0
$\begingroup$

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

$\endgroup$
2
  • 1
    $\begingroup$ 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 $|-\rangle$. $\endgroup$
    – Hans Wurst
    Commented Aug 18, 2021 at 6:52
  • 1
    $\begingroup$ Related : Understanding the Bloch sphere. $\endgroup$
    – Frobenius
    Commented Aug 18, 2021 at 12:02

1 Answer 1

2
$\begingroup$

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.

$\endgroup$
0

Not the answer you're looking for? Browse other questions tagged or ask your own question.