# Mutually Exclusive States

1. Dec 12, 2011

### Demon117

I understand that it is important for two eigenvectors to be orthogonal, but what is it exactly about mutually exclusive states that makes them orthogonal?

2. Dec 12, 2011

### facenian

they don't have to be orthogonal because if they were when you express one of then in the orthogonal base of the other observable you would get a null vector

3. Dec 12, 2011

### Demon117

Thats strange, maybe I just misunderstood but my professor said that mutually exclusive states must be represented by orthogonal kets. So, why would he say that if its not true?

4. Dec 12, 2011

### facenian

Well, maybe he is right and there's something wrong in my reasoning. Let's wait and see if someone else has something to say

5. Dec 12, 2011

### Truecrimson

It's the other way around. Orthogonal states, by the rule of probability in quantum mechanics, are mutually exclusive. Otherwise, you can't distinguish different outcomes of a measurement. (There is a generalization to this though.)

6. Dec 12, 2011

### Fredrik

Staff Emeritus
Eigenvectors corresponding to different eigenvalues of a self-adjoint operator are always orthogonal.

\begin{align}
&Ax=\lambda x\\
&Ay=\mu y\\
\\
& \lambda^*\langle x,y\rangle=\langle\lambda x,y\rangle=\langle Ax,y\rangle=\langle x,A^*y\rangle=\langle x,Ay\rangle\\
& \mu\langle x,y\rangle=\langle x,\mu y\rangle=\langle x,Ay\rangle\\
&\Rightarrow\ (\lambda^*-\mu)\langle x,y\rangle=\langle x,Ay\rangle-\langle x,Ay\rangle=0
\end{align}
This calculation shows that eigenvalues of self-adjoint operators are real (because if x=y, then $\mu=\lambda$, and the result we found says that $(\lambda^*-\lambda)\|x\|^2=0$, which implies that Im λ=0). This implies that our result can be written as $(\lambda-\mu)\langle x,y\rangle=0$, and if $\lambda\neq\mu$, this implies that $\langle x,y\rangle=0$.

Last edited: Dec 12, 2011
7. Dec 13, 2011

### facenian

After this other two contributions I think know I understand better what your professor might have said, he was not talking of general states he had in mind eingenstates of the same obsevable then his assertion is simply the theorem of linear algebra that Fredrik explained.

8. Dec 13, 2011

### Fredrik

Staff Emeritus
It's not entirely clear to me what he meant by "mutually exclusive". If it refers to the states corresponding to different results of the same measurement, then the theorem above is the answer. But he might have been talking about zero transition probability, and in that case, the statement is kind of trivial, once we understand what it says.

The probability that a system prepared in state $|\psi\rangle$ will end up in state $|\phi\rangle$ after a measurement of an observable that has $|\phi\rangle$ as the only eigenvector corresponding to some specific result, is $|\langle\phi|\psi\rangle|^2$. He could mean that $|\psi\rangle$ and $|\phi\rangle$ are mutually exclusive if that probability is 0. That would of course imply that these state vectors are orthogonal.

9. Dec 13, 2011

### Demon117

After looking at a few of these explanations it seems like you are explaining exactly what he had in mind, if I am looking at my notes correctly. It is a rather trivial statement, but once and a while you come across vocabulary which tricks you in to thinking it is a far more difficult concept than what is reality.