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enter image description hereI know that $\left|_x\langle+|-\rangle\right|^2 = 1/2$, so is it just as simple as taking the square root of $1/2$? Thanks.

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  • $\begingroup$ Why is there an $x$? $\endgroup$ – G. Smith Aug 22 at 0:54
  • $\begingroup$ It is supposed to represent the up state in the x axis; my textbook has it written this way (a subscript one). @G.Smith $\endgroup$ – mw.astronomy Aug 22 at 0:56
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    $\begingroup$ Inner products like that are complex numbers. The vertical bars mean its complex magnitude. There are an infinite number of complex numbers with a particular magnitude. $\endgroup$ – G. Smith Aug 22 at 1:00
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    $\begingroup$ This is an XY problem... the real answer is to use a different book. Just from the picture I can tell this one is amazingly unclear. $\endgroup$ – knzhou Aug 22 at 1:09
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$|+\rangle_x$ and $|-\rangle_x$ are the eigenvectors of the $\sigma_x$ Pauli matrix. In the $|\pm\rangle_z$ basis (which is the usual basis, so we will drop the $z$), it reads,

$$ \sigma_x = \begin{pmatrix} 0 & 1\\1 & 0 \end{pmatrix} $$

The eigenvectors are

$$ \begin{align} v_+ &= \alpha\begin{pmatrix} 1\\1 \end{pmatrix}\quad \text{with eigenvalue }\lambda = +1\\ v_- &= \beta\begin{pmatrix} 1\\-1 \end{pmatrix}\quad \text{with eigenvalue }\lambda = -1 \end{align} $$

The coefficients $\alpha$ and $\beta$ can be any number, since an eigenvector stays an eigenvector even after we multiply it with any scalar number. But let's say we want to normalize the eigenvectors. To do this, we can write them like this:

$$ \begin{align} v_+ &= \frac{\tilde \alpha}{\sqrt 2}\begin{pmatrix} 1\\1 \end{pmatrix}\quad \text{with eigenvalue }\lambda = +1\\ v_- &= \frac{\tilde \beta}{\sqrt 2}\begin{pmatrix} 1\\-1 \end{pmatrix}\quad \text{with eigenvalue }\lambda = -1 \end{align} $$

where now $\tilde \alpha$ and $\tilde\beta$ can only be a phase factor $\text{e}^{\text{i}\varphi_{\alpha,\beta}}$. Usually, one chooses that $\varphi_{\alpha,\beta}=0$, but let's keep them for now.

If we identify $v_\pm$ with $|\pm\rangle_x$, we have

$$ \begin{align} |+\rangle_x &= \frac{\tilde \alpha}{\sqrt 2}\big( |+\rangle + |-\rangle \big)\\ |-\rangle_x &= \frac{\tilde \beta}{\sqrt 2}\big( |+\rangle - |-\rangle \big)\\ \end{align} $$

As you can verify, if we take the inner product ${}_x\langle +|-\rangle$, we get

$$ {}_x\langle +|-\rangle = \frac{\tilde \alpha^*}{\sqrt 2}\big( \langle+| + \langle-| \big)\,|-\rangle = \frac{\tilde \alpha^*}{\sqrt 2}\langle -|-\rangle = \frac{\tilde \alpha^*}{\sqrt 2} $$

As you can see, if you only take the square root of $1/2$, you wouldn't get this phase factor $\tilde\alpha^* = \text{e}^{-\text{i}\varphi_\alpha}$. However, we are allowed to choose the phase such that $\varphi_\alpha=0$, which leads to

$$ {}_x\langle +|-\rangle = \frac{1}{\sqrt 2} $$

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No. Inner products like that are complex numbers. The vertical bars mean its complex magnitude. There are an infinite number of complex numbers with a particular magnitude. They all lie on the same circle in the complex plane. So you know the magnitude of the complex number but not its direction.

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  • $\begingroup$ Okay so what would you do to find the inner product? $\endgroup$ – mw.astronomy Aug 22 at 1:04
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    $\begingroup$ @mw.astronomy Express $|+\rangle$ and $|-\rangle$ in some basis and compute the inner product. $\endgroup$ – Aaron Stevens Aug 22 at 1:21

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