A hydrogen atom is prepared in the normalised state | = 1 3 |2,0,0 + 1 3 |2,1,0 + 1 3…

PHYSICS · Multi-part · Fermi University Physics · Quantum Mechanics · The formalism of quantum mechanics · The formalism of quantum mechanics

The Structure of Quantum Mechanics, Ch The postulates, measurement and collapse

Problem

A hydrogen atom is prepared in the normalised state ∣ψ⟩=13∣2,0,0⟩+13∣2,1,0⟩+13∣2,1,1⟩,|\psi\rangle = \tfrac{1}{\sqrt3}|2,0,0\rangle + \tfrac{1}{\sqrt3}|2,1,0\rangle + \tfrac{1}{\sqrt3}|2,1,1\rangle, where ∣n,l,m⟩|n,l,m\rangle denotes an energy eigenstate of hydrogen with principal quantum number nn, orbital quantum number ll and magnetic quantum number mm, and the three kets are mutually orthonormal. The energy of the n=2n=2 level is E2=−3.4 eVE_2=-3.4\,\mathrm{eV}. ((a)) Write the projector P^E2\hat P_{E_2} onto the whole n=2n=2 eigenspace (spin ignored), using all four kets ∣2,l,m⟩|2,l,m\rangle with n=2n=2, and say which of its terms act on ∣ψ⟩|\psi\rangle. ((b)) Use P(E2)=⟨ψ∣P^E2∣ψ⟩P(E_2)=\langle\psi|\hat P_{E_2}|\psi\rangle to find the probability that a measurement of energy on this atom returns E2E_2. ((c)) State, without recalculating, what the probability would be if the three coefficients were instead 1/21/2, 1/21/2, 1/21/\sqrt2 (still a normalised state built only from n=2n=2 kets).

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