Multiple sources, including the prodigal paper on the topic, "Theory of Hyperfine Structure" by Charles Schwartz, state that the hyperfine interaction can be written

$$ \hat{H}_{hfs} = \sum_k \mathbf{T}^{(k)}_n \cdot \mathbf{T}^{(k)}_e $$

where the first of these is a rank-$k$ tensor operator acting on nuclear degrees of freedom, and the second a same-rank operator acting on electronic degrees of freedom.

My feeling is that, if you're defining the hyperfine interaction to be all of the interactions between the nuclear and electronic degrees of freedom, this is the most general form of the interaction you can write and still get a scalar operator in the end. Is this correct, and is it the only justification for this form? If not, how can we derive this?

Further, we know that $k=1$ is the magnetic dipole term, $k=2$ the electric quadrupole term, and so on. But both the electric and magnetic multipole expansions in principle have all of these terms; how do we know that the electric dipole, magnetic quadrupole, etc. terms are zero? In a first-order approximation, can we treat this Hamiltonian as diagonal, take expectation values, and use parity arguments to say that this Hamiltonian is equivalent to one with only the nonzero terms? If so, what is the parity of the nuclear spin operator $\hat{I}$ and other relevant operators?


1 Answer 1


In the equation, $\hat{\boldsymbol T}^{(k)}$ is a $k$-order spherical tensor operator (with $2k+1$ components), subscript $e,n$ denotes electron and nucleu subsystem respectively. The dot $\cdot$ denotes the scalar product of the two tensor operators.

Answer to the 1st question

Is this correct, and is it the only justification for this form? If not, how can we derive this?

In each subsystem, a set of $k$-order tensor operators $\{\hat E{}^{(k)}_q\}$ can be choosen as bases superpositioned into components of $\hat{\boldsymbol{ T}}{}^{(k)}$, namely $\hat T{}^{(k)}_q=c^k_q \hat E{}^{(k)}_q$, where $c^k_q$ is $c$-number. Let $\{\hat E{}^{(k)}_q\},\{\hat N{}^{(k)}_q\}$ denote basis tensors in electron and nuclear systems, respectively, with coeffients $e_q^k,n_q^k$. Then the equation can be written as $$ \hat H_1=\sum_k \sum_{q=-k}^k(-)^q e^k_q n^k_{-q}\hat E{}^{(k)}_q \hat N{}^{(k)}_{-q}. $$ The set $\{\hat E^{(k)}_q\hat N{}^{(k)}_{-q}\}$ can be shown to be linearly independent scalar operators. It will be shown below that the set is all we need for $\hat H_1$, because the number of $\hat E{}^{(k)}_q\hat N{}^{(k)}_{-q}$ equals to that of the freedom of $\hat H_1$.

The dimension of electron and nuclear system is $(2j+1)$ and $(2i+1)$, respectively. For electron $k\leq 2j$ and for nuclear $k\leq 2i$. Therefore the summation over $k$ is from 0 to $2\min{(i,j)}$. And the number of $\hat E^{(k)}_q\hat N^{(k)}_{-q}$ is $2\min{(i,j)}+1$.

As a scalar operator, $\hat H_1$ is propotional to identity in each irreducible subspace of total angular momentum $\hat {\boldsymbol F}$. The eigenvalue of $\hat F_z$ ranges from $|i-j|$ to $i+j$, so there are $2\min{(i,j)}+1$ subspaces, or degrees of freedom for $\hat H_1$.

Remark: I did not test the above theory in cases where atoms are in superpostion sates of many $i$-s and $j$-s. These computation may be more complicated. Given that energy split caused by hyperfine structure is often very small, such simplification is resonable.

Answer to the 2nd question

how do we know that the electric dipole, magnetic quadrupole, etc. terms are zero?

Note.27 by Littlejohn provides two rules governing allowed multipole moments of nucleis, which are

  1. Electric multipoles of odd k and magnetic multipoles of even k are forbidden for the sake of parity and time-reversal violation.
  2. A $2^k$-pole can occur only if k ≤ 2i. This conclusion can be obsidianed from group theory.

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