So what is the one-body transition density?

NB: The text is incomplete and I have to go through it thoroughly!

I’ll dot down some thoughts about one-body transition operators and one-body transition densities here. Please do a fact-check on what I write here as these words are quick thoughts.

Preliminary

Before I start writing I want to define that a single-particle state $| \alpha \rangle$ is exactly what the wording says: It represents one particle (a proton or a neutron) in a specific quantum state labelled $\alpha$. In the $M$-scheme, this will be a specific $m$-substate of an orbital, for example the $j_z = -3/2$ in the $1d5/2$ orbital in the $sd$ major shell. A many-particle state $| \Psi \rangle$ however, is a state representing the ground state or some excited level of a nucleus. It is a linear combination of basis states

$$| \Psi \rangle = v_0 | b_0 \rangle + v_1 | b_1 \rangle + ... + v_{d_m - 1} | b_{d_m - 1} \rangle = \sum\limits_{i = 0}^{d_m - 1} v_i | b_i \rangle$$

where the basis states $| b_i \rangle$ are in a sense also many-particle states, but each basis state is one unique distribution of particles in the available $m$-substates of the system in question. One basis state might for example be: “One proton in $1d5/2$ with $j_z = -3/2$ and one proton in $1d5/2$ with $j_z = +3/2$”. In this example there are two valence particles, both protons, but more generally there are any number of protons and any number of neutrons. $d_m$ is called the $M$-scheme dimension and is equal to the number of basis states.

The one-body transition operator

Single-particle operators acting on a system of $N$ particles

Not to be confused with the one-body transition density (OBTD), the one-body transition operator (OBTO) is a generalisation of operators which act on single particles and extends a single-particle operator into a many-particle system. Assume a many-particle system of $N$-particles. A single-particle operator $\hat{o}_n$ which acts only on the $n$-th particle, generally is extended to act on all $N$ particles by

$$\hat{O}_1 = \sum_{n = 0}^{N-1} \hat{o}_n.$$

An example is the kinetic energy operator

$$\hat{E}_\text{kin} = \sum_{n = 0}^{N - 1} \frac{\hat{p}_n}{2m}$$

where $\hat{p}$ is the momentum operator. In this case, we simply sum the kinetic energy of each particle to get the total kinetic energy of the system. However, if we’re interested in properties or processes that involve changes to the state of the system, we have to remember the antisymmetry requirement to the wave function of a system of fermions. Recall that respecting antisymmetry is an intrinsic property of the second quantisation formalism, which comes from the anti-commutation relations of the annihilation and creation operators:

$$\{ \hat{c}_a, \hat{c}_b^\dagger \} = \hat{c}_a \hat{c}_b^\dagger + \hat{c}_b^\dagger \hat{c}_a = \delta_{ab},$$ $$\{ \hat{c}_a, \hat{c}_b \} = \hat{c}_a \hat{c}_b + \hat{c}_b \hat{c}_a = 0,$$ $$\{ \hat{c}_a^\dagger, \hat{c}_b^\dagger \} = \hat{c}_a^\dagger \hat{c}_b^\dagger + \hat{c}_b^\dagger \hat{c}_a^\dagger = 0,$$

where we see that changing the order of the operators generally brings along a sign. We will now use the convenient properties of second quantisation to generalise the representation of a single-particle operator $\hat{o}_n$ acting on a single particle, extended to acting on all $N$ particles. Keep in mind that the term single-particle operator refers to an operator which acts on a single particle at a time, but the operator can act on several particles consecutively, like $\hat{O}$. $\hat{o}$ however is defined to act only on one single particle.

Derivation of the OBTO

Assume that the operator $\hat{o}$ is diagonal in its orthonormal eigenbasis, the basis being

$$| b_i \rangle, i = 0, 1, ..., d_m - 1$$

where each basis state can be represented in second quantisation by letting $N$ creation operators act on the vacuum state, one example being

$$| b_i \rangle = \hat{c}_{\lambda_0}^\dagger \hat{c}_{\lambda_1}^\dagger ... \hat{c}_{\lambda_{N-1}}^\dagger | \text{vac} \rangle = | \lambda_0, \lambda_1, ..., \lambda_{N-1} \rangle.$$

When we say that an operator is diagonal in some basis, it means that, when represented in that particular basis, the operator takes the form of a diagonal matrix. When the operator is acting on one of the basis states, the basis state is only scaled by some factor, the factor being its eigenvalue. The diagonal matrix representation of the operator will thus have the eigenvalues of each basis vector in the diagonal. The operator $\hat{o}$ can then be represented as

$$\hat{o} = \sum_{i = 0}^{d_m - 1} o_i | b_i \rangle \langle b_i |,$$

where the eigenvalue $o_i$ is

$$o_i = \langle b_i | \hat{o} | b_i \rangle.$$

Note the similarity to the completeness relation

$$\sum_{i = 0}^{d_m - 1} | b_i \rangle \langle b_i | = \mathbb{1},$$

which in its matrix representation naturally is the identity matrix. Let’s now define a state… (the notation in the text I’m trying to follow is really confusing…)

OBTO

We now have the general definition of the one-body transition operator

$$\hat{O} = \sum_{\mu \nu} \langle \mu | \hat{o} | \nu \rangle \hat{c}_\mu^\dagger \hat{c}_\nu$$

which can be understood as the scattering of a particle from a state $\nu$ into a state $\mu$ with probability amplitude $\langle \mu | \hat{o} | \nu \rangle$. We can now substitute $\hat{o}$ for whatever one-body operator we wish to apply to a many-particle system, for example the $\hat{M}1$ operator.