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I'm currently doing a bit of quantum mechanics, and I can't figure out how to pick out eigenvectors. Let me explain through an example. An operator $A= \begin{bmatrix} 1 &0 &0 \\ 0&0 &-i \\ 0 &i &0 \end{bmatrix}$ acts on a state $|\psi(t) \rangle$. For the time being, the exact value of the state is not our concern. Quantum mechanics says that an action of an operator on a state vector forces the state vector to change to one of the Eigenstates of said operator. The eigenvalues turn out to be $a_1=-1$, and $a_2=a_3=1$, which happens to be doubly degenerate. This much I am clear. Now the method I use to figure out the Eigenstates is $A\psi=a_n \psi$. Let us consider $a_2=a_3=1$. If we take $\psi=\begin{bmatrix} x \\ y\\ z\\ \end{bmatrix}$ we get, $x=x$, $-iz=y$ and $iy=z$. This system of equations have an infinite number of solutions. A few of them being $\begin{bmatrix} 1\\ 0\\ 0\\ \end{bmatrix}$, $\dfrac{1}{\sqrt{2}}\begin{bmatrix} 0\\ -i\\ 1\\ \end{bmatrix}$ , $\dfrac{1}{\sqrt{2}}\begin{bmatrix} 0\\ 1\\ i\\ \end{bmatrix}$ and so on. The book I refer to takes the first 2 that I mentioned as the degenerate Eigenstates. I don't understand what the latter was not chosen? Am I missing something, or is the choice of Eigenstates completely arbritrary? Is there a reason why the first 2 are chosen and not the third?

P.S: The probability of $\psi$ to fall to any of these states are the same which makes me think that there is no real distinction between these Eigenstates.

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    $\begingroup$ Note that in this particular case, since the matrix is block diagonal with blocks of the form $\text{diag}(I_{1\times 1},\sigma_y)$ you can diagonalise each block in turn. The first block is trivial and the second is a Pauli matrix that is probably best to just get familiar with. $\endgroup$
    – jacob1729
    Oct 7, 2020 at 14:05
  • $\begingroup$ I believe your book may have actually missed out on a solution - see the edit to my answer for details. $\endgroup$
    – Drubbels
    Oct 7, 2020 at 14:46

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Okay, let's just do this from the beginning. We have a Hermitian matrix:

$$A = \left[ \begin{array}{c|c} I_{1\times 1} \\ \hline & \sigma_y \end{array} \right]$$

where $I_{1\times 1}$ is the $1\times 1$ identity matrix and $\sigma_y$ is the second Pauli matrix. The upper block is $1\times 1$ so is already diagonal: the eigenvalue is $1$ and the eigenvector is $(1,0,0)^T$. We need to diagonalise the Pauli matrix next. Lets label this eigenvector $|1\rangle$.

Now, at a certain point you end up just memorising the eigenvalues of $\sigma_x,\sigma_y$ but for now you can note that since $\sigma^2=I_{2\times 2}$ and $\text{Tr}(\sigma)=0$ the eigenvalues are $\pm 1$ and then you can solve for the eigenvectors as you did in the OP. We'll label these $|+\rangle,|-\rangle$. These eigenvectors are unique within this $2\times 2$ subspace up to multiplication by a complex number $\lambda$. Clearly if $\sigma_y |+\rangle = |+\rangle$ then we also have $\sigma_y \lambda |+\rangle = \lambda|+\rangle$.

Now the issue comes about because of the block diagonal structure: we have $A|1\rangle = |1\rangle$ but also $A|+\rangle = |+\rangle$ and $A|-\rangle = -|-\rangle$. You should be able to convince yourself that any linear combination of $|1\rangle,|+\rangle$ has eigenvalue $1$. So even though we had a unique choice of eigenvectors put upon us whilst working in the $1\times 1$ and $2\times 2$ subspaces, we now have an ambiguity.

But that's okay! There is indeed a $2D$ space of states with eigenvalue $1$. Your book picks the two most 'natural' ones in some sense (corresponding to our $|1\rangle$ and $|+\rangle$). but you could 'rotate' this basis within the 2D plane they lie in and get a different basis. Since this is a $2D$ subspace we sometimes say "there are two degenerate states with eigenvalue $1$". Technically there's infinitely many states but we only include basis states in that counting.


(Minor aside: block diagonalisability often happens because a space breaks up into different symmetry sectors. Thus it's probably best to work with your book's answers since this might respect the symmetry of the underlying situation best.)

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The third vector is equal to $-i$ times the second one. Thus, it doesn't matter which one you choose: the second and third are in fact 'the same' eigenvector, and you can choose either one, or any other (preferably unit-length) multiple of them. The matrix has only two (independent) eigenvectors.

EDIT: I just idly ran through the calculations myself, and I seem to get a different eigenvector for the eigenvalue $-1$ than you. Instead of $\dfrac{1}{\sqrt{2}}\begin{bmatrix} 0\\ 1\\ i\\ \end{bmatrix}$, I get $\dfrac{1}{\sqrt{2}}\begin{bmatrix} 0\\ i\\ 1\\ \end{bmatrix}$, which does give rise to three distinct (i.e. linearly independent) eigenvectors. Could it be that your book got it wrong?

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  • $\begingroup$ does it imply that it would only have 2 independent eigenvectors, and no more? $\endgroup$ Oct 7, 2020 at 14:01
  • $\begingroup$ It's been a while since I've actually done any QM or linear algebra, so I'm hesitant to answer confidently and definitively, but yes, I believe so. $\endgroup$
    – Drubbels
    Oct 7, 2020 at 14:04
  • $\begingroup$ Remember that any given matrix is not actually guaranteed to have any eigenvectors, much less to have the maximal number of independent ones. $\endgroup$
    – Drubbels
    Oct 7, 2020 at 14:09
  • $\begingroup$ Yes, but this particular one was given as an example of an observable. My question preassumes the existence of such eigenvectors. Also, in this particular case, each eigenstate satisfies $A\psi=a_n\psi$ $\endgroup$ Oct 7, 2020 at 14:13
  • $\begingroup$ Well as I said, it's been a while for me, so it's not out of the question that I might make some elementary mistakes - but $\begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & -i \\ 0 & i & 0\\ \end{bmatrix}$ $\begin{bmatrix} 0\\ i\\ 1\\ \end{bmatrix}$ = $\begin{bmatrix} 0\\ -i\\ -1\\ \end{bmatrix}$ = -1 $\begin{bmatrix} 0\\ i\\ 1\\ \end{bmatrix}$, no? $\endgroup$
    – Drubbels
    Oct 7, 2020 at 16:35
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Without going through all the calculations that you and other answers have done:

In quantum mechanics, if we obtain degenerate eigenvalues for a given operator, we construct/choose the eigenvectors such that they are mutually orthogonal to each other and eigenvectors of other non-degenerate eigenvalues.

Given your three choices for $\lambda=1$, $|\alpha\rangle = \begin{bmatrix}1\\ 0\\ 0 \end{bmatrix}$, $|\beta\rangle =\frac{1}{\sqrt{2}}\begin{bmatrix}0\\ -i\\ 1 \end{bmatrix}$, and $|\gamma\rangle =\frac{1}{\sqrt{2}}\begin{bmatrix}0\\ 1\\ i \end{bmatrix}$, only $\langle\alpha | \beta\rangle = 0$ and $\langle\alpha | \gamma\rangle = 0$.

$\langle\gamma | \beta\rangle = -i$ so those are not orthogonal. In fact $$|\gamma\rangle = e^{i\frac{\pi}{2}} | \beta\rangle,$$ so they are merely rotations of each other and are not linearly independent. So you can choose $| \alpha\rangle$ for one of the $\lambda=1$ and either $| \beta\rangle$ or $| \gamma\rangle$ for the other $\lambda=1$.

Eigenvectors of QM operators are constructed to be linearly independent, even if the eigenvalues are degenerate.

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We postulate that different quantum states represent same physics state, when they only differ by multiplying a complex constant (i.e. all wavefunctions are physically same if they have same "direction").

Your 2nd and 3rd are same physics state, as they only differ by a $i$.

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  • $\begingroup$ Yes. but there are more possible states. does it then imply that there are only 2 independent eigenstates? $\endgroup$ Oct 7, 2020 at 14:09
  • $\begingroup$ @AndrewMichealAnderson you mean how many eigenstates you get? Depends on how many eigenvalues you get. # of eigenstates can't exceed # of eigenvalues. $\endgroup$
    – Shing
    Oct 7, 2020 at 14:13
  • $\begingroup$ @Shing no, you can have multiple eigenvectors with the same eigenvalue. The relevent issue is linear independence. $\endgroup$
    – jacob1729
    Oct 7, 2020 at 14:16
  • $\begingroup$ @jacob1729 you mean repeated eigenvalues? I should clarify that I consider +1, +1 are two eigenvalues (once you solve the det, and find them). $\endgroup$
    – Shing
    Oct 7, 2020 at 14:20
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I think you are confusing there being infinite solutions with degeneracy. Let's label the vectors first to make things more clear. When I solved the eigensystem I got slightly different eigenvectors so I'm not sure what happened here. (I used Mathematica which is probably right) $$v_1=\pmatrix{0\\i\\1}\\v_2=\pmatrix{0\\-i\\1}\\v_3=\pmatrix{1\\0\\0}$$ This way the indices of the eigenvectors matches that of the eigenvalues: $Av_i=a_iv_i$. Keep in mind that you can always match an eigenvector with its corresponding eigenvalue. Now for any eigenvector you can rescale it by some (complex) constant and get the same eigenvector. If you take $v'=\alpha v$ then it still has the same eigenvalue so it is considered the same eigenvector. Often eigenvectors are normalised for convenience but this is not necessary.

Now degeneracy means that two or more eigenvectors share the same eigenvalue. Notably a linear combination of two degenerate eigenvectors is still an eigenvector with the same eigenvalue If you take two degenerate eigenvectors $v_i,v_j$ sharing eigenvalue $\lambda$ then $$A(\alpha v_i+\beta v_j)=(\lambda \alpha v_i+\lambda\beta v_j)=\lambda(\alpha v_i+\beta v_j)$$ So two (or more) degenerate eigenvectors span a basis and can form infinite other solutions. This means I could have replaced $v_2,v_3$ by two other vectors that are linear combinations of them. This is different from the rescaling I mentioned because the rescaling keeps the same direction as the original vector.

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