I've got an assignment with a given quantum state, denoted $\phi=\frac{1}{\sqrt{2}} \begin{bmatrix}0 \\ 1 \\ -1\end{bmatrix}$ and an operator for the observable B, given by $B=b\begin{bmatrix}0 & 2 & 0 \\ 2 & 0 & 0 \\ 0 & 0 & 2\end{bmatrix}$

And I'm asked to find the possible measurements of B, for the quantum state $\phi$.

Now; I've tried two different methods (given by two different instructors), and I'm unsure which is correct.

Method 1

The possible measurements of B should be given by the product $B|\phi\rangle$, which yields the vector

$B|\phi\rangle=\begin{bmatrix}b\sqrt{2} \\ 0 \\ -b\sqrt{2}\end{bmatrix}$

Which then indicates that the possible values of a measurement of B are $b$ and $-b$ and their respective chances are $(\sqrt{2})^2=\frac{1}{2}=50$% each.

Method 2 - and here there's a problem...

The possible measurements of B are its eigenvalues; I find the eigenvalues of B, given by the characteristic polynomial, and get $\lambda = -2b,2b,2b$ (note the degeneracy), with respective eigenvectors

$v_1 = \begin{bmatrix}-1 \\ 1 \\ 0\end{bmatrix}, v_2 = \begin{bmatrix}0 \\ 0 \\ 1\end{bmatrix}, v_3 = \begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix}$

I express the state $\phi$ as a linearcombination of these. It's $\phi=\frac{1}{\sqrt{2}}\bigg(\frac{1}{2} v_1+\frac{1}{2} v_3 - 1v_2\bigg)$

The chances for each individual eigenvalue $B=\lambda_n$ are the norm-squares of the coefficients for the corresponding eigenvector, in the linearcombination for $\phi$. I.e. The chance to get $B=\lambda_1=-b$ should be


And now you can probably already see my problem, because the chances don't seem to add up. $\frac{1}{8}+\frac{1}{8}+\frac{1}{2}\neq 1.0$

And the strangest thing is, for another, similar assignment, with the observable $L_z$ and a different quantum states, I used both methods and they yielded exact same results.

However; my idea is that the problem arises from the degeneracy of B's eigenvalues? The other problem (where both methods worked) did not have degeneracies in the eigenvalues for the matrix $[L_z]$

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    $\begingroup$ For staters in 2. you need to normalize your eigenvectors. $\endgroup$ Dec 21 '19 at 19:36

You failed to make Method 1 make any sense at al;, and, of course, the answer you provide is wrong. What you can ascertain from it is $\langle B\rangle=b$, which distinctly contradicts your unsound conclusion.

Method 2 is the standard one, if only you normalized your eigenvectors properly, as suggested by @ZeroTheHero:

The eigenvalues of B are $\lambda = -2b,2b,2b$, with respective normalized eigenvectors $$v_1 = \frac{1}{\sqrt{2}}\begin{bmatrix}-1 \\ 1 \\ 0\end{bmatrix}, \qquad v_2 = \begin{bmatrix}0 \\ 0 \\ 1\end{bmatrix},\qquad v_3 = \frac{1}{\sqrt{2}}\begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix}, $$ so that $\phi=\bigg(\frac{1}{2} v_1+\frac{1}{2} v_3 - \frac{1}{\sqrt{2}}v_2\bigg)$.

As a result, $| \frac{1}{2}|^2=\frac{1}{4}$ of the time you measure $-2b$, while $| \frac{1}{2}|^2 +| \frac{1}{\sqrt{2}}|^2=\frac{3}{4} $ of the time you measure 2b, so on the average b, as found above.

Degeneracy is innocent here.

  • $\begingroup$ Thanks for the answer. Must have been no more than a fortunate (or unfortunate, depending on how you look at it) coincidence that Method 1 got the right answer on that other, similar assignment I did. $\endgroup$ Dec 22 '19 at 21:30

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