This might seem obvious to some but I would like to know what is exactly the purpose of this? And if there is a change of basis...it's just simply taking the new basis and writing it in terms of the old basis right?

  • $\begingroup$ Can you give some more details about what you're actually asking? You can write the state and measurement operators with respext to whatever basis you want, this doesn't affect the results. $\endgroup$
    – Rammus
    Dec 13, 2020 at 21:37
  • $\begingroup$ But if you say are measuring sigma_z, why does it matter then which basis you are in? $\endgroup$
    – mikanim
    Dec 15, 2020 at 21:35
  • $\begingroup$ Well the crux is what does $\sigma_z$ mean? This probably means the operator corresponding to the matrix $\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$ when written in the computational basis. But if you choose another basis to work in then the matrix representing $\sigma_z$ will change. $\endgroup$
    – Rammus
    Dec 16, 2020 at 9:11

1 Answer 1


I'm guessing that what you mean is the following: if you have a state $|\psi\rangle$ to be measured in an orthonormal basis $|\phi_+\rangle,|\phi_-\rangle$, then you should express $|\psi\rangle$ as $$ |\psi\rangle=a|\phi_+\rangle+b|\phi_-\rangle. $$ (I assume this is what you mean by being "in the basis"?) If you can do it like this, then it makes a huge difference to being able to read of the probabilities of measurement outcomes: the corresponding probabilities are simply $|a|^2,|b|^2$.

But, really, it's not that important. You could have $|\psi\rangle$ expressed in any basis, because all you have to calculate is $|\langle\phi_+|\psi\rangle|^2$ and $|\langle\phi_-|\psi\rangle|^2$.

Equally, as you mention, you could just perform a basis transformation. Imagine that you had $$ |\psi\rangle=\alpha|0\rangle+\beta|1\rangle, $$ then if you construct $$ U=|0\rangle\langle\phi_+|+|1\rangle\langle\phi_-|, $$ calculating $$ U|\psi\rangle=a|0\rangle+b|1\rangle, $$ so reading off the amplitudes of the 0 and 1 components gives you exactly the components you need. To see this, note $$ \langle0|U|\psi\rangle=\langle\phi_+|\psi\rangle. $$


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