Feynman V. III - Ch. 12 - deriving eqn 12-47

Question

Working my way through Feynman V.III. Sec. 12-6 derives Eqn 12-47. I follow all the steps until the equation, but cannot see how he gets 12-47 from 12-46. Appears that he's using the rule: If $\langle+T|+S\rangle = a^2$ then $|+S\rangle = a^2|T\rangle$. Am I correct? If so, where does that rule come from?

http://www.feynmanlectures.caltech.edu/III_12.html

Answer 1

Think of evaluating a bra-ket as doing a dot product and |+S> and |+T> as vectors. Since |+S> is normalized (making it a "unit vector"), it follows <+T|+S> is akin to projecting |+T> along the |+S> direction. Since $\vec{a}\cdot \vec{b}=q$ does not imply $\vec{a}= q*\vec{b}$ unless they lie in the same direction you're statement also isn't always true.

What is happening here is that we are looking to represent a quantum state, $\rvert ++\rangle$, in some other inertial frame. Feynman defines the new basis vectors for our two spin states as $\rvert +'+'\rangle$, $\rvert +'-'\rangle$, $\rvert-'+'\rangle$, $\rvert -'-'\rangle$. Now one just proceeds as in normal linear algebra when changing a basis by projecting onto each. \begin{equation} \lvert ++\rangle=\langle ++ | +'+'\rangle| +'+'\rangle+\langle ++ | +'-'\rangle| +'-'\rangle + \langle ++ |-'+'\rangle|-'+'\rangle+\langle ++ | -'-'\rangle| -'-'\rangle \end{equation}

Now just plug in the values for all inner products above as found in the feynman lectures 12-46.