The simplest possible choice for many-body wavefunctions are product wavefunctions. That is $$ \Psi(x_1, x_2, x_3, \ldots, x_A) \approx \phi_1(x_1) \phi_2(x_2) \phi_3(x_3) \ldots $$ because we are really only good at thinking about one particle at a time. Such product wavefunctions, without correlations, are easy to work with; for example, if the single-particle states \( \phi_i(x) \) are orthonormal, then the product wavefunctions are easy to orthonormalize.
Similarly, computing matrix elements of operators are relatively easy, because the integrals factorize.
The price we pay is the lack of correlations, which we must build up by using many, many product wavefunctions. (Thus we have a trade-off: compact representation of correlations but difficult integrals versus easy integrals but many states required.)
Because we have fermions, we are required to have antisymmetric wavefunctions, e.g. $$ \Psi(x_1, x_2, x_3, \ldots, x_A) = - \Psi(x_2, x_1, x_3, \ldots, x_A) $$ etc. This is accomplished formally by using the determinantal formalism $$ \Psi(x_1, x_2, \ldots, x_A) = \frac{1}{\sqrt{A!}} \det \left | \begin{array}{cccc} \phi_1(x_1) & \phi_1(x_2) & \ldots & \phi_1(x_A) \\ \phi_2(x_1) & \phi_2(x_2) & \ldots & \phi_2(x_A) \\ \vdots & & & \\ \phi_A(x_1) & \phi_A(x_2) & \ldots & \phi_A(x_A) \end{array} \right | $$ Product wavefunction + antisymmetry = Slater determinant.
$$ \Psi(x_1, x_2, \ldots, x_A) = \frac{1}{\sqrt{N!}} \det \left | \begin{array}{cccc} \phi_1(x_1) & \phi_1(x_2) & \ldots & \phi_1(x_A) \\ \phi_2(x_1) & \phi_2(x_2) & \ldots & \phi_2(x_A) \\ \vdots & & & \\ \phi_A(x_1) & \phi_A(x_2) & \ldots & \phi_A(x_A) \end{array} \right | $$ Properties of the determinant (interchange of any two rows or any two columns yields a change in sign; thus no two rows and no two columns can be the same) lead to the Pauli principle:
As a practical matter, however, Slater determinants beyond \( N=4 \) quickly become unwieldy. Thus we turn to the occupation representation or second quantization to simplify calculations.
The occupation representation, using fermion creation and annihilation operators, is compact and efficient. It is also abstract and, at first encounter, not easy to internalize. It is inspired by other operator formalism, such as the ladder operators for the harmonic oscillator or for angular momentum, but unlike those cases, the operators do not have coordinate space representations.
Instead, one can think of fermion creation/annihilation operators as a game of symbols that compactly reproduces what one would do, albeit clumsily, with full coordinate-space Slater determinants.
We start with a set of orthonormal single-particle states \( \{ \phi_i(x) \} \). (Note: this requirement, and others, can be relaxed, but leads to a more involved formalism.) Any orthonormal set will do.
To each single-particle state \( \phi_i(x) \) we associate a creation operator \( \hat{a}^\dagger_i \) and an annihilation operator \( \hat{a}_i \).
When acting on the vacuum state \( | 0 \rangle \), the creation operator \( \hat{a}^\dagger_i \) causes a particle to occupy the single-particle state \( \phi_i(x) \): $$ \phi_i(x) \rightarrow \hat{a}^\dagger_i |0 \rangle $$
But with multiple creation operators we can occupy multiple states: $$ \phi_i(x) \phi_j(x^\prime) \phi_k(x^{\prime \prime}) \rightarrow \hat{a}^\dagger_i \hat{a}^\dagger_j \hat{a}^\dagger_k |0 \rangle. $$
Now we impose antisymmetry, by having the fermion operators satisfy anticommutation relations: $$ \hat{a}^\dagger_i \hat{a}^\dagger_j + \hat{a}^\dagger_j \hat{a}^\dagger_i = [ \hat{a}^\dagger_i ,\hat{a}^\dagger_j ]_+ = \{ \hat{a}^\dagger_i ,\hat{a}^\dagger_j \} = 0 $$ so that $$ \hat{a}^\dagger_i \hat{a}^\dagger_j = - \hat{a}^\dagger_j \hat{a}^\dagger_i $$
Because of this property, automatically \( \hat{a}^\dagger_i \hat{a}^\dagger_i = 0 \), enforcing the Pauli exclusion principle. Thus when writing a Slater determinant using creation operators, $$ \hat{a}^\dagger_i \hat{a}^\dagger_j \hat{a}^\dagger_k \ldots |0 \rangle $$ each index \( i,j,k, \ldots \) must be unique.
We have defined the ansatz for the ground state as $$ |\Phi_0\rangle = \left(\prod_{i\le F}\hat{a}_{i}^{\dagger}\right)|0\rangle, $$ where the index \( i \) defines different single-particle states up to the Fermi level. We have assumed that we have \( N \) fermions. A given one-particle-one-hole (\( 1p1h \)) state can be written as $$ |\Phi_i^a\rangle = \hat{a}_{a}^{\dagger}\hat{a}_i|\Phi_0\rangle, $$ while a \( 2p2h \) state can be written as $$ |\Phi_{ij}^{ab}\rangle = \hat{a}_{a}^{\dagger}\hat{a}_{b}^{\dagger}\hat{a}_j\hat{a}_i|\Phi_0\rangle, $$ and a general \( NpNh \) state as $$ |\Phi_{ijk\dots}^{abc\dots}\rangle = \hat{a}_{a}^{\dagger}\hat{a}_{b}^{\dagger}\hat{a}_{c}^{\dagger}\dots\hat{a}_k\hat{a}_j\hat{a}_i|\Phi_0\rangle. $$
We can then expand our exact state function for the ground state as $$ |\Psi_0\rangle=C_0|\Phi_0\rangle+\sum_{ai}C_i^a|\Phi_i^a\rangle+\sum_{abij}C_{ij}^{ab}|\Phi_{ij}^{ab}\rangle+\dots =(C_0+\hat{C})|\Phi_0\rangle, $$ where we have introduced the so-called correlation operator $$ \hat{C}=\sum_{ai}C_i^a\hat{a}_{a}^{\dagger}\hat{a}_i +\sum_{abij}C_{ij}^{ab}\hat{a}_{a}^{\dagger}\hat{a}_{b}^{\dagger}\hat{a}_j\hat{a}_i+\dots $$ Since the normalization of \( \Psi_0 \) is at our disposal and since \( C_0 \) is by hypothesis non-zero, we may arbitrarily set \( C_0=1 \) with corresponding proportional changes in all other coefficients. Using this so-called intermediate normalization we have $$ \langle \Psi_0 | \Phi_0 \rangle = \langle \Phi_0 | \Phi_0 \rangle = 1, $$ resulting in $$ |\Psi_0\rangle=(1+\hat{C})|\Phi_0\rangle. $$
We rewrite $$ |\Psi_0\rangle=C_0|\Phi_0\rangle+\sum_{ai}C_i^a|\Phi_i^a\rangle+\sum_{abij}C_{ij}^{ab}|\Phi_{ij}^{ab}\rangle+\dots, $$ in a more compact form as $$ |\Psi_0\rangle=\sum_{PH}C_H^P\Phi_H^P=\left(\sum_{PH}C_H^P\hat{A}_H^P\right)|\Phi_0\rangle, $$ where \( H \) stands for \( 0,1,\dots,n \) hole states and \( P \) for \( 0,1,\dots,n \) particle states. Our requirement of unit normalization gives $$ \langle \Psi_0 | \Phi_0 \rangle = \sum_{PH}|C_H^P|^2= 1, $$ and the energy can be written as $$ E= \langle \Psi_0 | \hat{H} |\Phi_0 \rangle= \sum_{PP'HH'}C_H^{*P}\langle \Phi_H^P | \hat{H} |\Phi_{H'}^{P'} \rangle C_{H'}^{P'}. $$
Normally $$ E= \langle \Psi_0 | \hat{H} |\Phi_0 \rangle= \sum_{PP'HH'}C_H^{*P}\langle \Phi_H^P | \hat{H} |\Phi_{H'}^{P'} \rangle C_{H'}^{P'}, $$ is solved by diagonalization setting up the Hamiltonian matrix defined by the basis of all possible Slater determinants. A diagonalization is equivalent to finding the variational minimum of $$ \langle \Psi_0 | \hat{H} |\Phi_0 \rangle-\lambda \langle \Psi_0 |\Phi_0 \rangle, $$ where \( \lambda \) is a variational multiplier to be identified with the energy of the system. The minimization process results in $$ \delta\left[ \langle \Psi_0 | \hat{H} |\Phi_0 \rangle-\lambda \langle \Psi_0 |\Phi_0 \rangle\right]= $$ $$ \sum_{P'H'}\left\{\delta[C_H^{*P}]\langle \Phi_H^P | \hat{H} |\Phi_{H'}^{P'} \rangle C_{H'}^{P'}+ C_H^{*P}\langle \Phi_H^P | \hat{H} |\Phi_{H'}^{P'} \rangle \delta[C_{H'}^{P'}]- \lambda( \delta[C_H^{*P}]C_{H'}^{P'}+C_H^{*P}\delta[C_{H'}^{P'}]\right\} = 0. $$ Since the coefficients \( \delta[C_H^{*P}] \) and \( \delta[C_{H'}^{P'}] \) are complex conjugates it is necessary and sufficient to require the quantities that multiply with \( \delta[C_H^{*P}] \) to vanish.
This leads to $$ \sum_{P'H'}\langle \Phi_H^P | \hat{H} |\Phi_{H'}^{P'} \rangle C_{H'}^{P'}-\lambda C_H^{P}=0, $$ for all sets of \( P \) and \( H \).
If we then multiply by the corresponding \( C_H^{*P} \) and sum over \( PH \) we obtain $$ \sum_{PP'HH'}C_H^{*P}\langle \Phi_H^P | \hat{H} |\Phi_{H'}^{P'} \rangle C_{H'}^{P'}-\lambda\sum_{PH}|C_H^P|^2=0, $$ leading to the identification \( \lambda = E \). This means that we have for all \( PH \) sets $$ \begin{equation} \sum_{P'H'}\langle \Phi_H^P | \hat{H} -E|\Phi_{H'}^{P'} \rangle = 0. \label{eq:fullci} \end{equation} $$
An alternative way to derive the last equation is to start from $$ (\hat{H} -E)|\Psi_0\rangle = (\hat{H} -E)\sum_{P'H'}C_{H'}^{P'}|\Phi_{H'}^{P'} \rangle=0, $$ and if this equation is successively projected against all \( \Phi_H^P \) in the expansion of \( \Psi \), then the last equation on the previous slide results. As stated previously, one solves this equation normally by diagonalization. If we are able to solve this equation exactly (that is numerically exactly) in a large Hilbert space (it will be truncated in terms of the number of single-particle states included in the definition of Slater determinants), it can then serve as a benchmark for other many-body methods which approximate the correlation operator \( \hat{C} \).
Suppose, as an example, that we have six fermions below the Fermi level. This means that we can make at most \( 6p-6h \) excitations. If we have an infinity of single particle states above the Fermi level, we will obviously have an infinity of say \( 2p-2h \) excitations. Each such way to configure the particles is called a configuration. We will always have to truncate in the basis of single-particle states. This gives us a finite number of possible Slater determinants. Our Hamiltonian matrix would then look like (where each block can have a large dimensionalities):
| \( 0p-0h \) | \( 1p-1h \) | \( 2p-2h \) | \( 3p-3h \) | \( 4p-4h \) | \( 5p-5h \) | \( 6p-6h \) | |
| \( 0p-0h \) | x | x | x | 0 | 0 | 0 | 0 |
| \( 1p-1h \) | x | x | x | x | 0 | 0 | 0 |
| \( 2p-2h \) | x | x | x | x | x | 0 | 0 |
| \( 3p-3h \) | 0 | x | x | x | x | x | 0 |
| \( 4p-4h \) | 0 | 0 | x | x | x | x | x |
| \( 5p-5h \) | 0 | 0 | 0 | x | x | x | x |
| \( 6p-6h \) | 0 | 0 | 0 | 0 | x | x | x |
with a two-body force. Why are there non-zero blocks of elements?
If we use a Hartree-Fock basis, this corresponds to a particular unitary transformation where matrix elements of the type \( \langle 0p-0h \vert \hat{H} \vert 1p-1h\rangle =\langle \Phi_0 | \hat{H}|\Phi_{i}^{a}\rangle=0 \) and our Hamiltonian matrix becomes
| \( 0p-0h \) | \( 1p-1h \) | \( 2p-2h \) | \( 3p-3h \) | \( 4p-4h \) | \( 5p-5h \) | \( 6p-6h \) | |
| \( 0p-0h \) | \( \tilde{x} \) | 0 | \( \tilde{x} \) | 0 | 0 | 0 | 0 |
| \( 1p-1h \) | 0 | \( \tilde{x} \) | \( \tilde{x} \) | \( \tilde{x} \) | 0 | 0 | 0 |
| \( 2p-2h \) | \( \tilde{x} \) | \( \tilde{x} \) | \( \tilde{x} \) | \( \tilde{x} \) | \( \tilde{x} \) | 0 | 0 |
| \( 3p-3h \) | 0 | \( \tilde{x} \) | \( \tilde{x} \) | \( \tilde{x} \) | \( \tilde{x} \) | \( \tilde{x} \) | 0 |
| \( 4p-4h \) | 0 | 0 | \( \tilde{x} \) | \( \tilde{x} \) | \( \tilde{x} \) | \( \tilde{x} \) | \( \tilde{x} \) |
| \( 5p-5h \) | 0 | 0 | 0 | \( \tilde{x} \) | \( \tilde{x} \) | \( \tilde{x} \) | \( \tilde{x} \) |
| \( 6p-6h \) | 0 | 0 | 0 | 0 | \( \tilde{x} \) | \( \tilde{x} \) | \( \tilde{x} \) |
If we do not make any truncations in the possible sets of Slater determinants (many-body states) we can make by distributing \( A \) nucleons among \( n \) single-particle states, we call such a calculation for Full configuration interaction theory
If we make truncations, we have different possibilities
Full configuration interaction theory calculations provide in principle, if we can diagonalize numerically, all states of interest. The dimensionality of the problem explodes however quickly.
The total number of Slater determinants which can be built with say \( N \) neutrons distributed among \( n \) single particle states is $$ \left (\begin{array}{c} n \\ N\end{array} \right) =\frac{n!}{(n-N)!N!}. $$
For a model space which comprises the first for major shells only \( 0s \), \( 0p \), \( 1s0d \) and \( 1p0f \) we have \( 40 \) single particle states for neutrons and protons. For the eight neutrons of oxygen-16 we would then have $$ \left (\begin{array}{c} 40 \\ 8\end{array} \right) =\frac{40!}{(32)!8!}\sim 10^{9}, $$ and multiplying this with the number of proton Slater determinants we end up with approximately witha dimensionality \( d \) of \( d\sim 10^{18} \).
This number can be reduced if we look at specific symmetries only. However, the dimensionality explodes quickly!
To see this, we look at the contributions arising from $$ \langle \Phi_H^P | = \langle \Phi_0| $$ in Eq. \eqref{eq:fullci}, that is we multiply with \( \langle \Phi_0 | \) from the left in $$ (\hat{H} -E)\sum_{P'H'}C_{H'}^{P'}|\Phi_{H'}^{P'} \rangle=0. $$ If we assume that we have a two-body operator at most, Slater's rule gives then an equation for the correlation energy in terms of \( C_i^a \) and \( C_{ij}^{ab} \) only. We get then $$ \langle \Phi_0 | \hat{H} -E| \Phi_0\rangle + \sum_{ai}\langle \Phi_0 | \hat{H} -E|\Phi_{i}^{a} \rangle C_{i}^{a}+ \sum_{abij}\langle \Phi_0 | \hat{H} -E|\Phi_{ij}^{ab} \rangle C_{ij}^{ab}=0, $$ or $$ E-E_0 =\Delta E=\sum_{ai}\langle \Phi_0 | \hat{H}|\Phi_{i}^{a} \rangle C_{i}^{a}+ \sum_{abij}\langle \Phi_0 | \hat{H}|\Phi_{ij}^{ab} \rangle C_{ij}^{ab}, $$ where the energy \( E_0 \) is the reference energy and \( \Delta E \) defines the so-called correlation energy. The single-particle basis functions could be the results of a Hartree-Fock calculation or just the eigenstates of the non-interacting part of the Hamiltonian.
To see this, we look at the contributions arising from $$ \langle \Phi_H^P | = \langle \Phi_0| $$ in Eq. \eqref{eq:fullci}, that is we multiply with \( \langle \Phi_0 | \) from the left in $$ (\hat{H} -E)\sum_{P'H'}C_{H'}^{P'}|\Phi_{H'}^{P'} \rangle=0. $$
If we assume that we have a two-body operator at most, Slater's rule gives then an equation for the correlation energy in terms of \( C_i^a \) and \( C_{ij}^{ab} \) only. We get then $$ \langle \Phi_0 | \hat{H} -E| \Phi_0\rangle + \sum_{ai}\langle \Phi_0 | \hat{H} -E|\Phi_{i}^{a} \rangle C_{i}^{a}+ \sum_{abij}\langle \Phi_0 | \hat{H} -E|\Phi_{ij}^{ab} \rangle C_{ij}^{ab}=0, $$ or $$ E-E_0 =\Delta E=\sum_{ai}\langle \Phi_0 | \hat{H}|\Phi_{i}^{a} \rangle C_{i}^{a}+ \sum_{abij}\langle \Phi_0 | \hat{H}|\Phi_{ij}^{ab} \rangle C_{ij}^{ab}, $$ where the energy \( E_0 \) is the reference energy and \( \Delta E \) defines the so-called correlation energy. The single-particle basis functions could be the results of a Hartree-Fock calculation or just the eigenstates of the non-interacting part of the Hamiltonian.
In our notes on Hartree-Fock calculations, we have already computed the matrix \( \langle \Phi_0 | \hat{H}|\Phi_{i}^{a}\rangle \) and \( \langle \Phi_0 | \hat{H}|\Phi_{ij}^{ab}\rangle \). If we are using a Hartree-Fock basis, then the matrix elements \( \langle \Phi_0 | \hat{H}|\Phi_{i}^{a}\rangle=0 \) and we are left with a correlation energy given by $$ E-E_0 =\Delta E^{HF}=\sum_{abij}\langle \Phi_0 | \hat{H}|\Phi_{ij}^{ab} \rangle C_{ij}^{ab}. $$
Inserting the various matrix elements we can rewrite the previous equation as $$ \Delta E=\sum_{ai}\langle i| \hat{f}|a \rangle C_{i}^{a}+ \sum_{abij}\langle ij | \hat{v}| ab \rangle C_{ij}^{ab}. $$ This equation determines the correlation energy but not the coefficients \( C \).
We need more equations. Our next step is to set up $$ \langle \Phi_i^a | \hat{H} -E| \Phi_0\rangle + \sum_{bj}\langle \Phi_i^a | \hat{H} -E|\Phi_{j}^{b} \rangle C_{j}^{b}+ \sum_{bcjk}\langle \Phi_i^a | \hat{H} -E|\Phi_{jk}^{bc} \rangle C_{jk}^{bc}+ \sum_{bcdjkl}\langle \Phi_i^a | \hat{H} -E|\Phi_{jkl}^{bcd} \rangle C_{jkl}^{bcd}=0, $$ as this equation will allow us to find an expression for the coefficents \( C_i^a \) since we can rewrite this equation as $$ \langle i | \hat{f}| a\rangle +\langle \Phi_i^a | \hat{H}|\Phi_{i}^{a} \rangle C_{i}^{a}+ \sum_{bj\ne ai}\langle \Phi_i^a | \hat{H}|\Phi_{j}^{b} \rangle C_{j}^{b}+ \sum_{bcjk}\langle \Phi_i^a | \hat{H}|\Phi_{jk}^{bc} \rangle C_{jk}^{bc}+ \sum_{bcdjkl}\langle \Phi_i^a | \hat{H}|\Phi_{jkl}^{bcd} \rangle C_{jkl}^{bcd}=EC_i^a. $$
We see that on the right-hand side we have the energy \( E \). This leads to a non-linear equation in the unknown coefficients. These equations are normally solved iteratively ( that is we can start with a guess for the coefficients \( C_i^a \)). A common choice is to use perturbation theory for the first guess, setting thereby $$ C_{i}^{a}=\frac{\langle i | \hat{f}| a\rangle}{\epsilon_i-\epsilon_a}. $$
The observant reader will however see that we need an equation for \( C_{jk}^{bc} \) and \( C_{jkl}^{bcd} \) as well. To find equations for these coefficients we need then to continue our multiplications from the left with the various \( \Phi_{H}^P \) terms.
For \( C_{jk}^{bc} \) we need then $$ \langle \Phi_{ij}^{ab} | \hat{H} -E| \Phi_0\rangle + \sum_{kc}\langle \Phi_{ij}^{ab} | \hat{H} -E|\Phi_{k}^{c} \rangle C_{k}^{c}+ $$ $$ \sum_{cdkl}\langle \Phi_{ij}^{ab} | \hat{H} -E|\Phi_{kl}^{cd} \rangle C_{kl}^{cd}+\sum_{cdeklm}\langle \Phi_{ij}^{ab} | \hat{H} -E|\Phi_{klm}^{cde} \rangle C_{klm}^{cde}+\sum_{cdefklmn}\langle \Phi_{ij}^{ab} | \hat{H} -E|\Phi_{klmn}^{cdef} \rangle C_{klmn}^{cdef}=0, $$ and we can isolate the coefficients \( C_{kl}^{cd} \) in a similar way as we did for the coefficients \( C_{i}^{a} \).
A standard choice for the first iteration is to set $$ C_{ij}^{ab} =\frac{\langle ij \vert \hat{v} \vert ab \rangle}{\epsilon_i+\epsilon_j-\epsilon_a-\epsilon_b}. $$ At the end we can rewrite our solution of the Schroedinger equation in terms of \( n \) coupled equations for the coefficients \( C_H^P \). This is a very cumbersome way of solving the equation. However, by using this iterative scheme we can illustrate how we can compute the various terms in the wave operator or correlation operator \( \hat{C} \). We will later identify the calculation of the various terms \( C_H^P \) as parts of different many-body approximations to full CI. In particular, we can relate this non-linear scheme with Coupled Cluster theory and many-body perturbation theory.
If we can diagonalize large matrices, FCI is the method of choice since:
The correlation energy is defined as, with a two-body Hamiltonian, $$ \Delta E=\sum_{ai}\langle i| \hat{f}|a \rangle C_{i}^{a}+ \sum_{abij}\langle ij | \hat{v}| ab \rangle C_{ij}^{ab}. $$ The coefficients \( C \) result from the solution of the eigenvalue problem. The energy of say the ground state is then $$ E=E_{ref}+\Delta E, $$ where the so-called reference energy is the energy we obtain from a Hartree-Fock calculation, that is $$ E_{ref}=\langle \Phi_0 \vert \hat{H} \vert \Phi_0 \rangle. $$
However, as we have seen, even for a small case like the four first major shells and a nucleus like oxygen-16, the dimensionality becomes quickly intractable. If we wish to include single-particle states that reflect weakly bound systems, we need a much larger single-particle basis. We need thus approximative methods that sum specific correlations to infinite order.
Popular methods are
Here we will discuss how we can set up a single-particle basis which we can use in the various parts of our projects, from the simple pairing model to infinite nuclear matter. We will use here the simple pairing model to illustrate in particular how to set up a single-particle basis. We will also use this do discuss standard FCI approaches like:
An important step in an FCI code is to construct the many-body basis.
While the formalism is independent of the choice of basis, the effectiveness of a calculation will certainly be basis dependent.
Furthermore there are common conventions useful to know.
First, the single-particle basis has angular momentum as a good quantum number. You can imagine the single-particle wavefunctions being generated by a one-body Hamiltonian, for example a harmonic oscillator. Modifications include harmonic oscillator plus spin-orbit splitting, or self-consistent mean-field potentials, or the Woods-Saxon potential which mocks up the self-consistent mean-field. For nuclei, the harmonic oscillator, modified by spin-orbit splitting, provides a useful language for describing single-particle states.
Each single-particle state is labeled by the following quantum numbers:
In practice the single-particle space has to be severely truncated. This truncation is typically based upon the single-particle energies, which is the effective energy from a mean-field potential.
Sometimes we freeze the core and only consider a valence space. For example, one may assume a frozen \( {}^{4}\mbox{He} \) core, with two protons and two neutrons in the \( 0s_{1/2} \) shell, and then only allow active particles in the \( 0p_{1/2} \) and \( 0p_{3/2} \) orbits.
Another example is a frozen \( {}^{16}\mbox{O} \) core, with eight protons and eight neutrons filling the \( 0s_{1/2} \), \( 0p_{1/2} \) and \( 0p_{3/2} \) orbits, with valence particles in the \( 0d_{5/2}, 1s_{1/2} \) and \( 0d_{3/2} \) orbits.
Sometimes we refer to nuclei by the valence space where their last nucleons go. So, for example, we call \( {}^{12}\mbox{C} \) a \( p \)-shell nucleus, while \( {}^{26}\mbox{Al} \) is an \( sd \)-shell nucleus and \( {}^{56}\mbox{Fe} \) is a \( pf \)-shell nucleus.
There are different kinds of truncations.
Here one works in a harmonic oscillator basis, with each major oscillator shell assigned a principal quantum number \( N=0,1,2,3,... \). The \( N\hbar\Omega \) or \( N_{max} \) truncation: Any configuration is given an noninteracting energy, which is the sum of the single-particle harmonic oscillator energies. (Thus this ignores spin-orbit splitting.)
Excited state are labeled relative to the lowest configuration by the number of harmonic oscillator quanta.
This truncation is useful because if one includes all configuration up to some \( N_{max} \), and has a translationally invariant interaction, then the intrinsic motion and the center-of-mass motion factor. In other words, we can know exactly the center-of-mass wavefunction.
In almost all cases, the many-body Hamiltonian is rotationally invariant. This means it commutes with the operators \( \hat{J}^2, \hat{J}_z \) and so eigenstates will have good \( J,M \). Furthermore, the eigenenergies do not depend upon the orientation \( M \).
Therefore we can choose to construct a many-body basis which has fixed \( M \); this is called an \( M \)-scheme basis.
Alternately, one can construct a many-body basis which has fixed \( J \), or a \( J \)-scheme basis.
The Hamiltonian matrix will have smaller dimensions (a factor of 10 or more) in the \( J \)-scheme than in the \( M \)-scheme. On the other hand, as we'll show in the next slide, the \( M \)-scheme is very easy to construct with Slater determinants, while the \( J \)-scheme basis states, and thus the matrix elements, are more complicated, almost always being linear combinations of \( M \)-scheme states. \( J \)-scheme bases are important and useful, but we'll focus on the simpler \( M \)-scheme.
The quantum number \( m \) is additive (because the underlying group is Abelian): if a Slater determinant \( \hat{a}_i^\dagger \hat{a}^\dagger_j \hat{a}^\dagger_k \ldots | 0 \rangle \) is built from single-particle states all with good \( m \), then the total $$ M = m_i + m_j + m_k + \ldots $$ This is not true of \( J \), because the angular momentum group SU(2) is not Abelian.
The upshot is that
Example: two \( j=1/2 \) orbits
| Index | \( n \) | \( l \) | \( j \) | \( m_j \) |
|---|---|---|---|---|
| 1 | 0 | 0 | 1/2 | -1/2 |
| 2 | 0 | 0 | 1/2 | 1/2 |
| 3 | 1 | 0 | 1/2 | -1/2 |
| 4 | 1 | 0 | 1/2 | 1/2 |
Note that the order is arbitrary.
There are \( \left ( \begin{array}{c} 4 \\ 2 \end{array} \right) = 6 \) two-particle states, which we list with the total \( M \):
| Occupied | \( M \) |
|---|---|
| 1,2 | 0 |
| 1,3 | -1 |
| 1,4 | 0 |
| 2,3 | 0 |
| 2,4 | 1 |
| 3,4 | 0 |
As another example, consider using only single particle states from the \( 0d_{5/2} \) space. They have the following quantum numbers
| Index | \( n \) | \( l \) | \( j \) | \( m_j \) |
|---|---|---|---|---|
| 1 | 0 | 2 | 5/2 | -5/2 |
| 2 | 0 | 2 | 5/2 | -3/2 |
| 3 | 0 | 2 | 5/2 | -1/2 |
| 4 | 0 | 2 | 5/2 | 1/2 |
| 5 | 0 | 2 | 5/2 | 3/2 |
| 6 | 0 | 2 | 5/2 | 5/2 |
There are \( \left ( \begin{array}{c} 6 \\ 2 \end{array} \right) = 15 \) two-particle states, which we list with the total \( M \):
| Occupied | \( M \) | Occupied | \( M \) | Occupied | \( M \) |
|---|---|---|---|---|---|
| 1,2 | -4 | 2,3 | -2 | 3,5 | 1 |
| 1,3 | -3 | 2,4 | -1 | 3,6 | 2 |
| 1,4 | -2 | 2,5 | 0 | 4,5 | 2 |
| 1,5 | -1 | 2,6 | 1 | 4,6 | 3 |
| 1,6 | 0 | 3,4 | 0 | 5,6 | 4 |
There are 3 states with \( M= 0 \), 2 with \( M = 1 \), and so on.
The basic goal of this project is for you to build your own configuration-interaction shell-model code. The code will be fairly basic; it will assume that we have a single species of particles, e.g. only neutrons, and you could, if you wish to, read in uncoupled two-body matrix elements. Furthermore the pieces of the code will not be the most efficient. Nonetheless it will be usable; most importantly, you will gain a good idea of what goes into a many-body shell-model code.
The first step is to construct the \( M \)-scheme basis of Slater determinants. Here \( M \)-scheme means the total \( J_z \) of the many-body states is fixed.
The steps could be:
The format of a possible input file could be
| Index | \( n \) | \( l \) | \( 2j \) | \( 2m_j \) |
|---|---|---|---|---|
| 1 | 1 | 0 | 1 | -1 |
| 2 | 1 | 0 | 1 | 1 |
| 3 | 0 | 2 | 3 | -3 |
| 4 | 0 | 2 | 3 | -1 |
| 5 | 0 | 2 | 3 | 1 |
| 6 | 0 | 2 | 3 | 3 |
| 7 | 0 | 2 | 5 | -5 |
| 8 | 0 | 2 | 5 | -3 |
| 9 | 0 | 2 | 5 | -1 |
| 10 | 0 | 2 | 5 | 1 |
| 11 | 0 | 2 | 5 | 3 |
| 12 | 0 | 2 | 5 | 5 |
This represents the \( 1s_{1/2}0d_{3/2}0d_{5/2} \) valence space, or just the \( sd \)-space. There are twelve single-particle states, labeled by an overall index, and which have associated quantum numbers the number of radial nodes, the orbital angular momentum \( l \), and the angular momentum \( j \) and third component \( j_z \). To keep everything as integers, we could store \( 2 \times j \) and \( 2 \times j_z \).
To read in the single-particle states you need to:
The next step is to read in the number of particles \( N \) and the fixed total \( M \) (or, actually, \( 2 \times M \)). For this project we assume only a single species of particles, say neutrons, although this can be relaxed. Note: Although it is often a good idea to try to write a more general code, given the short time alloted we would suggest you keep your ambition in check, at least in the initial phases of the project.
You should probably write an error trap to make sure \( N \) and \( M \) are congruent; if \( N \) is even, then \( 2 \times M \) should be even, and if \( N \) is odd then \( 2\times M \) should be odd.
The final step is to generate the set of \( N \)-particle Slater determinants with fixed \( M \). The Slater determinants will be stored in occupation representation. Although in many codes this representation is done compactly in bit notation with ones and zeros, but for greater transparency and simplicity we will list the occupied single particle states.
Hence we can store the Slater determinant basis states as \( sd(i,j) \), that is an array of dimension \( N_{SD} \), the number of Slater determinants, by \( N \), the number of occupied state. So if for the 7th Slater determinant the 2nd, 3rd, and 9th single-particle states are occupied, then \( sd(7,1) = 2 \), \( sd(7,2) = 3 \), and \( sd(7,3) = 9 \).
We can construct an occupation representation of Slater determinants by the odometer method. Consider \( N_{sp} = 12 \) and \( N=4 \). Start with the first 4 states occupied, that is:
When we restrict ourselves to an \( M \)-scheme basis, we could choose two paths. The first is simplest (and simplest is often best, at least in the first draft of a code): generate all possible Slater determinants, and then extract from this initial list a list of those Slater determinants with a given \( M \). (You will need to write a short function or routine that computes \( M \) for any given occupation.)
Alternately, and not too difficult, is to run the odometer routine twice: each time, as as a Slater determinant is calculated, compute \( M \), but do not store the Slater determinants except the current one. You can then count up the number of Slater determinants with a chosen \( M \). Then allocated storage for the Slater determinants, and run the odometer algorithm again, this time storing Slater determinants with the desired \( M \) (this can be done with a simple logical flag).
Some example solutions: Let's begin with a simple case, the \( 0d_{5/2} \) space containing six single-particle states
| Index | \( n \) | \( l \) | \( j \) | \( m_j \) |
|---|---|---|---|---|
| 1 | 0 | 2 | 5/2 | -5/2 |
| 2 | 0 | 2 | 5/2 | -3/2 |
| 3 | 0 | 2 | 5/2 | -1/2 |
| 4 | 0 | 2 | 5/2 | 1/2 |
| 5 | 0 | 2 | 5/2 | 3/2 |
| 6 | 0 | 2 | 5/2 | 5/2 |
For two particles, there are a total of 15 states, which we list here with the total \( M \):
You should try by hand to show that in this same single-particle space, that for \( N=3 \) there are 3 states with \( M=1/2 \) and for \( N= 4 \) there are also only 3 states with \( M=0 \).
To test your code, confirm the above.
Also, for the \( sd \)-space given above, for \( N=2 \) there are 14 states with \( M=0 \), for \( N=3 \) there are 37 states with \( M=1/2 \), for \( N=4 \) there are 81 states with \( M=0 \).
For our project, we will only consider the pairing model. A simple space is the \( (1/2)^2 \) space with four single-particle states
| Index | \( n \) | \( l \) | \( s \) | \( m_s \) |
|---|---|---|---|---|
| 1 | 0 | 0 | 1/2 | -1/2 |
| 2 | 0 | 0 | 1/2 | 1/2 |
| 3 | 1 | 0 | 1/2 | -1/2 |
| 4 | 1 | 0 | 1/2 | 1/2 |
For \( N=2 \) there are 4 states with \( M=0 \); show this by hand and confirm your code reproduces it.
Another, slightly more challenging space is the \( (1/2)^4 \) space, that is, with eight single-particle states we have
| Index | \( n \) | \( l \) | \( s \) | \( m_s \) |
|---|---|---|---|---|
| 1 | 0 | 0 | 1/2 | -1/2 |
| 2 | 0 | 0 | 1/2 | 1/2 |
| 3 | 1 | 0 | 1/2 | -1/2 |
| 4 | 1 | 0 | 1/2 | 1/2 |
| 5 | 2 | 0 | 1/2 | -1/2 |
| 6 | 2 | 0 | 1/2 | 1/2 |
| 7 | 3 | 0 | 1/2 | -1/2 |
| 8 | 3 | 0 | 1/2 | 1/2 |
For \( N=2 \) there are 16 states with \( M=0 \); for \( N=3 \) there are 24 states with \( M=1/2 \), and for \( N=4 \) there are 36 states with \( M=0 \).
In the shell-model context we can interpret this as 4 \( s_{1/2} \) levels, with \( m = \pm 1/2 \), we can also think of these are simple four pairs, \( \pm k, k = 1,2,3,4 \). Later on we will assign single-particle energies, depending on the radial quantum number \( n \), that is, \( \epsilon_k = |k| \delta \) so that they are equally spaced.
For application in the pairing model we can go further and consider only states with no "broken pairs," that is, if \( +k \) is filled (or \( m = +1/2 \), so is \( -k \) (\( m=-1/2 \)). If you want, you can write your code to accept only these, and obtain the following six states:
Hints for coding
Our recommended occupation representation, e.g. \( | 1,2,4,8 \rangle \), is easy to code, but numerically inefficient when one has hundreds of millions of Slater determinants.
In state-of-the-art shell-model codes, one generally uses bit representation, i.e. \( |1101000100... \rangle \) where one stores the Slater determinant as a single (or a small number of) integer.
This is much more compact, but more intricate to code with considerable more overhead. There exist bit-manipulation functions. This is left as a challenge for those of you who would like to study this topic further for the final project to be presented for the oral examination.
We consider a space with \( 2\Omega \) single-particle states, with each state labeled by \( k = 1, 2, 3, \Omega \) and \( m = \pm 1/2 \). The convention is that the state with \( k>0 \) has \( m = + 1/2 \) while \( -k \) has \( m = -1/2 \).
The Hamiltonian we consider is $$ \hat{H} = -G \hat{P}_+ \hat{P}_-, $$ where $$ \hat{P}_+ = \sum_{k > 0} \hat{a}^\dagger_k \hat{a}^\dagger_{-{k}}. $$ and \( \hat{P}_- = ( \hat{P}_+)^\dagger \).
This problem can be solved using what is called the quasi-spin formalism to obtain the exact results. Thereafter we will try again using the explicit Slater determinant formalism.
One can show (and this is part of the project) that $$ \left [ \hat{P}_+, \hat{P}_- \right ] = \sum_{k> 0} \left( \hat{a}^\dagger_k \hat{a}_k + \hat{a}^\dagger_{-{k}} \hat{a}_{-{k}} - 1 \right) = \hat{N} - \Omega. $$ Now define $$ \hat{P}_z = \frac{1}{2} ( \hat{N} -\Omega). $$ Finally you can show $$ \left [ \hat{P}_z , \hat{P}_\pm \right ] = \pm \hat{P}_\pm. $$ This means the operators \( \hat{P}_\pm, \hat{P}_z \) form a so-called \( SU(2) \) algebra, and we can use all our insights about angular momentum, even though there is no actual angular momentum involved (this is similar to project 1).
So we rewrite the Hamiltonian to make this explicit: $$ \hat{H} = -G \hat{P}_+ \hat{P}_- = -G \left( \hat{P}^2 - \hat{P}_z^2 + \hat{P}_z\right) $$
Because of the SU(2) algebra, we know that the eigenvalues of \( \hat{P}^2 \) must be of the form \( p(p+1) \), with \( p \) either integer or half-integer, and the eigenvalues of \( \hat{P}_z \) are \( m_p \) with \( p \geq | m_p| \), with \( m_p \) also integer or half-integer.
But because \( \hat{P}_z = (1/2)(\hat{N}-\Omega) \), we know that for \( N \) particles the value \( m_p = (N-\Omega)/2 \). Furthermore, the values of \( m_p \) range from \( -\Omega/2 \) (for \( N=0 \)) to \( +\Omega/2 \) (for \( N=2\Omega \), with all states filled).
We deduce the maximal \( p = \Omega/2 \) and for a given \( n \) the values range of \( p \) range from \( |N-\Omega|/2 \) to \( \Omega/2 \) in steps of 1 (for an even number of particles)
Following Racah we introduce the notation \( p = (\Omega - v)/2 \) where \( v = 0, 2, 4,..., \Omega - |N-\Omega| \) With this it is easy to deduce that the eigenvalues of the pairing Hamiltonian are $$ -G(N-v)(2\Omega +2-N-v)/4 $$ This also works for \( N \) odd, with \( v= 1,3,5, \dots \).
Let's take a specific example: \( \Omega = 3 \) so there are 6 single-particle states, and \( N = 3 \), with \( v= 1,3 \). Therefore there are two distinct eigenvalues, $$ E = -2G, 0 $$ Now let's work this out explicitly. The single particle degrees of freedom are defined as
| Index | \( k \) | \( m \) |
|---|---|---|
| 1 | 1 | -1/2 |
| 2 | -1 | 1/2 |
| 3 | 2 | -1/2 |
| 4 | -2 | 1/2 |
| 5 | 3 | -1/2 |
| 6 | -3 | 1/2 |
There are \( \left( \begin{array}{c}6 \\ 3 \end{array} \right) = 20 \) three-particle states, but there are 9 states with \( M = +1/2 \), namely \( | 1,2,3 \rangle, |1,2,5\rangle, | 1,4,6 \rangle, | 2,3,4 \rangle, |2,3,6 \rangle, | 2,4,5 \rangle, | 2, 5, 6 \rangle, |3,4,6 \rangle, | 4,5,6 \rangle \).
In this basis, the operator $$ \hat{P}_+ = \hat{a}^\dagger_1 \hat{a}^\dagger_2 + \hat{a}^\dagger_3 \hat{a}^\dagger_4 + \hat{a}^\dagger_5 \hat{a}^\dagger_6 $$ From this we can determine that $$ \hat{P}_- | 1, 4, 6 \rangle = \hat{P}_- | 2, 3, 6 \rangle = \hat{P}_- | 2, 4, 5 \rangle = 0 $$ so those states all have eigenvalue 0.
Now for further example, $$ \hat{P}_- | 1,2,3 \rangle = | 3 \rangle $$ so $$ \hat{P}_+ \hat{P}_- | 1,2,3\rangle = | 1,2,3\rangle+ | 3,4,3\rangle + | 5,6,3\rangle $$ The second term vanishes because state 3 is occupied twice, and reordering the last term we get $$ \hat{P}_+ \hat{P}_- | 1,2,3\rangle = | 1,2,3\rangle+ |3, 5,6\rangle $$ without picking up a phase.
Continuing in this fashion, with the previous ordering of the many-body states ( \( | 1,2,3 \rangle, |1,2,5\rangle, | 1,4,6 \rangle, | 2,3,4 \rangle, |2,3,6 \rangle, | 2,4,5 \rangle, | 2, 5, 6 \rangle, |3,4,6 \rangle, | 4,5,6 \rangle \)) the Hamiltonian matrix of this system is $$ H = -G\left( \begin{array}{ccccccccc} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \end{array} \right ) $$ This is useful for our project. One can by hand confirm that there are 3 eigenvalues \( -2G \) and 6 with value zero.
Another example Using the \( (1/2)^4 \) single-particle space, resulting in eight single-particle states
| Index | \( n \) | \( l \) | \( s \) | \( m_s \) |
|---|---|---|---|---|
| 1 | 0 | 0 | 1/2 | -1/2 |
| 2 | 0 | 0 | 1/2 | 1/2 |
| 3 | 1 | 0 | 1/2 | -1/2 |
| 4 | 1 | 0 | 1/2 | 1/2 |
| 5 | 2 | 0 | 1/2 | -1/2 |
| 6 | 2 | 0 | 1/2 | 1/2 |
| 7 | 3 | 0 | 1/2 | -1/2 |
| 8 | 3 | 0 | 1/2 | 1/2 |
and then taking only 4-particle, \( M=0 \) states that have no `broken pairs', there are six basis Slater determinants:
Now we take the following Hamiltonian $$ \hat{H} = \sum_n n \delta \hat{N}_n - G \hat{P}^\dagger \hat{P} $$ where $$ \hat{N}_n = \hat{a}^\dagger_{n, m=+1/2} \hat{a}_{n, m=+1/2} + \hat{a}^\dagger_{n, m=-1/2} \hat{a}_{n, m=-1/2} $$ and $$ \hat{P}^\dagger = \sum_{n} \hat{a}^\dagger_{n, m=+1/2} \hat{a}^\dagger_{n, m=-1/2} $$ We can write down the $ 6 \times 6$ Hamiltonian in the basis from the prior slide: $$ H = \left ( \begin{array}{cccccc} 2\delta -2G & -G & -G & -G & -G & 0 \\ -G & 4\delta -2G & -G & -G & -0 & -G \\ -G & -G & 6\delta -2G & 0 & -G & -G \\ -G & -G & 0 & 6\delta-2G & -G & -G \\ -G & 0 & -G & -G & 8\delta-2G & -G \\ 0 & -G & -G & -G & -G & 10\delta -2G \end{array} \right ) $$ (You should check by hand that this is correct.)
For \( \delta = 0 \) we have the closed form solution of the g.s. energy given by \( -6G \).
The goal is to compute the matrix elements of the Hamiltonian, specifically matrix elements between many-body states (Slater determinants) of two-body operators $$ \sum_{p < q, r < s}V_{pqr} \hat{a}^\dagger_p \hat{a}^\dagger_q\hat{a}_s \hat{a}_r $$ In particular we will need to compute $$ \langle \beta | \hat{a}^\dagger_p \hat{a}^\dagger_q\hat{a}_s \hat{a}_r |\alpha \rangle $$ where \( \alpha, \beta \) are indices labeling Slater determinants and \( p,q,r,s \) label single-particle states.
Note: there are other, more efficient ways to do this than the method we describe, but you will be able to produce a working code quickly.
As we coded in the first step, a Slater determinant \( | \alpha \rangle \) with index \( \alpha \) is a list of \( N \) occupied single-particle states \( i_1 < i_2 < i_3 \ldots i_N \).
Furthermore, for the two-body matrix elements \( V_{pqrs} \) we normally assume \( p < q \) and \( r < s \). For our specific project, the interaction is much simpler and you can use this to simplify considerably the setup of a shell-model code for project 2.
What follows here is a more general, but still brute force, approach.
Write a function that:
This is equivalent to computing $$ \langle \beta | \hat{a}^\dagger_p \hat{a}^\dagger_q\hat{a}_s \hat{a}_r |\alpha \rangle $$
The first step can take as input an initial Slater determinant (whose position in the list of basis Slater determinants is \( \alpha \)) written as an ordered listed of occupied single-particle states, e.g. \( 1,2,5,8 \), and the indices \( p,q,r,s \) from the two-body operator.
It will return another final Slater determinant if the single-particle states \( r \) and \( s \) are occupied, else it will return an empty Slater determinant (all zeroes).
If \( r \) and \( s \) are in the list of occupied single particle states, then replace the initial single-particle states \( ij \) as \( i \rightarrow r \) and \( j \rightarrow r \).
The second step will take the final Slater determinant from the first step (if not empty), and then order by pairwise permutations (i.e., if the Slater determinant is \( i_1, i_2, i_3, \ldots \), then if \( i_n > i_{n+1} \), interchange \( i_n \leftrightarrow i_{n+1} \).
It will also output a phase. If any two single-particle occupancies are repeated, the phase is 0. Otherwise it is +1 for an even permutation and -1 for an odd permutation to bring the final Slater determinant into ascending order, \( j_1 < j_2 < j_3 \ldots \).
Example: Suppose in the \( sd \) single-particle space that the initial Slater determinant is \( 1,3,9,12 \). If \( p,q,r,s = 2,8,1,12 \), then after the first step the final Slater determinant is \( 2,3,9,8 \). The second step will return \( 2,3,8,9 \) and a phase of -1, because an odd number of interchanges is required.
Example: Suppose in the \( sd \) single-particle space that the initial Slater determinant is \( 1,3,9,12 \). If \( p,q,r,s = 3,8,1,12 \), then after the first step the final Slater determinant is \( 3,3,9,8 \), but after the second step the phase is 0 because the single-particle state 3 is occupied twice.
Lastly, the final step takes the ordered final Slater determinant and we search through the basis list to determine its index in the many-body basis, that is, \( \beta \).
The Hamiltonian is then stored as an \( N_{SD} \times N_{SD} \) array of real numbers, which can be allocated once you have created the many-body basis and know \( N_{SD} \).
You will also need to include the single-particle energies. This is easy: they only contribute to diagonal matrix elements, that is, \( H(\alpha,\alpha) \). Simply find the occupied single-particle states \( i \) and add the corresponding \( \epsilon(i) \).