What is the collapse state after measuring Sx?

[Lengalicious]

Homework Statement

So say I have a 1 qubit normalised state |ψ> = 1/√N(c₁|z⁺> - c₂|z^->) and I make the measurement Sx and the measurement yields +hbar/2
as |x⁺> is a superposition of both z⁺ & z^- basis eigenstates does the collapsed state end up the same as the original state? Or would I need to take the inner product of <x⁺|ψ> to find the collapsed state? Confused. .

Homework Equations

The Attempt at a Solution

Its not really a direct homework problem its more of an understanding thing so my attempt is above in 1.

 

[TSny]

It doesn't matter what the original state is. If you know that the outcome of a measurement is "spin up" along the x axis, then the collapsed wave function is simply |x⁺>, independent of the original state |ψ>.

The complex number <x⁺|ψ> is the probability amplitude that the outcome of the measurement yields |x⁺>. If <x⁺|ψ> happened to be zero, then it would have been impossible for the measurement outcome to be spin up along x.

As long as <x⁺|ψ> is nonzero, then there is a nonzero probability |<x⁺|ψ>|² that the outcome is spin up along x. If it does turn out that the outcome is spin up along x, then the measurement has collapsed |ψ> to |x⁺>

 

[Lengalicious]

Ok great, that clears things up a lot, thanks!