The proof is from Principles of Quantum Mechanics by Shankar. The theorem is:
If $\Omega$ and $\Lambda$ are two commuting Hermitian operators, there exists (at least) a basis of common eigenvectors that diagonalizes them both.
The proof is: Consider first the case where at least one of the operators is nondegenerate, i.e. to a given eigenvalue, there is just one eigenvector, up to a scale. Let us assume $\Omega$ is nondegenerate. Consider any one of its eigenvectors:
$$\Omega\left|\omega_i\right\rangle=\omega_i\left|\omega_i\right\rangle$$ $$\Lambda\Omega\left|\omega_i\right\rangle=\omega_i\Lambda\left|\omega_i\right\rangle$$ Since $[\Omega,\Lambda]=0$ $$\Omega\Lambda\left|\omega_i\right\rangle=\omega_i\Lambda\left|\omega_i\right\rangle$$
i.e., $\Lambda\left|\omega_i\right\rangle$ is an eigenvector with eigenvalue $\omega_i$. Since this vector is unique up to a scale,
$$\Lambda\left|\omega_i\right\rangle=\lambda_i\left|\omega_i\right\rangle$$
Thus $\left|\omega_i\right\rangle$ is also an eigenvector of $\Lambda$ with eigenvalue $\lambda_i$...
What I do not understand is the statement/argument "Since this vector is unique up to a scale." I do not see how the argument allows to state the equation following it. What axiom or what other theorem is he using when he states "since this vector is unique up to a scale"?
Answer
When $\lambda_1$ is an eigenvalue of a matrix and $v_1$ and $v_2$ are the components of the corresponding eigenvector, then the following equation holds:
$\begin{pmatrix} a-\lambda_1 & b \\ c &d-\lambda_1 \end{pmatrix}\begin{pmatrix} v_1 \\ v_2 \end{pmatrix}=\begin{pmatrix} 0 \\ 0 \end{pmatrix}$
Now when you scale up the eigenvector (say by three) it looks like this:
$\begin{pmatrix} a-\lambda_1 & b \\ c &d-\lambda_1 \end{pmatrix}\begin{pmatrix} 3v_1 \\ 3v_2 \end{pmatrix}=\begin{pmatrix} 0 \\ 0 \end{pmatrix}$
This you can write as
$3 \begin{pmatrix} a-\lambda_1 & b \\ c &d-\lambda_1 \end{pmatrix}\begin{pmatrix} v_1 \\ v_2 \end{pmatrix}=\begin{pmatrix} 0 \\ 0 \end{pmatrix}$
But the matrix multiplied with the eigenvector still yields the zero vector!
This is what he meant when he said "Since this vector is unique up to a scale.": any scaled up eigenvector of a matrix is still an eigenvector. And how does the last equation follow from it?
When you write $$\Omega\Lambda\left|\omega_i\right\rangle=\omega_i\Lambda\left|\omega_i\right\rangle$$ then you know that $\Lambda\left|\omega_i\right\rangle$ gives you a vector which is an eigenvector of $\Omega$. But you said that $\Omega$ is nondegenerate, so for any $\omega_i$ there is a unique $\left|\omega_i\right\rangle$. What this means is that this eigenvector you get by applying $\Lambda\left|\omega_i\right\rangle$ must be $\left|\omega_i\right\rangle$. Luckily, any eigenvector which is scaled up (here by $\lambda_i$) is still an eigenvector, so that you get
$$\Omega\lambda_i\left|\omega_i\right\rangle=\omega_i \lambda_i \left|\omega_i\right\rangle$$
or
$$\Lambda\left|\omega_i\right\rangle=\lambda_i\left|\omega_i\right\rangle $$
No comments:
Post a Comment