Introduction to second quantization
Contents
2. Introduction to second quantization¶
We introduce the time-independent operators \(a_\alpha^{\dagger}\) and \(a_\alpha\) which create and annihilate, respectively, a particle in the single-particle state \(\varphi_\alpha\). We define the fermion creation operator \(a_\alpha^{\dagger}\)
and
In Eq. (1) the operator \(a_\alpha^{\dagger}\) acts on the vacuum state \(|0\rangle\), which does not contain any particles. Alternatively, we could define a closed-shell nucleus or atom as our new vacuum, but then we need to introduce the particle-hole formalism, see the discussion to come.
In Eq. (2) \(a_\alpha^{\dagger}\) acts on an antisymmetric \(n\)-particle state and creates an antisymmetric \((n+1)\)-particle state, where the one-body state \(\varphi_\alpha\) is occupied, under the condition that \(\alpha \ne \alpha_1, \alpha_2, \dots, \alpha_n\). It follows that we can express an antisymmetric state as the product of the creation operators acting on the vacuum state.
It is easy to derive the commutation and anticommutation rules for the fermionic creation operators \(a_\alpha^{\dagger}\). Using the antisymmetry of the states (3)
we obtain
Using the Pauli principle
it follows that
If we combine Eqs. (5) and (7), we obtain the well-known anti-commutation rule
The hermitian conjugate of \(a_\alpha^{\dagger}\) is
If we take the hermitian conjugate of Eq. (8), we arrive at
What is the physical interpretation of the operator \(a_\alpha\) and what is the effect of \(a_\alpha\) on a given state \(|\alpha_1\alpha_2\dots\alpha_n\rangle_{\mathrm{AS}}\)? Consider the following matrix element
where both sides are antisymmetric. We distinguish between two cases. The first (1) is when \(\alpha \in \{\alpha_i\}\). Using the Pauli principle of Eq. (6) it follows
The second (2) case is when \(\alpha \notin \{\alpha_i\}\). It follows that an hermitian conjugation
Eq. (13) holds for case (1) since the lefthand side is zero due to the Pauli principle. We write Eq. (11) as
Here we must have \(m = n+1\) if Eq. (14) has to be trivially different from zero.
For the last case, the minus and plus signs apply when the sequence \(\alpha ,\alpha_1, \alpha_2, \dots, \alpha_n\) and \(\alpha_1', \alpha_2', \dots, \alpha_{n+1}'\) are related to each other via even and odd permutations. If we assume that \(\alpha \notin \{\alpha_i\}\) we obtain
when \(\alpha \in \{\alpha_i'\}\). If \(\alpha \notin \{\alpha_i'\}\), we obtain
and in particular
If \(\{\alpha\alpha_i\} = \{\alpha_i'\}\), performing the right permutations, the sequence \(\alpha ,\alpha_1, \alpha_2, \dots, \alpha_n\) is identical with the sequence \(\alpha_1', \alpha_2', \dots, \alpha_{n+1}'\). This results in
and thus
The action of the operator \(a_\alpha\) from the left on a state vector is to to remove one particle in the state \(\alpha\). If the state vector does not contain the single-particle state \(\alpha\), the outcome of the operation is zero. The operator \(a_\alpha\) is normally called for a destruction or annihilation operator.
The next step is to establish the commutator algebra of \(a_\alpha^{\dagger}\) and \(a_\beta\).
The action of the anti-commutator \(\{a_\alpha^{\dagger}\),\(a_\alpha\}\) on a given \(n\)-particle state is
if the single-particle state \(\alpha\) is not contained in the state.
If it is present we arrive at
From Eqs. (20) and (21) we arrive at
The action of \(\left\{a_\alpha^{\dagger}, a_\beta\right\}\), with \(\alpha \ne \beta\) on a given state yields three possibilities. The first case is a state vector which contains both \(\alpha\) and \(\beta\), then either \(\alpha\) or \(\beta\) and finally none of them.
The first case results in
while the second case gives
Finally if the state vector does not contain \(\alpha\) and \(\beta\)
For all three cases we have
We can summarize our findings in Eqs. (22) and (26) as
with \(\delta_{\alpha\beta}\) is the Kroenecker \(\delta\)-symbol.
The properties of the creation and annihilation operators can be summarized as (for fermions)
and
from which follows
The hermitian conjugate has the folowing properties
Finally we found
and
and the corresponding commutator algebra
2.1. One-body operators in second quantization¶
A very useful operator is the so-called number-operator. Most physics cases we will study in this text conserve the total number of particles. The number operator is therefore a useful quantity which allows us to test that our many-body formalism conserves the number of particles. In for example \((d,p)\) or \((p,d)\) reactions it is important to be able to describe quantum mechanical states where particles get added or removed. A creation operator \(a_\alpha^{\dagger}\) adds one particle to the single-particle state \(\alpha\) of a give many-body state vector, while an annihilation operator \(a_\alpha\) removes a particle from a single-particle state \(\alpha\).
Let us consider an operator proportional with \(a_\alpha^{\dagger} a_\beta\) and \(\alpha=\beta\). It acts on an \(n\)-particle state resulting in
Summing over all possible one-particle states we arrive at
The operator
is called the number operator since it counts the number of particles in a give state vector when it acts on the different single-particle states. It acts on one single-particle state at the time and falls therefore under category one-body operators. Next we look at another important one-body operator, namely \(\hat{H}_0\) and study its operator form in the occupation number representation.
We want to obtain an expression for a one-body operator which conserves the number of particles. Here we study the one-body operator for the kinetic energy plus an eventual external one-body potential. The action of this operator on a particular \(n\)-body state with its pertinent expectation value has already been studied in coordinate space. In coordinate space the operator reads
and the anti-symmetric \(n\)-particle Slater determinant is defined as
Defining
we can easily evaluate the action of \(\hat{H}_0\) on each product of one-particle functions in Slater determinant. From Eq. (32) we obtain the following result without permuting any particle pair
If we interchange particles \(1\) and \(2\) we obtain
We can continue by computing all possible permutations. We rewrite also our Slater determinant in its second quantized form and skip the dependence on the quantum numbers \(x_i.\) Summing up all contributions and taking care of all phases \((-1)^p\) we arrive at
In Eq. (35) we have expressed the action of the one-body operator of Eq. (31) on the \(n\)-body state in its second quantized form. This equation can be further manipulated if we use the properties of the creation and annihilation operator on each primed quantum number, that is
Inserting this in the right-hand side of Eq. (35) results in
In the number occupation representation or second quantization we get the following expression for a one-body operator which conserves the number of particles
Obviously, \(\hat{H}_0\) can be replaced by any other one-body operator which preserved the number of particles. The stucture of the operator is therefore not limited to say the kinetic or single-particle energy only.
The opearator \(\hat{H}_0\) takes a particle from the single-particle state \(\beta\) to the single-particle state \(\alpha\) with a probability for the transition given by the expectation value \(\langle \alpha|\hat{h}_0|\beta\rangle\).
It is instructive to verify Eq. (38) by computing the expectation value of \(\hat{H}_0\) between two single-particle states
Using the commutation relations for the creation and annihilation operators we have
which results in
and
2.2. Two-body operators in second quantization¶
Let us now derive the expression for our two-body interaction part, which also conserves the number of particles. We can proceed in exactly the same way as for the one-body operator. In the coordinate representation our two-body interaction part takes the following expression
where the summation runs over distinct pairs. The term \(V\) can be an interaction model for the nucleon-nucleon interaction or the interaction between two electrons. It can also include additional two-body interaction terms.
The action of this operator on a product of two single-particle functions is defined as
We can now let \(\hat{H}_I\) act on all terms in the linear combination for \(|\alpha_1\alpha_2\dots\alpha_n\rangle\). Without any permutations we have
where on the rhs we have a term for each distinct pairs.
For the other terms on the rhs we obtain similar expressions and summing over all terms we obtain
We introduce second quantization via the relation
Inserting this in (46) gives
Here we let \(\sum'\) indicate that the sums running over \(\alpha\) and \(\beta\) run over all single-particle states, while the summations \(\gamma\) and \(\delta\) run over all pairs of single-particle states. We wish to remove this restriction and since
we get
where we have used the anti-commutation rules.
Changing the summation indices \(\alpha\) and \(\beta\) in (51) we obtain
From this it follows that the restriction on the summation over \(\gamma\) and \(\delta\) can be removed if we multiply with a factor \(\frac{1}{2}\), resulting in
where we sum freely over all single-particle states \(\alpha\), \(\beta\), \(\gamma\) og \(\delta\).
With this expression we can now verify that the second quantization form of \(\hat{H}_I\) in Eq. (53) results in the same matrix between two anti-symmetrized two-particle states as its corresponding coordinate space representation. We have
Using the commutation relations we get
The vacuum expectation value of this product of operators becomes
Insertion of Eq. (56) in Eq. (54) results in
The two-body operator can also be expressed in terms of the anti-symmetrized matrix elements we discussed previously as
The factors in front of the operator, either \(\frac{1}{4}\) or \(\frac{1}{2}\) tells whether we use antisymmetrized matrix elements or not.
We can now express the Hamiltonian operator for a many-fermion system in the occupation basis representation as
This is the form we will use in the rest of these lectures, assuming that we work with anti-symmetrized two-body matrix elements.
2.3. Proof of Wick’s theorem¶
Discuss also Wick’s generalized theorem
2.4. Interaction, Schroedinger and Heisenberg pictures¶
2.5. Time dependent wick’s theorem¶
2.6. Gell-Man and Low’s theorem and Adiabatic switching¶
2.7. Particle-hole formalism¶
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 \(|0\rangle \). Following Eq. (3) we have
If we use Eq. (60) as our new reference state, we can simplify considerably the representation of this state
The new reference states for the \(n+1\) and \(n-1\) states can then be written as
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. (65)
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
We assume also that the new reference state is properly normalized
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
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
resulting in
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
We can now rewrite our one-body and two-body operators in terms of the new creation and annihilation operators. The number operator becomes
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
Here \(n=n_p +n_c - n_h\) is the total number of particles in the quasi-particle state of Eq. (72). Note that \(\hat{N}\) counts the total number of particles present
gives us the number of quasi-particles as can be seen by computing
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
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
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
Using anti-symmetrized matrix elements, bthe term \(\hat{H}_I^{(a)}\) is
The next term \(\hat{H}_I^{(b)}\) reads
This term conserves the number of quasiparticles but creates or removes a three-particle-one-hole state. For \(\hat{H}_I^{(c)}\) we have
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
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
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.
2.8. Summarizing and defining a normal-ordered Hamiltonian¶
which is equivalent with \(|\alpha_1 \dots \alpha_A\rangle= a_{\alpha_1}^{\dagger} \dots a_{\alpha_A}^{\dagger} |0\rangle\). We have also
and
with \(i,j,\ldots \leq \alpha_F, \quad a,b,\ldots > \alpha_F, \quad p,q, \ldots - \textrm{any}\)
and
The one-body operator is defined as
while the two-body opreator is defined as
where we have defined the antisymmetric matrix elements
We can also define a three-body operator
with the antisymmetrized matrix element
2.9. Operators in second quantization¶
In the build-up of an FCI code that is meant to tackle large dimensionalities is the action of the Hamiltonian \(\hat{H}\) on a Slater determinant represented in second quantization as
The time consuming part stems from the action of the Hamiltonian on the above determinant,
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 orbits \(\alpha_0,\alpha_2,\dots,\alpha_{n-1}\) and that we can distribute among these orbits \(N\le n\) 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
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
which translates into the decimal number
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
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
in a more compact way as
The action of a creation operator is thus
which becomes
Similarly
which becomes
This gives a simple recipe:
If one of the bits \(b_j\) is \(1\) and we act with a creation operator on this bit, we return a null vector
If \(b_j=0\), we set it to \(1\) and return a sign factor \((-1)^l\), where \(l\) is the number of bits set before bit \(j\).
Consider the action of \(a^{\dagger}_{\alpha_2}\) on various slater determinants:
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
in a more compact way as
The action of
which becomes
The action
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
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 orbital \(\alpha_4\) is not already occupied.
It is trickier however to get the phase \((-1)^l\). One possibility is as follows
Let \(S_1\) be a word that represents the \(1-\)bit to be removed and all others set to zero.
In the previous example \(S_1=|1000000000000000\rangle\)
Define \(S_2\) as the similar word that represents the bit to be added, that is in our case
\(S_2=|0000100000000000\rangle\).
Compute then \(S=S_1-S_2\), which here becomes
Perform then the logical AND operation of \(S\) with the word containing
which results in \(|0001000000000000\rangle\). Counting the number of \(1-\)bits gives the phase. Here you need however an algorithm for bitcounting. Several efficient ones available.
2.10. Schr”odinger picture¶
The time-dependent Schr”odinger equation (or equation of motion) reads
where the subscript \(S\) stands for Schr”odinger here. A formal solution is given by
The Hamiltonian \(\hat{H}\) is hermitian and the exponent represents a unitary operator with an operation carried ut on the wave function at a time \(t_0\).
2.11. Interaction picture¶
Our Hamiltonian is normally written out as the sum of an unperturbed part \(\hat{H}_0\) and an interaction part \(\hat{H}_I\), that is
In general we have \([\hat{H}_0,\hat{H}_I]\ne 0\) since \([\hat{T},\hat{V}]\ne 0\). We wish now to define a unitary transformation in terms of \(\hat{H}_0\) by defining
which is again a unitary transformation carried out now at the time \(t\) on the wave function in the Schr”odinger picture.
We can easily find the equation of motion by taking the time derivative
Using the definition of the Schr”odinger equation, we can rewrite the last equation as
which gives us
with
The order of the operators is important since \(\hat{H}_0\) and \(\hat{H}_I\) do generally not commute. The expectation value of an arbitrary operator in the interaction picture can now be written as
and using the definition
we obtain
stating that a unitary transformation does not change expectation values!
If we take the time derivative of the operator in the interaction picture we arrive at the following equation of motion
Here we have used the time-independence of the Schr”odinger equation together with the observation that any function of an operator commutes with the operator itself.
In order to solve the equation of motion equation in the interaction picture, we define a unitary operator time-development operator \(\hat{U}(t,t')\). Later we will derive its connection with the linked-diagram theorem, which yields a linked expression for the actual operator. The action of the operator on the wave function is
with the obvious value \(\hat{U}(t_0,t_0)=1\).
The time-development operator \(U\) has the properties that
which implies that \(U\) is unitary
Further,
and
which leads to
Using our definition of Schr”odinger’s equation in the interaction picture, we can then construct the operator \(\hat{U}\). We have defined
which can be rewritten as
or
From the last expression we can define
It is then easy to convince oneself that the properties defined above are satisfied by the definition of \(\hat{U}\).
We derive the equation of motion for \(\hat{U}\) using the above definition. This results in
which we integrate from \(t_0\) to a time \(t\) resulting in
which can be rewritten as
We can solve this equation iteratively keeping in mind the time-ordering of the of the operators
The third term can be written as
We obtain this expression by changing the integration order in the second term via a change of the integration variables \(t'\) and \(t''\) in
We can rewrite the terms which contain the double integral as
with \(\Theta(t''-t')\) being the standard Heavyside or step function. The step function allows us to give a specific time-ordering to the above expression.
With the \(\Theta\)-function we can rewrite the last expression as
where \(\Hat{T}\) is the so-called time-ordering operator.
With this definition, we can rewrite the expression for \(\hat{U}\) as
The above time-evolution operator in the interaction picture will be used to derive various contributions to many-body perturbation theory.
2.12. Heisenberg picture¶
We wish now to define a unitary transformation in terms of \(\hat{H}\) by defining
which is again a unitary transformation carried out now at the time \(t\) on the wave function in the Schr”odinger picture. If we combine this equation with Schr”odinger’s equation we obtain the following equation of motion
meaning that \(|\Psi_H(t)\rangle\) is time independent. An operator in this picture is defined as
The time dependence is then in the operator itself, and this yields in turn the following equation of motion
We note that an operator in the Heisenberg picture can be related to the corresponding operator in the interaction picture as
With our definition of the time evolution operator we see that
which in turn implies that \(\hat{O}_S=\hat{O}_I(0)=\hat{O}_H(0)\), all operators are equal at \(t=0\). The wave function in the Heisenberg formalism is related to the other pictures as
since the wave function in the Heisenberg picture is time independent. We can relate this wave function to that a given time \(t\) via the time evolution operator as
2.13. Adiabatic hypothesis¶
We assume that the interaction term is switched on gradually. Our wave function at time \(t=-\infty\) and \(t=\infty\) is supposed to represent a non-interacting system given by the solution to the unperturbed part of our Hamiltonian \(\hat{H}_0\). We assume the ground state is given by \(|\Phi_0\rangle\), which could be a Slater determinant.
We define our Hamiltonian as
where \(\varepsilon\) is a small number. The way we write the Hamiltonian and its interaction term is meant to simulate the switching of the interaction.
The time evolution of the wave function in the interaction picture is then
with
In the limit \(t_0\rightarrow -\infty\), the solution ot Schr”odinger’s equation is \(|\Phi_0\rangle\), and the eigenenergies are given by
meaning that
with the corresponding interaction picture wave function given by
The solution becomes time independent in the limit \(t_0\rightarrow -\infty\). The same conclusion can be reached by looking at
and taking the limit \(t\rightarrow \pm\infty\). We can rewrite the equation for the wave function at a time \(t=0\) as
2.14. Wigner-Jordan transformation and second quantization¶
2.15. Baker-Campbell-Hausdorf¶
Discuss also Suzuki-Trotter as an approximation to BCH
2.16. Exercise 1: Relation between basis functions¶
This exercise serves to convince you about the relation between two different single-particle bases, where one could be our new Hartree-Fock basis and the other a harmonic oscillator basis.
Consider a Slater determinant built up of single-particle orbitals \(\psi_{\lambda}\), with \(\lambda = 1,2,\dots,A\). The unitary transformation
brings us into the new basis.
The new basis has quantum numbers \(a=1,2,\dots,A\).
Show that the new basis is orthonormal.
Show that the new Slater determinant constructed from the new single-particle wave functions can be
written as the determinant based on the previous basis and the determinant of the matrix \(C\).
Show that the old and the new Slater determinants are equal up to a complex constant with absolute value unity.
(Hint, \(C\) is a unitary matrix).
Starting with the second quantization representation of the Slater determinant
use Wick’s theorem to compute the normalization integral \(\langle\Phi_{0}|\Phi_{0}\rangle\).
2.17. Exercise 2: Matrix elements¶
Calculate the matrix elements
and
with
and
Compare these results with those from exercise 3c).
2.18. Exercise 3: Normal-ordered one-body operator¶
Show that the onebody part of the Hamiltonian
can be written, using standard annihilation and creation operators, in normal-ordered form as
Explain the meaning of the various symbols. Which reference vacuum has been used?
2.19. Exercise 4: Normal-ordered two-body operator¶
Show that the twobody part of the Hamiltonian
can be written, using standard annihilation and creation operators, in normal-ordered form as
Explain again the meaning of the various symbols.
This exercise is optional: Derive the normal-ordered form of the threebody part of the Hamiltonian.
and specify the contributions to the twobody, onebody and the scalar part.
2.20. Exercise 5: Matrix elements using the Slater-Condon rule¶
The aim of this exercise is to set up specific matrix elements that will turn useful when we start our discussions of the nuclear shell model. In particular you will notice, depending on the character of the operator, that many matrix elements will actually be zero.
Consider three \(N\)-particle Slater determinants \(|\Phi_0\), \(|\Phi_i^a\rangle\) and \(|\Phi_{ij}^{ab}\rangle\), where the notation means that Slater determinant \(|\Phi_i^a\rangle\) differs from \(|\Phi_0\rangle\) by one single-particle state, that is a single-particle state \(\psi_i\) is replaced by a single-particle state \(\psi_a\). It is often interpreted as a so-called one-particle-one-hole excitation. Similarly, the Slater determinant \(|\Phi_{ij}^{ab}\rangle\) differs by two single-particle states from \(|\Phi_0\rangle\) and is normally thought of as a two-particle-two-hole excitation. We assume also that \(|\Phi_0\rangle\) represents our new vacuum reference state and the labels \(ijk\dots\) represent single-particle states below the Fermi level and \(abc\dots\) represent states above the Fermi level, so-called particle states. We define thereafter a general onebody normal-ordered (with respect to the new vacuum state) operator as
with
and a general normal-ordered two-body operator
with for example the direct matrix element given as
with \(g\) being invariant under the interchange of the coordinates of two particles. The single-particle states \(\psi_i\) are not necessarily eigenstates of \(\hat{f}\). The curly brackets mean that the operators are normal-ordered with respect to the new vacuum reference state.
How would you write the above Slater determinants in a second quantization formalism, utilizing the fact that we have defined a new reference state?
Use thereafter Wick’s theorem to find the expectation values of
and
Find thereafter
and
Finally, find
and
What happens with the two-body operator if we have a transition probability of the type
where the Slater determinant to the right of the operator differs by more than two single-particle states?
2.21. Exercise 6: Program to set up Slater determinants¶
Write a program which sets up all possible Slater determinants given \(N=4\) eletrons which can occupy the atomic single-particle states defined by the \(1s\), \(2s2p\) and \(3s3p3d\) shells. How many single-particle states \(n\) are there in total? Include the spin degrees of freedom as well.
2.22. Exercise 7: Using sympy to compute matrix elements¶
Compute the matrix element
using Wick’s theorem and express the two-body operator \(G\) in the occupation number (second quantization) representation.
2.23. Exercise 8: Using sympy to compute matrix elements¶
The last exercise can be solved using the symbolic Python package called SymPy. SymPy is a Python package for general purpose symbolic algebra. There is a physics module with several interesting submodules. Among these, the submodule called secondquant, contains several functionalities that allow us to test our algebraic manipulations using Wick’s theorem and operators for second quantization.
from sympy import *
from sympy.physics.secondquant import *
i, j = symbols('i,j', below_fermi=True)
a, b = symbols('a,b', above_fermi=True)
p, q = symbols('p,q')
print simplify(wicks(Fd(i)*F(a)*Fd(p)*F(q)*Fd(b)*F(j), keep_only_fully_contracted=True))
Input In [1]
print simplify(wicks(Fd(i)*F(a)*Fd(p)*F(q)*Fd(b)*F(j), keep_only_fully_contracted=True))
^
SyntaxError: invalid syntax
The code defines single-particle states above and below the Fermi level, in addition to the genereal symbols \(pq\) which can refer to any type of state below or above the Fermi level. Wick’s theorem is implemented between the creation and annihilation operators Fd and F, respectively. Using the simplify option, one can lump together several Kronecker-\(\delta\) functions.
2.24. Exercise 9: Using sympy to compute matrix elements¶
We can expand the above Python code by defining one-body and two-body operators using the following SymPy code
# This code sets up a two-body Hamiltonian for fermions
from sympy import symbols, latex, WildFunction, collect, Rational
from sympy.physics.secondquant import F, Fd, wicks, AntiSymmetricTensor, substitute_dummies, NO
# setup hamiltonian
p,q,r,s = symbols('p q r s',dummy=True)
f = AntiSymmetricTensor('f',(p,),(q,))
pr = NO((Fd(p)*F(q)))
v = AntiSymmetricTensor('v',(p,q),(r,s))
pqsr = NO(Fd(p)*Fd(q)*F(s)*F(r))
Hamiltonian=f*pr + Rational(1)/Rational(4)*v*pqsr
print "Hamiltonian defined as:", latex(Hamiltonian)
Here we have used the AntiSymmetricTensor functionality, together with normal-ordering defined by the NO function. Using the latex option, this program produces the following output
2.25. Exercise 10: Using sympy to compute matrix elements¶
We can now use this code to compute the matrix elements between two two-body Slater determinants using Wick’s theorem.
from sympy import symbols, latex, WildFunction, collect, Rational, simplify
from sympy.physics.secondquant import F, Fd, wicks, AntiSymmetricTensor, substitute_dummies, NO, evaluate_deltas
# setup hamiltonian
p,q,r,s = symbols('p q r s',dummy=True)
f = AntiSymmetricTensor('f',(p,),(q,))
pr = NO((Fd(p)*F(q)))
v = AntiSymmetricTensor('v',(p,q),(r,s))
pqsr = NO(Fd(p)*Fd(q)*F(s)*F(r))
Hamiltonian=f*pr + Rational(1)/Rational(4)*v*pqsr
c,d = symbols('c, d',above_fermi=True)
a,b = symbols('a, b',above_fermi=True)
expression = wicks(F(b)*F(a)*Hamiltonian*Fd(c)*Fd(d),keep_only_fully_contracted=True, simplify_kronecker_deltas=True)
expression = evaluate_deltas(expression)
expression = simplify(expression)
print "Hamiltonian defined as:", latex(expression)
The result is as expected,
2.26. Exercise 11: Using sympy to compute matrix elements¶
We can continue along these lines and define a normal-ordered Hamiltonian with respect to a given reference state. In our first step we just define the Hamiltonian
from sympy import symbols, latex, WildFunction, collect, Rational, simplify
from sympy.physics.secondquant import F, Fd, wicks, AntiSymmetricTensor, substitute_dummies, NO, evaluate_deltas
# setup hamiltonian
p,q,r,s = symbols('p q r s',dummy=True)
f = AntiSymmetricTensor('f',(p,),(q,))
pr = Fd(p)*F(q)
v = AntiSymmetricTensor('v',(p,q),(r,s))
pqsr = Fd(p)*Fd(q)*F(s)*F(r)
#define the Hamiltonian
Hamiltonian = f*pr + Rational(1)/Rational(4)*v*pqsr
#define indices for states above and below the Fermi level
index_rule = {
'below': 'kl',
'above': 'cd',
'general': 'pqrs'
}
Hnormal = substitute_dummies(Hamiltonian,new_indices=True, pretty_indices=index_rule)
print "Hamiltonian defined as:", latex(Hnormal)
which results in
2.27. Exercise 12: Using sympy to compute matrix elements¶
In our next step we define the reference energy \(E_0\) and redefine the Hamiltonian by subtracting the reference energy and collecting the coefficients for all normal-ordered products (given by the NO function).
from sympy import symbols, latex, WildFunction, collect, Rational, simplify
from sympy.physics.secondquant import F, Fd, wicks, AntiSymmetricTensor, substitute_dummies, NO, evaluate_deltas
# setup hamiltonian
p,q,r,s = symbols('p q r s',dummy=True)
f = AntiSymmetricTensor('f',(p,),(q,))
pr = Fd(p)*F(q)
v = AntiSymmetricTensor('v',(p,q),(r,s))
pqsr = Fd(p)*Fd(q)*F(s)*F(r)
#define the Hamiltonian
Hamiltonian=f*pr + Rational(1)/Rational(4)*v*pqsr
#define indices for states above and below the Fermi level
index_rule = {
'below': 'kl',
'above': 'cd',
'general': 'pqrs'
}
Hnormal = substitute_dummies(Hamiltonian,new_indices=True, pretty_indices=index_rule)
E0 = wicks(Hnormal,keep_only_fully_contracted=True)
Hnormal = Hnormal-E0
w = WildFunction('w')
Hnormal = collect(Hnormal, NO(w))
Hnormal = evaluate_deltas(Hnormal)
print latex(Hnormal)
which gives us
again as expected, with the reference energy to be subtracted.
2.28. Exercise 13: Using sympy to compute matrix elements¶
We can now go back to exercise 7 and define the Hamiltonian and the second-quantized representation of a three-body Slater determinant.
from sympy import symbols, latex, WildFunction, collect, Rational, simplify
from sympy.physics.secondquant import F, Fd, wicks, AntiSymmetricTensor, substitute_dummies, NO, evaluate_deltas
# setup hamiltonian
p,q,r,s = symbols('p q r s',dummy=True)
v = AntiSymmetricTensor('v',(p,q),(r,s))
pqsr = NO(Fd(p)*Fd(q)*F(s)*F(r))
Hamiltonian=Rational(1)/Rational(4)*v*pqsr
a,b,c,d,e,f = symbols('a,b, c, d, e, f',above_fermi=True)
expression = wicks(F(c)*F(b)*F(a)*Hamiltonian*Fd(d)*Fd(e)*Fd(f),keep_only_fully_contracted=True, simplify_kronecker_deltas=True)
expression = evaluate_deltas(expression)
expression = simplify(expression)
print latex(expression)
resulting in nine terms (as expected),
2.29. Exercise 14: Diagrammatic representation of Hartree-Fock equations¶
What is the diagrammatic representation of the HF equation?
(Represent \((-u^{HF})\) by the symbol \(---\)X .)
2.30. Exercise 15: Derivation of Hartree-Fock equations¶
Consider the ground state \(|\Phi\rangle\) of a bound many-particle system of fermions. Assume that we remove one particle from the single-particle state \(\lambda\) and that our system ends in a new state \(|\Phi_{n}\rangle\). Define the energy needed to remove this particle as
where \(E_{0}\) and \(E_{n}\) are the ground state energies of the states \(|\Phi\rangle\) and \(|\Phi_{n}\rangle\), respectively.
Show that
where \(H\) is the Hamiltonian of this system.
If we assume that \(\Phi\) is the Hartree-Fock result, find the
relation between \(E_{\lambda}\) and the single-particle energy \(\varepsilon_{\lambda}\) for states \(\lambda \leq F\) and \(\lambda >F\), with
and
We have assumed an antisymmetrized matrix element here. Discuss the result.
The Hamiltonian operator is defined as