hyperfine splitting of hydrogen
type $\bracket{+}{\sigma_z}{+}$âwere useful for describing the behavior
\end{equation*}, With that clue you can immediately see another solution from the last
The hyperfine structure of atomic hydrogen is derived in a simple and self-contained way that makes the theory accessible to advanced undergraduates in … the four states:
particlesâsame idea.
Now in the most general case we could have more complex things. For example, if we have $\sigmap_x$ acting on
in (12.15) as a simple operator equation:
\Hop\,\ket{+\,+}=A\{\ket{-\,-}-\ket{-\,-}+\ket{+\,+}\}=A\,\ket{+\,+}.
way. proton alone. \sigmae_y\sigmap_y+
fields are easy to understand, and rather interesting. negative, and about $1000$ times larger than $\mu_{\text{p}}$âwhich
The Z3-increase leads to a transition energy in the UV-region of the optical spectrum for the case of Bi82+. in a more complicated situation than we are considering, the base
\kern{-6em}\text{with}\\ % ebook remove
\label{Eq:III:12:39}
where the $+$ and $-$ represent the components of angular momentum
\label{Eq:III:12:25}
\frac{E+A-(\mu_{\text{e}}-\mu_{\text{p}})B}{2A}\notag\\
states in (12.1). \begin{aligned}
light. A\{-1+2\sqrt{1+(\mu_{\text{e}}-\mu_{\text{p}})^2B^2/4A^2}\},\\[1ex]
\label{Eq:III:12:3}
ground state. \sigmae_x\sigmap_z\,\ket{-\,-}&=-\,\ket{+\,-}. \label{Eq:III:12:9}
\begin{align}
f=\omega/2\pi=(1{,}420{,}405{,}751.800\pm0.028)\text{ cycles per second}. For
âthe ground state.â The details of the configurationâthe
been calculated to an accuracy of about $30$ parts in one million. Fig.
G. M. Zinov’ev, B. V. Struminskii, R. N. Faustov et al, Yad. We get only the one term. than $\mu_{\text{e}}$, and has the opposite sign). therefore, the Hamiltonian matrix in particularâcan be written as a
handy rule for figuring out $\FLPsigmae\cdot\FLPsigmap$. \begin{equation}
what happens next. The effect of decreasing r0 (i.e., compressing the sphere isotropically) on the hyperfine splitting A of the ground state is nonlinear, with A increasing approximately as r−10. \ket{-S}\\
The mixtures of the base states $\ket{+\,-}$ and $\ket{-\,+}$
\ket{++};\;\text{electron}\;\textit{up},\;\text{proton}\;\textit{up}\\
\label{Eq:III:12:33}
… there is an energy proportional to the cosine of the angle between the
We can
E_{\slIII}\;&=-A+\mu'B,&\quad
and $\ket{-S}$ for the states of a spin-one particle. \sigmae_x(-\,\ket{+\,-})=
But they are both just the
\end{equation*}
\end{equation}
On the other hand, the proton can
\braket{+,\FLPp}{\psi}\quad
of a single particle of spin one-half. We have to do as we did for
\end{alignat}
\begin{equation}
Rev.
will, the rate of change in time is given by the operator $\Hop$. (12.47)
The FAMU experiment aims to measure for the first time the hyperfine splitting of the muonic hydrogen ground state. \end{equation*}
horizontal slopes. important thing is that Eq. That is all there is
\label{Eq:III:12:25}
On the
six transitions shown by the vertical arrows in Fig. H_{21}&\to2A,\\[1ex]
The first question we have to answer is: What are the base
5-6âthe
\end{align}, \begin{equation}
All the operator does is interchange the spin directions of the two
&\textit{State 4:}\;\small
(1{,}420{,}405{,}751.800\pm0.028)\text{ cycles per second}. want to find those special states $\ket{\psi}$ for which each
You have probably heard before about the â$21$-centimeter
E_{\slIV}&=-3A. in the last two terms. we can use them to get:
They are, in fact, just the amplitude we
the hydrogen can be calculated from a more detailed theory. +\,&\ket{-\,+}C_3+\ket{-\,-}C_4. \frac{ad+bc}{\sqrt{2}}\,\{\ket{+'\,-'}+\ket{-'\,+'}\}+
\label{Eq:III:12:48}
a_1=1,\quad a_2=a_3=a_4=0,
find an analogous device to describe a system with two spins? Its three components $\sigmap_x$,
have only certain values always $\hbar$ apart. then they are all quite similar. There is no such thing as âtheâ base states,
Even our labeling of the states agrees with
\end{equation}, Before solving these equations we canât resist telling you about a
G. T. Bodwin and D. R. Yennie, Phys. So the only possibility for a Hamiltonian with the
state $\ketsl{\slIV}$ pick up energy and go into one of the upper statesâbut
So the question is not what is
\ketsl{\slIII}&=\frac{1}{\sqrt{2}}\,(\ketsl{\slTwo}+\ketsl{\slThree})\\[1ex]
and $\braket{-'}{+}_{\text{p}}$ of having spin âupâ or spin âdownâ in
worked out by the atom so to speakâit has worked itself out by
P_{\text{spin exch}}\,\ket{-\,+}&=\ket{+\,-},\\[1ex]
\end{alignat}
E_{\slIV}&=-3A.
all of the energies change in a different way. You will remember that in the last chapter we were able to describe
electron, but does nothing to the proton and multiplies the result
physics of the situation. So we have found the energies of the four stationary states of a
way the atom works. If these four amplitudes change with time, as they
One solution is the state $\ketsl{\slI}$ for which $a_1=1$, $a_2=$ $a_3=$ $a_4=$ $0$, or
that the product $\sigmae_x\sigmap_z$ gives the following results for
astronomy. for any particle its component of angular momentum along any axis can
In our frame the electron in the $\ket{+\,+}$ state
states for zero field in (12.41) and (12.42)
\sigmae_x\sigmap_x\,&\ket{+\,+}=+\,\ket{-\,-}\\[.5ex]
start with a base set which is physically the clearest. Or, if we shine microwaves on hydrogen gas,
condition of the atom. each of the four base states, we getâalways using
\right\}
\notag \ketsl{\slII}&=\ketsl{\slFour}=\ket{-\,-},\\
\end{equation}
field is determined completely by the interaction of the two magnetic
As
i\hbar\,dC_3/dt&=
The four spin states do not all have exactly
\braket{-'\,-'}{+\,+}&=
Then they begin to curve, and for large $B$
straight, we summarize the new notation in Table 12-4. $\FLPmu_{\text{e}}$ and $\FLPmu_{\text{p}}$, the mutual energy will
Although in Chapter 9 we used to call these energies
Now,
\;\text{spin $1$}\;
\label{Eq:III:12:52}
12-4. complicated that once you see how this one is handled you can get
we have
all four states, energy can be absorbedâor emittedâin any one of the
the energy $+A$âwhich has spin one. combination is too hard at this level. positions of the electron and proton because that has all been
J_z=m\hbar,
He will be referring his states to what we will call
second term tells us all we need to know to find the level splitting
Ea_3&=2Aa_2-\{A-(\mu_{\text{e}}-\mu_{\text{p}})B\}a_3,\\[1ex]
\left\{
\ket{\OS}=\sqrt{2}\,ac\,\ket{+T}+(ad+bc)\,\ket{\OT}+
\begin{equation}
\sigmae_x\sigmap_z\,\ket{+\,-}=
\sigmae_z\sigmap_z\}\,\ket{+\,+}. numberâ ($j$) and âmagnetic quantum numberâ ($m$)]. amplitudes for all the momentum base statesâcan be calculated, but
\ket{+\,+}&=&\phantom{ab}&a^2&\,&\ket{+'\,+'}\notag\\[.5ex]
Now we are concerned only with the effects of
So you see that to write down the
(Of course, taking all the $a$âs equal to zero also gives a
Now we will work through the general case for all the states. electron, but to describe the general Hamiltonian of such a
workâno force on them in a magnetic field gradient.) Suppose we run it through a
to the steady strong field $\FLPB$). \end{equation*}, \begin{alignat*}{3}
\end{equation}
\end{equation}
\begin{equation}
we could also have taken the amplitudes that go with them. But
\label{Eq:III:12:6}
therefore, only diagonal elementsâwe can just add the
took $A$ as positive because the theory we spoke of says it should be, and
It has, in fact,
by $-i$.) \begin{equation}
gives a slightly different magnetic energy for each spin state. at very low magnetic fields. to $21$-cm waves (or $1420$ megacycles approximately) we can observe the
The total splitting of the 1S state is 182.725 meV; this value can be used as a reliable estimate in conducting a corresponding experiment with an accuracy of 30 ppm. first spin symbolâthat is, on the electron spin. \begin{equation*}
We invent a thing which we
only at the frequency $\omega=4A/\hbar$. \ket{\OS}&=&&\phantom{a}\frac{2}{\sqrt{2}}\,ac&\,&\ket{+'\,+'}\\[.75ex]
12.2.1 below), and to describe it theoretically we need to consider additional contributions to HFS connected with the bound state nature of the proton. The magnetic dipole moment due to the nuclear spin is much smaller than that of the electron because the mass appears in the denominator. We can, then, write the state $\ket{+\,+}$ as the following linear
\begin{equation*}
\end{equation}
Next we define the corresponding operator âsigma protonâ for the proton spin.
For $\ket{\OS}$ itâs a little more complicated, because
time, people usually write
(12.5), we can use it
It would be ridiculous,
There is still the small additional term $A$
Hyperfine Splitting. \end{gather*}
Its energy is
\text{The electron is âdownâ and the proton is âup.â}\\[3pt]
Then we have to add to our Hamiltonian
\ket{+\,-}&&=ac\,&\ket{+'\,+'}+ad\,\ket{+'\,-'}\notag\\[.5ex]
It follows, from the above analysis, that spin-spin coupling breaks the degeneracy of the two states in hydrogen, lifting the energy of the triplet configuration, and lowering that of the singlet. for writing down the Hamiltonian. state will have the energy $E=\hbar\omega$. \Hop\,\ket{+\,+}=A\FLPsigmae\cdot\FLPsigmap\,\ket{+\,+}=
the rest, so the same technique works again. \end{alignat*}
\ket{\text{electron âupâ with momentum $\FLPp$}}
\sigmae_y\sigmap_y\,&\ket{+\,-}=+\,\ket{-\,+}\\[.5ex]
(9.24), which is
\sigmae_x(\sigmap_z\,\ket{+\,-})=
The sum of the two
there if there were no magnetic field. \label{Eq:III:12:12}
&=A\{\sigmae_x\sigmap_x+\sigmae_y\sigmap_y+\sigmae_z\sigmap_z\}\,\ket{+\,+}. E_{\slI}&=A+\mu B,\\[1.3ex]
\end{align*}
\end{equation*}
-\sigmae_x\,\ket{+\,-}=
(When $\sigmae_y$ acts on the combined state it flips over the
This frequency has been
\end{align*}
vectors, we get the following Hamiltonian matrix, $H_{ij}$:
\begin{equation*}
\end{equation}
E_{\slIV}&=-A\{1+2\sqrt{1+\mu'^2B^2/4A^2}\}. of axes for a coordinate system. calculate in this chapter. \label{Eq:III:12:51}
\end{equation}
A\{\sigmae_x\sigmap_x+
scalar invariant combination of two vectors is the dot product,
what we called them before. \{A-(\mu_{\text{e}}+\mu_{\text{p}})B\}C_1,\\[1ex]
Now the difference in energy between state $\ketsl{\slIV}$ and any one
amplitudes,
Such a classical qualitative
(12.5) is the correct quantum
\sigma_y\,&\ket{-}=-i\,\ket{+}
and (12.51); we get that
If the proton is also âdown,â the two
do with quantum mechanics. \begin{aligned}
In the same way, the last
In the first place, the energies for large
\begin{equation}
solution, but thatâs no state at all!) $\ketsl{\slIII}$ goes straight through. The theoretical knowledge of the two-photon exchange (TPE) correction to the hyperfine splitting (HFS) of the S energy levels in muonic hydrogen exceeds by two orders of magnitude the expected ppm level of the experimental accuracy in the forthcoming measurements of 1S HFS by CREMA Pohl:2016tqq and FAMU Dupays:2003zz (); Adamczak:2016pdb collaborations as well as at J-PARC … of $R_y(\alpha)$ in Table 6-2. state of hydrogen. absorbs or emits radio waves at $21$ centimetersâwe have to know
matricesâor the exactly equivalent sigma operators. \Hop\,\ket{+\,+}&=A\FLPsigmae\cdot\FLPsigmap\,\ket{+\,+}\notag\\[1ex]