# 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.

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