Commutation and Measurement of Observables

[Mr_Allod]

Homework Statement

Let the Hilbert space be $\mathcal H = \mathbb C^3$. Consider the two observables:
$$\hat B = \begin{pmatrix}
2 & 0 & 0 \\
0 & 0 & 1 \\
0 & 1 & 0
\end{pmatrix}$$
$$\hat C = \begin{pmatrix}
-1 & 0 & 0\\
0 & 0 & 0\\
0 & 0 & 1
\end{pmatrix}
$$

a. Do $\hat B$ and $\hat C$ have common eigenvectors?
b. Assume that we have done the measurement characterized by $\hat C$ and we are measuring the value -1. Then we immediately do the measurement characterized by $\hat B$ right after measuring $\hat C$. What is the expectation value of that measurement?

Relevant Equations

Commutator Relation: $\left[ \hat B, \hat C\right] = \hat B \hat C - \hat C \hat B$
Expectation value: $\langle \hat B \rangle = \langle \psi | \hat B \psi \rangle$

Hello there, I am having trouble with part b. of this problem. I've solved part a. by calculating the commutator of the two observables and found it to be non-zero, which should mean that $\hat B$ and $\hat C$ do not have common eigenvectors. Although calculating the eigenvectors for each one actually yields that they do have one in common ($\vec v = (1, 0, 0)$), I chose to interpret the question as asking if they have a common set of eigenvectors, in which case my answer would be that they do not.

Now I don't really know how to approach part b. I feel like I am missing information even though I'm sure this isn't the case and I'm just not seeing something that's right in front of me. I'd appreciate it if someone could explain the concept behind part b. to help me reach a solution, thank you.

 

[DrClaude]

What is the state of the system after the measurement of $\hat{C}$ if the result was -1?

 

[Mr_Allod]

What is the state of the system after the measurement of $\hat{C}$ if the result was -1?

Would it be the state given by the eigenvector corresponding to the eigenvalue $\lambda = -1$?

 

[DrClaude]

Would it be the state given by the eigenvector corresponding to the eigenvalue $\lambda = -1$?

Correct. Now that you have the state, you can calculate the expectation value of $\hat{B}$.

 

[vanhees71]

Hello there, I am having trouble with part b. of this problem. I've solved part a. by calculating the commutator of the two observables and found it to be non-zero, which should mean that $\hat B$ and $\hat C$ do not have common eigenvectors.

This is not a valid argument!

Example: Due to the angular-momentum commutation relations $[\hat{J}_x,\hat{J}_y]=\mathrm{i} \hat{J}_z$ the two angular momentum components $\hat{J}_x$ and $\hat{J}_y$ don't commute, but they have a common eigenvector, namely the one with $j=0$, $j_z=0$. This is a common eigenvector of all three angular-momentum components with the eigenvalue $0$. Of course there's no common complete orthonormal set of eigenvectors, because the operators don't commute, but there can be special cases of common eigenvectors although two self-adjoint operators don't commute. So you have to check it carefully for any given example!

In your example the two matrices have the obvious common eigenvector $(1,0,0)^{\text{T}}$!