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 Exercise [22.10] 
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Joined: 12 Jul 2010, 07:44
Posts: 154
Post Exercise [22.10]
Look at the last sentence of ยง22.3 on page 535. We have:


\overline{\left\langle\phi\middle{|}\boldsymbol{L}\middle{|}\psi\right\rangle} = \left\langle\psi\middle{|}\boldsymbol{L^*}\middle{|}\phi\right\rangle


For any \phi, \psi, and operator \boldsymbol{L}.

Let us take \phi=\psi, and let \boldsymbol{L} be the hermitian operator \boldsymbol{Q}, i.e. \boldsymbol{L}=\boldsymbol{Q}=\boldsymbol{L^*}. Then the above becomes:


\overline{\left\langle\psi\middle{|}\boldsymbol{Q}\middle{|}\psi\right\rangle} = \left\langle\psi\middle{|}\boldsymbol{Q}\middle{|}\psi\right\rangle


i.e. \left\langle\psi\middle{|}\boldsymbol{Q}\middle{|}\psi\right\rangle \in \mathbb{R}


If \psi is an eigenfunction of \boldsymbol{Q} with eigenvalue \lambda, then


\left\langle\psi\middle{|}\boldsymbol{Q}\middle{|}\psi\right\rangle = \left\langle\psi\middle{|}\lambda\psi\right\rangle = \lambda\left\langle\psi\middle{|}\psi\right\rangle = \lambda\|\psi\| \in \mathbb{R}


Since \|\psi\| is always a real number, that means that the eigenvalue \lambda must always be real, as well.

\blacksquare


07 Oct 2012, 15:05
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