Fund Quantum Mechanics Lect &amp; HW Solutions 68

Fund Quantum Mechanics Lect &amp; HW Solutions 68 - x...

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50 CHAPTER 4. SINGLE-PARTICLE SYSTEMS For the z -angular momentum, the expectation value is zero but the two states have eigenvalues ¯ h and ¯ h , so σ L z = r 1 2 h 0) 2 + 1 2 ( ¯ h 0) 2 = ¯ h. Whether ¯ h or ¯ h is measured, the deviation from zero has magnitude ¯ h . 4.3.3 Simplifed expressions 4.3.3.1 Solution esdb2-a Question: The 2p x pointer state of the hydrogen atom was de±ned as 1 2 ( ψ 211 + ψ 21 1 ) . where both ψ 211 and ψ 21 1 are eigenfunctions of the total energy Hamiltonian H with eigen- value E 2 and of square angular momentum h L 2 with eigenvalue 2¯ h 2 ; however, ψ 211 is an eigen- function of z -angular momentum h L z with eigenvalue ¯ h , while ψ 21 1 is one with eigenvalue ¯ h . Evaluate the expectation values of energy, square angular momentum, and z -angular momentum in the 2p x state using inner products. (Of course, since 2p
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Unformatted text preview: x is already written out in terms of the eigenfunctions, there is no simpli±cation in this case.) Answer: For energy you have, a E A = 1 2 a− ψ 211 + ψ 21 − 1 | H | − ψ 211 + ψ 21 − 1 A . By the de±nition of eigenfunction, the products with H simplify: a E A = 1 2 a− ψ 211 + ψ 21 − 1 | − E 2 ψ 211 + E 2 ψ 21 − 1 A . Multiplying out further, while noting that on account of orthonormality of the eigenstates, a ψ 211 | ψ 211 A = a ψ 21 − 1 | ψ 21 − 1 A = 1 , a ψ 211 | ψ 21 − 1 A = a ψ 21 − 1 | ψ 211 A = 0 , you get a E A = E 2 . Similarly, for the square angular momentum, a L 2 A = 1 2 a− ψ 211 + ψ 21 − 1 | h L 2 | − ψ 211 + ψ 21 − 1 A ....
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