Question

Lin. algebra

Answer 1

@ganeshie8 @zepdrix

Answer 2

the results are eigenvalues of the operator
the probabilities are the coefficients of the eigenfunctions
squared

Answer 3

So like a) e and probability is 1

Answer 4

Give it your best shot and I'll help you out. I'm sorta in the middle of helping someone else so I'll be here to help more in a minute

Answer 5

Alright so we started out with
\[| \psi \rangle = 1 | 1 \rangle + 0 | 2 \rangle\]
the probabilities add up to 1 like we expect, \(1^2+0^2=1\) so that's good.

So your answer is right, we get \(\varepsilon\) with probability 1 since the vectors \(|1\rangle\) and \(|2 \rangle\) are eigenvectors of the \(\hat Z\) operator. However these are not eigenvectors of the Y operator, this is important in a minute.

So they want you to measure Y after Z, so we do that:
\[\hat Y \hat Z |\psi\rangle\]
What do you get? You must now represent your state vector in the new \(\hat Y\) basis if you want to find out what is observed, since the observables are eigenvalues of this operator. It wouldn't make sense to use the \(\hat Z\) eigenvectors since they have no corresponding eigenvalues here. Kinda weird to explain maybe, so ask if you need me toclarify anything.

For now, do the multiplicaton as best as you can, and then express your new state vector in the Y eigenbasis.

Answer 6

do you want me to multiply YZ?

Answer 7

It doesn't matter since matrix and vector multiplication is associative, you can multiply:

\[Y(Z|\psi \rangle ) = (YZ)|\psi \rangle \]

Remember, \(|\psi\rangle\) is just a 2x1 column vector, it just looks fancy but it's nothing special really.

Answer 8

I think I forgot to answer your question directly, I want you to multiply all 3 of those things together, not just he two matrices haha.

Answer 9

Okay,