Exercise [22.10]

deant · 07 Oct 2012, 15:05

Look at the last sentence of §22.3 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.

$\blacksquare$