# Using Variational Principle to solve ground state energy

• 1missing

#### 1missing

Homework Statement
A certain Hamiltonian can be expressed as: ##\hat H = | \varphi_1 \rangle \langle \varphi_1| + 2 | \varphi_2 \rangle \langle \varphi_2| + 3 | \varphi_3 \rangle \langle \varphi_3| ##

where, ##| \varphi_1 \rangle## , ##| \varphi_2 \rangle## , ##| \varphi_3 \rangle## are normalized but not fully orthogonal.

##\langle \varphi_1 | \varphi_2 \rangle = \langle \varphi_2 | \varphi_1 \rangle = \langle \varphi_2 | \varphi_3 \rangle = \langle \varphi_3 | \varphi_2 \rangle = \frac {1} {2}##

##\langle \varphi_1 | \varphi_3 \rangle = \langle \varphi_3 | \varphi_1 \rangle = 0##
Relevant Equations
1: ##E_\phi = \frac { \langle \phi | \hat H | \phi \rangle } { \langle \phi | \phi \rangle } ##

2: ##\sum_{j=1}^n (H_{ij} - ES_{ij} ) C_j = 0## , i = 1, 2,..., n
First I picked an arbitrary state ##|ϕ⟩=C_1|φ_1⟩+C_2|φ_2⟩+C_3|φ_3⟩## and went to use equation 1. Realizing my answer was a mess of constants and not getting me closer to a ground state energy, I abandoned that approach and went with equation two.

I proceeded to calculate the following matrix:
##\begin{pmatrix}

( \frac {3} {2} - E) & ( \frac {3} {2} - \frac {E} {2} ) & ( \frac {1} {2} ) \\

( \frac {3} {2} - \frac {E} {2} ) & ( 3 - E ) & ( \frac {5} {2} - \frac {E} {2} ) \\

( \frac {1} {2} ) & ( \frac {5} {2} - \frac {E} {2} ) & ( \frac {7} {2} - E )

\end{pmatrix}\vec C = 0##

Calculating the determinant, I end up with ##E^3 - 6E^2 + 9E - 3 = 0## which doesn't seem to have a rational solution. I must've gone through the calculation half a dozen times now, and confirmed with my professor that equation 2 was the one I should use, but I'm just not seeing how I'm supposed to find a solution.

I end up with ##E^3 - 6E^2 + 9E - 3 = 0## which doesn't seem to have a rational solution.
Why do you expect the solutions to be rational? Or did you mean real instead of rational? The equation does have 3 real solutions.

I don't see how any variational method is being used here.

Why do you expect the solutions to be rational? Or did you mean real instead of rational? The equation does have 3 real solutions.

I don't see how any variational method is being used here.
He usually models his exam questions off the homework problems he assigns. If he gave this problem on the exam there wouldn't be any way for me to solve for those roots, which made me think I messed up somewhere. If there were at least one integer solution, that term could be factored out by polynomial long division and I'd be left with a solvable quadratic for the other two.

He called the method "trial states as linear combinations of basis states". He starts with the variational principle, takes the derivative with respect to Ci, and sets it to zero. From that he derives equation 2.

OK, that sounds good. I checked the entries for your matrix and the equation coming from setting the determinant equal to zero. I get the same results. I don't see anything wrong with your work.

• Einstein44