Complex conjugate of an inner product

[Pablo315]

Hi everyone.

Yesterday I had an exam, and I spent half the exam trying to solve this question.

Show that $\left\langle\Psi\left(\vec{r}\right)\right|\hat{p_{y}^{2}}\left|\phi\left(\vec{r}\right)\right\rangle =\left\langle \phi\left(\vec{r}\right)\right|\hat{p_{y}^{2}}\left|\Psi\left(\vec{r}\right)\right\rangle$

However, I don't see how that is true. If we choose $\Psi\left(\vec{r}\right)=i\phi\left(\vec{r}\right)$

then

$\left\langle \Psi\left(\vec{r}\right)\right|\hat{p_{y}^{2}}\left|\phi\left(\vec{r}\right)\right\rangle =\left\langle i\phi\left(\vec{r}\right)\right|\hat{p_{y}^{2}}\left|\phi\left(\vec{r}\right)\right\rangle =-i\left\langle \phi\left(\vec{r}\right)\right|\hat{p_{y}^{2}}\left|\phi\left(\vec{r}\right)\right\rangle$

and

$\left\langle \phi\left(\vec{r}\right)\right|\hat{p_{y}^{2}}\left|\Psi\left(\vec{r}\right)\right\rangle =\left\langle \phi\left(\vec{r}\right)\right|\hat{p_{y}^{2}}\left|i\phi\left(\vec{r}\right)\right\rangle =i\left\langle \phi\left(\vec{r}\right)\right|\hat{p_{y}^{2}}\left|\phi\left(\vec{r}\right)\right\rangle$

which clearly aren't the same. Am I missing something here?

Thank you!

 

[DrClaude]

Indeed, that should be
$$
\left\langle\Psi\right|\hat{p_{y}^{2}}\left|\phi\right\rangle =\left\langle \phi\right|\hat{p}_{y}^{2}\left|\Psi\right\rangle^*
$$
Only if $|\Psi\rangle = |\phi\rangle$ do you get
$$
\left\langle\Psi\right|\hat{p_{y}^{2}}\left|\phi\right\rangle =\left\langle \phi\right|\hat{p}_{y}^{2}\left|\Psi\right\rangle
$$
as $\hat{p}_{y}^{2}$ is Hermitian.

Note also that the notation $\left|\phi\left(\vec{r}\right)\right\rangle$ doesn't make sense. A ket is an abstract vector, not a wave function. You then have
$$
\phi(\mathbf{r}) = \langle \mathbf{r} | \phi \rangle
$$

 

[Pablo315]

Yes, that is exactly what I thought.

Thank you very much.