Variational Principle in Quantum Physics

Normalize the wave function $\psi=\phi_0+a*\phi_1$ and then use the variational principle to determine the parameter $a$ such that the energy corresponding to the the normalized $\psi$ is minimal when $\phi_0$ and $\phi_1$ are orthogonal on each other and normalized eigenfunctions of the Hamiltonian $\hat H$ of the system and when they are

a) non degenerate eigenfunctions of $\hat H$ with energies $E_0 < E_1$

b) degenerate eigenfunctions of $\hat H$ both with energy $E_0$.

If the energy cannot be minimized in this way, explain why.

To normalize make the inner product of ψ

$<ψ|ψ>=<φ_0 | φ_0>+a<φ_0 |φ_1>+a<φ_1 | φ_0>+a^2<φ_0 |φ_0>=1+0+0+a^2=1+a^2$

So that

$ψ_n=1/\sqrt{<ψ|ψ>}*(φ_0+aφ_1 )=1/\sqrt{1+a^2 }*(φ_0+aφ_1 )$

Non degenerate eigenfunctions

Write Schroedinger equation for $ψ_n$

$Hψ_n=Eψ_n$   so that

$E=\frac {(<ψ_n |H| ψ_n>)}{(<ψ_n |ψ_n>)}=\frac{1}{(1+a^2 )}*\frac{(<φ_0+aφ_1 |H| φ_0+aφ_1>)}{1}$

$E=\frac {1}{(1+a^2 )}*(<φ_0 |H| φ_0>+a^2<φ_1 |H| φ_1>)=\frac{1}{(1+a^2 )}*(E_0+a^2 E_1 )$

Variational principle says

$\frac {d E}{d a}=0$  so that $\frac{2aE_1 (1+a^2 )-2a(E_0+a^2 E_1 )} {(1+a^2 )^2} =0$

$2aE_1+2a^3 E_1-2aE_0+2a^3 E_1=0 or 4a^3 E_1+2a(E_1-E_0 )=0$

or $2a(2a^2 E_1+E_1-E_0 )=0$

$a_1=0$ a and $2a^2=E_0/E_1 -1$   or $a_{2,3}=±\sqrt{(E_0-E_1)/(2E_1 )}$

Degenerate eigenfunctions

$Hψ_n=Eψ_n$   so that

$E=\frac {(<ψ_n |H| ψ_n>)}{(<ψ_n |ψ_n>)}=\frac {1} {(1+a^2 )}*\frac{(<φ_0+aφ_1 |H| φ_0+aφ_1>)}{1}$

$E=\frac {1}{(1+a^2 )}*(<φ_0 |H| φ_0>+a^2<φ_1 |H| φ_1>)=\frac{1}{(1+a^2 )} (E_0+a^2 E_0 )=\frac {E_0}{(1+a^2 )}$

$d E/d a=0$  so that $-E_0/(1+a^2 )^2 *2a=0$  so that $a=0$

The energy cannot be minimized in this case because there is only one energy in the spectrum E0 (so there is nothing to minimize). In the case above a) there were two energies in the spectrum (E1 and E0 and the state was a mix of these two states)