I have the following non-stationary state of the QSHO:(adsbygoogle = window.adsbygoogle || []).push({});

Ψ = [ e^{−3iωt/2}+ [ ]e^{−5iωt/2}]

where β = mω/ħ

in calculating <E^{2}>

The answer I see in the textbook is 6.17 ħ^{2}ω^{2}.

This answer suggests that in calculating <E^{2}> = ∫ Ψ*Ĥ^{2}Ψ dx

where Ψ is composed of the above two terms Ψ_{1}and Ψ_{2}(a linear combination of E_{1}and E_{2}of the QSHO) the cross terms need to be zero to eliminate the time dependence as reflected in the answer in the text book.

This leaves us with the inside and outside terms

∫ Ψ_{1}*Ĥ^{2}Ψ_{1}dx + ∫ Ψ_{2}*Ĥ^{2}Ψ_{2}dx and calculating the sum of these inside and outside terms producing the answer in the text book.

However, as I calculated the cross terms ∫ Ψ_{1}*Ĥ^{2}Ψ_{2}dx is zero as it is an odd function, but ∫ Ψ_{2}*Ĥ^{2}Ψ_{1}dx is not zero as it is an even function. This would not eliminate the time dependence and would not result in a non oscillating result as reflected in the text book.

Am I missing something here?

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# I Solving for <E^2> of a non-stationary state of the QSHO

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