8.2: Excited States

The variational method can be adapted to give bounds on the energies of excited states, under certain conditions. Suppose we choose a trial function $\Phi_1 (\beta_n)$ with variational parameters $\beta_n$. which is made orthogonal to the ground state $\phi_0$, by imposing the condition $\langle \phi_0|\phi_1 \rangle = 0$.

If we know $|\phi_0 \rangle = |i_0 \rangle$, then similar to the above

$$E[a_n] = \frac{\langle \Phi_1|\hat{H} |\Phi_1 \rangle}{\langle \Phi_1|\Phi_1 \rangle} = \sum_{ij} \langle \phi_1|i \rangle \langle i|\hat{H} |j \rangle \langle j|\phi_1 \rangle = \sum_i |\langle \phi_1|i \rangle |^2 E_i = 0+E_1+ \sum_{i = 2} |\langle \phi_1|i \rangle|^2 (E_i−E_1) \geq E_1 \nonumber$$

So the variational method gives an upper bound on the first excited-state energy, and so on. We can satisfy $\langle i_0|\phi_1 \rangle = 0$ if $|i_0 \rangle$ is known, or if it has a known symmetry from which we can exploit (e.g. if $|i_0 \rangle$ has even parity, chosing $|\Phi_1 \rangle$ to be odd.)

In general, though, we only have a variational estimate of the ground state $\phi_0(\alpha_n)$. In this case the expression above, subject to the constraint $\langle \phi_1(\beta_n)|\phi_0(\alpha_n) \rangle = 0$, gives an estimate of $E_1$. However, the error in this approach will be larger than for $E_0$ because not only is the wavefunction incorrect, but also the constraint $\langle \phi_1|\phi_0 \rangle = 0$ is not quite correct; using an approximate ground state does not guarantee that we get an upper bound for the excited states.

If the excited state has different symmetry from those of the lower-lying levels, and we choose trial functions with the correct symmetries, orthogonality is guaranteed and we get an upper bound to the energy of the lowest-lying level with those symmetries, which is the excited state.