# Rebasing half spin from $|u\rangle$ and $|d\rangle$ to $|+\rangle$ and $|-\rangle$ [closed]

I want to know if my logic in rebasing the pure $$|u\rangle$$ and $$|d\rangle$$ to $$|+\rangle$$ and $$|-\rangle$$ is correct:

$$|+\rangle=\frac{|u\rangle+|d\rangle}{\sqrt{2}}\implies|u\rangle=\sqrt{2}|+\rangle-|d\rangle\tag1$$
$$|-\rangle=\frac{|u\rangle-|d\rangle}{\sqrt{2}}\implies|d\rangle=\sqrt{2}|-\rangle+|u\rangle\tag2$$

substituting $$(2)$$ into $$(1)$$ we get: $$|u\rangle=\frac{\sqrt{2}}{2}(|+\rangle+|-\rangle)$$

substituting $$(1)$$ into $$(2)$$ we get: $$|d\rangle=\frac{\sqrt{2}}{2}(|+\rangle-|-\rangle)$$

• Equation (2) looks wrong to me. It should read $|d\rangle = |u\rangle - \sqrt 2|-\rangle$. Also note that this is in principle a linear algebra problem. Using pure substitutions won't solve it. You have to add/subtract equations to eliminate variables, a method to solve such systems manually is for example the en.wikipedia.org/wiki/Gaussian_elimination. Or you can invert the matrix that defines the relation between both bases. Your results are correct despite your wrong setup, which is easily seen by adding/subtracting your initial equations for $|+\rangle$ and $|-\rangle$. Commented Aug 18, 2021 at 6:52
• Related : Understanding the Bloch sphere. Commented Aug 18, 2021 at 12:02

This is a simple linear algebra problem. We can rephrase it in vector notation as such: take the first basis $$|u\rangle, |d\rangle$$ and consider a vector associated to it $$\vec{u} = (u\quad d)^T$$. Then do the same for the $$|+\rangle, |-\rangle$$ like this $$\vec{v} = (+\quad -)^T$$. Then the change of basis is just a linear transformation to which we associate a square matrix $$\vec{v} = A\vec{u}$$ The matrix $$A$$ can be easily found by the transformation you gave. When you have your matrix, then it is just a matter of inverting it, so that $$\vec{u} = A^{-1}\vec{v}$$
The matrix is just $$A=\frac{1}{\sqrt{2}}\left(\begin{matrix} 1& 1\\ 1&-1 \end{matrix}\right)$$ which is really easy to invert.