{\prime }/(1+z).\,\!$

Denoting the lower $1S$ level by 0, the higher by 1, one can define the “spin temperature $T_{S}$” to characterize the relative abundance of these two states:

${\frac {n_{1}}{n_{0}}}={\frac {g_{1}}{g_{0}}}e^{-h\nu /k_{B}T_{S}}=3\exp(-T_{*}/T_{S})\approx 3(1-T_{*}/T_{S}).\,\!$

During the dark ages till the end of reionization, several processes control the relative abundances. Among them are direct radiative transitions (Einstein Coefficients) [1], collisional excitation $C_{01}$, $C_{10}$ and the Wouthuysen-Field (WF) Effect $W_{01}$ and $W_{10}$. The WF effect will be explained in detail. For now let’s just note that the overall statistical balance gives

$n_{1}(A_{10}+B_{10}I_{CMB}+C_{10}+W_{10})=n_{0}(B_{01}I_{CMB}+C_{01}+W_{01}),\,\!$

where $A_{10}=2.869\times 10^{-15}{\mbox{s}}^{-1}$ and $B$ are the Einstein Coefficients of the 21cm transition. Recall that in thermodynamic equilibrium, we had

$B_{01}={\frac {g_{1}}{g_{0}}}B_{10}={\frac {3c^{2}}{2h\nu ^{3}}}A_{10}\,\!$

In the Rayleigh-Jeans limit, radiative coefficients satisfy

${\begin{aligned}B_{01}I_{CMB}&={\frac {3kT_{\gamma }}{h\nu }}A_{10}={\frac {3T_{\gamma }}{T_{*}}}A_{10},\\B_{10}I_{CMB}&={\frac {T_{\gamma }}{T_{*}}}A_{10},\end{aligned}}\,\!$

Here $T_{\gamma }=2.74(1+z){\mbox{ K }}$ is the temperature of the CMB.

Remember that for collisional excitations of the hyperfine transition, in thermal equilibrium, the gas temperature is coupled directly to the spin temperature. In other words, collisions are the dominant effect that sets the hyperfine level population. This means the transition rates $C_{01}$ and $C_{10}$ are given by:

${\begin{aligned}{\frac {C_{01}}{C_{10}}}&={\frac {g_{1}}{g_{0}}}e^{-T_{*}/T_{K}}\approx 3\left(1-{\frac {T_{*}}{T_{K}}}\right).\end{aligned}}\,\!$

Similarly, WF effect rates are given by the “color temperature” $T_{W}$:

${\begin{aligned}{\frac {W_{01}}{W_{10}}}&={\frac {g_{1}}{g_{0}}}e^{-T_{*}/T_{W}}\approx 3\left(1-{\frac {T_{*}}{T_{W}}}\right).\end{aligned}}\,\!$

Plugging the above results into the balance equation eq:balance, we have the relation of the temperatures:

${T_{S}^{-1}={\frac {T_{\gamma }^{-1}+x_{c}T_{K}^{-1}+x_{W}T_{W}^{-1}}{1+x_{c}+x_{W}}},}\,\!$

where

${\begin{aligned}x_{c}&={\frac {C_{10}}{A_{10}}}{\frac {T_{*}}{T_{\gamma }}},\\x_{W}&={\frac {W_{10}}{A_{10}}}{\frac {T_{*}}{T_{\gamma }}},\end{aligned}}\,\!$

represents the relative rates. It remains to determine these coefficients.

The rest of this article shall concentrate on the WF effect, and in particular the determination of $x_{W}.$ The WF effect involves absorption of an Lyman-$\alpha$ photon from the $1_{0}S_{1/2}$ state and subsequent decay into the $1_{1}S_{1/2}$ state. The CMB temperature during the dark ages have temperatures $3000K>T_{\gamma }>15K$, and is much smaller than the Lyman-$\alpha$ transition temperature of $T=13.6eV\sim 1.6\times 10^{5}{\mbox{K}}$, and thus we expect WF effect to be most important during the epoch of reionization, when the first stars provide the abundance of ${\mbox{Ly-}}\alpha$ photons. To study the WF effect, we first look at the rules for allowed transitions.

Parity and Selection Rules

Now we need to look at the selection rules for dipole transitions. First of all, there is the selection rule that $\Delta L=0$ is forbidden by parity, a symmetry of electromagnetic interactions, so only transitions between $1S$ and $2P$ will matter. (In other words, nevermind about the $2S$ states we wrote down above.) This is the selection rule that forbids the 21cm transition, which is why the WF effect can dominate over the direct 21cm transition so long as enough Ly-$\alpha$ photons are around.

Rotational symmetry is where the rest of our selection rules come from. There is a fancy thing in quantum called the Wigner-Eckart theorem which applies to “irreducible tensor operators” in general, and can generate selection rules for any such operator. The selection rule goes like this: a transition from $j$ to $j^{/}prime$ is only allowed if $j^{/}prime$ is in the range $|j-k|$ to $j+k$, where $k$ is the “order” of the tensor operator. For dipole transitions, $k=1$: the dipole operator is an “order 1 irreducible tensor operator”, so the rule becomes $j^{/}prime=|j-1|,...,j+1$. (Lowercase $j$ is a totally generic angular momentum quantum number, not the same as $J$.)

To put illustrate all of these in simpler language, a photon is a vector $A^{\mu }$ and carries one unit of angular momentum. This means that a single photon must be circularly polarized (left or right) [2]. Since the vector does not has any spin dependence, it commutes with the electron spin operator:

$[A^{\mu },\ S]=0.\,\!$

Here $S=|{\vec {S}}_{e}+{\vec {S}}_{p}|$ is the vector sum of the electron and proton spins.

This implies the first dipole selection rule:

$\Delta S=0.\,\!$

The spin along given directions ($m_{s}$), however, are not “good” quantum numbers, and can change.

To consider orbital angular momentum, recall the parity operator $P$. It reverses all physical space directions:

$P:x_{i}\rightarrow -x_{i}.\,\!$

If we fix the phase such that $P^{2}=1$, we have parity as a Hermitian and hence observable:

$P^{-1}=P^{\dagger }=P.\,\!$

Parity thus acts on other operators via conjugation. For example

$PA^{\mu }(t,x)P^{-1}=-A^{\mu }(t,-x).\,\!$

More generally, it is clear that quantities such as vector, pseudo-scalar change sign under parity while a pseudo-vector (such as angular momentum or magnetic field) or a true scalar do not change sign. For example, parity commutes with spin $[P,S]=0$, while it anti-commutes with momentum $\{P,p\}=0$. This means that while a spin state is a good quantum number when paired with parity, a momentum state is not.

So how do states in a hydrogen atom behave under parity? To answer this question recall that such a state is proportional to the spherical harmonics:

$|nlm\rangle \propto R_{lm}(r)Y_{l}^{m}(\Omega )\propto R_{lm}(r)P_{l}^{m}(\cos \theta )e^{im\phi }.\,\!$

Under parity, $\phi \rightarrow \pi +\phi$, $\theta \rightarrow \pi -\theta$, and since $P_{l}^{m}(-x)=(-1)^{l-m}P_{l}^{m}(x),$ we have

$P(|nlm\rangle )=(-1)^{l}|nlm\rangle .\,\!$

The parity of a state $|nlm\rangle$ does not depend on $m$, as expected since an intrinsic property such as parity should not depend on the orientation of the system.

Thus the matrix element of the vector can be written as

${\begin{aligned}M&=\langle n^{/}primel^{/}primem^{/}prime|A^{\mu }|nlm\rangle ,\\&=\langle n^{/}primel^{/}primem^{/}prime|P^{\dagger }PA^{\mu }P^{\dagger }P|nlm\rangle ,\\&=(-1)^{l-l^{/}prime+1}\langle n^{/}primel^{/}primem^{/}prime|A^{\mu }|nlm\rangle .\end{aligned}}\,\!$

Since a matrix element is a scalar and has even parity, we see that the end state must have different parity from the initial state. In other words, $l^{/}prime-l$ must be odd!

Furthermore, since the photon carries only one unit of angular momentum, any angular momentum change must be of order unity or less:

$|\Delta L|,|\Delta F|,|\Delta J|\leq 1.\,\!$

Thus we have the rule

$\Delta l=\pm 1.\,\!$

To see this more explicitly, one can also write out the full expression

$|nlm\rangle =R_{lm}(r)P_{l}^{m}(\cos \theta )e^{im\phi }{\frac {1}{\sqrt {2\pi }}}.\,\!$

Since a dipole interaction, along a direction called $z$ has the form

${\vec {d}}=E{\vec {z}}=Er\cos \theta {\hat {z}},\,\!$

The matrix element can be written as

${\begin{aligned}\langle n^{/}primel^{/}primem^{/}prime|d|nlm\rangle &\propto \int _{0}^{\pi }P_{l^{/}prime}^{m^{/}prime}P_{l}^{m}\cos \theta \sin \theta d\theta \times {\frac {1}{2\pi }}\int _{0}^{2\pi }e^{i(m-m^{/}prime)\phi }d\phi .\end{aligned}}\,\!$

From properties of associated Legendre polynomials, we can decompose

$P_{l}^{m}(\cos \theta )\cos \theta =bP_{l+1}^{m}+cP_{l-1}^{m},\,\!$

for some constants $b$ and $c$. And since they also satisfy the orthogonality rules

$\int _{-1}^{1}P_{l^{/}prime}^{m}(x)P_{l}^{m}(x)dx={\frac {2(l+m)!}{(2l+1)(l-m)!}}\delta _{l^{/}primel},\,\!$

the above matrix element vanishes unless

${\begin{cases}m^{/}prime&=m,\\l^{/}prime&=l\pm 1.\end{cases}}\,\!$

In particular, the angular momentum along the dipole direction is unchanged. This is expected since the spin of the photon is pointed perpendicular to the field (a transverse wave).

A difference between $l$ and $L$ is in place when there are more than one electron in the atom. We have seen that a single electron contributes a factor of $(-1)^{l}$ when acted on by parity. In the multi-electron case, this generalizes to

$P(|l_{1}l_{2}...\rangle )=(-1)^{\sum l_{i}}|l_{1}l_{2}...\rangle .\,\!$

Since in general the total orbital angular momentum $L\neq \sum l_{i}$, two states with the same $L$ can have different parity, and so one cannot conclude that $\Delta L=0$ is not allowed. One can conclude, however, that $(L=0)\rightarrow (L=0)$ transitions are not allowed. This is because the $S$ orbital has complete spherical symmetry, a symmetry broken by the vector photon, and thus an $S$ orbital cannot absorb a photon and stay an $S$ orbital. Similarly, considering L-S coupling, we have corresponding selection rules of forbidden $(J=0)\rightarrow (J=0)$ or $(F=0)\rightarrow (F=0)$ transitions. Physically, these last two rules amounts to the combined effect of angular momentum conservation, and the first rule, that $\Delta S=0.$ For example, the transition from $1_{1}S_{1/2}$ to $2_{1}P_{1/2}$ is allowed, even though $\Delta F=0$, because in this transition the photon brought $\Delta L=1$, electron flips its spin to keep $\Delta J=0$, and proton also flips its spin, to preserve $\Delta S=0$ and $\Delta F=0$, in the process $m_{F}$ is flipped. The transition $1_{0}S_{1/2}$ to $2_{0}P_{1/2}$ on the other hand, is forbidden, because as $\Delta L=1$, the two conditions $\Delta S=0$ and $\Delta F=0$ cannot be simultaneously satisfied.

In summary, we have the following selection rules for dipole transitions in a neutral hydrogen atom

${\begin{cases}\Delta S&=0,\\\Delta L&=\pm 1,\\\Delta J&=0,\pm 1,{\mbox{(except 0 to 0),}}\\\Delta F&=0,\pm 1,{\mbox{(except 0 to 0). }}\end{cases}}\,\!$

The WF effect

Coupling coefficient

We would like to find the coupling coefficient $x_{W}$ defined at the end of the Cosmological Context subsection. The allowed transitions between the lowest levels are shown in Fig. 2. The transitions relevant to the WF effect are traced with solid lines. Since we have denoted the $1S$ states by 0 and 1, we shall by convenience denote the 2$P$ states by 2,3,4,5, in order of increasing energy.

Energy level diagram with allowed transitions
Energy level diagram of WF effect

Evidently, the (de-)excitation rate due to WF effect is given by:

${\begin{aligned}W_{01}&=B_{03}J_{03}{\frac {A_{31}}{A_{31}+A_{30}}}+B_{04}J_{04}{\frac {A_{41}}{A_{41}+A_{40}}},\\&={\frac {3T_{\gamma }^{03}}{T_{*}^{03}}}{\frac {A_{30}A_{31}}{A_{31}+A_{30}}}+{\frac {3T_{\gamma }^{04}}{T_{*}^{04}}}{\frac {A_{40}A_{41}}{A_{41}+A_{40}}},\\W_{10}&=B_{13}J_{13}{\frac {A_{30}}{A_{31}+A_{30}}}+B_{14}J_{14}{\frac {A_{40}}{A_{41}+A_{40}}},\\&={\frac {T_{\gamma }^{13}}{T_{*}^{13}}}{\frac {A_{30}A_{31}}{A_{31}+A_{30}}}+{\frac {T_{\gamma }^{14}}{T_{*}^{14}}}{\frac {A_{40}A_{41}}{A_{41}+A_{40}}}.\end{aligned}}\,\!$

In the second and fourth lines I have used the relation from above:

${\begin{aligned}B_{01}I_{CMB}&={\frac {3kT_{\gamma }}{h\nu }}A_{10}={\frac {3T_{\gamma }}{T_{*}}}A_{10},\\B_{10}I_{CMB}&={\frac {T_{\gamma }}{T_{*}}}A_{10},\end{aligned}}\,\!$

We would like to relate this rate to the total Ly$\alpha$ scattering rate

$P_{\alpha }=4\pi \chi _{\alpha }\int d\nu J_{\nu }\phi _{\alpha }(\nu ),\,\!$

where $\sigma _{\nu }=\chi _{\nu }\phi _{\alpha }$ is the absorption cross section and $\chi _{\alpha }=f_{\alpha }{\frac {\pi e^{2}}{m_{e}c^{2}}}$, with $f_{\alpha }=0.4162$ the oscillator strength. The line $\phi _{\alpha }$ can be assumed to be a Voigt profile. Thermal broadening leads to Doppler width

$\Delta \nu _{D}={\sqrt {\frac {2k_{B}T_{K}}{m_{H}c^{2}}}}\nu _{\alpha },\,\!$

where $\nu _{\alpha }=2.47\times 10^{15}{\mbox{Hz}}$ is the Lyman $\alpha$ line center frequency.

Since the dipole operator commutes with spin, the transition of $|nJFm_{F}\rangle$ to $|n^{/}primeJ^{/}primeF^{/}primem^{/}prime_{F}\rangle$ is independent of $F$ and $m_{F}$. This means that the emission intensity of transition from $|nJF\rangle$, summed over all $m_{F}$, to $|n^{/}primeJ^{/}prime\rangle$, summed over $m^{/}prime_{F}$ and $F^{/}prime$, is the intensity of transition from a particular $|nJFm_{F}\rangle$, times $2F+1$, the degeneracy of the initial state. This is called the “sum rule” of the transition.

The sum rules, together with the selected transitions in Fig. 2, immediately lead to

${\begin{aligned}{\frac {I_{51}}{I_{41}+I_{40}}}={\frac {5}{3}},\ &{\frac {I_{41}+I_{40}}{I_{31}+I_{30}}}=1,\ &{\frac {I_{21}}{I_{31}+I_{30}}}={\frac {1}{3}},\\{\frac {I_{40}}{I_{41}+I_{51}}}={\frac {1}{3}},\ &{\frac {I_{30}}{I_{21}+I_{31}}}={\frac {1}{3}}.\end{aligned}}\,\!$

The second line follow from the transition into excited states.

Next we shall neglect radiative transfer effect and make the assumption that background intensities and temperatures are constant across all the hyperfine lines, and are given by the CMB values. Under this assumption

$P_{\alpha }=n_{(n=1)}B_{Ly\alpha }={\frac {3T_{\gamma }}{T_{*}}}A_{Ly\alpha }.\,\!$

Let $I_{tot}=I_{51}+I_{40}+I_{41}+I_{31}+I_{30}+I_{21}$ be the total intensity of de-excitation. Then from the relations on $I$ above we get

${\begin{aligned}I_{30}=I_{41}&={\frac {1}{12}}I_{tot},\\I_{31}=I_{40}&={\frac {1}{6}}I_{tot}.\end{aligned}}\,\!$

Furthermore, the intensities are related to the Einstein coefficients by

${\frac {I_{ki}}{I_{\alpha }}}={\frac {g_{k}}{g_{tot}}}{\frac {A_{kj}}{A_{\alpha }}},\,\!$

where $g_{k}=2F_{k}+1$ and $g_{tot}=1+3+3+5=12$ is the total degeneracy of $n=2$ level.

Thus we have

${\begin{aligned}A_{30}&=A_{41}=A_{\alpha }/3,\\A_{31}&=A_{40}=2A_{\alpha }/3.\end{aligned}}\,\!$

Thus finally $P_{10}=4P_{\alpha }/27$ and the coupling coefficient becomes

${x_{W}={\frac {4P_{\alpha }}{27A_{10}}}{\frac {T_{*}}{T_{\gamma }}}}.\,\!$

Color temperature and corrections

The effect of Ly$\alpha$ absorption on spin temperature also depends on the color temperature $T_{W}$. For an environment that’s optically thick, such as the high-redshift IGM, we may take

$T_{W}=T_{K}.\,\!$

One commonly considered correction is due to energy loss in spin-exchange collisions. In these collisions, the electron spins of the two colliding hydrogen atoms exchange, while the total spin remains unchanged. This correction to the color temperature is worked out in for example [, section 2.3.3. We shall here just quote the result:

${T_{W}=T_{K}\left({\frac {1+T_{se}/T_{K}}{1+T_{se}/T_{S}}}\right),}\,\!$

where the spin exchange temperature

$T_{se}={\frac {2T_{K}\nu _{se}^{2}}{9\Delta \nu _{D}^{2}}}\sim 0.40{\mbox{K}}.\,\!$

To use Eq. eq:colort, the spin temperature must be determined iteratively. The iteration is shown to converge quickly.

Another commonly considered correction are radiative transfer effects. Previously we assumed that the background intensity is constant and given by the CMB intensity. The Ly$\alpha$ absorption would decrease the background intensity and hence the estimated scattering rate. The effect is more important for lower kinetic temperature. This leads to a correction to the WF effect coupling coefficient $S_{\alpha }$ such that

$x_{W}=S_{\alpha }{\frac {J_{\alpha }}{J_{\nu }^{c}}},\,\!$

where

$J_{\nu }^{c}=1.165\times 10^{-10}\left({\frac {1+z}{20}}\right){\mbox{cm}}^{-2}{\mbox{s}}^{-1}{\mbox{Hz}}^{-1}{\mbox{sr}}^{-1}.\,\!$

Neglecting spin exchange, the suppression factor is given by

$S_{\alpha }\sim \exp \left[-0.803T_{K}^{-2/3}(10^{-6}\tau _{GP})^{1/3}\right],\,\!$

where the Gunn-Peterson optical depth can be written for overdensity $\delta$ as roughly:

$\tau _{GP}=7\times 10^{5}\left({\frac {\Omega _{b}h_{100}}{0.03}}\right)\left({\frac {\Omega _{m}}{0.25}}\right)^{-1/2}\left({\frac {1+z}{10}}\right)(1+\delta ).\,\!$

[1] Direct transitions are disfavored by selection rules, as we shall explain later.

[2] In general, a tensor of rank $k$ carries $k$ units of angular momentum, and has $2k+1$ possible spin states. Due to the lack of mass of a photon however, the electromagnetic wave is transverse in all reference frames (easily seen with Maxwell’s equations), and thus only has two polarization states, with angular momentum (anti-)parallel to the direction of motion.