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When learning measurement basis, my teacher told us $|0\rangle=\frac{1}{\sqrt{2}}(|+\rangle+|-\rangle)$ and said that we can derive it ourselves. Along this, he also mentioned $|+\rangle=\frac{1}{\sqrt{2}}(|0\rangle+|1\rangle)$.

I understand that when we visualize those vectors on a bloch sphere, $|0\rangle$ lies in between $|+\rangle$ and $|-\rangle$, and if we normalize the coefficient, we would get $\frac{1}{\sqrt{2}}$. However, I'm confused how we know that the phase is + ($|+\rangle+|-\rangle$) instead of -? Is this just a definition for $|0\rangle$ or is it backed by a deeper reason?

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You can do it via use of substitution, or via the expansion into vectors and comparison. However, for this and other expansions, I find the use of the identity operator, which can be diagonalised in all bases, is the most informative:

$$|0\rangle=I|0\rangle=(|+\rangle\langle+|+|-\rangle\langle-|)|0\rangle$$

$$=\langle+|0\rangle|+\rangle+\langle-|0\rangle|-\rangle=\frac{1}{\sqrt{2}}|+\rangle + \frac{1}{\sqrt{2}}|-\rangle$$

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    $\begingroup$ This is an interesting approach! Nice $\endgroup$ Jul 29 at 16:06
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A way to see it is by writing the column representation of each state you mentioned:

$$ |0\rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix}, \; |1\rangle = \begin{pmatrix} 0 \\ 1 \end{pmatrix}, \; |+\rangle =\frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 1 \end{pmatrix}, \; |-\rangle =\frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ -1 \end{pmatrix} $$

This way, it is easy to see why the phase of $|-\rangle$ in the state $|0\rangle$ needs to be positive ($+1$) and negative ($-1$) in the representation of $|1\rangle$.

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This can be done entirely in Dirac notation by substituting the definitions

$$ |+\rangle = \frac{1}{\sqrt2}(|0\rangle + |1\rangle) \\ |-\rangle = \frac{1}{\sqrt2}(|0\rangle - |1\rangle) $$

into $\frac{1}{\sqrt2}(|+\rangle + |-\rangle)$. We get

$$ \begin{align} \frac{1}{\sqrt2}(|+\rangle + |-\rangle) &= \frac{1}{\sqrt2}\left(\frac{1}{\sqrt2}(|0\rangle + |1\rangle) + \frac{1}{\sqrt2}(|0\rangle - |1\rangle)\right)\\ &=\frac12|0\rangle + \frac12|1\rangle + \frac12|0\rangle - \frac12|1\rangle \\ &= |0\rangle. \end{align} $$

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