Difference between revisions of "Wouthuysen Field effect"

From AstroBaki
Jump to navigationJump to search
Line 28: Line 28:
 
the dominant effect until reionization is complete.
 
the dominant effect until reionization is complete.
  
\subsection*{Spectroscopic Notation}
+
\subsection*{Fine structure of hydrogen}
To understand how the WF effect works, we need to have a reasonable
 
notation for all of the angular momentum quantum numbers of the
 
hydrogen atom. There are 3 sources of angular momentum:
 
\begin{itemize}
 
\item orbital angular momentum ($L = 0, ..., n-1$)
 
\item electron spin ($\frac{1}{2}$)
 
\item proton spin ($\frac{1}{2}$)
 
\end{itemize}
 
Spectroscopic notation uses the following quantum numbers to describe
 
the energy states of the hydrogen atom, with fine structure and
 
hyperfine structure:
 
\begin{itemize}
 
\item orbital angular momentum ($L$)
 
\item orbital combined with electron spin ($J = L + \frac{1}{2}$)
 
\item total angular momentum ($F = J + \frac{1}{2}$)
 
\end{itemize}
 
The orbital angular momentum is denoted by letters, as you probably
 
remember vaguely from chemistry. For the purposes of this discussion,
 
we'll only be talking about $L=0$ and $L=1$, which are denoted by $S$
 
and $P$.
 
  
The format for spectroscopic notation is $n~_FL_J$. So, let's list all
+
Energy levels in hydrogen atoms are split due to spin orbit interaction
of the ground state and first excited state energies in hydrogen using
+
(fine splitting), and the smaller effect of electron-proton spin interaction
this notation.\footnote{This list comes from knowing all the possible
+
(hyperfine splitting). Anti-aligned spins lead to lower energy levels.
  ways to combine angular momenta. For example, $J$ is a combination
+
The splittings of the lowest energy levels are shown in Fig. 1:
  of $L$ and $\frac{1}{2}$, so it can range from $|L-\frac{1}{2}|$ to
 
  $L+\frac{1}{2}$ in integer steps. This means for $L=0$ you can only have
 
  $J=\frac{1}{2}$, and for $L=1$ you can have $J=\frac{1}{2}$ or
 
  $J=\frac{3}{2}$. The same rule applies when you add in proton spin;
 
  $F$ can range from $|J-\frac{1}{2}|$ to $J + \frac{1}{2}$. This
 
  means for $J=\frac{1}{2}$ you can have $F=0$ or $F=1$, and for
 
  $J=\frac{3}{2}$ you can have $F=1$ or $F=2$.}
 
\begin{itemize}
 
\item $n=1$, $L=0$: $1~_0S_\frac{1}{2}$, $1~_1S_\frac{1}{2}$
 
\item $n=2$, $L=0$: $2~_0S_\frac{1}{2}$, $2~_1S_\frac{1}{2}$
 
\item $n=2$, $L=1$: $2~_0P_\frac{1}{2}$, $2~_1P_\frac{1}{2}$,
 
  $2~_1P_\frac{3}{2}$, $2~_2P_\frac{3}{2}$
 
\end{itemize}
 
 
 
For better visualization, figure \ref{all levels} is an energy level
 
diagram labeled with the same notation. It's formatted with increasing
 
energy going up (but not to scale), and with $S$ levels $(L=0)$ in the
 
first column and $P$ levels $(L=1)$ in the second column.
 
  
 
\begin{figure}
 
\begin{figure}
Line 79: Line 41:
 
\end{figure}
 
\end{figure}
  
\subsection*{Selection Rules}
+
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 %
 +
\footnote{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 $T_{\gamma}=2.74(1+z)\mbox{ K}$.
 +
 
 +
\subsection*{Cosmological context}
 +
 
 +
After recombination $(z\sim1100)$, 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
 +
\[
 +
T_{b}=T'_{b}/(1+z).
 +
\]
 +
 
 +
 
 +
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:
 +
\[
 +
\frac{n_{1}}{n_{0}}=\frac{g_{1}}{g_{0}}e^{-h\nu/k_{B}T_{S}}=3\exp(-T_{*}/T_{S})\approx3(1-T_{*}/T_{S}).
 +
\]
 +
During the dark ages till the end of reionization, several processes
 +
control the relative abundances. Among them are direct radiative transitions
 +
(Einstein coefficients) %
 +
\footnote{Direct transitions are disfavored by selection rules, as we shall
 +
explain later. %
 +
}, 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
 +
\begin{equation}
 +
n_{1}(A_{10}+B_{10}I_{CMB}+C_{10}+W_{10})=n_{0}(B_{01}I_{CMB}+C_{01}+W_{01}),\label{eq:balance}
 +
\end{equation}
 +
where $A_{10}=2.869\times10^{-15}\mbox{s}^{-1}$ 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{equation}
 +
\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}
 +
\label{eq:BI}
 +
\end{equation}
 +
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:
 +
 
 +
\[
 +
\begin{aligned}\frac{C_{01}}{C_{10}} & =\frac{g_{1}}{g_{0}}e^{-T_{*}/T_{K}}\approx3\left(1-\frac{T_{*}}{T_{K}}\right).\end{aligned}
 +
\]
 +
Similarly, WF effect rates are given by the ``color temperature''
 +
$T_{W}$:
 +
\[
 +
\begin{aligned}\frac{W_{01}}{W_{10}} & =\frac{g_{1}}{g_{0}}e^{-T_{*}/T_{W}}\approx3\left(1-\frac{T_{*}}{T_{W}}\right).\end{aligned}
 +
\]
 +
Plugging the above results into the balance equation \eqref{eq:balance},
 +
we have the relation of the temperatures:
 +
\[
 +
\boxed{T_{S}^{-1}=\frac{T_{\gamma}^{-1}+x_{c}T_{K}^{-1}+x_{W}T_{W}^{-1}}{1+x_{c}+x_{W}},}
 +
\]
 +
where
 +
\begin{equation}
 +
\boxed{\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}
 +
}\label{eq:coeff}
 +
\end{equation}
 +
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 $T=13.6eV\sim1.6\times10^{5}\mbox{K}$,
 +
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.
 +
 
 +
\subsection*{Parity and Selection Rules}
 
Now we need to look at the selection rules for dipole
 
Now we need to look at the selection rules for dipole
 
transitions. First of all, there is the selection rule that $\Delta L
 
transitions. First of all, there is the selection rule that $\Delta L
= 0$ is forbidden by parity, so only transitions between $1S$ and $2P$
+
= 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
 
will matter. (In other words, nevermind about the $2S$ states we wrote
 
down above.) This is the selection rule that forbids the 21cm
 
down above.) This is the selection rule that forbids the 21cm
Line 99: Line 167:
 
momentum quantum number, not the same as $J$.)
 
momentum quantum number, not the same as $J$.)
  
Notice that this is the same rule as if you were combining angular
+
To put illustrate all of these in simpler language, a photon is a vector $A^{\mu}$ and carries one unit
momentum $j$ with another angular momentum $k=1$. It can be summarized
+
of angular momentum. This means that a single photon must be circularly
as $\Delta j = 0, \pm 1$, except that a transition from $j = 0$ to $j' =
+
polarized (left or right) %
0$ is not allowed. (Combining $j=0$ and $k=1$ can only give $j'=1$.)
+
\footnote{In general, a tensor of rank $k$ carries $k$ units of angular momentum,
This selection rule applies to every angular momentum quantum
+
and has $2k+1$ possible spin states. Due to the lack of mass of a
number, $L$ and $J$ and $F$. In contrast, the rule that $\Delta L = 0$
+
photon however, the electromagnetic wave is transverse in all reference
is forbidden by parity only applies to $L$. So in total, our selection
+
frames (easily seen with Maxwell's equations), and thus only has two
rules are:
+
polarization states, with angular momentum (anti-)parallel to the
\begin{itemize}
+
direction of motion. %
\item $\Delta L = \pm 1$ (see footnote)\footnote{Remember $\Delta L =
+
}. Since the vector does not has any spin dependence, it commutes with
    0$ is forbidden by parity}
+
the electron spin operator:
\item $\Delta J = 0, \pm 1$ (see footnote)\footnote{$J=0$ is
+
\[
    impossible since $J$ is always half integers, so that exception
+
[A^{\mu},\ S]=0.
    for $J = J' = 0$ doesn't apply.}
+
\]
\item $\Delta F = 0, \pm 1$, except $F = 0$ to $F' = 0$ is forbidden.  
+
Here $S=|\vec{S}_{e}+\vec{S}_{p}|$ is the vector sum of the electron
\end{itemize}
+
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
 +
\[
 +
\begin{aligned}M & =\langle n'l'm'|A^{\mu}|nlm\rangle,\\
 +
& =\langle n'l'm'|P^{\dagger}PA^{\mu}P^{\dagger}P|nlm\rangle,\\
 +
& =(-1)^{l-l'+1}\langle n'l'm'|A^{\mu}|nlm\rangle.
 +
\end{aligned}
 +
\]
 +
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, $l'-l$ 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:
 +
\[
 +
|\Delta L|,|\Delta F|,|\Delta J|\le1.
 +
\]
 +
Thus we have the rule
 +
\[
 +
\Delta l=\pm1.
 +
\]
 +
 
  
Figure \ref{all transitions} is another copy of the energy level
+
To see this more explicitly, one can also write out the full expression
diagram with the allowed transitions marked. The $2S$ states are left
+
\[
off the diagram completely this time.
+
|nlm\rangle=R_{lm}(r)P_{l}^{m}(\cos\theta)e^{im\phi}\frac{1}{\sqrt{2\pi}}.
 +
\]
 +
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
 +
\[
 +
\begin{aligned}\langle n'l'm'|d|nlm\rangle & \propto\int_{0}^{\pi}P_{l'}^{m'}P_{l}^{m}\cos\theta\sin\theta d\theta\times\frac{1}{2\pi}\int_{0}^{2\pi}e^{i(m-m')\phi}d\phi.\end{aligned}
 +
\]
 +
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
 +
\[
 +
\int_{-1}^{1}P_{l'}^{m}(x)P_{l}^{m}(x)dx=\frac{2(l+m)!}{(2l+1)(l-m)!}\delta_{l'l},
 +
\]
 +
the above matrix element vanishes unless
 +
\[
 +
\begin{cases}
 +
m' & =m,\\
 +
l' & =l\pm1.
 +
\end{cases}
 +
\]
 +
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 $L\ne\sum l_{i}$,
 +
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
 +
\[
 +
\begin{cases}
 +
\Delta S & =0,\\
 +
\Delta L & =\pm1,\\
 +
\Delta J & =0,\pm1,\mbox{(except 0 to 0),}\\
 +
\Delta F & =0,\pm1,\mbox{(except 0 to 0). }
 +
\end{cases}
 +
\]
 +
 
 +
 
 +
\subsection*{The WF effect}
 +
 
 +
 
 +
\subsubsection*{Coupling coefficient}
 +
 
 +
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.  
 
\begin{figure}
 
\begin{figure}
 
\center{\includegraphics{WFfig2.jpg}}
 
\center{\includegraphics{WFfig2.jpg}}
Line 125: Line 338:
 
\end{figure}
 
\end{figure}
  
But we care only about transitions that can cause an effective
 
hyperfine transition, so the diagram really looks like figure \ref{wf
 
  transitions}.
 
  
 
\begin{figure}
 
\begin{figure}
Line 135: Line 345:
 
\end{figure}
 
\end{figure}
  
Thus we can see there are 2 possible pathways for the WF effect to
 
cause a hyperfine excitation or de-excitation. It's possible to get
 
into the details of transition rates for each intermediate transition,
 
transition rates for each pathway, etc, but it's not really worth the
 
effort. Since Ly-$\alpha$ radiation will usually be coupled to the
 
kinetic temperature of the gas anyway, we'll go straight into
 
discussing temperatures.
 
  
\subsection*{Spin Temperature}
+
Evidently, the (de-)excitation rate due to WF effect is given by:
The Ly-$\alpha$ line is so optically thick that we expect the
 
radiation field to be coupled to the kinetic gas temperature. In other
 
words, the Ly-$\alpha$ radiation temperature will be the same as the
 
gas temperature $T_K$. The other important temperature is $T_s$, the
 
``spin temperature'' or ``excitation temperature'', which is
 
defined by the following equation. $T_s$ only describes the hyperfine
 
level populations, and may or may not be related to any physical
 
temperatures in the problem.
 
 
\begin{equation}
 
\begin{equation}
\frac{n_1}{n_0} = \frac{g_1}{g_0} e^{-h\nu/kT_s}
+
\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}
 +
\label{eq:W01}
 
\end{equation}
 
\end{equation}
 +
In the second and fourth lines I have used Eq. \eqref{eq:BI}.
 +
 +
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 $\nu_{\alpha}=2.47\times10^{15}\mbox{Hz}$ is the Lyman $\alpha$
 +
line center frequency.
 +
 +
Since the dipole operator commutes with spin, the transition of $|nJFm_{F}\rangle$
 +
to $|n'J'F'm'_{F}\rangle$ is independent of $F$ and $m_{F}$. This
 +
means that the emission intensity of transition from $|nJF\rangle$,
 +
summed over all $m_{F}$, to $|n'J'\rangle$, summed over $m'_{F}$
 +
and $F'$, 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
  
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 $q_{01}$ and $q_{10}$ are given by:
 
 
\begin{equation}
 
\begin{equation}
\frac{q_{01}}{q_{10}} = \frac{g_1}{g_0} e^{-h\nu/kT_K}
+
\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}
 +
\label{eq:int}
 
\end{equation}
 
\end{equation}
 +
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 Eq. \eqref{eq:int} 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
 +
in Eq. \eqref{eq:coeff} becomes
 +
\[
 +
\boxed{x_{W}=\frac{4P_{\alpha}}{27A_{10}}\frac{T_{*}}{T_{\gamma}}}.
 +
\]
 +
 +
 +
 +
\subsubsection*{Color temperature and corrections}
  
So, the WF effect is very similar, since it couples $T_s$ and $T_K$
+
The effect of Ly$\alpha$ absorption on spin temperature also depends
(indirectly through Ly-$\alpha$). If you define the transition rates due
+
on the color temperature $T_{W}$. For an environment that's optically
to the WF effect as $P_{01}$ and $P_{10}$, you have:
+
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 \cite{key-1},
 +
section 2.3.3. We shall here just quote the result:
 
\begin{equation}
 
\begin{equation}
\frac{P_{01}}{P_{10}} = \frac{g_1}{g_0} e^{-h\nu/kT_K}
+
\boxed{T_{W}=T_{K}\left(\frac{1+T_{se}/T_{K}}{1+T_{se}/T_{S}}\right),}\label{eq:colort}
 
\end{equation}
 
\end{equation}
 +
where the spin exchange temperature
 +
\[
 +
T_{se}=\frac{2T_{K}\nu_{se}^{2}}{9\Delta\nu_{D}^{2}}\sim0.40\mbox{K}.
 +
\]
 +
To use Eq. \eqref{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
 +
\[
 +
J_{\nu}^{c}=1.165\times10^{-10}\left(\frac{1+z}{20}\right)\mbox{cm}^{-2}\mbox{s}^{-1}\mbox{Hz}^{-1}\mbox{sr}^{-1}.
 +
\]
 +
 +
 +
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:
  
In order to get an exact expression for $P_{01}$ or $P_{10}$, you
+
\[
would need to look at the Ly-$\alpha$ radiation field, the einstein
+
\tau_{GP}=7\times10^{5}\left(\frac{\Omega_{b}h_{100}}{0.03}\right)\left(\frac{\Omega_{m}}{0.25}\right)^{-1/2}\left(\frac{1+z}{10}\right)(1+\delta).
coefficients for each transition in the WF effect diagram, and all the
+
\]
other nasty details. All we need to know is that once the first stars
 
and AGN turn on, the WF effect dominates the hyperfine level
 
populations.
 
  
 
\end{document}
 
\end{document}
 
</latex>
 
</latex>

Revision as of 16:18, 30 April 2015

Reference Materials

The Wouthuysen-Field Effect

The Wouthuysen-Field effect is a coupling of the 21cm hyperfine transition to Ly- 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 state, and back down to the other hyperfine state are not forbidden. So, if there is sufficient Ly- 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 shown in Fig. 1:

WFfig1.jpg
Energy level diagram with spectroscopic notation

Here we have used spectroscopic notation , where is the principal quantum number, are the electron orbital angular momentum. () shall denote the electron (proton) spin (not to be confused with the orbital). Then is the electron total angular momentum, and is the hydrogen total angular momentum [1].

The splitting between the two levels is of particular interest. It has energy difference , corresponding to wavelength . We also define . In the regime, the Rayleigh-Jeans limit holds and we can define the “brightness temperature”

only serves to redefine , and should be distinguished from the ambient temperature of the CMB .

Cosmological context

After recombination , 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 . Studying the 21cm absorption of neutral hydrogen can thus potentially probe the universe during the dark ages and epoch of reionization. The frequency of 21cm undergoes cosmological redshift and hence the actual observed brightness temperature is

Failed to parse (syntax error): {\displaystyle T_{b}=T’_{b}/(1+z). \,\!}

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

During the dark ages till the end of reionization, several processes control the relative abundances. Among them are direct radiative transitions (Einstein coefficients) [2], collisional excitation , and the Wouthuysen-Field (WF) Effect and . The WF effect will be explained in detail. For now let’s just note that the overall statistical balance gives

where and are the Einstein coefficients of the 21cm transition. Recall that in thermodynamic equilibrium, we had

In the Rayleigh-Jeans limit, radiative coefficients satisfy

Here 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 and are given by:

Similarly, WF effect rates are given by the “color temperature” :

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

where

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 The WF effect involves absorption of an Lyman- photon from the state and subsequent decay into the state. The CMB temperature during the dark ages have temperatures , and is much smaller than the Lyman- transition temperature of , and thus we expect WF effect to be most important during the epoch of reionization, when the first stars provide the abundance of 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 is forbidden by parity, a symmetry of electromagnetic interactions, so only transitions between and will matter. (In other words, nevermind about the 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- 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 to Failed to parse (syntax error): {\displaystyle j’} is only allowed if Failed to parse (syntax error): {\displaystyle j’} is in the range to , where is the “order” of the tensor operator. For dipole transitions, : the dipole operator is an “order 1 irreducible tensor operator”, so the rule becomes Failed to parse (syntax error): {\displaystyle j’ = |j-1|, ..., j+1} . (Lowercase is a totally generic angular momentum quantum number, not the same as .)

To put illustrate all of these in simpler language, a photon is a vector and carries one unit of angular momentum. This means that a single photon must be circularly polarized (left or right) [3]. Since the vector does not has any spin dependence, it commutes with the electron spin operator:

Here is the vector sum of the electron and proton spins.

This implies the first dipole selection rule:

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

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

If we fix the phase such that , we have parity as a Hermitian and hence observable:

Parity thus acts on other operators via conjugation. For example

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 , while it anti-commutes with momentum . 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:

Under parity, , , and since we have

The parity of a state does not depend on , 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

Failed to parse (unknown function "\begin{align}"): {\displaystyle \begin{align} M & =\langle n’l’m’|A^{\mu }|nlm\rangle ,\\ & =\langle n’l’m’|P^{\dagger }PA^{\mu }P^{\dagger }P|nlm\rangle ,\\ & =(-1)^{l-l'+1}\langle n’l’m’|A^{\mu }|nlm\rangle . \end{align} \,\!}

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, Failed to parse (syntax error): {\displaystyle l’-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:

Thus we have the rule

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

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

The matrix element can be written as

Failed to parse (unknown function "\begin{align}"): {\displaystyle \begin{align} \langle n’l’m’|d|nlm\rangle & \propto \int _{0}^{\pi }P_{l'}^{m'}P_{l}^{m}\cos \theta \sin \theta d\theta \times \frac{1}{2\pi }\int _{0}^{2\pi }e^{i(m-m')\phi }d\phi .\end{align} \,\!}

From properties of associated Legendre polynomials, we can decompose

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

the above matrix element vanishes unless

Failed to parse (unknown function "\begin{cases}"): {\displaystyle \begin{cases} m’ & =m,\\ l’ & =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 and is in place when there are more than one electron in the atom. We have seen that a single electron contributes a factor of when acted on by parity. In the multi-electron case, this generalizes to

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

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

The WF effect

Coupling coefficient

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 states by 0 and 1, we shall by convenience denote the 2 states by 2,3,4,5, in order of increasing energy.

WFfig2.jpg
Energy level diagram with allowed transitions
WFfig3a.jpg
Energy level diagram of WF effect

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

In the second and fourth lines I have used Eq. eq:BI.

We would like to relate this rate to the total Ly scattering rate

where is the absorption cross section and , with the oscillator strength. The line can be assumed to be a Voigt profile. Thermal broadening leads to Doppler width

where is the Lyman line center frequency.

Since the dipole operator commutes with spin, the transition of to Failed to parse (syntax error): {\displaystyle |n’J’F’m’_{F}\rangle } is independent of and . This means that the emission intensity of transition from , summed over all , to Failed to parse (syntax error): {\displaystyle |n’J’\rangle } , summed over Failed to parse (syntax error): {\displaystyle m’_{F}} and Failed to parse (syntax error): {\displaystyle F’} , is the intensity of transition from a particular , times , 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

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

Let be the total intensity of de-excitation. Then from Eq. eq:int we get

Furthermore, the intensities are related to the Einstein coefficients by

where and is the total degeneracy of level.

Thus we have

Thus finally and the coupling coefficient in Eq. eq:coeff becomes

Color temperature and corrections

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

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:

where the spin exchange temperature

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 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 such that

where

Neglecting spin exchange, the suppression factor is given by

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

[1] Note these are vector sums.

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

[3] In general, a tensor of rank carries units of angular momentum, and has 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.