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View Full Version : quantum mechanics, way of writing the eigenstates


Chronos000
Apr1-11, 03:05 PM
1. The problem statement, all variables and given/known data

I'm having some trouble understanding exactly how the eigenstates of this matrix are being presented.

A= ( 0 -i
i 0 ) <- matrix

I can find the eigenvalues to be +/- 1 which gives the eigenvectors to be x=-iy or x=iy.

The eigenvectors are then being presented as:

|A+> = (|1> +i|2> )/root2

|A-> = (|1> -i|2> )/root2

I thought to get this you take the action 2 of A away from action 1 and the resolve it.(for A-). But I cant seem to get this.

I really just need some clarification on this as I have a further complicated matrix to do

grey_earl
Apr1-11, 03:52 PM
What they give you as |1> is simply the vector (1, 0), and |2> is the vector (0,1), so |A+> = (1,i)/sqrt(2) which satisfies x = -i y, as you got. Similar for |A->.

Chronos000
Apr1-11, 04:24 PM
thanks for your reply. I think I may be blind here- but I don't see how A=(1,i)/sqrt2 satisfies x=-iy. - if x=1 and y equals 1 from the vectors you spoke of

JaWiB
Apr1-11, 07:38 PM
|A+> = (1,i) = |1>+i|2>

i.e., x=1 and y=i, which satisfies x=-i*y

similarly for the |A-> eigenvector.