Dirac Bra-ket Notation

A state with definite momentum $\bgroup\color{black}$p$\egroup$ . $\bgroup\color{black}$\vert p\rangle$\egroup$
A state with definite position $\bgroup\color{black}$x$\egroup$ . $\bgroup\color{black}$\vert x\rangle$\egroup$
The ``dot product'' between two abstract states $\bgroup\color{black}$\psi_1$\egroup$ and $\bgroup\color{black}$\psi_2$\egroup$ .

$\bgroup\color{black}$\displaystyle \langle\psi_1\vert\psi_2\rangle=\int\limits_{-\infty}^\infty\psi_1^*\psi_2 dx$\egroup$

This dot product projects the state $\bgroup\color{black}$\psi_2$\egroup$ onto $\bgroup\color{black}$\psi_1$\egroup$ and represents the amplitude to go from $\bgroup\color{black}$\psi_2$\egroup$ to $\bgroup\color{black}$\psi_1$\egroup$ .

To find the probability amplitude for our particle to by at any position $\bgroup\color{black}$x$\egroup$ , we dot the state of definite $\bgroup\color{black}$x$\egroup$ into our state $\bgroup\color{black}$\psi$\egroup$ . $\bgroup\color{black}$\psi(x)=\langle x\vert\psi\rangle$\egroup$

To find the probability amplitude for our particle to have a momentum $\bgroup\color{black}$p$\egroup$ , we dot the state of definite $\bgroup\color{black}$x$\egroup$ into our state $\bgroup\color{black}$\psi$\egroup$ . $\bgroup\color{black}$\phi(p)=\langle p\vert\psi\rangle$\egroup$

Jim Branson 2013-04-22