# An eigenstates, eigenvectors and eigenvalues question

1. May 11, 2013

### pigletbear

Good evening :-)

I have an exam on Wednesday and am working through some past papers. My uni doesn't give the model answers out, and I have come a bit stuck with one question. I have done part one, but not sure where to go from here, would be great if someone could point me in the right direction:

S2) Show that the state vectors |Sx+> = $\frac{1}{\sqrt{}2}$ times a 2x1 matrix (1,1) and |Sy+> = $\frac{1}{\sqrt{}2}$ times a 2x1 matrix (1,-1) are eigenvectors of Sx = h/2 times a 2x2 matrix (0 1, 1 0) with respective eigenvalues plus and minus h/2...

This I can do by using |A - λI| = 0, finding the eigenvalues, then using A.v=λv and setting up simutaneous quations to find the eigenvalues

Part two... Of what operator is the state $\frac{1}{sqrt{}2$}[/itex]/$(|Sx+> + |Sy+>) and eigenstate, and with what eigenvalue... Any help would be great and much appreciated 2. May 11, 2013 ### Ilmrak Hello! My suggestion is to try to explicitly add up the two states in matrix notation... The answer should then be obvious to you :) Edit: The answer you gave to the first question is right. Nevertheless it is more time consuming then it was necessary and time is precious during exams :) You have to show that those are egenvectors with given eigenvalue, so you could simply show that [itex]S_x |S_x ± \rangle = ± \frac{h}{2}|S_x ± \rangle$,

without solving the egenvalue equation

$|S_x-\lambda I|=0$.

Ilm

Last edited: May 11, 2013
3. May 11, 2013

### Dreak

$|S_x-\lambda I|=0$. is a lot easier and faster to solve ;)

about the second question, it works in the same way:

A(|Sx+> + |Sy+>) = λ (|Sx+> + |Sy+>)

=> (A-λI)(|Sx+> + |Sy+>) = 0

A-λI = 0; Fill in λ = 1/sqrt(2) and the matrix will be (1/sqrt(2) 0; 0 1/sqrt(2) )

It's been a while since I did this, I could be wrong ofcourse.. (and it looks a bit to easy, but hey?)

edit: I think I'm wrong...

4. May 11, 2013

### Ilmrak

Yes it is easy but no, it isn't faster.
And if it were a bigger matrix (or worst a differential operator) it wouldn't be so easy to solve that equation while it would still be easy to let a matrix (or a differential operator) act on a vector (or a function).
To solve an equation is (almost) always more difficult than checking one solution ^^

Ilm