I want to understand why there is freedom in choosing entries of an eigenvectors on some instances. I will take up a particular Hamiltonian to explain this. $$H=H_0 \left[ {\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & -i \\ 0 & i & 0 \end{array} } \right]$$ This Hamiltonian has $H_0,H_0,-H_0$ as its eigenvalues. When we go ahead and solve for the eigenvectors for one of the eigenvalue, say for $+H_0$. Assuming that the entries of the eigenvector are $\left[ {\begin{array}{ccc} a & b & c\\ \end{array} } \right]$, we get the following equations $$a=a$$ $$-ic=b$$ $$ib=c$$ The last equation is same as the second one. Giving us $a=a$, $-ic=b$ as the final equations. At this point we make assumptions,
- We can assume, $a=1$. The next assumption could be $c=1$, giving us $b=-i$. Eigenvector being $\left[ {\begin{array}{ccc} 1 & -i & 1\\ \end{array} } \right]$.
- Or we could also assume, $a=0$. Now $c=i$, giving $b=1$. The eigenvector now changes to $\left[ {\begin{array}{ccc} 0 & 1 & i\\ \end{array} } \right]$
There are other possible combinations giving us different eigenvectors, with most of them working fine. I want to understand why is there such a freedom in choosing the entries of the eigenvectors. I am particularly interested in understanding this in the context of quantum mechanics. I do know that irrespective of our choice of the elements, somehow the completeness of the eigenvectors is satisfied; so is the property of eigenvectors being orthonormal(although sometimes this needs to be put in there by hand). Even though the coefficients of the eigenvectors are somewhat influenced by the choice we make, the coefficient squared which gives us the probability is irrespective of the choices we make.
So, why is there such a freedom? Are there any implications of this freedom?
Edit: I am only slightly worried about the coefficient, my major concern is about the freedom in choosing the elements of the eigenvector. I do understand that changing the phase does not really change the essence of the state as the inner product still remains the same. But if we see the two choices of the eigenvectors corresponding to the same eigenvalue, the inner product of those both choices is different.