# Eigenvalues and operators, step the involves switching and substitutin

1. Sep 15, 2014

### rwooduk

Operator C = I+><-I + I-><+I

Wavefunction PSI = Q I+> +V I->

C PSI = Q I-> + V I+>

note the I is just a straight line (BRAKET vectors), the next step is where I get confused, p is subbed in and the ket vectors switch places...

C PSI = pQ I+> + pV I-> <---- why??

therefore

V = pQ and Q = pV

which gives p^2 Q = Q therefore p = +/- 1

why can p be subbed in and the ket vectors switch around, i.e. I-> no longer with Q and I+> no longer with V

thanks

2. Sep 15, 2014

### kith

What is p? What are you even trying to do? What source of information are you using?

3. Sep 15, 2014

### rwooduk

Hi, im reviewing last semesters quantum mechanics, the complete question is:

Determine the eigenvalues of the operator C = I+><-I + I-><+I. You may use PSI = Q I+> + V I-> as an eigenvector.

so p would be the eigenvalues of the operator C which are +1 and -1

the solution introduces p to C PSI = Q I-> + V I+> by switching the ket vectors around and putting p with each term, I don't understand why this is.

4. Sep 15, 2014

### Fredrik

Staff Emeritus
Please don't write I. There's a vertical line on the keyboard that you can use. Or you can copy and paste this one: |

I assume that Q and V are just numbers. What is p and how did it enter the calculation?

Oh wait, p is the eigenvalue, right? Then by definition of eigenvalue and eigenvector, you have $C|\psi\rangle =p|\psi\rangle$. Now you can just use the definition of $|\psi\rangle$ to get
$$C|\psi\rangle =p|\psi\rangle =p(Q|+\rangle+V|-\rangle)=pQ|+\rangle+pV|-\rangle.$$ Since you already know that
$$C|\psi\rangle=Q|-\rangle+V|+\rangle,$$ this implies that pQ=V and pV=Q.

Last edited: Sep 15, 2014
5. Sep 15, 2014

### rwooduk

oh dear I missed something in the solution:

Recall the eigenvalue equation: C |PSI> = p |PSI>

that would explain where the p comes from, but it doesn't explain why the ket vectors switch around... or does it?

Yes Q and V are just numbers.

Apologies on the I.

6. Sep 15, 2014

### Fredrik

Staff Emeritus
If p is the eigenvalue, then by definition of eigenvalue and eigenvector, you have $C|\psi\rangle =p|\psi\rangle$. Now you can just use the definition of $|\psi\rangle$ to get
$$C|\psi\rangle =p|\psi\rangle =p\big(Q|+\rangle+V|-\rangle\big)=pQ|+\rangle+pV|-\rangle.$$ Since you already know that
$$C|\psi\rangle=Q|-\rangle+V|+\rangle,$$ this implies that
$$(pQ-V)|+\rangle +(pV-Q)|-\rangle=0.$$ Since the basis vectors are linearly independent, this implies that pQ-V=0 and pV-Q=0.

7. Sep 15, 2014

### rwooduk

Ahhh I see now, that's great!! Thanks very much for your time, really appreciated!