We start with a reminder on determinants in the number representation.
- - -The simplest possible choice for many-body wavefunctions are product wavefunctions. -That is -
-$$ -\Psi(x_1, x_2, x_3, \ldots, x_N) \approx \phi_1(x_1) \phi_2(x_2) \phi_3(x_3) \ldots -$$ - -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.
-$$ -\Psi(x_1, x_2, x_3, \ldots, x_N) = - \Psi(x_2, x_1, x_3, \ldots, x_N) -$$ - -etc. This is accomplished formally by using the determinantal formalism
-$$ -\Psi(x_1, x_2, \ldots, x_N) -= \frac{1}{\sqrt{N!}} -\det \left | -\begin{array}{cccc} -\phi_1(x_1) & \phi_1(x_2) & \ldots & \phi_1(x_N) \\ -\phi_2(x_1) & \phi_2(x_2) & \ldots & \phi_2(x_N) \\ - \vdots & & & \\ -\phi_N(x_1) & \phi_N(x_2) & \ldots & \phi_N(x_N) -\end{array} -\right | -$$ - -Product wavefunction + antisymmetry = Slater determinant.
-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 or number 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, -
+We have defined the ansatz for the ground state as
$$ -\hat{a}^\dagger_i \hat{a}^\dagger_j \hat{a}^\dagger_k \ldots |0 \rangle +|\Phi_0\rangle = \left(\prod_{i\le F}\hat{a}_{i}^{\dagger}\right)|0\rangle, $$ -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.
-where the index \( i \) defines different single-particle states up to the Fermi level. We have assumed that we have \( N \) fermions.
-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 -
+A given one-particle-one-hole (\( 1p1h \)) state can be written as
$$ |\Phi_i^a\rangle = \hat{a}_{a}^{\dagger}\hat{a}_i|\Phi_0\rangle, $$ @@ -1156,15 +957,11 @@We can then expand our exact state function for the ground state as
@@ -1178,6 +975,9 @@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
@@ -1189,15 +989,11 @@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, @@ -1208,9 +1004,12 @@where \( H \) stands for \( 0,1,\dots,n \) hole states and \( P \) for \( 0,1,\dots,n \) particle states. -We have introduced the operator \( \hat{A}_H^P \) which contains an equal number of creation and annihilation operators. -
+where \( H \) stands for \( 0,1,\dots,n \) hole states and \( P \) for \( 0,1,\dots,n \) particle states.
+ + +We have introduced the operator \( \hat{A}_H^P \) which contains an equal number of creation and annihilation operators.
Our requirement of unit normalization gives
$$ @@ -1221,15 +1020,11 @@Normally
$$ E= \langle \Psi_0 | \hat{H} |\Psi_0 \rangle= \sum_{PP'HH'}C_H^{*P}\langle \Phi_H^P | \hat{H} |\Phi_{H'}^{P'} \rangle C_{H'}^{P'}, @@ -1242,9 +1037,12 @@where \( \lambda \) is a variational multiplier to be identified with the energy of the system. -The minimization process results in -
+where \( \lambda \) is a variational multiplier to be identified with the energy of the system.
+ + +The minimization process results in
$$ \delta\left[ \langle \Psi_0 | \hat{H} |\Psi_0 \rangle-\lambda \langle \Psi_0 |\Psi_0 \rangle\right]=0, $$ @@ -1254,16 +1052,10 @@This leads to
$$ \sum_{P'H'}\langle \Phi_H^P | \hat{H} |\Phi_{H'}^{P'} \rangle C_{H'}^{P'}-\lambda C_H^{P}=0, @@ -1277,15 +1069,10 @@leading to the identification \( \lambda = E \).
-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, @@ -1297,15 +1084,10 @@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
@@ -1319,9 +1101,6 @@and multiplying this with the number of proton Slater determinants we end up with approximately with a dimensionality \( d \) of \( d\sim 10^{18} \).
-To see this, we look at the contributions arising from
$$ \langle \Phi_H^P | = \langle \Phi_0|, @@ -1355,7 +1132,10 @@If we assume that we have a two-body operator at most, Slater's rule gives then an equation for the + + +
If we assume that we have a two-body operator at most, using the Condon-Slater rule gives then an equation for the correlation energy in terms of \( C_i^a \) and \( C_{ij}^{ab} \) only. We get then
$$ @@ -1372,15 +1152,10 @@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|, @@ -1392,24 +1167,24 @@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
$$ \langle \Phi_0 | \hat{H} -E| \Phi_0\rangle + \sum_{ai}\langle \Phi_0 | \hat{H} -E|\Phi_{i}^{a} \rangle C_{i}^{a}+ -\sum_{abij}\langle \Phi_0 | \hat{H} -E|\Phi_{ij}^{ab} \rangle C_{ij}^{ab}=0, +\sum_{abij}\langle \Phi_0 | \hat{H} -E|\Phi_{ij}^{ab} \rangle C_{ij}^{ab}=0. $$ -or
+ + +Which we can rewrite
$$ 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}, @@ -1418,17 +1193,14 @@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 +
In our discussions of the Hartree-Fock method planned for week 39, +we are going to compute the elements \( \langle \Phi_0 | \hat{H}|\Phi_{i}^{a}\rangle \) and \( \langle \Phi_0 | \hat{H}|\Phi_{ij}^{ab}\rangle \). If we are using a Hartree-Fock basis, then these quantities result in \( \langle \Phi_0 | \hat{H}|\Phi_{i}^{a}\rangle=0 \) and we are left with a correlation energy given by
$$ @@ -1456,28 +1228,27 @@We need more equations. Our next step is to set up
$$ \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, +\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
+ + +This equation will allow us to find an expression for the coefficents \( C_i^a \) since we can rewrite this equation as
$$ \langle i | \hat{f}| a\rangle +\langle \Phi_i^a | \hat{H}|\Phi_{i}^{a} \rangle C_{i}^{a}+ \sum_{bj\ne ai}\langle \Phi_i^a | \hat{H}|\Phi_{j}^{b} \rangle C_{j}^{b}+ \sum_{bcjk}\langle \Phi_i^a | \hat{H}|\Phi_{jk}^{bc} \rangle C_{jk}^{bc}+ \sum_{bcdjkl}\langle \Phi_i^a | \hat{H}|\Phi_{jkl}^{bcd} \rangle C_{jkl}^{bcd}=EC_i^a. $$ -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 -
+The coefficients \( C \) result from the solution of the eigenvalue problem.
+ + +The energy of say the ground state is then
$$ E=E_{ref}+\Delta E, $$ @@ -1574,8 +1345,6 @@We start with a reminder on determinants in the number representation.
--
The simplest possible choice for many-body wavefunctions are product wavefunctions. -That is -
-
-$$
-\Psi(x_1, x_2, x_3, \ldots, x_N) \approx \phi_1(x_1) \phi_2(x_2) \phi_3(x_3) \ldots
-$$
-
-
-
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.
-
-$$
-\Psi(x_1, x_2, x_3, \ldots, x_N) = - \Psi(x_2, x_1, x_3, \ldots, x_N)
-$$
-
-
-
etc. This is accomplished formally by using the determinantal formalism
-
-$$
-\Psi(x_1, x_2, \ldots, x_N)
-= \frac{1}{\sqrt{N!}}
-\det \left |
-\begin{array}{cccc}
-\phi_1(x_1) & \phi_1(x_2) & \ldots & \phi_1(x_N) \\
-\phi_2(x_1) & \phi_2(x_2) & \ldots & \phi_2(x_N) \\
- \vdots & & & \\
-\phi_N(x_1) & \phi_N(x_2) & \ldots & \phi_N(x_N)
-\end{array}
-\right |
-$$
-
-
-
Product wavefunction + antisymmetry = Slater determinant.
--
-$$
-\Psi(x_1, x_2, \ldots, x_N)
-= \frac{1}{\sqrt{N!}}
-\det \left |
-\begin{array}{cccc}
-\phi_1(x_1) & \phi_1(x_2) & \ldots & \phi_1(x_N) \\
-\phi_2(x_1) & \phi_2(x_2) & \ldots & \phi_2(x_N) \\
- \vdots & & & \\
-\phi_N(x_1) & \phi_N(x_2) & \ldots & \phi_N(x_N)
-\end{array}
-\right |
-$$
-
-
-
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 or number 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, -
+We have defined the ansatz for the ground state as
$$
-\hat{a}^\dagger_i \hat{a}^\dagger_j \hat{a}^\dagger_k \ldots |0 \rangle
+|\Phi_0\rangle = \left(\prod_{i\le F}\hat{a}_{i}^{\dagger}\right)|0\rangle,
$$
-
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.
-where the index \( i \) defines different single-particle states up to the Fermi level. We have assumed that we have \( N \) fermions.
-
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 -
+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,
@@ -1134,14 +936,11 @@
-
+
We can then expand our exact state function for the ground state as
@@ -1158,7 +957,10 @@ 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
+
+Intermediate normalization
Full Configuration Interaction Th
|\Psi_0\rangle=(1+\hat{C})|\Phi_0\rangle.
$$
-
+
We rewrite
$$
@@ -1196,9 +995,13 @@
-
where \( H \) stands for \( 0,1,\dots,n \) hole states and \( P \) for \( 0,1,\dots,n \) particle states. -We have introduced the operator \( \hat{A}_H^P \) which contains an equal number of creation and annihilation operators. -
+where \( H \) stands for \( 0,1,\dots,n \) hole states and \( P \) for \( 0,1,\dots,n \) particle states.
+ + +We have introduced the operator \( \hat{A}_H^P \) which contains an equal number of creation and annihilation operators.
Our requirement of unit normalization gives
@@ -1213,14 +1016,11 @@
-
+
Normally
$$
@@ -1237,9 +1037,13 @@
-
where \( \lambda \) is a variational multiplier to be identified with the energy of the system. -The minimization process results in -
+where \( \lambda \) is a variational multiplier to be identified with the energy of the system.
+ + +The minimization process results in
$$
\delta\left[ \langle \Psi_0 | \hat{H} |\Psi_0 \rangle-\lambda \langle \Psi_0 |\Psi_0 \rangle\right]=0,
@@ -1253,15 +1057,10 @@
-
-
This leads to
$$
@@ -1279,14 +1078,11 @@
leading to the identification \( \lambda = E \).
-+
An alternative way to derive the last equation is to start from
$$
@@ -1300,14 +1096,11 @@
+
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
@@ -1325,7 +1118,6 @@
and multiplying this with the number of proton Slater determinants we end up with approximately with a dimensionality \( d \) of \( d\sim 10^{18} \).
-+
To see this, we look at the contributions arising from
$$
@@ -1362,8 +1152,11 @@
+
-
If we assume that we have a two-body operator at most, Slater's rule gives then an equation for the
+ If we assume that we have a two-body operator at most, using the Condon-Slater rule gives then an equation for the
correlation energy in terms of \( C_i^a \) and \( C_{ij}^{ab} \) only. We get then
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.
Using the Condon-Slater rule
+
@@ -1384,14 +1177,11 @@ A non-practical w
+
To see this, we look at the contributions arising from
$$
@@ -1407,25 +1197,26 @@
-
+
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
$$
\langle \Phi_0 | \hat{H} -E| \Phi_0\rangle + \sum_{ai}\langle \Phi_0 | \hat{H} -E|\Phi_{i}^{a} \rangle C_{i}^{a}+
-\sum_{abij}\langle \Phi_0 | \hat{H} -E|\Phi_{ij}^{ab} \rangle C_{ij}^{ab}=0,
+\sum_{abij}\langle \Phi_0 | \hat{H} -E|\Phi_{ij}^{ab} \rangle C_{ij}^{ab}=0.
$$
+
-
or
+Which we can rewrite
$$
E-E_0 =\Delta E=\sum_{ai}\langle \Phi_0 | \hat{H}|\Phi_{i}^{a} \rangle C_{i}^{a}+
@@ -1436,7 +1227,6 @@
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 +
In our discussions of the Hartree-Fock method planned for week 39, +we are going to compute the elements \( \langle \Phi_0 | \hat{H}|\Phi_{i}^{a}\rangle \) and \( \langle \Phi_0 | \hat{H}|\Phi_{ij}^{ab}\rangle \). If we are using a Hartree-Fock basis, then these quantities result in \( \langle \Phi_0 | \hat{H}|\Phi_{i}^{a}\rangle=0 \) and we are left with a correlation energy given by
@@ -1475,19 +1265,20 @@
+
We need more equations. Our next step is to set up
$$
\langle \Phi_i^a | \hat{H} -E| \Phi_0\rangle + \sum_{bj}\langle \Phi_i^a | \hat{H} -E|\Phi_{j}^{b} \rangle C_{j}^{b}+
\sum_{bcjk}\langle \Phi_i^a | \hat{H} -E|\Phi_{jk}^{bc} \rangle C_{jk}^{bc}+
-\sum_{bcdjkl}\langle \Phi_i^a | \hat{H} -E|\Phi_{jkl}^{bcd} \rangle C_{jkl}^{bcd}=0,
+\sum_{bcdjkl}\langle \Phi_i^a | \hat{H} -E|\Phi_{jkl}^{bcd} \rangle C_{jkl}^{bcd}=0.
$$
+
-
as this equation will allow us to find an expression for the coefficents \( C_i^a \) since we can rewrite this equation as
+This equation will allow us to find an expression for the coefficents \( C_i^a \) since we can rewrite this equation as
$$
\langle i | \hat{f}| a\rangle +\langle \Phi_i^a | \hat{H}|\Phi_{i}^{a} \rangle C_{i}^{a}+ \sum_{bj\ne ai}\langle \Phi_i^a | \hat{H}|\Phi_{j}^{b} \rangle C_{j}^{b}+
@@ -1495,11 +1286,10 @@
-
@@ -1580,9 +1370,7 @@
+
The correlation energy is defined as, with a two-body Hamiltonian,
$$
@@ -1591,9 +1379,12 @@
-
The coefficients \( C \) result from the solution of the eigenvalue problem. -The energy of say the ground state is then -
+The coefficients \( C \) result from the solution of the eigenvalue problem.
+ + +The energy of say the ground state is then
$$
E=E_{ref}+\Delta E,
@@ -1606,7 +1397,6 @@
-
We start with a reminder on determinants in the number representation.
- --
The simplest possible choice for many-body wavefunctions are product wavefunctions. -That is -
-$$ -\Psi(x_1, x_2, x_3, \ldots, x_N) \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.
-$$ -\Psi(x_1, x_2, x_3, \ldots, x_N) = - \Psi(x_2, x_1, x_3, \ldots, x_N) -$$ - -etc. This is accomplished formally by using the determinantal formalism
-$$ -\Psi(x_1, x_2, \ldots, x_N) -= \frac{1}{\sqrt{N!}} -\det \left | -\begin{array}{cccc} -\phi_1(x_1) & \phi_1(x_2) & \ldots & \phi_1(x_N) \\ -\phi_2(x_1) & \phi_2(x_2) & \ldots & \phi_2(x_N) \\ - \vdots & & & \\ -\phi_N(x_1) & \phi_N(x_2) & \ldots & \phi_N(x_N) -\end{array} -\right | -$$ - -Product wavefunction + antisymmetry = Slater determinant.
--$$ -\Psi(x_1, x_2, \ldots, x_N) -= \frac{1}{\sqrt{N!}} -\det \left | -\begin{array}{cccc} -\phi_1(x_1) & \phi_1(x_2) & \ldots & \phi_1(x_N) \\ -\phi_2(x_1) & \phi_2(x_2) & \ldots & \phi_2(x_N) \\ - \vdots & & & \\ -\phi_N(x_1) & \phi_N(x_2) & \ldots & \phi_N(x_N) -\end{array} -\right | -$$ - -
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 or number 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, -
+We have defined the ansatz for the ground state as
$$ -\hat{a}^\dagger_i \hat{a}^\dagger_j \hat{a}^\dagger_k \ldots |0 \rangle +|\Phi_0\rangle = \left(\prod_{i\le F}\hat{a}_{i}^{\dagger}\right)|0\rangle, $$ -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.
-where the index \( i \) defines different single-particle states up to the Fermi level. We have assumed that we have \( N \) fermions.
-
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 -
+A given one-particle-one-hole (\( 1p1h \)) state can be written as
$$ |\Phi_i^a\rangle = \hat{a}_{a}^{\dagger}\hat{a}_i|\Phi_0\rangle, $$ @@ -1092,14 +900,11 @@+
We can then expand our exact state function for the ground state as
@@ -1113,6 +918,9 @@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
@@ -1124,14 +932,11 @@+
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, @@ -1142,9 +947,12 @@where \( H \) stands for \( 0,1,\dots,n \) hole states and \( P \) for \( 0,1,\dots,n \) particle states. -We have introduced the operator \( \hat{A}_H^P \) which contains an equal number of creation and annihilation operators. -
+where \( H \) stands for \( 0,1,\dots,n \) hole states and \( P \) for \( 0,1,\dots,n \) particle states.
+ +We have introduced the operator \( \hat{A}_H^P \) which contains an equal number of creation and annihilation operators.
Our requirement of unit normalization gives
$$ @@ -1155,14 +963,11 @@+
Normally
$$ E= \langle \Psi_0 | \hat{H} |\Psi_0 \rangle= \sum_{PP'HH'}C_H^{*P}\langle \Phi_H^P | \hat{H} |\Phi_{H'}^{P'} \rangle C_{H'}^{P'}, @@ -1175,9 +980,12 @@where \( \lambda \) is a variational multiplier to be identified with the energy of the system. -The minimization process results in -
+where \( \lambda \) is a variational multiplier to be identified with the energy of the system.
+ +The minimization process results in
$$ \delta\left[ \langle \Psi_0 | \hat{H} |\Psi_0 \rangle-\lambda \langle \Psi_0 |\Psi_0 \rangle\right]=0, $$ @@ -1187,15 +995,10 @@-
This leads to
$$ \sum_{P'H'}\langle \Phi_H^P | \hat{H} |\Phi_{H'}^{P'} \rangle C_{H'}^{P'}-\lambda C_H^{P}=0, @@ -1209,14 +1012,10 @@leading to the identification \( \lambda = E \).
-+
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, @@ -1228,14 +1027,10 @@+
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
@@ -1249,8 +1044,6 @@and multiplying this with the number of proton Slater determinants we end up with approximately with a dimensionality \( d \) of \( d\sim 10^{18} \).
-+
To see this, we look at the contributions arising from
$$ \langle \Phi_H^P | = \langle \Phi_0|, @@ -1283,7 +1074,10 @@If we assume that we have a two-body operator at most, Slater's rule gives then an equation for the
+
+
+
If we assume that we have a two-body operator at most, using the Condon-Slater rule gives then an equation for the correlation energy in terms of \( C_i^a \) and \( C_{ij}^{ab} \) only. We get then
$$ @@ -1300,14 +1094,10 @@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|, @@ -1319,23 +1109,24 @@+
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
$$ \langle \Phi_0 | \hat{H} -E| \Phi_0\rangle + \sum_{ai}\langle \Phi_0 | \hat{H} -E|\Phi_{i}^{a} \rangle C_{i}^{a}+ -\sum_{abij}\langle \Phi_0 | \hat{H} -E|\Phi_{ij}^{ab} \rangle C_{ij}^{ab}=0, +\sum_{abij}\langle \Phi_0 | \hat{H} -E|\Phi_{ij}^{ab} \rangle C_{ij}^{ab}=0. $$ -or
+ +Which we can rewrite
$$ 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}, @@ -1344,16 +1135,14 @@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 +
In our discussions of the Hartree-Fock method planned for week 39, +we are going to compute the elements \( \langle \Phi_0 | \hat{H}|\Phi_{i}^{a}\rangle \) and \( \langle \Phi_0 | \hat{H}|\Phi_{ij}^{ab}\rangle \). If we are using a Hartree-Fock basis, then these quantities result in \( \langle \Phi_0 | \hat{H}|\Phi_{i}^{a}\rangle=0 \) and we are left with a correlation energy given by
$$ @@ -1379,27 +1168,27 @@+
We need more equations. Our next step is to set up
$$ \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, +\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
+ +This equation will allow us to find an expression for the coefficents \( C_i^a \) since we can rewrite this equation as
$$ \langle i | \hat{f}| a\rangle +\langle \Phi_i^a | \hat{H}|\Phi_{i}^{a} \rangle C_{i}^{a}+ \sum_{bj\ne ai}\langle \Phi_i^a | \hat{H}|\Phi_{j}^{b} \rangle C_{j}^{b}+ \sum_{bcjk}\langle \Phi_i^a | \hat{H}|\Phi_{jk}^{bc} \rangle C_{jk}^{bc}+ \sum_{bcdjkl}\langle \Phi_i^a | \hat{H}|\Phi_{jkl}^{bcd} \rangle C_{jkl}^{bcd}=EC_i^a. $$ -@@ -1472,18 +1261,18 @@
+
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 -
+The coefficients \( C \) result from the solution of the eigenvalue problem.
+ +The energy of say the ground state is then
$$ E=E_{ref}+\Delta E, $$ @@ -1492,7 +1281,6 @@We start with a reminder on determinants in the number representation.
- --
The simplest possible choice for many-body wavefunctions are product wavefunctions. -That is -
-$$ -\Psi(x_1, x_2, x_3, \ldots, x_N) \approx \phi_1(x_1) \phi_2(x_2) \phi_3(x_3) \ldots -$$ - -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.
-$$ -\Psi(x_1, x_2, x_3, \ldots, x_N) = - \Psi(x_2, x_1, x_3, \ldots, x_N) -$$ - -etc. This is accomplished formally by using the determinantal formalism
-$$ -\Psi(x_1, x_2, \ldots, x_N) -= \frac{1}{\sqrt{N!}} -\det \left | -\begin{array}{cccc} -\phi_1(x_1) & \phi_1(x_2) & \ldots & \phi_1(x_N) \\ -\phi_2(x_1) & \phi_2(x_2) & \ldots & \phi_2(x_N) \\ - \vdots & & & \\ -\phi_N(x_1) & \phi_N(x_2) & \ldots & \phi_N(x_N) -\end{array} -\right | -$$ - -Product wavefunction + antisymmetry = Slater determinant.
--$$ -\Psi(x_1, x_2, \ldots, x_N) -= \frac{1}{\sqrt{N!}} -\det \left | -\begin{array}{cccc} -\phi_1(x_1) & \phi_1(x_2) & \ldots & \phi_1(x_N) \\ -\phi_2(x_1) & \phi_2(x_2) & \ldots & \phi_2(x_N) \\ - \vdots & & & \\ -\phi_N(x_1) & \phi_N(x_2) & \ldots & \phi_N(x_N) -\end{array} -\right | -$$ - -
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 or number 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, -
+We have defined the ansatz for the ground state as
$$ -\hat{a}^\dagger_i \hat{a}^\dagger_j \hat{a}^\dagger_k \ldots |0 \rangle +|\Phi_0\rangle = \left(\prod_{i\le F}\hat{a}_{i}^{\dagger}\right)|0\rangle, $$ -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.
-where the index \( i \) defines different single-particle states up to the Fermi level. We have assumed that we have \( N \) fermions.
-
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 -
+A given one-particle-one-hole (\( 1p1h \)) state can be written as
$$ |\Phi_i^a\rangle = \hat{a}_{a}^{\dagger}\hat{a}_i|\Phi_0\rangle, $$ @@ -1169,14 +977,11 @@+
We can then expand our exact state function for the ground state as
@@ -1190,6 +995,9 @@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
@@ -1201,14 +1009,11 @@+
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, @@ -1219,9 +1024,12 @@where \( H \) stands for \( 0,1,\dots,n \) hole states and \( P \) for \( 0,1,\dots,n \) particle states. -We have introduced the operator \( \hat{A}_H^P \) which contains an equal number of creation and annihilation operators. -
+where \( H \) stands for \( 0,1,\dots,n \) hole states and \( P \) for \( 0,1,\dots,n \) particle states.
+ +We have introduced the operator \( \hat{A}_H^P \) which contains an equal number of creation and annihilation operators.
Our requirement of unit normalization gives
$$ @@ -1232,14 +1040,11 @@+
Normally
$$ E= \langle \Psi_0 | \hat{H} |\Psi_0 \rangle= \sum_{PP'HH'}C_H^{*P}\langle \Phi_H^P | \hat{H} |\Phi_{H'}^{P'} \rangle C_{H'}^{P'}, @@ -1252,9 +1057,12 @@where \( \lambda \) is a variational multiplier to be identified with the energy of the system. -The minimization process results in -
+where \( \lambda \) is a variational multiplier to be identified with the energy of the system.
+ +The minimization process results in
$$ \delta\left[ \langle \Psi_0 | \hat{H} |\Psi_0 \rangle-\lambda \langle \Psi_0 |\Psi_0 \rangle\right]=0, $$ @@ -1264,15 +1072,10 @@-
This leads to
$$ \sum_{P'H'}\langle \Phi_H^P | \hat{H} |\Phi_{H'}^{P'} \rangle C_{H'}^{P'}-\lambda C_H^{P}=0, @@ -1286,14 +1089,10 @@leading to the identification \( \lambda = E \).
-+
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, @@ -1305,14 +1104,10 @@+
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
@@ -1326,8 +1121,6 @@and multiplying this with the number of proton Slater determinants we end up with approximately with a dimensionality \( d \) of \( d\sim 10^{18} \).
-+
To see this, we look at the contributions arising from
$$ \langle \Phi_H^P | = \langle \Phi_0|, @@ -1360,7 +1151,10 @@If we assume that we have a two-body operator at most, Slater's rule gives then an equation for the
+
+
+
If we assume that we have a two-body operator at most, using the Condon-Slater rule gives then an equation for the correlation energy in terms of \( C_i^a \) and \( C_{ij}^{ab} \) only. We get then
$$ @@ -1377,14 +1171,10 @@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|, @@ -1396,23 +1186,24 @@+
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
$$ \langle \Phi_0 | \hat{H} -E| \Phi_0\rangle + \sum_{ai}\langle \Phi_0 | \hat{H} -E|\Phi_{i}^{a} \rangle C_{i}^{a}+ -\sum_{abij}\langle \Phi_0 | \hat{H} -E|\Phi_{ij}^{ab} \rangle C_{ij}^{ab}=0, +\sum_{abij}\langle \Phi_0 | \hat{H} -E|\Phi_{ij}^{ab} \rangle C_{ij}^{ab}=0. $$ -or
+ +Which we can rewrite
$$ 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}, @@ -1421,16 +1212,14 @@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 +
In our discussions of the Hartree-Fock method planned for week 39, +we are going to compute the elements \( \langle \Phi_0 | \hat{H}|\Phi_{i}^{a}\rangle \) and \( \langle \Phi_0 | \hat{H}|\Phi_{ij}^{ab}\rangle \). If we are using a Hartree-Fock basis, then these quantities result in \( \langle \Phi_0 | \hat{H}|\Phi_{i}^{a}\rangle=0 \) and we are left with a correlation energy given by
$$ @@ -1456,27 +1245,27 @@+
We need more equations. Our next step is to set up
$$ \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, +\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
+ +This equation will allow us to find an expression for the coefficents \( C_i^a \) since we can rewrite this equation as
$$ \langle i | \hat{f}| a\rangle +\langle \Phi_i^a | \hat{H}|\Phi_{i}^{a} \rangle C_{i}^{a}+ \sum_{bj\ne ai}\langle \Phi_i^a | \hat{H}|\Phi_{j}^{b} \rangle C_{j}^{b}+ \sum_{bcjk}\langle \Phi_i^a | \hat{H}|\Phi_{jk}^{bc} \rangle C_{jk}^{bc}+ \sum_{bcdjkl}\langle \Phi_i^a | \hat{H}|\Phi_{jkl}^{bcd} \rangle C_{jkl}^{bcd}=EC_i^a. $$ -@@ -1549,18 +1338,18 @@
+
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 -
+The coefficients \( C \) result from the solution of the eigenvalue problem.
+ +The energy of say the ground state is then
$$ E=E_{ref}+\Delta E, $$ @@ -1569,7 +1358,6 @@