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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: $$|d\rangle=\frac{\sqrt{2}}{2}(|+\rangle-|-\rangle)$$

jng224
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al pal
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1 Answers1

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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 $$\vec{u} = A^{-1}\vec{v}$$

Obviously in the case of a system of 2 equations, this method is a bit too much. You can easily get the inverse transformation by substitution, as you did. But in general one can have many equations and by then, even just for three, inverting a matrix is just faster and easier.

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.

Davide Morgante
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