Help understanding proof in simultaneous diagonalization

Question

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 1

Note that he explains above: "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." So he uses the assumption that the operator is nondegenerate and the definition of nondegeneracy (or a statement equivalent to the definition of nondegeneracy, if you use a different definition).

The phrase "to a given eigenvalue, there is just one eigenvector, up to a scale" means that any two eigenvectors with the same eigenvalue $\omega_i$ coincide up to a factor. Then $\lambda_i$ is such a factor.

Answer 2

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

$$\Lambda\left|\omega_i\right\rangle=\lambda_i\left|\omega_i\right\rangle $$

Answer 3

Since the vector $\Lambda | \omega_i \rangle $ has the same eigenvalue as $| \omega_i \rangle $, it must be in the same invariant subspace as $| \omega_i \rangle $, which Shankar takes to be one dimensional.

Answer 4

I don't know exsactly what it means with scalar but remember that:

A vector $|\Psi\rangle$ is invariant up to a phase because a global phase is always ruled out when you calculate, for example, with the state $e^{i\theta}|\Psi\rangle$, the expected value of and observable $A$ using this state is

$$\langle A \rangle = \langle\Psi | e^{-i\theta} A e^{i\theta} |\Psi \rangle = \langle\Psi | A | \Psi\rangle,$$

which is the same result as if you calculate the expected value by just using the vector $|\Psi\rangle$. So for all practical purposes, you can use just a state $|\Psi\rangle$ in your punch line of your proof.