Derivator
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Hi folks,
I'm just reading Ballentine's book on quantum mechanics and was wondering whether he really made a mistake. It's about the variational principle.
In chapter 10.6 (p. 296 in the current edition) he says:
Shouldn't
read
<\psi|\mathcal{H}|\psi> {\color{red}\geq} E_m \sum_n |<\psi|\Psi_n>|^2 = E_m <\psi|\psi>
?
--derivator
I'm just reading Ballentine's book on quantum mechanics and was wondering whether he really made a mistake. It's about the variational principle.
In chapter 10.6 (p. 296 in the current edition) he says:
Although the variational theorem applies to the lowest eigenvalue, it is possible to generalize it to calculate low-lying excited states. In proving that theorem, we formally express the trial function as a linear combination of eigenvectors of \mathcal{H}, so that <\psi|\mathcal{H}|\psi> = \sum_n E_n |<\psi|\Psi_n>|^2. Suppose that we want to calculate the excited state eigenvalue E_m. If we constrain the trial function |\psi> to satisfy <\psi|\Psi_{n'}> = 0 for all n' such that E_{n'} \leq E_m, then it will follow that <\psi|\mathcal{H}|\psi>\leq E_m \sum_n |<\psi|\Psi_n>|^2 = E_m <\psi|\psi>. Hence we can calculateE_m by minimizig <\mathcal{H}> \equiv <\psi|\mathcal{H}|\psi>/<\psi|\psi> subject to the constraint that |\psi> be orthogonal to all state functions and energies lower than E_m.
Shouldn't
<\psi|\mathcal{H}|\psi>\leq E_m \sum_n |<\psi|\Psi_n>|^2 = E_m <\psi|\psi>
read
<\psi|\mathcal{H}|\psi> {\color{red}\geq} E_m \sum_n |<\psi|\Psi_n>|^2 = E_m <\psi|\psi>
?
--derivator
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