Fund Quantum Mechanics Lect &amp; HW Solutions 69

# Fund Quantum Mechanics Lect &amp; HW Solutions 69 - 4.3...

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Unformatted text preview: 4.3. EXPECTATION VALUE AND STANDARD DEVIATION 51 or multiplying out L2 = 1 −ψ211 + ψ21−1 | − 2¯ 2 ψ211 + 2¯ 2 ψ21−1 . h h 2 multiplying out further to L2 = 2¯ 2 . h For the z -angular momentum, Lz = or multiplying out Lz = multiplying out further to Lz 4.3.3.2 1 −ψ211 + ψ21−1 |Lz | − ψ211 + ψ21−1 . 2 1 −ψ211 + ψ21−1 | − hψ211 − hψ21−1 . ¯ ¯ 2 = 0. Solution esdb2-b Question: Continuing the previous question, evaluate the standard deviations in energy, square angular momentum, and z -angular momentum in the 2px state using inner products. Answer: For energy you have, 2 σE = (H − E2 )2 = 1 −ψ211 + ψ21−1 |(H − E2 )2 | − ψ211 + ψ21−1 . 2 or multiplying out, noting that Hψ21±1 = E2 ψ21±1 , so that (H − E − 2)ψ21±1 = 0, 2 σE = 1 −ψ211 + ψ21−1 | − 0ψ211 + 0ψ21−1 . 2 which is zero. The same way, σL2 = 0. For z -angular momentum, you have, since the expectation value is zero, 2 σLz = (Lz − 0)2 = or multiplying out, 1 −ψ211 + ψ21−1 |(Lz − 0)2 | − ψ211 + ψ21−1 . 2 1 −ψ211 + ψ21−1 | − h2 ψ211 + h2 ψ21−1 ¯ ¯ 2 which multiplies out to h2 , so σLz itself is h. ¯ ¯ 2 σLz = 4.3.4 Some examples ...
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