# What is the collapse state after measuring Sx?

1. Dec 7, 2013

### Lengalicious

1. The problem statement, all variables and given/known data

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. .

2. Relevant equations

3. 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.

2. Dec 7, 2013

### 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+>

3. Dec 7, 2013

### Lengalicious

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