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From what I understand, spin in the x direction takes the form $$ |+\rangle_x = a|+\rangle + b |-\rangle $$ (in the z basis), where $a$ and $b$ are complex. You can work out the value of the constants $a$ and $b$ (which is $ \frac{1}{\sqrt{2}} $) from using the results attained in the Stern Gerlach experiment.

During the derivation of spin in the x direction, I understand that a phase factor is added since the constants can be complex, however why is this added to only one of the terms (in this case the $ |-\rangle $ co-efficients).

$$ |+\rangle_x = \frac{1}{\sqrt{2}}(|+\rangle+e^{i\alpha}|-\rangle)$$ $$ |-\rangle_x = \frac{1}{\sqrt{2}}(|+\rangle-e^{i\beta}|-\rangle)$$

(where $ \alpha $ and $ \beta $ are constants)

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  • $\begingroup$ Mixing bra and ket vectors doesn't make much sense. You probably mean $|+\rangle_x$ and $|-\rangle_x$ instead of $\langle+|_x$ and $\langle-|_x$. $\endgroup$ – Thomas Fritsch Sep 3 '19 at 18:59
  • $\begingroup$ Apologises, I have corrected it. $\endgroup$ – Lol Lolling Sep 3 '19 at 19:04
  • $\begingroup$ I'm not sure I understand your question correctly. $|+\rangle_x$ and $|-\rangle_x$ must be orthogonal to each other. Therefore $e^{i\alpha}$ and $e^{i\beta}$ cannot be the same. $\endgroup$ – Thomas Fritsch Sep 3 '19 at 19:17
  • $\begingroup$ Yes they are not, they are different values. $\endgroup$ – Lol Lolling Sep 3 '19 at 19:26
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Adding an overall phase to a quantum state doesn't change anything.

$\frac{1}{\sqrt{2}}\left(|+\rangle+ e^{i \alpha}|-\rangle\right)$ is the same as $\frac{1}{\sqrt{2}}\left(e^{i \delta}|+\rangle+ e^{i (\alpha + \delta)}|-\rangle\right)$, up to an overall phase. So we can freely set $\delta = 0$.

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It is kind of arbitrary if you apply the factors to the $|+\rangle$ or to the $|-\rangle$ components.

Instead of defining $$\begin{align} |+\rangle_x = \frac{1}{\sqrt{2}}(|+\rangle+e^{i\alpha}|-\rangle) \\ |-\rangle_x = \frac{1}{\sqrt{2}}(|+\rangle+e^{i\beta}|-\rangle) \end{align} \tag{1}$$ you could as well define $$\begin{align} |+\rangle_x = \frac{1}{\sqrt{2}}(e^{-i\alpha}|+\rangle+|-\rangle) \\ |-\rangle_x = \frac{1}{\sqrt{2}}(e^{-i\beta}|+\rangle+|-\rangle) \end{align} \tag{2}$$

The vectors $|+\rangle_x$ from (1) and (2) differ only by an over-all phase factor $e^{2i\alpha}$. This is irrelevant to the physics. They are physically the same state.

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