# Wavefunction collapse for non-commuting operators

jekeckel
Science 20 Mar '10 05:43
1. 20 Mar '10 05:43
For two commuting operators, one can find a common eigenbasis. For simplicity, assume the eigenvalues here are nondegenerate. In this case, after applying the first operator, the state collapses into one of the basis eigenvectors. Since this is also an eigenvector of the second operator, we essentially know which eigenvalue will be returned when the second operator is applied to this collapsed state. This much makes sense to me.

But what about for non-commuting operators, where no common eigenbasis can be found? Lets say the original, uncollapsed state vector is a linear combination of the eigenvectors for the first operator. After applying the first operator to the state, it collapses into one of its eigenvectors. What exactly happens after applying the second, non-commuting operator into this collapsed state? Do we now find some basis comprised of eigenvectors for the second operator, which superpose into the collapsed state, and then make the measurement as usual? Or is there some deeper, more intricate process going on?
2. 20 Mar '10 05:43
Omg, I'm terrible about logging off of her account. Sorry guys.
Baby Gauss
20 Mar '10 12:27
Originally posted by jekeckel
But what about for non-commuting operators, where no common eigenbasis can be found? Lets say the original, uncollapsed state vector is a linear combination of the eigenvectors for the first operator. After applying the first operator to the state, it collapses into one of its eigenvectors. What exactly happens after applying the second, non-commuting op ...[text shortened]... d then make the measurement as usual? Or is there some deeper, more intricate process going on?
If both operators are observables, let us call them A and B, we know that we can span the full space on some set of eigenvectors, a_n and b_n, and that there is no common set of eigenvectors, c_n, to both operators that can span the whole space.

Think about it this way. Imagine that you're measuring the spin of an electron in the x and y directions. These are observables and they don't commute. Let us denote the operators as X and Y. If you first apply X and measure it as -h/2 (sorry there is no h bar here) you don't have the flimsiest idea of what you'll get when you apply Y to the previous state (except of course that it will be -h/2 or h/2).
On the other hand if you're measuring X and the square of the total spin, L^2, there is always absolute certainty of what you're going to get in the final state. It doesn't matter if you first measure X and then L^2 or if you first measure L^2 and then X.
The thing is that once you've measured one of them you already have enough information to know the other.

On the other hand if you measure X first there isn't enough information for you to know what Y or Z would be like.

If the operators in question aren't observables things get a little bit more complicated and I don't remember enough of it... ðŸ˜›
4. 20 Mar '10 21:13
Thanks a lot for the explanation!
5. 21 Mar '10 02:03
Originally posted by amolv06
Omg, I'm terrible about logging off of her account. Sorry guys.
she'd probably have a 1600 rating if you didn't accidentally move for her! ðŸ˜
6. Palynka
Upward Spiral
21 Mar '10 12:40
Originally posted by amolv06
Omg, I'm terrible about logging off of her account. Sorry guys.
Show off. ðŸ˜›