Our goal is to find the ground state energy of a system. To do so, we simply have to review the Schrödinger equation
H∣Ψ0⟩=E0∣Ψ0⟩
Multiplying the bra vector of the ground state to both side will give us
⟨Ψ0∣H∣Ψ0⟩=E0⟨Ψ0∣Ψ0⟩
The inner product is just a number, so we can divide both sides by it to get an expression for the ground state energy
E0=⟨Ψ0∣Ψ0⟩⟨Ψ0∣H∣Ψ0⟩
Suppose our state vector is normalized, the ground state energy of a system can be calculated as such
E0=⟨Ψ0∣H∣Ψ0⟩
This is a simple way to find the ground state energy if we know what the corresponding state vector is
Unfortunately, whenever we need to employ the variational methods, it is unlikely that the Schrödinger equation of that particular system is solvable
Fortunately, we can proceed simply by taking a guess of what the state vector may be
∣Ψtrial⟩
The trial solution must be normalized, well-behaved and satisfies the boundary conditions
We can then find the corresponding trial energy by plugging it into the expectation value of the Hamiltonian
Etrial=⟨Ψtrial∣H∣Ψtrial⟩
If our trial energy is close to that of the actual energy, then our trial solution is also a good approximation. Hence, we need set up a relation between the trial energy and the actual energy
The actual state vectors of the Hamiltonian constitutes a complete basis, so we can write down the completeness relation
I=j=0∑∣Ψj⟩⟨Ψj∣
We can insert the completeness relation next to the Hamiltonian without altering the expectation value
The goal of the variational method is to approximate the ground state energy, so we shall rewrite our relation in terms of the actual ground state energy
By definition, all other energy levels in the spectrum are higher than that of the ground state energy
We can then factorize out the actual ground state energy and rewrite the expression back as a summation and the
Etrial≥E0j=0∑⟨Ψtrial∣Ψj⟩⟨Ψj∣Ψtrial⟩
We can eliminate the summation by acknowledging the completeness relation and the fact that the inner product between the normalized trial solutions must be 1
We can therefore conclude that the trial energy will never be smaller than the ground state energy
Etrial≥E0
The only case where this becomes an equality is when the trial solution is the same as the actual solution
Apart from our mathematical proof, there is a good reason why the trial energy cannot be smaller than that of the actual ground state energy
Nature will adjust the distribution of the particles so as to minimize the total energy, so the distribution of the particle predicted the actual ground state wavefunction must yield the lowest possible energy
Any other distributions of the particle will be less stable, and thus produce a higher energy
Since our trial energy provides an upper limit to the ground state energy, we have a way to make our trial solution better and better
We simply need to insert parameters into our trial solution and then adjust said parameters to achieve the minimum trial energy, in the knowledge that, as we make the trial energy smaller and smaller, we must be approaching the exact ground state energy
To do so, we need to find out how each of the parameter varies with the trial energy and choose the value of the parameter that corresponds to the lowest trial energy
Since we are varying the parameters to produce a minimum trial energy, the computed approximate energy is called the variational energy
Moreover, if we can make the variation energy exactly the same as the ground state energy simply by adjusting the parameter, then we can make our trial solution the exact solution to the Hamiltonian of the system
The Rayleigh-Ritz variational principle tells us that we can find a good approximation to the wavefunction simply by varying the parameters
These parameters can be anything, the proportionality constant, the power of an exponential etc.
However, when we employ the variational methods to complex systems, the presence of a nonlinear variational parameter can get very messy
If we can only allow for linear variational parameters, meaning that we are restricting them to be the multiplicative coefficients to the first power, the situation will be simplified to some algebraic equations, which is a lot easier to solve
To ensure that all the variational parameters are linear, we can write the trial solution as a linear combination
∣Ψtrial⟩=i∑nλi∣κi⟩
The ∣κi⟩s are some arbitrary basis vectors of our choice and the λis are the linear variational parameters
The basis vectors can be anything as long as they do not contain variational parameters within them
Just like before, we shall express the variational energy as such
Etrial=⟨Ψtrial∣Ψtrial⟩⟨Ψtrial∣H∣Ψtrial⟩
Previously, we omitted the denominator because the inner product of the same normalized trial solution must be 1
Here, we cannot ignore it even if the trial solution is normalized. This is because we are not interested in the trial solution, but its linear combination
We shall expand the variational energy in terms of the basis chosen
To save time writing, we denote each inner product as Hij or Sij respectively
Etrial=ij∑nλiλj∗Sijij∑nλiλj∗Hij
Having expanded the numerator and the denominator separately, our next step is to create a simple equation such that the variational parameters can be evaluated
Recall that at minimum value of the variational energy, the derivatives with respect to each of the variational parameter must be zero
Using the fact that the variational parameters are independent of each other and the skew-property of inner products and Hermitian operator, we can further simplify the situation
One possible solution is that all the variational parameters are equal to zero, but that corresponds to a wavefunction being zero everywhere, which is impossible. To avoid the trivial solution, where all the variational parameter has a value of zero, we must set our determinant to zero
To solve for the determinant, we must first evaluate the value of each Hik and Sik respectively
Hik and Sik are presumably knowable since the Hamiltonian and the basis set are known
When expanded, the determinant yields an nth order polynomial equation in Etrial that has n roots
When arranged in ascending order, the roots are the best approximation and the upper limit for the energy of each state
By substituting the calculated variational energy back to the system of linear equations, we will be able to solve for the variational parameters
The system of linear equations are capable of telling us the ratios of the variational parameters, but not their absolute values. To find their absolute values, we shall invoke the requirement that the wavefunction must be normalized
Each value of the variational energy has its own associated unique set of variational parameters
Trial wavefunctions for different states can be found by obtaining different sets of variational parameters
Assumptions:
normalized wavefunctions: SAA=SBB=1
atomic wavefunctions can be made real: SAB=SBA and HAB=HBA
the two wavefunctions are not necessarily the same H_{AA}≠H_
