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.
$$
\Psi(x_1, x_2, \ldots, x_A)
= \frac{1}{\sqrt{A!}}
\det \left |
\begin{array}{cccc}
\phi_1(x_1) & \phi_1(x_2) & \ldots & \phi_1(x_A) \\
\phi_2(x_1) & \phi_2(x_2) & \ldots & \phi_2(x_A) \\
\vdots & & & \\
\phi_A(x_1) & \phi_A(x_2) & \ldots & \phi_A(x_A)
\end{array}
\right |
$$
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.
For some relevant exercises with solutions see chapter 8 of Lecture Notes in Physics, volume 936.
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
$$
\begin{equation}
\sum_{P'H'}\langle \Phi_H^P | \hat{H} -E|\Phi_{H'}^{P'} \rangle = 0. \tag{1}
\end{equation}
$$
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
What happens if we have a three-body interaction and a Hartree-Fock basis?
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 with a 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. (1), 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. (1), 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
All these methods start normally with a Hartree-Fock basis as the calculational basis.
In the steps that lead to the development of clean code you should think of
Many of the above aspetcs will save you a lot of time when you incrementally extend software over time from simpler to more complicated problems. In particular, you will benefit from many good habits:
Unit Testing is the practice of testing the smallest testable parts, called units, of an application individually and independently to determine if they behave exactly as expected.
Unit tests (short code fragments) are usually written such that they can be preformed at any time during the development to continually verify the behavior of the code.
In this way, possible bugs will be identified early in the development cycle, making the debugging at later stages much easier.
There are many benefits associated with Unit Testing, such as
Look up the guide on how to install unit tests for c++ at course webpage. This is the version with classes.
#include <unittest++/UnitTest++.h>
class MyMultiplyClass{
public:
double multiply(double x, double y) {
return x * y;
}
};
TEST(MyMath) {
MyMultiplyClass my;
CHECK_EQUAL(56, my.multiply(7,8));
}
int main()
{
return UnitTest::RunAllTests();
}
And without classes
#include <unittest++/UnitTest++.h>
double multiply(double x, double y) {
return x * y;
}
TEST(MyMath) {
CHECK_EQUAL(56, multiply(7,8));
}
int main()
{
return UnitTest::RunAllTests();
}
For Fortran users, the link at http://sourceforge.net/projects/fortranxunit/ contains a similar software for unit testing. For Python go to https://docs.python.org/2/library/unittest.html.
There are many types of unit test libraries. One which is very popular with C++ programmers is Catch
Catch is header only. All you need to do is drop the file(s) somewhere reachable from your project - either in some central location you can set your header search path to find, or directly into your project tree itself!
This is a particularly good option for other Open-Source projects that want to use Catch for their test suite.
Computing factorials
inline unsigned int Factorial( unsigned int number ) {
return number > 1 ? Factorial(number-1)*number : 1;
}
Simple test where we put everything in a single file
#define CATCH_CONFIG_MAIN // This tells Catch to provide a main()
#include "catch.hpp"
inline unsigned int Factorial( unsigned int number ) {
return number > 1 ? Factorial(number-1)*number : 1;
}
TEST_CASE( "Factorials are computed", "[factorial]" ) {
REQUIRE( Factorial(0) == 1 );
REQUIRE( Factorial(1) == 1 );
REQUIRE( Factorial(2) == 2 );
REQUIRE( Factorial(3) == 6 );
REQUIRE( Factorial(10) == 3628800 );
}
This will compile to a complete executable which responds to command line arguments. If you just run it with no arguments it will execute all test cases (in this case there is just one), report any failures, report a summary of how many tests passed and failed and return the number of failed tests.
#define
one identifier and
#include
one header and we got everything - even an implementation of main() that will respond to command line arguments. Once you have more than one file with unit tests in you'll just need to
#include "catch.hpp"
and go. Usually it's a good idea to have a dedicated implementation file that just has
#define CATCH_CONFIG_MAIN
#include "catch.hpp".
You can also provide your own implementation of main and drive Catch yourself.
TEST_CASE
macro.
The test name must be unique. You can run sets of tests by specifying a wildcarded test name or a tag expression. All we did was define one identifier and include one header and we got everything.
We write our individual test assertions using the
REQUIRE
macro.
Three levels of tests
Computing is understanding and your understanding is reflected in your abilities to write clear and clean code.
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 this format one labels states by \( n(l)_j \), with \( (l) \) replaced by a letter: \( s \) for \( l=0 \), \( p \) for \( l=1 \), \( d \) for \( l=2 \), \( f \) for \( l=3 \), and thenceforth alphabetical.
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
Note that the individual \( M \)-scheme basis states will not, in general, have good total \( J \). Because the Hamiltonian is rotationally invariant, however, the eigenstates will have good \( J \). (The situation is muddied when one has states of different \( J \) that are nonetheless degenerate.)
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:
Now increase the last occupancy recursively:
Then start over with
and again increase the rightmost digit
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 \( M \):
Of these, there are only 3 states with \( M=0 \).
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
The number one mistake is using a too complex a system to test. For example , if you are computing particles in a potential in a box, try removing the potential–you should get particles in a box. And start with one particle, then two, then three... Don't start with eight particles.
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. We will discuss these in more detail at the beginning of the third week.
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.
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:
$$
| \beta \rangle = \pm \hat{a}^\dagger_p \hat{a}^\dagger_q\hat{a}_s \hat{a}_r |\alpha \rangle
$$
as well as the phase
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 should force the resulting matrix \( H \) to be symmetric. To do this, when updating \( H(\alpha,\beta) \), if \( \alpha \neq \beta \), also update \( H(\beta,\alpha) \).
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 \( \epsilon(i) \).
Consider the many-body state \( \Psi_{\lambda} \) expressed as linear combinations of Slater determinants (\( SD \)) of orthonormal single-particle states \( \phi({\bf r}) \):
$$
\begin{equation}
\Psi_{\lambda} = \sum_i C_{\lambda i} SD_i
\tag{2}
\end{equation}
$$
Using the Slater-Condon rules the matrix elements of any one-body
(\( {\cal O}_1 \)) or two-body (\( {\cal O}_2 \)) operator expressed in the
determinant space have simple expressions involving one- and two-fermion
integrals in our given single-particle basis.
The diagonal elements are given by:
$$
\begin{eqnarray}
\langle SD | {\cal O}_1 | SD \rangle & = & \sum_{i \in SD} \langle \phi_i | {\cal O}_1 | \phi_i \rangle \\
\langle SD | {\cal O}_2 | SD \rangle & = & \frac{1}{2} \sum_{(i,j) \in SD}
\langle \phi_i \phi_j | {\cal O}_2 | \phi_i \phi_j \rangle - \nonumber \\
& &
\langle \phi_i \phi_j | {\cal O}_2 | \phi_j \phi_i \rangle \nonumber
\end{eqnarray}
$$
For two determinants which differ only by the substitution of single-particle states \( i \) with a single-particle state \( j \):
$$
\begin{eqnarray}
\langle SD | {\cal O}_1 | SD_i^j \rangle & = & \langle \phi_i | {\cal O}_1 | \phi_j \rangle \\
\langle SD | {\cal O}_2 | SD_i^j \rangle & = & \sum_{k \in SD}
\langle \phi_i \phi_k | {\cal O}_2 | \phi_j \phi_k \rangle -
\langle \phi_i \phi_k | {\cal O}_2 | \phi_k \phi_j \rangle \nonumber
\end{eqnarray}
$$
For two determinants which differ by two single-particle states
$$
\begin{eqnarray}
\langle SD | {\cal O}_1 | SD_{ik}^{jl} \rangle & = & 0 \\
\langle SD | {\cal O}_2 | SD_{ik}^{jl} \rangle & = &
\langle \phi_i \phi_k | {\cal O}_2 | \phi_j \phi_l \rangle -
\langle \phi_i \phi_k | {\cal O}_2 | \phi_l \phi_j \rangle \nonumber
\end{eqnarray}
$$
All other matrix elements involving determinants with more than two
substitutions are zero.
An efficient implementation of these rules requires
We can solve this problem using our odometric approach or alternatively using a bit representation as discussed below and in more detail in
We recommend in particular the article by Simen Kvaal. It contains nice general classes for creation and annihilation operators as well as the calculation of the phase (see below).
When we diagonalize the Hamiltonian matrix, the eigenvectors are the coefficients \( C_{\lambda i} \) used to express the many-body state \( \Psi_{\lambda} \) in terms of a linear combinations of Slater determinants (\( SD \)) of orthonormal single-particle states \( \phi({\bf r}) \).
With these eigenvectors we can compute say the transition likelyhood of a one-body operator as
$$
\langle \Psi_{\lambda} \vert {\cal O}_1 \vert \Psi_{\sigma} \rangle =
\sum_{ij}C_{\lambda i}^*C_{\sigma j} \langle SD_i | {\cal O}_1 | SD_j \rangle .
$$
Writing the one-body operator in second quantization as
$$
{\cal O}_1 = \sum_{pq} \langle p \vert {\cal o}_1 \vert q\rangle a_p^{\dagger} a_q,
$$
we have
$$
\langle \Psi_{\lambda} \vert {\cal O}_1 \vert \Psi_{\sigma} \rangle =
\sum_{pq}\langle p \vert {\cal o}_1 \vert q\rangle \sum_{ij}C_{\lambda i}^*C_{\sigma j} \langle SD_i |a_p^{\dagger} a_q | SD_j \rangle .
$$
The terms we need to evalute then are just the elements
$$
\langle SD_i |a_p^{\dagger} a_q | SD_j \rangle,
$$
which can be rewritten in terms of spectroscopic factors by inserting a complete set of Slater determinats as
$$
\langle SD_i |a_p^{\dagger} a_q | SD_j \rangle = \sum_{l}\langle SD_i \vert a_p^{\dagger}\vert SD_l\rangle \langle SD_l \vert a_q \vert SD_j \rangle,
$$
where \( \langle SD_l\vert a_q(a_p^{\dagger})\vert SD_j\rangle \) are the spectroscopic factors. These can be easily evaluated in \( m \)-scheme. Using the Wigner-Eckart theorem we can transform these to a \( J \)-coupled scheme through so-called reduced matrix elements.
In the build-up of a shell-model or FCI code that is meant to tackle large dimensionalities we need to deal with the action of the Hamiltonian \( \hat{H} \) on a Slater determinant represented in second quantization as
$$
|\alpha_1\dots \alpha_n\rangle = a_{\alpha_1}^{\dagger} a_{\alpha_2}^{\dagger} \dots a_{\alpha_n}^{\dagger} |0\rangle.
$$
The time consuming part stems from the action of the Hamiltonian
on the above determinant,
$$
\left(\sum_{\alpha\beta} \langle \alpha|t+u|\beta\rangle a_\alpha^{\dagger} a_\beta + \frac{1}{4} \sum_{\alpha\beta\gamma\delta}
\langle \alpha \beta|\hat{v}|\gamma \delta\rangle a_\alpha^{\dagger} a_\beta^{\dagger} a_\delta a_\gamma\right)a_{\alpha_1}^{\dagger} a_{\alpha_2}^{\dagger} \dots a_{\alpha_n}^{\dagger} |0\rangle.
$$
A practically useful way to implement this action is to encode a Slater determinant as a bit pattern.
Assume that we have at our disposal \( n \) different single-particle states \( \alpha_0,\alpha_2,\dots,\alpha_{n-1} \) and that we can distribute among these states \( N\le n \) particles.
A Slater determinant can then be coded as an integer of \( n \) bits. As an example, if we have \( n=16 \) single-particle states \( \alpha_0,\alpha_1,\dots,\alpha_{15} \) and \( N=4 \) fermions occupying the states \( \alpha_3 \), \( \alpha_6 \), \( \alpha_{10} \) and \( \alpha_{13} \) we could write this Slater determinant as
$$
\Phi_{\Lambda} = a_{\alpha_3}^{\dagger} a_{\alpha_6}^{\dagger} a_{\alpha_{10}}^{\dagger} a_{\alpha_{13}}^{\dagger} |0\rangle.
$$
The unoccupied single-particle states have bit value \( 0 \) while the occupied ones are represented by bit state \( 1 \).
In the binary notation we would write this 16 bits long integer as
$$
\begin{array}{cccccccccccccccc}
{\alpha_0}&{\alpha_1}&{\alpha_2}&{\alpha_3}&{\alpha_4}&{\alpha_5}&{\alpha_6}&{\alpha_7} & {\alpha_8} &{\alpha_9} & {\alpha_{10}} &{\alpha_{11}} &{\alpha_{12}} &{\alpha_{13}} &{\alpha_{14}} & {\alpha_{15}} \\
{0} & {0} &{0} &{1} &{0} &{0} &{1} &{0} &{0} &{0} &{1} &{0} &{0} &{1} &{0} & {0} \\
\end{array}
$$
which translates into the decimal number
$$
2^3+2^6+2^{10}+2^{13}=9288.
$$
We can thus encode a Slater determinant as a bit pattern.
With \( N \) particles that can be distributed over \( n \) single-particle states, the total number of Slater determinats (and defining thereby the dimensionality of the system) is
$$
\mathrm{dim}(\mathcal{H}) = \left(\begin{array}{c} n \\N\end{array}\right).
$$
The total number of bit patterns is \( 2^n \).
We assume again that we have at our disposal \( n \) different single-particle orbits \( \alpha_0,\alpha_2,\dots,\alpha_{n-1} \) and that we can distribute among these orbits \( N\le n \) particles. The ordering among these states is important as it defines the order of the creation operators. We will write the determinant
$$
\Phi_{\Lambda} = a_{\alpha_3}^{\dagger} a_{\alpha_6}^{\dagger} a_{\alpha_{10}}^{\dagger} a_{\alpha_{13}}^{\dagger} |0\rangle,
$$
in a more compact way as
$$
\Phi_{3,6,10,13} = |0001001000100100\rangle.
$$
The action of a creation operator is thus
$$
a^{\dagger}_{\alpha_4}\Phi_{3,6,10,13} = a^{\dagger}_{\alpha_4}|0001001000100100\rangle=a^{\dagger}_{\alpha_4}a_{\alpha_3}^{\dagger} a_{\alpha_6}^{\dagger} a_{\alpha_{10}}^{\dagger} a_{\alpha_{13}}^{\dagger} |0\rangle,
$$
which becomes
$$
-a_{\alpha_3}^{\dagger} a^{\dagger}_{\alpha_4} a_{\alpha_6}^{\dagger} a_{\alpha_{10}}^{\dagger} a_{\alpha_{13}}^{\dagger} |0\rangle=-|0001101000100100\rangle.
$$
Similarly
$$
a^{\dagger}_{\alpha_6}\Phi_{3,6,10,13} = a^{\dagger}_{\alpha_6}|0001001000100100\rangle=a^{\dagger}_{\alpha_6}a_{\alpha_3}^{\dagger} a_{\alpha_6}^{\dagger} a_{\alpha_{10}}^{\dagger} a_{\alpha_{13}}^{\dagger} |0\rangle,
$$
which becomes
$$
-a^{\dagger}_{\alpha_4} (a_{\alpha_6}^{\dagger})^ 2 a_{\alpha_{10}}^{\dagger} a_{\alpha_{13}}^{\dagger} |0\rangle=0!
$$
This gives a simple recipe:
Consider the action of \( a^{\dagger}_{\alpha_2} \) on various slater determinants:
$$
\begin{array}{ccc}
a^{\dagger}_{\alpha_2}\Phi_{00111}& = a^{\dagger}_{\alpha_2}|00111\rangle&=0\times |00111\rangle\\
a^{\dagger}_{\alpha_2}\Phi_{01011}& = a^{\dagger}_{\alpha_2}|01011\rangle&=(-1)\times |01111\rangle\\
a^{\dagger}_{\alpha_2}\Phi_{01101}& = a^{\dagger}_{\alpha_2}|01101\rangle&=0\times |01101\rangle\\
a^{\dagger}_{\alpha_2}\Phi_{01110}& = a^{\dagger}_{\alpha_2}|01110\rangle&=0\times |01110\rangle\\
a^{\dagger}_{\alpha_2}\Phi_{10011}& = a^{\dagger}_{\alpha_2}|10011\rangle&=(-1)\times |10111\rangle\\
a^{\dagger}_{\alpha_2}\Phi_{10101}& = a^{\dagger}_{\alpha_2}|10101\rangle&=0\times |10101\rangle\\
a^{\dagger}_{\alpha_2}\Phi_{10110}& = a^{\dagger}_{\alpha_2}|10110\rangle&=0\times |10110\rangle\\
a^{\dagger}_{\alpha_2}\Phi_{11001}& = a^{\dagger}_{\alpha_2}|11001\rangle&=(+1)\times |11101\rangle\\
a^{\dagger}_{\alpha_2}\Phi_{11010}& = a^{\dagger}_{\alpha_2}|11010\rangle&=(+1)\times |11110\rangle\\
\end{array}
$$
What is the simplest way to obtain the phase when we act with one annihilation(creation) operator
on the given Slater determinant representation?
We have an SD representation
$$
\Phi_{\Lambda} = a_{\alpha_0}^{\dagger} a_{\alpha_3}^{\dagger} a_{\alpha_6}^{\dagger} a_{\alpha_{10}}^{\dagger} a_{\alpha_{13}}^{\dagger} |0\rangle,
$$
in a more compact way as
$$
\Phi_{0,3,6,10,13} = |1001001000100100\rangle.
$$
The action of
$$
a^{\dagger}_{\alpha_4}a_{\alpha_0}\Phi_{0,3,6,10,13} = a^{\dagger}_{\alpha_4}|0001001000100100\rangle=a^{\dagger}_{\alpha_4}a_{\alpha_3}^{\dagger} a_{\alpha_6}^{\dagger} a_{\alpha_{10}}^{\dagger} a_{\alpha_{13}}^{\dagger} |0\rangle,
$$
which becomes
$$
-a_{\alpha_3}^{\dagger} a^{\dagger}_{\alpha_4} a_{\alpha_6}^{\dagger} a_{\alpha_{10}}^{\dagger} a_{\alpha_{13}}^{\dagger} |0\rangle=-|0001101000100100\rangle.
$$
The action
$$
a_{\alpha_0}\Phi_{0,3,6,10,13} = |0001001000100100\rangle,
$$
can be obtained by subtracting the logical sum (AND operation) of \( \Phi_{0,3,6,10,13} \) and
a word which represents only \( \alpha_0 \), that is
$$
|1000000000000000\rangle,
$$
from \( \Phi_{0,3,6,10,13}= |1001001000100100\rangle \).
This operation gives \( |0001001000100100\rangle \).
Similarly, we can form \( a^{\dagger}_{\alpha_4}a_{\alpha_0}\Phi_{0,3,6,10,13} \), say, by adding \( |0000100000000000\rangle \) to \( a_{\alpha_0}\Phi_{0,3,6,10,13} \), first checking that their logical sum is zero in order to make sure that the state \( \alpha_4 \) is not already occupied.
It is trickier however to get the phase \( (-1)^l \). One possibility is as follows
In the previous example \( S_1=|1000000000000000\rangle \)
\( S_2=|0000100000000000\rangle \).
$$
S=|0111000000000000\rangle
$$
$$
\Phi_{0,3,6,10,13} = |1001001000100100\rangle,
$$
which results in \( |0001000000000000\rangle \). Counting the number of 1-bits gives the phase. Here you need however an algorithm for bitcounting.
We include here a python program which may aid in this direction. It uses bit manipulation functions from http://wiki.python.org/moin/BitManipulation.
import math
"""
A simple Python class for Slater determinant manipulation
Bit-manipulation stolen from:
http://wiki.python.org/moin/BitManipulation
"""
# bitCount() counts the number of bits set (not an optimal function)
def bitCount(int_type):
""" Count bits set in integer """
count = 0
while(int_type):
int_type &= int_type - 1
count += 1
return(count)
# testBit() returns a nonzero result, 2**offset, if the bit at 'offset' is one.
def testBit(int_type, offset):
mask = 1 << offset
return(int_type & mask) >> offset
# setBit() returns an integer with the bit at 'offset' set to 1.
def setBit(int_type, offset):
mask = 1 << offset
return(int_type | mask)
# clearBit() returns an integer with the bit at 'offset' cleared.
def clearBit(int_type, offset):
mask = ~(1 << offset)
return(int_type & mask)
# toggleBit() returns an integer with the bit at 'offset' inverted, 0 -> 1 and 1 -> 0.
def toggleBit(int_type, offset):
mask = 1 << offset
return(int_type ^ mask)
# binary string made from number
def bin0(s):
return str(s) if s<=1 else bin0(s>>1) + str(s&1)
def bin(s, L = 0):
ss = bin0(s)
if L > 0:
return '0'*(L-len(ss)) + ss
else:
return ss
class Slater:
""" Class for Slater determinants """
def __init__(self):
self.word = int(0)
def create(self, j):
print "c^+_" + str(j) + " |" + bin(self.word) + "> = ",
# Assume bit j is set, then we return zero.
s = 0
# Check if bit j is set.
isset = testBit(self.word, j)
if isset == 0:
bits = bitCount(self.word & ((1<<j)-1))
s = pow(-1, bits)
self.word = setBit(self.word, j)
print str(s) + " x |" + bin(self.word) + ">"
return s
def annihilate(self, j):
print "c_" + str(j) + " |" + bin(self.word) + "> = ",
# Assume bit j is not set, then we return zero.
s = 0
# Check if bit j is set.
isset = testBit(self.word, j)
if isset == 1:
bits = bitCount(self.word & ((1<<j)-1))
s = pow(-1, bits)
self.word = clearBit(self.word, j)
print str(s) + " x |" + bin(self.word) + ">"
return s
# Do some testing:
phi = Slater()
phi.create(0)
phi.create(1)
phi.create(2)
phi.create(3)
print
s = phi.annihilate(2)
s = phi.create(7)
s = phi.annihilate(0)
s = phi.create(200)
Let us consider the matrix \( \mathbf{A} \) of dimension \( n \). The eigenvalues of \( \mathbf{A} \) are defined through the matrix equation
$$
\mathbf{A}\mathbf{x}^{(\nu)} = \lambda^{(\nu)}\mathbf{x}^{(\nu)},
$$
where \( \lambda^{(\nu)} \) are the eigenvalues and \( \mathbf{x}^{(\nu)} \) the
corresponding eigenvectors.
Unless otherwise stated, when we use the wording eigenvector we mean the
right eigenvector. The left eigenvalue problem is defined as
$$
\mathbf{x}^{(\nu)}_L\mathbf{A} = \lambda^{(\nu)}\mathbf{x}^{(\nu)}_L
$$
The above right eigenvector problem is equivalent to a set of \( n \) equations with \( n \) unknowns
\( x_i \).
The eigenvalue problem can be rewritten as
$$
\left( \mathbf{A}-\lambda^{(\nu)} \mathbf{I} \right) \mathbf{x}^{(\nu)} = 0,
$$
with \( \mathbf{I} \) being the unity matrix. This equation provides
a solution to the problem if and only if the determinant
is zero, namely
$$
\left| \mathbf{A}-\lambda^{(\nu)}\mathbf{I}\right| = 0,
$$
which in turn means that the determinant is a polynomial
of degree \( n \) in \( \lambda \) and in general we will have
\( n \) distinct zeros.
The eigenvalues of a matrix \( \mathbf{A}\in {\mathbb{C}}^{n\times n} \) are thus the \( n \) roots of its characteristic polynomial
$$
P(\lambda) = det(\lambda\mathbf{I}-\mathbf{A}),
$$
or
$$
P(\lambda)= \prod_{i=1}^{n}\left(\lambda_i-\lambda\right).
$$
The set of these roots is called the spectrum and is denoted as
\( \lambda(\mathbf{A}) \).
If \( \lambda(\mathbf{A})=\left\{\lambda_1,\lambda_2,\dots ,\lambda_n\right\} \) then we have
$$
det(\mathbf{A})= \lambda_1\lambda_2\dots\lambda_n,
$$
and if we define the trace of \( \mathbf{A} \) as
$$
Tr(\mathbf{A})=\sum_{i=1}^n a_{ii}$$
then
$$
Tr(\mathbf{A})=\lambda_1+\lambda_2+\dots+\lambda_n.
$$
The Abel-Ruffini theorem (also known as Abel's impossibility theorem) states that there is no general solution in radicals to polynomial equations of degree five or higher.
The content of this theorem is frequently misunderstood. It does not assert that higher-degree polynomial equations are unsolvable. In fact, if the polynomial has real or complex coefficients, and we allow complex solutions, then every polynomial equation has solutions; this is the fundamental theorem of algebra. Although these solutions cannot always be computed exactly with radicals, they can be computed to any desired degree of accuracy using numerical methods such as the Newton-Raphson method or Laguerre method, and in this way they are no different from solutions to polynomial equations of the second, third, or fourth degrees.
The theorem only concerns the form that such a solution must take. The content of the theorem is that the solution of a higher-degree equation cannot in all cases be expressed in terms of the polynomial coefficients with a finite number of operations of addition, subtraction, multiplication, division and root extraction. Some polynomials of arbitrary degree, of which the simplest nontrivial example is the monomial equation \( ax^n = b \), are always solvable with a radical.
The Abel-Ruffini theorem says that there are some fifth-degree equations whose solution cannot be so expressed. The equation \( x^5 - x + 1 = 0 \) is an example. Some other fifth degree equations can be solved by radicals, for example \( x^5 - x^4 - x + 1 = 0 \). The precise criterion that distinguishes between those equations that can be solved by radicals and those that cannot was given by Galois and is now part of Galois theory: a polynomial equation can be solved by radicals if and only if its Galois group is a solvable group.
Today, in the modern algebraic context, we say that second, third and fourth degree polynomial equations can always be solved by radicals because the symmetric groups \( S_2, S_3 \) and \( S_4 \) are solvable groups, whereas \( S_n \) is not solvable for \( n \ge 5 \).
In the present discussion we assume that our matrix is real and symmetric, that is \( \mathbf{A}\in {\mathbb{R}}^{n\times n} \). The matrix \( \mathbf{A} \) has \( n \) eigenvalues \( \lambda_1\dots \lambda_n \) (distinct or not). Let \( \mathbf{D} \) be the diagonal matrix with the eigenvalues on the diagonal
$$
\mathbf{D}= \left( \begin{array}{ccccccc} \lambda_1 & 0 & 0 & 0 & \dots &0 & 0 \\
0 & \lambda_2 & 0 & 0 & \dots &0 &0 \\
0 & 0 & \lambda_3 & 0 &0 &\dots & 0\\
\dots & \dots & \dots & \dots &\dots &\dots & \dots\\
0 & \dots & \dots & \dots &\dots &\lambda_{n-1} & \\
0 & \dots & \dots & \dots &\dots &0 & \lambda_n
\end{array} \right).
$$
If \( \mathbf{A} \) is real and symmetric then there exists a real orthogonal matrix \( \mathbf{S} \) such that
$$
\mathbf{S}^T \mathbf{A}\mathbf{S}= \mathrm{diag}(\lambda_1,\lambda_2,\dots ,\lambda_n),
$$
and for \( j=1:n \) we have \( \mathbf{A}\mathbf{S}(:,j) = \lambda_j \mathbf{S}(:,j) \).
To obtain the eigenvalues of \( \mathbf{A}\in {\mathbb{R}}^{n\times n} \), the strategy is to perform a series of similarity transformations on the original matrix \( \mathbf{A} \), in order to reduce it either into a diagonal form as above or into a tridiagonal form.
We say that a matrix \( \mathbf{B} \) is a similarity transform of \( \mathbf{A} \) if
$$
\mathbf{B}= \mathbf{S}^T \mathbf{A}\mathbf{S}, \hspace{1cm} \mathrm{where} \hspace{1cm} \mathbf{S}^T\mathbf{S}=\mathbf{S}^{-1}\mathbf{S} =\mathbf{I}.
$$
The importance of a similarity transformation lies in the fact that
the resulting matrix has the same
eigenvalues, but the eigenvectors are in general different.
To prove this we start with the eigenvalue problem and a similarity transformed matrix \( \mathbf{B} \).
$$
\mathbf{A}\mathbf{x}=\lambda\mathbf{x} \hspace{1cm} \mathrm{and}\hspace{1cm}
\mathbf{B}= \mathbf{S}^T \mathbf{A}\mathbf{S}.
$$
We multiply the first equation on the left by \( \mathbf{S}^T \) and insert
\( \mathbf{S}^{T}\mathbf{S} = \mathbf{I} \) between \( \mathbf{A} \) and \( \mathbf{x} \). Then we get
$$
\begin{equation}
(\mathbf{S}^T\mathbf{A}\mathbf{S})(\mathbf{S}^T\mathbf{x})=\lambda\mathbf{S}^T\mathbf{x} ,
\tag{3}
\end{equation}
$$
which is the same as
$$
\mathbf{B} \left ( \mathbf{S}^T\mathbf{x} \right ) = \lambda \left (\mathbf{S}^T\mathbf{x}\right ).
$$
The variable \( \lambda \) is an eigenvalue of \( \mathbf{B} \) as well, but with
eigenvector \( \mathbf{S}^T\mathbf{x} \).
The basic philosophy is to
$$
\begin{equation}
\mathbf{S}_N^T\dots \mathbf{S}_1^T\mathbf{A}\mathbf{S}_1\dots \mathbf{S}_N=\mathbf{D} ,
\tag{4}
\end{equation}
$$
One speaks normally of two main approaches to solving the eigenvalue problem.
Direct or non-iterative methods require for matrices of dimensionality \( n\times n \) typically \( O(n^3) \) operations. These methods are normally called standard methods and are used for dimensionalities \( n \sim 10^5 \) or smaller. A brief historical overview
| Year | \( n \) | |
|---|---|---|
| 1950 | \( n=20 \) | (Wilkinson) |
| 1965 | \( n=200 \) | (Forsythe et al.) |
| 1980 | \( n=2000 \) | Linpack |
| 1995 | \( n=20000 \) | Lapack |
| This decade | \( n\sim 10^5 \) | Lapack |
shows that in the course of 60 years the dimension that direct diagonalization methods can handle has increased by almost a factor of \( 10^4 \) (note this is for serial versions). However, it pales beside the progress achieved by computer hardware, from flops to petaflops, a factor of almost \( 10^{15} \). We see clearly played out in history the \( O(n^3) \) bottleneck of direct matrix algorithms.
Sloppily speaking, when \( n\sim 10^4 \) is cubed we have \( O(10^{12}) \) operations, which is smaller than the \( 10^{15} \) increase in flops.
If the matrix to diagonalize is large and sparse, direct methods simply become impractical, also because many of the direct methods tend to destroy sparsity. As a result large dense matrices may arise during the diagonalization procedure. The idea behind iterative methods is to project the $n-$dimensional problem in smaller spaces, so-called Krylov subspaces. Given a matrix \( \mathbf{A} \) and a vector \( \mathbf{v} \), the associated Krylov sequences of vectors (and thereby subspaces) \( \mathbf{v} \), \( \mathbf{A}\mathbf{v} \), \( \mathbf{A}^2\mathbf{v} \), \( \mathbf{A}^3\mathbf{v},\dots \), represent successively larger Krylov subspaces.
| Matrix | \( \mathbf{A}\mathbf{x}=\mathbf{b} \) | \( \mathbf{A}\mathbf{x}=\lambda\mathbf{x} \) |
|---|---|---|
| \( \mathbf{A}=\mathbf{A}^* \) | Conjugate gradient | Lanczos |
| \( \mathbf{A}\ne \mathbf{A}^* \) | GMRES etc | Arnoldi |
Basic features with a real symmetric matrix (and normally huge \( n> 10^6 \) and sparse) \( \hat{A} \) of dimension \( n\times n \):
$$
\hat{T}= \hat{Q}^{T}\hat{A}\hat{Q},
$$
with the first vector \( \hat{Q}\hat{e}_1=\hat{q}_1 \).
We are going to solve iteratively
$$
\hat{T}= \hat{Q}^{T}\hat{A}\hat{Q},
$$
with the first vector \( \hat{Q}\hat{e}_1=\hat{q}_1 \).
We can write out the matrix \( \hat{Q} \) in terms of its column vectors
$$
\hat{Q}=\left[\hat{q}_1\hat{q}_2\dots\hat{q}_n\right].
$$
The matrix
$$
\hat{T}= \hat{Q}^{T}\hat{A}\hat{Q},
$$
can be written as
$$
\hat{T} = \left(\begin{array}{cccccc}
\alpha_1& \beta_1 & 0 &\dots & \dots &0 \\
\beta_1 & \alpha_2 & \beta_2 &0 &\dots &0 \\
0& \beta_2 & \alpha_3 & \beta_3 & \dots &0 \\
\dots& \dots & \dots &\dots &\dots & 0 \\
\dots& & &\beta_{n-2} &\alpha_{n-1}& \beta_{n-1} \\
0& \dots &\dots &0 &\beta_{n-1} & \alpha_{n} \\
\end{array} \right)
$$
Using the fact that
$$
\hat{Q}\hat{Q}^T=\hat{I},
$$
we can rewrite
$$
\hat{T}= \hat{Q}^{T}\hat{A}\hat{Q},
$$
as
$$
\hat{Q}\hat{T}= \hat{A}\hat{Q}.
$$
If we equate columns
$$
\hat{T} = \left(\begin{array}{cccccc}
\alpha_1& \beta_1 & 0 &\dots & \dots &0 \\
\beta_1 & \alpha_2 & \beta_2 &0 &\dots &0 \\
0& \beta_2 & \alpha_3 & \beta_3 & \dots &0 \\
\dots& \dots & \dots &\dots &\dots & 0 \\
\dots& & &\beta_{n-2} &\alpha_{n-1}& \beta_{n-1} \\
0& \dots &\dots &0 &\beta_{n-1} & \alpha_{n} \\
\end{array} \right)
$$
we obtain
$$
\hat{A}\hat{q}_k=\beta_{k-1}\hat{q}_{k-1}+\alpha_k\hat{q}_k+\beta_k\hat{q}_{k+1}.
$$
We have thus
$$
\hat{A}\hat{q}_k=\beta_{k-1}\hat{q}_{k-1}+\alpha_k\hat{q}_k+\beta_k\hat{q}_{k+1},
$$
with \( \beta_0\hat{q}_0=0 \) for \( k=1:n-1 \). Remember that the vectors \( \hat{q}_k \) are orthornormal and this implies
$$
\alpha_k=\hat{q}_k^T\hat{A}\hat{q}_k,
$$
and these vectors are called Lanczos vectors.
We have thus
$$
\hat{A}\hat{q}_k=\beta_{k-1}\hat{q}_{k-1}+\alpha_k\hat{q}_k+\beta_k\hat{q}_{k+1},
$$
with \( \beta_0\hat{q}_0=0 \) for \( k=1:n-1 \) and
$$
\alpha_k=\hat{q}_k^T\hat{A}\hat{q}_k.
$$
If
$$
\hat{r}_k=(\hat{A}-\alpha_k\hat{I})\hat{q}_k-\beta_{k-1}\hat{q}_{k-1},
$$
is non-zero, then
$$
\hat{q}_{k+1}=\hat{r}_{k}/\beta_k,
$$
with \( \beta_k=\pm ||\hat{r}_{k}||_2 \).