Second quantization is a useful and elegant formalism for constructing many-body states and quantum mechanical operators. One can express and translate many physical processes into simple pictures such as Feynman diagrams. Expecation values of many-body states are also easily calculated.
However, although the equations are seemingly easy to set up, from a practical point of view, that is the solution of Schroedinger's equation, there is no particular gain. The many-body equation is equally hard to solve, irrespective of representation. The cliche that there is no free lunch brings us down to earth again. Note however that a transformation to a particular basis, for cases where the interaction obeys specific symmetries, can ease the solution of Schroedinger's equation.
But there is at least one important case where second quantization comes to our rescue. It is namely easy to introduce another reference state than the pure vacuum \( |0\rangle \), where all single-particle states are active. With many particles present it is often useful to introduce another reference state than the vacuum state$|0\rangle $. We will label this state \( |c\rangle \) (\( c \) for core) and as we will see it can reduce considerably the complexity and thereby the dimensionality of the many-body problem. It allows us to sum up to infinite order specific many-body correlations. The particle-hole representation is one of these handy representations.
In the original particle representation these states are products of the creation operators \( a_{\alpha_i}^\dagger \) acting on the true vacuum \( |0\rangle \). We have
$$ \begin{align} |\alpha_1\alpha_2\dots\alpha_{n-1}\alpha_n\rangle &=& a_{\alpha_1}^\dagger a_{\alpha_2}^\dagger \dots a_{\alpha_{n-1}}^\dagger a_{\alpha_n}^\dagger |0\rangle \label{eq:2-47a} \\ |\alpha_1\alpha_2\dots\alpha_{n-1}\alpha_n\alpha_{n+1}\rangle &=& a_{\alpha_1}^\dagger a_{\alpha_2}^\dagger \dots a_{\alpha_{n-1}}^\dagger a_{\alpha_n}^\dagger a_{\alpha_{n+1}}^\dagger |0\rangle \label{eq:2-47b} \\ |\alpha_1\alpha_2\dots\alpha_{n-1}\rangle &=& a_{\alpha_1}^\dagger a_{\alpha_2}^\dagger \dots a_{\alpha_{n-1}}^\dagger |0\rangle \label{eq:2-47c} \end{align} $$If we use Eq. \eqref{eq:2-47a} as our new reference state, we can simplify considerably the representation of this state
$$ \begin{equation} |c\rangle \equiv |\alpha_1\alpha_2\dots\alpha_{n-1}\alpha_n\rangle = a_{\alpha_1}^\dagger a_{\alpha_2}^\dagger \dots a_{\alpha_{n-1}}^\dagger a_{\alpha_n}^\dagger |0\rangle \label{eq:2-48a} \end{equation} $$The new reference states for the \( n+1 \) and \( n-1 \) states can then be written as
$$ \begin{align} |\alpha_1\alpha_2\dots\alpha_{n-1}\alpha_n\alpha_{n+1}\rangle &=& (-1)^n a_{\alpha_{n+1}}^\dagger |c\rangle \equiv (-1)^n |\alpha_{n+1}\rangle_c \label{eq:2-48b} \\ |\alpha_1\alpha_2\dots\alpha_{n-1}\rangle &=& (-1)^{n-1} a_{\alpha_n} |c\rangle \equiv (-1)^{n-1} |\alpha_{n-1}\rangle_c \label{eq:2-48c} \end{align} $$The first state has one additional particle with respect to the new vacuum state \( |c\rangle \) and is normally referred to as a one-particle state or one particle added to the many-body reference state. The second state has one particle less than the reference vacuum state \( |c\rangle \) and is referred to as a one-hole state. When dealing with a new reference state it is often convenient to introduce new creation and annihilation operators since we have from Eq. \eqref{eq:2-48c}
$$ \begin{equation} a_\alpha |c\rangle \neq 0 \label{eq:2-49} \end{equation} $$since \( \alpha \) is contained in \( |c\rangle \), while for the true vacuum we have \( a_\alpha |0\rangle = 0 \) for all \( \alpha \).
The new reference state leads to the definition of new creation and annihilation operators which satisfy the following relations
$$ \begin{align} b_\alpha |c\rangle &=& 0 \label{eq:2-50a} \\ \{b_\alpha^\dagger , b_\beta^\dagger \} = \{b_\alpha , b_\beta \} &=& 0 \nonumber \\ \{b_\alpha^\dagger , b_\beta \} &=& \delta_{\alpha \beta} \label{eq:2-50c} \end{align} $$We assume also that the new reference state is properly normalized
$$ \begin{equation} \langle c | c \rangle = 1 \label{eq:2-51} \end{equation} $$The physical interpretation of these new operators is that of so-called quasiparticle states. This means that a state defined by the addition of one extra particle to a reference state \( |c\rangle \) may not necesseraly be interpreted as one particle coupled to a core. We define now new creation operators that act on a state \( \alpha \) creating a new quasiparticle state
$$ \begin{equation} b_\alpha^\dagger|c\rangle = \Bigg\{ \begin{array}{ll} a_\alpha^\dagger |c\rangle = |\alpha\rangle, & \alpha > F \\ \\ a_\alpha |c\rangle = |\alpha^{-1}\rangle, & \alpha \leq F \end{array} \label{eq:2-52} \end{equation} $$where \( F \) is the Fermi level representing the last occupied single-particle orbit of the new reference state \( |c\rangle \).
The annihilation is the hermitian conjugate of the creation operator
$$ b_\alpha = (b_\alpha^\dagger)^\dagger, $$resulting in
$$ \begin{equation} b_\alpha^\dagger = \Bigg\{ \begin{array}{ll} a_\alpha^\dagger & \alpha > F \\ \\ a_\alpha & \alpha \leq F \end{array} \qquad b_\alpha = \Bigg\{ \begin{array}{ll} a_\alpha & \alpha > F \\ \\ a_\alpha^\dagger & \alpha \leq F \end{array} \label{eq:2-54} \end{equation} $$With the new creation and annihilation operator we can now construct many-body quasiparticle states, with one-particle-one-hole states, two-particle-two-hole states etc in the same fashion as we previously constructed many-particle states. We can write a general particle-hole state as
$$ \begin{equation} |\beta_1\beta_2\dots \beta_{n_p} \gamma_1^{-1} \gamma_2^{-1} \dots \gamma_{n_h}^{-1}\rangle \equiv \underbrace{b_{\beta_1}^\dagger b_{\beta_2}^\dagger \dots b_{\beta_{n_p}}^\dagger}_{>F} \underbrace{b_{\gamma_1}^\dagger b_{\gamma_2}^\dagger \dots b_{\gamma_{n_h}}^\dagger}_{\leq F} |c\rangle \label{eq:2-56} \end{equation} $$We can now rewrite our one-body and two-body operators in terms of the new creation and annihilation operators.
The number operator becomes
$$ \begin{equation} \hat{N} = \sum_\alpha a_\alpha^\dagger a_\alpha= \sum_{\alpha > F} b_\alpha^\dagger b_\alpha + n_c - \sum_{\alpha \leq F} b_\alpha^\dagger b_\alpha \label{eq:2-57b} \end{equation} $$where \( n_c \) is the number of particle in the new vacuum state \( |c\rangle \). The action of \( \hat{N} \) on a many-body state results in
$$ \begin{equation} N |\beta_1\beta_2\dots \beta_{n_p} \gamma_1^{-1} \gamma_2^{-1} \dots \gamma_{n_h}^{-1}\rangle = (n_p + n_c - n_h) |\beta_1\beta_2\dots \beta_{n_p} \gamma_1^{-1} \gamma_2^{-1} \dots \gamma_{n_h}^{-1}\rangle \label{2-59} \end{equation} $$Here \( n=n_p +n_c - n_h \) is the total number of particles in the quasi-particle state of Eq. \eqref{eq:2-56}. Note that \( \hat{N} \) counts the total number of particles present
$$ \begin{equation} N_{qp} = \sum_\alpha b_\alpha^\dagger b_\alpha, \label{eq:2-60} \end{equation} $$gives us the number of quasi-particles as can be seen by computing
$$ \begin{equation} N_{qp}= |\beta_1\beta_2\dots \beta_{n_p} \gamma_1^{-1} \gamma_2^{-1} \dots \gamma_{n_h}^{-1}\rangle = (n_p + n_h)|\beta_1\beta_2\dots \beta_{n_p} \gamma_1^{-1} \gamma_2^{-1} \dots \gamma_{n_h}^{-1}\rangle \label{eq:2-61} \end{equation} $$where \( n_{qp} = n_p + n_h \) is the total number of quasi-particles.
We express the one-body operator \( \hat{H}_0 \) in terms of the quasi-particle creation and annihilation operators, resulting in
$$ \begin{align} \hat{H}_0 &=& \sum_{\alpha\beta > F} \langle \alpha|\hat{h}_0|\beta\rangle b_\alpha^\dagger b_\beta + \sum_{\alpha > F, \beta \leq F } \left[\langle \alpha|\hat{h}_0|\beta\rangle b_\alpha^\dagger b_\beta^\dagger + \langle \beta|\hat{h}_0|\alpha\rangle b_\beta b_\alpha \right] \nonumber \\ &+& \sum_{\alpha \leq F} \langle \alpha|\hat{h}_0|\alpha\rangle - \sum_{\alpha\beta \leq F} \langle \beta|\hat{h}_0|\alpha\rangle b_\alpha^\dagger b_\beta \label{eq:2-63b} \end{align} $$The first term gives contribution only for particle states, while the last one contributes only for holestates. The second term can create or destroy a set of quasi-particles and the third term is the contribution from the vacuum state \( |c\rangle \).
Before we continue with the expressions for the two-body operator, we introduce a nomenclature we will use for the rest of this text. It is inspired by the notation used in quantum chemistry. We reserve the labels \( i,j,k,\dots \) for hole states and \( a,b,c,\dots \) for states above \( F \), viz. particle states. This means also that we will skip the constraint \( \leq F \) or \( > F \) in the summation symbols. Our operator \( \hat{H}_0 \) reads now
$$ \begin{align} \hat{H}_0 &=& \sum_{ab} \langle a|\hat{h}|b\rangle b_a^\dagger b_b + \sum_{ai} \left[ \langle a|\hat{h}|i\rangle b_a^\dagger b_i^\dagger + \langle i|\hat{h}|a\rangle b_i b_a \right] \nonumber \\ &+& \sum_{i} \langle i|\hat{h}|i\rangle - \sum_{ij} \langle j|\hat{h}|i\rangle b_i^\dagger b_j \label{eq:2-63c} \end{align} $$The two-particle operator in the particle-hole formalism is more complicated since we have to translate four indices \( \alpha\beta\gamma\delta \) to the possible combinations of particle and hole states. When performing the commutator algebra we can regroup the operator in five different terms
$$ \begin{equation} \hat{H}_I = \hat{H}_I^{(a)} + \hat{H}_I^{(b)} + \hat{H}_I^{(c)} + \hat{H}_I^{(d)} + \hat{H}_I^{(e)} \label{eq:2-65} \end{equation} $$Using anti-symmetrized matrix elements, bthe term \( \hat{H}_I^{(a)} \) is
$$ \begin{equation} \hat{H}_I^{(a)} = \frac{1}{4} \sum_{abcd} \langle ab|\hat{V}|cd\rangle b_a^\dagger b_b^\dagger b_d b_c \label{eq:2-66} \end{equation} $$The next term \( \hat{H}_I^{(b)} \) reads
$$ \begin{equation} \hat{H}_I^{(b)} = \frac{1}{4} \sum_{abci}\left(\langle ab|\hat{V}|ci\rangle b_a^\dagger b_b^\dagger b_i^\dagger b_c +\langle ai|\hat{V}|cb\rangle b_a^\dagger b_i b_b b_c\right) \label{eq:2-67b} \end{equation} $$This term conserves the number of quasiparticles but creates or removes a three-particle-one-hole state. For \( \hat{H}_I^{(c)} \) we have
$$ \begin{align} \hat{H}_I^{(c)}& =& \frac{1}{4} \sum_{abij}\left(\langle ab|\hat{V}|ij\rangle b_a^\dagger b_b^\dagger b_j^\dagger b_i^\dagger + \langle ij|\hat{V}|ab\rangle b_a b_b b_j b_i \right)+ \nonumber \\ && \frac{1}{2}\sum_{abij}\langle ai|\hat{V}|bj\rangle b_a^\dagger b_j^\dagger b_b b_i + \frac{1}{2}\sum_{abi}\langle ai|\hat{V}|bi\rangle b_a^\dagger b_b. \label{eq:2-68c} \end{align} $$The first line stands for the creation of a two-particle-two-hole state, while the second line represents the creation to two one-particle-one-hole pairs while the last term represents a contribution to the particle single-particle energy from the hole states, that is an interaction between the particle states and the hole states within the new vacuum state.
The fourth term reads
$$ \begin{align} \hat{H}_I^{(d)}& = &\frac{1}{4} \sum_{aijk}\left(\langle ai|\hat{V}|jk\rangle b_a^\dagger b_k^\dagger b_j^\dagger b_i+ \langle ji|\hat{V}|ak\rangle b_k^\dagger b_j b_i b_a\right)+\nonumber \\ &&\frac{1}{4}\sum_{aij}\left(\langle ai|\hat{V}|ji\rangle b_a^\dagger b_j^\dagger+ \langle ji|\hat{V}|ai\rangle - \langle ji|\hat{V}|ia\rangle b_j b_a \right). \label{eq:2-69d} \end{align} $$The terms in the first line stand for the creation of a particle-hole state interacting with hole states, we will label this as a two-hole-one-particle contribution. The remaining terms are a particle-hole state interacting with the holes in the vacuum state. Finally we have
$$ \begin{equation} \hat{H}_I^{(e)} = \frac{1}{4} \sum_{ijkl} \langle kl|\hat{V}|ij\rangle b_i^\dagger b_j^\dagger b_l b_k+ \frac{1}{2}\sum_{ijk}\langle ij|\hat{V}|kj\rangle b_k^\dagger b_i +\frac{1}{2}\sum_{ij}\langle ij|\hat{V}|ij\rangle \label{eq:2-70d} \end{equation} $$The first terms represents the interaction between two holes while the second stands for the interaction between a hole and the remaining holes in the vacuum state. It represents a contribution to single-hole energy to first order. The last term collects all contributions to the energy of the ground state of a closed-shell system arising from hole-hole correlations.
which is equivalent with \( |\alpha_1 \dots \alpha_N\rangle= a_{\alpha_1}^{\dagger} \dots a_{\alpha_N}^{\dagger} |0\rangle \). We have also
$$ a_p^\dagger|0\rangle = |p\rangle, \quad a_p |q\rangle = \delta_{pq}|0\rangle $$ $$ \delta_{pq} = \left\{a_p, a_q^\dagger \right\}, $$and
$$ 0 = \left\{a_p^\dagger, a_q \right\} = \left\{a_p, a_q \right\} = \left\{a_p^\dagger, a_q^\dagger \right\} $$ $$ |\Phi_0\rangle = |\alpha_1 \dots \alpha_N\rangle, \quad \alpha_1, \dots, \alpha_N \leq \alpha_F $$with \( i,j,\ldots \leq \alpha_F, \quad a,b,\ldots > \alpha_F, \quad p,q, \ldots - \textrm{any} \)
$$ a_i|\Phi_0\rangle = |\Phi_i\rangle, \hspace{0.5cm} a_a^\dagger|\Phi_0\rangle = |\Phi^a\rangle $$and
$$ a_i^\dagger|\Phi_0\rangle = 0 \hspace{0.5cm} a_a|\Phi_0\rangle = 0 $$The one-body operator is defined as
$$ \hat{F} = \sum_{pq} \langle p|\hat{f}|q\rangle a_p^\dagger a_q $$while the two-body opreator is defined as
$$ \hat{V} = \frac{1}{4} \sum_{pqrs} \langle pq|\hat{v}|rs\rangle_{AS} a_p^\dagger a_q^\dagger a_s a_r $$where we have defined the antisymmetric matrix elements
$$ \langle pq|\hat{v}|rs\rangle_{AS} = \langle pq|\hat{v}|rs\rangle - \langle pq|\hat{v}|sr\rangle. $$We can also define a three-body operator
$$ \hat{V}_3 = \frac{1}{36} \sum_{pqrstu} \langle pqr|\hat{v}_3|stu\rangle_{AS} a_p^\dagger a_q^\dagger a_r^\dagger a_u a_t a_s $$with the antisymmetrized matrix element
$$ \begin{align} \langle pqr|\hat{v}_3|stu\rangle_{AS} = \langle pqr|\hat{v}_3|stu\rangle + \langle pqr|\hat{v}_3|tus\rangle + \langle pqr|\hat{v}_3|ust\rangle- \langle pqr|\hat{v}_3|sut\rangle - \langle pqr|\hat{v}_3|tsu\rangle - \langle pqr|\hat{v}_3|uts\rangle. \label{_auto1} \end{align} $$For a spin-\( 1/2 \) one-dimensional quantum spin-chain a fermionization procedure exists which allows the mapping between spin operators and fermionic creation-annihilation operators. The algebra governing the spin chain is the \( SU(2) \) algebra, represented by the \( \sigma \)-matrices. The Jordan-Wigner transformation is a transformation from fermionic annihilation and creation operators to the \( \sigma \)-matrices of a spin-\( 1/2 \) chain.
There is an isomorphism between the two systems, meaning that any \( a \) or \( a^\dag \) operator can be transformed into a tensor product of \( \sigma \)-matrices operating on a set of qubits. The authors demonstrated, with an emphasis on single-particle fermionic operators, that the Jordan-Wigner transformation ensures efficient, i.e., not exponential complexity, simulations of a fermionic system on a quantum computer. Similar transformations must be found for other systems, in order to efficiently simulate many-body systems.
We present here the various ingredients needed in order to transform a given Hamiltonian into a practical form suitable for quantum mechanical simulations.
We begin with the fermionic creation and annihilation operators, which satisfy the following anticommutation relations
$$ \begin{equation} \label{eq:anticommutationrelations} \{a_k, a_l\}=\{a_k^\dag, a_l^\dag\}= 0, \quad \{a_k^\dag, a_l\} = \delta_{kl}. \end{equation} $$Thereafter we define the three traceless and Hermitian generators of the \( SU(2) \) group, the \( \sigma \)-matrices \( \sigma_x \), \( \sigma_y \) and \( \sigma_z \). Together with the identity matrix \( {\bf 1} \) they form a complete basis for all Hermitian \( 2\times2 \) matrices. They can be used to write all Hamiltonians on a spin \( 1/2 \) chain when taking sums of tensor products of these, in other words they form a product basis for the operators on the qubits.
The three \( \sigma \)-matrices are
$$ \begin{equation} \sigma_x = \begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}, \sigma_y = \begin{pmatrix} 0 & -i\\ i & 0 \end{pmatrix}, \quad \sigma_z = \begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix}. \label{_auto2} \end{equation} $$We define the raising and lowering matrices as
$$ \sigma_+ = \frac{1}{2}(\sigma_x + i\sigma_y)= \begin{pmatrix} 0 & 1\\ 0 & 0 \end{pmatrix}, $$ $$ \begin{equation} \label{eq:raisingAndLowerin} \sigma_- = \frac{1}{2}(\sigma_x - i\sigma_y) =\begin{pmatrix} 0 & 0\\ 1 & 0 \end{pmatrix}. \end{equation} $$The transformation is based on the fact that for each possible quantum state of the fermion system, there can be either one or zero fermions. Therefore we need \( n \) qubits for a system with \( n \) possible fermion states. A qubit in state $ |0\rangle ^i=a^\dag_i|vacuum\rangle$ represents a state with a fermion, while $ |1\rangle ^i=|vacuum\rangle$ represents no fermions. Then the raising operator \( \sigma_+ \) changes $ |1\rangle $ into $ |0\rangle $ when
$$ \begin{equation} |0\rangle \equiv \begin{pmatrix} 1 \\ 0 \end{pmatrix}, \quad |1\rangle \equiv \begin{pmatrix} 0 \\ 1 \end{pmatrix}. \label{_auto3} \end{equation} $$This means that \( \sigma_+ \) acts as a creation operator, and \( \sigma_- \) acts as an annihilation operator. In addition, because of the anticommutation of creation(annihilation) operators for different states we have \( a_1^\dag a_2^\dag |vacuum\rangle = - a_2^\dag a_1^\dag |vacuum\rangle \), meaning that for creation and annihilation operators for states higher than the state corresponding to the first qubit, we need to multiply with a \( \sigma_z \)-matrix on all the qubits leading up to the one in question, in order to get the correct sign in the final operation. This leads us to the Jordan-Wigner transformation
$$ \begin{equation} \label{eq:JWtransformation} a^\dag_n = \left(\prod_{k=1}^{n-1} \sigma_z^k\right) \sigma_+^n, \quad a_n = \left(\prod_{k=1}^{n-1} \sigma_z^k\right) \sigma_-^n. \end{equation} $$The notation \( \sigma_z^i\sigma_+^j \) means a tensor product of the identity matrix on all qubits other than \( i \) and \( j \), \( {\bf 1}\otimes \sigma_z \otimes {\bf 1} \otimes \sigma_+\otimes{\bf 1} \), if \( i < j \), with \( {\bf 1} \) being the identity matrices of appropriate dimension.
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.
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
$$ \langle \Psi_0 | \Phi_0 \rangle = \sum_{PH}|C_H^P|^2= 1, $$and the energy can be written as
$$ E= \langle \Psi_0 | \hat{H} |\Psi_0 \rangle= \sum_{PP'HH'}C_H^{*P}\langle \Phi_H^P | \hat{H} |\Phi_{H'}^{P'} \rangle C_{H'}^{P'}. $$Normally
$$ E= \langle \Psi_0 | \hat{H} |\Psi_0 \rangle= \sum_{PP'HH'}C_H^{*P}\langle \Phi_H^P | \hat{H} |\Phi_{H'}^{P'} \rangle C_{H'}^{P'}, $$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
$$ \langle \Psi_0 | \hat{H} |\Psi_0 \rangle-\lambda \langle \Psi_0 |\Psi_0 \rangle, $$where \( \lambda \) is a variational multiplier to be identified with the energy of the system.
The minimization process results in
$$ \delta\left[ \langle \Psi_0 | \hat{H} |\Psi_0 \rangle-\lambda \langle \Psi_0 |\Psi_0 \rangle\right]=0, $$and 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. Varying the latter coefficients we have then
$$ \sum_{P'H'}\left\{\delta[C_H^{*P}]\langle \Phi_H^P | \hat{H} |\Phi_{H'}^{P'} \rangle C_{H'}^{P'}- \lambda( \delta[C_H^{*P}]C_{H'}^{P'}]\right\} = 0. $$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 \( \lambda = E \).
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 \( \hat{C} \).
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
$$ \langle \Phi_H^P | = \langle \Phi_0|, $$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, 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
$$ \langle \Phi_0 | \hat{H} -E| \Phi_0\rangle + \sum_{ai}\langle \Phi_0 | \hat{H} -E|\Phi_{i}^{a} \rangle C_{i}^{a}+ \sum_{abij}\langle \Phi_0 | \hat{H} -E|\Phi_{ij}^{ab} \rangle C_{ij}^{ab}=0, $$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
$$ \langle \Phi_H^P | = \langle \Phi_0|, $$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
$$ \langle \Phi_0 | \hat{H} -E| \Phi_0\rangle + \sum_{ai}\langle \Phi_0 | \hat{H} -E|\Phi_{i}^{a} \rangle C_{i}^{a}+ \sum_{abij}\langle \Phi_0 | \hat{H} -E|\Phi_{ij}^{ab} \rangle C_{ij}^{ab}=0. $$Which we can rewrite
$$ E-E_0 =\Delta E=\sum_{ai}\langle \Phi_0 | \hat{H}|\Phi_{i}^{a} \rangle C_{i}^{a}+ \sum_{abij}\langle \Phi_0 | \hat{H}|\Phi_{ij}^{ab} \rangle C_{ij}^{ab}, $$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 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
$$ E-E_0 =\Delta E^{HF}=\sum_{abij}\langle \Phi_0 | \hat{H}|\Phi_{ij}^{ab} \rangle C_{ij}^{ab}. $$
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
$$ \langle \Phi_i^a | \hat{H} -E| \Phi_0\rangle + \sum_{bj}\langle \Phi_i^a | \hat{H} -E|\Phi_{j}^{b} \rangle C_{j}^{b}+ \sum_{bcjk}\langle \Phi_i^a | \hat{H} -E|\Phi_{jk}^{bc} \rangle C_{jk}^{bc}+ \sum_{bcdjkl}\langle \Phi_i^a | \hat{H} -E|\Phi_{jkl}^{bcd} \rangle C_{jkl}^{bcd}=0. $$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
$$ E_{ref}=\langle \Phi_0 \vert \hat{H} \vert \Phi_0 \rangle. $$