# Manipulation with braket notation

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

• 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. – Elvex May 16 '14 at 2:07
• ^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. – zzz May 16 '14 at 2:49

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 $$\frac{-\hbar^2}{2m}\int_{-\infty}^{\infty}\frac{\partial^2\phi_n^*}{\partial z^2} \phi\,dz = \int_{-\infty}^{\infty}\left(\frac{-\hbar^2}{2m} \frac{\partial^2}{\partial z^2}\right)\phi_n^* \phi\,dz\\ = \int_{-\infty}^{\infty} \langle \phi_n|\hat{K}|z \rangle \langle z|\phi \rangle \,dz= \langle \phi_n|\hat{K} \left( \int_{-\infty}^{\infty}|z \rangle \langle z| \,dz\right) \phi \rangle\\ = \langle \phi_n|\hat{K} |\phi \rangle$$