2.8. Variational theorem
Here, we derive the variational theorem, which states that the energy of any wavefunction \(\psi\) is always larger or equal to the true ground-state energy \(E_0\)
We start with a general wavefunction \(\psi\) that is not necessarily normalized. We write it as a linear combination of the complete orthonormal set of eigenfunctions \(\psi_i\) of the Hamiltonian (\(\hat{H}\psi_i = E_i\psi_i\) and \(\langle\psi_i|\psi_j\rangle=\delta_{i,j}\)):
\(c_i\) are the linear-combination coefficients. The expectation value of the energy of \(\psi\) is
\(\hat{H}\psi_i\) equals to \(E_i\psi_i\), so that we get
The set of \(\psi_i\) are orthonormal, i.e. \(\langle\psi_j|\psi_i\rangle = \delta_{j,i}\). With this, all terms in the two double sums except those with \(j=i\) vanish:
Now comes the key step of the proof: The energy of any energy eigenstate is always larger or equal to that of the ground state, \(E_i\ge E_0\). Therefore, we can replace all \(E_i\) by \(E_0\) and, at the same time, replace the equal sign by \(\ge\). This yields
After pulling \(E_0\) out of the sum, the sums in the enumerator and denominator cancel, so that the final result is
This proves the variational theorem.
Now comes the key step of the proof: The energy of any energy eigenstate is always larger or equal to that of the ground state, \(E_i\ge E_0\). Therefore, we can replace all \(E_i\) by \(E_0\) and, at the same time, replace the equal sign by \(\ge\). This yields