# States of spin of a quantum mechanical particle

Assuming a spin is prepared in the positive $$x$$-direction ($$|r\rangle$$) and a measuring apparatus is oriented on the $$z$$ axis, does this equation apply?

$$|r\rangle=\frac{1}{\sqrt{2}}|u\rangle+\frac{1}{\sqrt{2}}|d\rangle$$

What would the equation for $$|l\rangle$$ be? Explain how the equation was obtained.

If your axis of measurement aligns with the z-axis, then it's convention to choose $$|u\rangle = \begin{pmatrix}1\\0\end{pmatrix}, |d\rangle = \begin{pmatrix}0\\1\end{pmatrix},$$ representing spin up and spin down along the z-axis. Then your eigenstates along the x direction would be$$^1$$: $$|r\rangle = \frac{1}{\sqrt 2}\begin{pmatrix}1\\1\end{pmatrix}, |l\rangle = \frac{1}{\sqrt 2}\begin{pmatrix}1\\-1\end{pmatrix}$$ So your equation is correct. Furthermore, $$|l\rangle = \frac{1}{\sqrt 2}|u\rangle - \frac{1}{\sqrt 2}|d\rangle$$
$$^1$$ See, for example: Spin-½ - Wikipedia
• Plug in $|u\rangle$ and $|d\rangle$, if the sign is positive you won't get $|l\rangle$ – Hanting Zhang Nov 25 '18 at 17:56