# How to derive $|0\rangle=\frac{1}{\sqrt{2}}(|+\rangle+|-\rangle)$?

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?

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:

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

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

• This is an interesting approach! Nice Jul 29 at 16:06

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 $$|1\rangle$$.

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}