Self-consistent Hartree-Fock approach to Phase Transitions

Hartree-Fock approach to Phase Transitions

Introduction to self consistent Hartree-Fock theory

The self consistent Hartree-Fock method is a way to approximately find the ground state of a many body system made of n interacting fermions. We start by writing the many particle wavefunction such that it respects the antisymmetry property of fermions. This is achieved by taking Slater determinant of one particle wave functions. The one particle wave functions $\left|i\right\rangle$ can themselves be expressed in terms of the basis functions $\left|\chi _{k}\right\rangle$ as

$\left|i\right\rangle ={\underset {k}{\sum }}c_{ik}\left|\chi _{k}\right\rangle$

So, the Slater determinant is a function of the coefficients $c_{ik}$.

In self consistent field theory, we make a guess for the form of the many electron ground state wave function and then use variation principle to calculate the best approximation for the ground state energy as a function of the coefficients. This leads to a set of non linear Hartree -Fock equations in terms of the coefficients. We then solve the equations iteratively to find the right values of the coefficients upto a certain accuracy. Thus, we obtain the self consistent solution for the ground state wave function of the n particle system and the ground state energy.

With the general introduction self consistent Hartree-Fock method mentioned above, let us now elaborate on the topic with more mathematical details.

Consider a system of n interacting fermions. The Hamiltonian is expressed as

$H={\underset {\alpha \alpha '}{\sum }}b_{\alpha }^{\dagger }\left\langle \alpha \left|H_{0}\right|\alpha '\right\rangle b_{\alpha '}+{\frac {1}{2}}{\underset {\alpha \alpha '\beta \beta '}{\sum }}b_{\alpha }^{\dagger }b_{\beta }^{\dagger }\left\langle \alpha \beta \left|V\right|\alpha '\beta '\right\rangle b_{\beta '}b_{\alpha '}$

The independent particle state can be written as

$\left|\psi _{0}\right\rangle =a_{n}^{\dagger }a_{n-1}^{\dagger }...a_{2}^{\dagger }a_{1}^{\dagger }\left|0\right\rangle$

In this basis,

$H={\underset {i,j}{\sum }}a_{i}^{\dagger }\left\langle i\left|H_{0}\right|j\right\rangle a_{j}+{\frac {1}{2}}{\underset {qrst}{\sum a_{q}^{\dagger }a_{r}^{\dagger }\left\langle qr|V|ts\right\rangle a_{s}a_{t}}}$

The variation of state $\left|\psi _{0}\right\rangle$ is given by a linear combination

$\delta \psi _{jk}=\varepsilon _{jk}a_{j}^{\dagger }a_{k}\left|\psi _{0}\right\rangle$

The “best” state $\left|\psi _{0}\right\rangle$ should minimise the ground state energy. Also, since H is Hermitian, therefore

$\left\langle \psi _{0}|\psi _{0}\right\rangle \left\langle \delta \psi |H|\psi _{0}\right\rangle -\left\langle \psi _{0}|H|\psi _{0}\right\rangle \left\langle \delta \psi |\psi _{0}\right\rangle =0$

Further, since the variation preserves the fact that $\left\langle \psi _{0}|\psi _{0}\right\rangle =1$, this makes $\left\langle \delta \psi _{0}|\psi _{0}\right\rangle =0$. Then the above equation reduces to

$\left\langle \delta \psi |H|\psi _{0}\right\rangle =0$

Substituting the form of the variation and the Hamiltonian, we get

$\left\langle j\left|H_{0}\right|k\right\rangle +{\frac {1}{2}}{\underset {t}{\sum }}[\left\langle jt\left|V\right|kt\right\rangle -\left\langle jt\left|V\right|tk\right\rangle ]=0$

Or, $H_{0}\left|m\right\rangle +{\underset {p}{\sum }}\left|p\right\rangle {\underset {t}{\sum }}[\left\langle pt\left|V\right|mt\right\rangle -\left\langle pt\left|V\right|tm\right\rangle ]=\varepsilon _{m}\left|m\right\rangle$

If $\left\langle \alpha \beta \left|V\right|\alpha '\beta '\right\rangle =V_{\alpha \beta }\delta _{\alpha \alpha '}\delta _{\beta \beta '}$ then, we get

$H_{0}\left|m\right\rangle +{\underset {t}{\sum }}{\underset {\alpha ,\beta }{\sum }}\left|\alpha \right\rangle \left\langle t\right|\left.\beta \right\rangle V_{\alpha \beta }[\left\langle \beta \right|\left.t\right\rangle \left\langle \alpha \right|\left.m\right\rangle -\left\langle \alpha \right|\left.t\right\rangle \left\langle \beta \right|\left.m\right\rangle ]=\varepsilon _{m}\left|m\right\rangle$

In matrix form, this can be written as

${\underset {\beta }{\sum }}[\left\langle \alpha \right|H_{0}\left|\beta \right\rangle \left\langle \beta \right|\left.m\right\rangle +{\underset {t}{\sum }}\left\langle t\right|\left.\beta \right\rangle V_{\alpha \beta }[\left\langle \beta \right|\left.t\right\rangle \left\langle \alpha \right|\left.m\right\rangle -\left\langle \alpha \right|\left.t\right\rangle \left\langle \beta \right|\left.m\right\rangle ]]=\varepsilon _{k}\left\langle \alpha \right|\left.m\right\rangle$

This equation is known as Hartree-Fock equation. The first term in the summation over t on the left hand side of the above equation is known as the Hartree term or direct energy term. The last term in the sum is called the Fock term and it constitutes the exchange energy of the interaction between electrons.

The Hartree-Fock equations cannot be computed without knowing the n eigensolutions $\left|m\right\rangle$ since the matrix elements of interaction term V depend on them. We solve these equations iteratively starting by guessing a set of one particle states $\left|m\right\rangle$ and then solve the Hartree-Fock equations to find the next set of eigensolutions. We compare this set with previous set and if they don't match within error then we repeat the process till we get a self consistent set of eigensolutions.

Hartree-Fock in second quantized form

The interaction energy operator for a two body system with pairwise interactions is given by

$\nu ={\frac {1}{2}}\sum _{\sigma \sigma ^{\prime }}\int d^{3}rd^{3}r^{\prime }v({\vec {r}}-{\vec {r}}^{\prime })\psi _{\sigma }^{\dagger }({\vec {r}})\psi _{\sigma ^{\prime }}^{\dagger }({\vec {r}}^{\prime })\psi _{\sigma ^{\prime }}({\vec {r}}^{\prime })\psi _{\sigma }({\vec {r}})$

In the above operator, it is important to note the order of the operators. Including the kinetic energy term we can write the Hamiltonian for particles of mass m with pairwise interactions in second quantized formalism as

$H=\sum _{\sigma \sigma ^{\prime }}\int d^{3}r{\frac {1}{2m}}\nabla \psi _{\sigma }^{\dagger }({\vec {r}})\cdot \nabla \psi _{\sigma }({\vec {r}})+{\frac {1}{2}}\sum _{\sigma \sigma ^{\prime }}\int d^{3}rd^{3}r^{\prime }v({\vec {r}}-{\vec {r}}^{\prime })\psi _{\sigma }^{\dagger }({\vec {r}})\psi _{\sigma ^{\prime }}^{\dagger }({\vec {r}}^{\prime })\psi _{\sigma ^{\prime }}({\vec {r}}^{\prime })\psi _{\sigma }({\vec {r}})$

The first-order correction to the energy induced by the interaction is

$E^{(1)}={\frac {1}{2}}\int d^{3}rd^{3}r^{\prime }v({\vec {r}}-{\vec {r}}^{\prime })\sum _{\sigma \sigma ^{\prime }}<\Phi _{0}|\psi _{\sigma }^{\dagger }({\vec {r}})\psi _{\sigma ^{\prime }}^{\dagger }({\vec {r}}^{\prime })\psi _{\sigma ^{\prime }}({\vec {r}}^{\prime })\psi _{\sigma }({\vec {r}})|\Phi _{0}>$

Define $G_{\sigma }({\vec {r}}-{\vec {r}}^{\prime })=<\Phi _{0}|\psi _{\sigma }^{\dagger }({\vec {r}})\psi _{\sigma }({\vec {r}}^{\prime })|\Phi _{0}>$

Then, we find that

$E^{(1)}={\frac {1}{2}}\int d^{3}rd^{3}r^{\prime }v({\vec {r}}-{\vec {r}}^{\prime })[n^{2}-\sum _{\sigma }G_{\sigma }({\vec {r}}-{\vec {r}}^{\prime })^{2}]$

(You can find intermediate steps in "Lectures on Quantum Mechanics" by Gordon Baym).

The first term in $E^{(1)}$ is called the Hartree energy or direct energy. The second term is called the Fock term and it corresponds to the energy due to exchange interactions.