Why are mutually exclusive states orthogonal?

[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?

 

[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

 

[Demon117]

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

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?

 

[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

 

[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.)

 

[Fredrik]

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 I am 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.

 

[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.

 

[Fredrik]

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.

 

[Demon117]

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.

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 into thinking it is a far more difficult concept than what is reality.