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I am still getting used to the braket notation. Is this manipulation correct? $$ \frac{-\hbar^2}{2m}\int_{-\infty}^{\infty}\frac{\partial^2\phi_n^*}{\partial z^2} \phi\,dz = \int_{-\infty}^{\infty}\phi\left(\frac{-\hbar^2}{2m} \frac{\partial^2}{\partial z^2}\right)\phi_n^*\,dz = \int_{-\infty}^{\infty} \phi \hat{K}\phi_n^*\,dz = \left( \int_{-\infty}^{\infty} \phi^* \hat{K}\phi_n\,dz \right)^{*} = \langle\phi|\hat{K}|\phi_n\rangle^{*} $$

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  • $\begingroup$ I didn't look over everything but you don't need to add the * at the end of Braket product, that information is inherent into the nature of Bras and Kets. $\endgroup$ – Elvex May 16 '14 at 2:07
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    $\begingroup$ ^I didn't look over it either but what @Elvex means is that to take the complex conjugate of a bra-ket scalar just flip the bra and the ket. $\endgroup$ – zzz May 16 '14 at 2:49
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Your conclusion is correct, your steps shows maybe you're not quite sure what you're doing, so let me for clarity sake do the problem below. We know $$\phi(z)=\langle z|\phi \rangle$$ and $$\left(\frac{-\hbar^2}{2m} \frac{\partial^2}{\partial z^2}\right)\phi(z) =\langle z|\hat{K}|\phi \rangle $$ So in particular: $$\left(\frac{-\hbar^2}{2m} \frac{\partial^2}{\partial z^2}\right)\phi_n(z)^* =\langle z|\hat{K}|\phi_n \rangle^* =\langle \phi_n|\hat{K}|z \rangle $$ Therefore $$ \frac{-\hbar^2}{2m}\int_{-\infty}^{\infty}\frac{\partial^2\phi_n^*}{\partial z^2} \phi\,dz = \int_{-\infty}^{\infty}\left(\frac{-\hbar^2}{2m} \frac{\partial^2}{\partial z^2}\right)\phi_n^* \phi\,dz\\ = \int_{-\infty}^{\infty} \langle \phi_n|\hat{K}|z \rangle \langle z|\phi \rangle \,dz= \langle \phi_n|\hat{K} \left( \int_{-\infty}^{\infty}|z \rangle \langle z| \,dz\right) \phi \rangle\\ = \langle \phi_n|\hat{K} |\phi \rangle$$

So your conclusion is correct.

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