# Quantum Eigenvalues, measurements, chances not matching up

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

$$|\frac{1}{\sqrt{2}}\frac{1}{2}|^2=\frac{1}{8}$$

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 $$[L_z]$$

• For staters in 2. you need to normalize your eigenvectors. 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.
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.