Supersymmetry in Quantum Mechanics Josh Powell
نویسنده
چکیده
Hence, there is an eigenstate |ψ〉 = A|φ〉 of H with this same energy eigenvalue. We could equally well have begun with the statement that H|ψ〉 = E|ψ〉 and then found a state of H given by |φ〉 = A†|ψ〉 with the same energy. Note that a Hamiltonian of the two given forms will have only non-negative eigenvalues: 〈ψ|A†A|ψ〉 = ∣∣|Aψ〉∣∣2 ≥ 0. Equality holds only if the state is annihilated by A. Analogous statements hold for A†. In order to have both |φ〉 and |ψ〉 properly normalized, we need |φ〉 = E−1/2A†|ψ〉 |ψ〉 = E−1/2A|φ〉
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