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 Exercise [22.14] 
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Joined: 12 Jul 2010, 07:44
Posts: 154
Post Exercise [22.14]
Given \boldsymbol{E}^2 = \boldsymbol{E} = \boldsymbol{E}^*, we wish to show that \boldsymbol{E}\psi and (\boldsymbol{I}-\boldsymbol{E})\psi are orthogonal,

i.e. that \left\langle\left(\boldsymbol{E}\psi\right)\middle|\left(\left(\boldsymbol{I}-\boldsymbol{E}\right)\psi\right)\right\rangle = 0.

We have:


\left\langle\left(\boldsymbol{E}\psi\right)\middle|\left(\left(\boldsymbol{I}-\boldsymbol{E}\right)\psi\right)\right\rangle<br />= \left\langle\psi\middle|\boldsymbol{E}^*\left(\boldsymbol{I}-\boldsymbol{E}\right)\middle|\psi\right\rangle<br />= \left\langle\psi\middle|\boldsymbol{E}\left(\boldsymbol{I}-\boldsymbol{E}\right)\middle|\psi\right\rangle

= \left\langle\psi\middle|\left(\boldsymbol{E}-\boldsymbol{E}^2\right)\middle|\psi\right\rangle<br />= \left\langle\psi\middle|\left(\boldsymbol{E}-\boldsymbol{E}\right)\middle|\psi\right\rangle<br />= \left\langle\psi\middle|0\middle|\psi\right\rangle<br />= 0


\blacksquare


08 Oct 2012, 08:11

Joined: 12 Jul 2010, 07:44
Posts: 154
Post Re: Exercise [22.14]
Ah, I see Roberto actually has already posted the identical solution as a footnote to his exercise [22.15] solution...

I'll leave this here, though just because it's posted under the correct [22.14] heading.


08 Oct 2012, 08:16
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