I'm trying to understand how Hamiltonian matrices are built for optical applications. In the excerpts below, from the book "Optically polarized atoms: understanding light-atom interaction", what I don't understand is: Why are the μB parts not diagonal? If the Hamiltonian is →μ⋅→B, why aren't all the components just diagonal? How is this matrix built systematically? Can someone please explain?
We now consider the effect of a uniform magnetic field B=Bˆz on the hyperfine levels of the 2S1/2 ground state of hydrogen. Initially, we will neglect the effect of the nuclear (proton) magnetic moment. The energy eigenstates for the Hamiltonian describing the hyperfine interaction are also eigenstates of the operators {F2,Fz,I2,S2}. Therefor if we write out a matrix for the hyperfine Hamiltonian Hhf in the coupled basis |FmF⟩, it is diagonal. However, the Hamiltonian HB for the interaction of the magnetic moment of the electron with the external magnetic field,
HB=−μe⋅B=2μBBSz/ℏ,
is diagonal in the uncoupled basis |(SI)mS,mI⟩, made up of eigenstates of the operators {I2,Iz,S2,Sz}. We can write the matrix elements of the Hamiltonian in the coupled basis by relating the uncoupled to the coupled basis. (We could also carry out the analysis in the uncoupled basis, if we so chose.)
The relationship between the coupled |FmF⟩ and uncoupled |(SI)mSmI⟩ bases (see the discussion of the Clebsch-Gordan expansions in Chapter 3) is
|1,1⟩=|(1212)1212⟩,|1,0⟩=1√2(|(1212)12,−12⟩+|(1212),−1212⟩),|1,−1⟩=|(1212),−12,−12⟩,|0,0⟩=1√2(|(1212)12,−12⟩−|(1212),−1212⟩),
Employing the hyperfine energy shift formula (2.28) and Eq. (4.20), one finds for the matrix of the overall Hamiltonian Hhf+HB in the coupled basis
H=(A4+μBB0000A4−μBB0000A4μBB00μBB−3A4),
where we order the states (|1,1⟩,|1,−1⟩,|1,0⟩,|0,0⟩).
And for Eq. (2.28) the other part is
ΔEF=12AK+B32K(K+1)−2I(I+1)J(J+1)2I(2I−1)2J(2J−1),
where K=F(F+1)−I(I+1)−J(J+1). Here the constants A and B characterize the strengths of the magnetic-dipole and the electric-quadrupole interaction, respectively. B is zero unless I and J are both greater than 1/2.
Answer
Let me give it a shot:
If I interpret this correctly, F will be the operator for the full spin of the coupled system, S will be the operator of the electron spin (usually, one would consider J, the spin containing also spin-orbit coupling, but we are on the S-shell, hence no angular momentum) and I will be the nuclear spin. Then it should hold that F=S+I, right?
First, let's have a look at the hyperfine structure Hamiltonian Hhf. By construction of F, the eigenstates of Hhf will be eigenstates of F2 and Fz. This is just the same as for angular momentum and electron spin (and we construct F to have this property - this lets us label the eigenstates by the quantum number corresponding to F). Hence the Hamiltonian must be diagonal in the |F2,mF⟩-basis. One can also see that F2 commutes with I2 and S2 (and so does Fz), since F=I+S.
Now we have a look at HB, the interaction Hamiltonian with a constant magnetic field. We can see that (up to some prefactor) HB=Sz. Hence the eigenstates of HB must be eigenstates of Sz and thus also of S2 and, since the two operators are independent (they relate to two different types of spins, hence the operators should better commute) also to I2 and Iz, if you want.
The crucial problem is that Sz and F2 do not commute. Why? Well: F=I+S hence F2=S2+I2+2S⋅I. Now Sz and S do not commute, because Sz does not commute with e.g. Sx, which is part of S. Since F2 commutes with Hhf and Sz commutes with HB, but not with F2, we have that Hhf does not commute with HB. This means that HB and Hhf cannot be diagonal in the same basis, hence you need to have off-diagonal elements.
In order to see how the matrix representing HB looks like in the |F2,mF⟩-basis, you can express the |mI,mS⟩-basis (in which HB is diagonal) in terms of the other basis. This is exactly what equations (4.21) do. These are obtained by ordinary addition of angular momenta. From there, you can construct the unitary transforming the basis |mI,mS⟩ into |F2,mF⟩ and HB will be the diagonal matrix in the basis |mI,mS⟩ conjugated with this unitary.
EDIT: I'm not quite sure whether I understand correctly what your problem is, but let me elaborate: We want to find the Hamiltonian HB in the |mImS⟩ basis. In this basis, it is diagonal, because HB is essentially Sz (hence commutes with Sz) and it must also commute with Iz since Sz and Iz are independent.
If we order the basis according to |12,12⟩,|−12,−12⟩,|12,−12⟩,|−12,12⟩, then, we can just read off the Hamiltonian: The first and fourth vector are eigenvectors to eigenvalue μB, the others of −μB (by definition of Sz, since the second component in |mImS⟩=|(SI)Iz,Sz⟩ tells us the eigenvalue of Sz that the basis vector corresponds to), i.e. HB=(μB0000−μB0000−μB0000μB) Now, as I said, you just have to change the basis. The matrix transforming the above basis into the new basis is given by eqn. (4.21a-d): U:=(1000010000121200−1212) where the ordering of the |FmF⟩-basis is as for H in your text.
Now calculate UHBU† and that should give you the part of H coming from HB in the |F,mF⟩-basis and this will be exactly what is written in your book.
EDIT 2: I sort of suspected this, so here is some more linear algebra for the problem. I'll use Dirac notion since I suspect you are more familiar with this:
Now suppose you have given two bases |ei⟩ and |fi⟩ and suppose they are orthonormal bases. What we want is a matrix U that transforms one basis into the other (I'll call it U, since it'll be a unitary - if the bases are not orthonormal, it'll only be an invertible matrix). So we want a matrix such that |fi⟩:=U|ei⟩∀i How to construct this matrix? Well, given an equation for |fi⟩ in terms of the |ei⟩ will give you the i-th row of the matrix. You can also see the matrix elements in Dirac notation: ⟨ej|U|ei⟩=⟨ej|fi⟩
In your case, |ei⟩=|mImS⟩ and |fi⟩=|F2,mF⟩. Hence equation (4.21a) will give you the first row of the matrix (the ordering of the basis vectors |mImS⟩ as I proposed above), (4.21c) the second (notice the basis ordering in the matrix H!) (4.21b) the second and (4.21d) the last row of the matrix. Using the equation for the matrix elements above, you should be able to check that with not too much trouble. You can also easily check that U is indeed a unitary (i.e. UU†=U†U=1.
Then we can calculate the matrix elements: ⟨ei|H|ej⟩=⟨ei|U†UHU†U|ej⟩=⟨fi|UHU†|fj⟩, which tells you how the matrix looks like in the other basis.
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