Freedom in choosing elements/entries of an eigenvector

Question

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,

1. We can assume, $a=1$. The next assumption could be $c=1$, giving us $b=-i$. Eigenvector being $\left[ {\begin{array}{ccc} 1 & -i & 1\\ \end{array} } \right]$.
2. 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.

Answer 1

The freedom in choosing eigenvectors comes from the fact that the eigenvectors of $H$ with an eigenvalue $E$ form a vector-space. This means that if a solution exists for the eigenvalue problem $Hv = Ev$ for some $E$, then there exist infinitely many solutions.

One example would be to rescale the vector $v$: a vector $\alpha v$ would also be a solution (you may check that this is generally true). This leads to a freesom in choosing the components- we can rescale them how we wish and still get an eigenvector. Another freedom may come from the fact that another independent solution exists (or many).

The num'ber of independent solutions $v$ to $Hv = Ev$ is called the dimension of the eigenspace, or the geometric multiplicity of $E$.

Answer 2

The eigenspace of $H$ (a subspace of $\mathbb{C}^3$ here) corresponding to the eigenvalue 1 is the following vector space.

$$\ker (1\cdot \mathbb{I}_3 - H)= \left\{v=\begin{pmatrix}a\\b\\c \end{pmatrix}\in \mathbb{C}^3 \hspace{0.1cm} ; \hspace{0.1cm}\underbrace{\begin{pmatrix} 0 & 0 & 0 \\ 0 & 1 & -i \\ 0 & i & 1 \end{pmatrix} }_{=\mathbb{I}_3 -H}\begin{pmatrix}a\\b\\c\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix}\right\}$$

Then, $\ker(1\cdot \mathbb{I}_3-H) = \{(a,b,c)\hspace{0.1cm};\hspace{0.1cm} a,b,c\in\mathbb{C}\hspace{0.1cm}\text{and}\hspace{0.1cm}c=ib \}$ so the eigenvectors are in the set $\ker(\mathbb{I}_3-H)\backslash\{0_{\mathbb{C}^3}\}$ because we don't define the 0-vector as an eigenvector as it is always in $\ker(\lambda \mathbb{I}_3 - H)$, for any eigenvalue $\lambda$. If you like, you could then decompose the eigenspace, $$\ker(1\cdot\mathbb{I}_3 - H) = \mathbb{C}\oplus\{(b,c)\hspace{0.1cm};\hspace{0.1cm}b,c\in\mathbb{C}\hspace{0.1cm}\text{and}\hspace{0.1cm}c=ib\},$$ which would allow you to do some computations with eigenstates of this Hamiltonian (corresponding to $\lambda = 1$) by using only 2-vectors without changing the physics. This seems like a good example of an eigenstate having a nice symmetry at a particular energy.

Somewhat analogously, when using the tight-binding Hamiltonian (used lots in solid state physics), there are certain aperiodic lattices for which at a distinguished energy you get a nice symmetry (a fractal state) and every other energy corresponds to a multifractal state. You can check some papers of Kohmoto, Sutherland, Tang, et al. from the 80's, if you're interested; some mathematicians are still interested in formalizing aspects of such things.