Number of eigenvalues of this Hermitian

[kuchenvater]

Hi. I'm trying to study QM from Shankar on my own. Asking this here because I don't really have a teacher to help me with this:

Homework Statement

I'm trying to solve problem 1.8.9 -part 3 of "The Principles of Quantum Mechanics" by R Shankar. Here's the problem:
Given the values of M_ij (see next section for equations) which is Hermitian, argue that there exist three directions such that the angular velocity and angular momentum will be parallel.

Homework Equations

Given
M_ij=∑_a=m_a[ r²_a δ_ij - (r_a)_i(r_a)_j ]
here i denotes the i^th Cartesian component.

|l>=M|ω>

The Attempt at a Solution

It's known that M is Hermitian. l and ω will be parallel if ω is an eigenvector of l. So I have to prove that the number of eigenvalues of M is 3. Somehow I can't figure out a way to do that. It's obvious that all eigenvalues will be real and the number of eigenvalues is equal to or less than the rank of a matrix. I was thinking I could consider only the diagonal elements of a diagonalized version of M or just the trace of M(which yields the sum of all eigenvalues) by setting i=j and then I'd have i as the index of sigma. If I can count the number of values of i, I'd have the number of eigenvalues. And since M is Hermitian and has real eigenvalues and this describes a rotating system of mass(i.e. it's moment of inertia), it must be defined in the real plane and that has 3 axes-x,y and z, one for each value of i. So i varies across 3 values, hence the diagonalized matrix has 3 diagonal elements and hence 3 eigenvalues. The question also mentions the mass's cartesian component, so that automatically means x, y, z.

Is that correct reasoning?

 

[D H]

That would be correct if you prove that your matrix M has three distinct eigenvalues. That is not always the case. Consider a set of 8 masses placed on the corners of a cube, with origin at the center of the cube. The moment of inertia tensor M for this is a diagonal matrix with all diagonal elements equal to one another. There's only one eigenvalue.

What you need to show is that one can always at least three directions along which angular velocity and angular momentum are parallel, regardless of the number of unique eigenvalues.

Aside: A eigenvalue of zero is a bit problematic. This means zero angular momentum for rotation along that eigenvector. You'd have to interpret the zero vector as being parallel to all vectors.

 

[kuchenvater]

What you need to show is that one can always at least three directions along which angular velocity and angular momentum are parallel, regardless of the number of unique eigenvalues.

Any hints on how to do that?

 

[D H]

The matrix is Hermitian. What does the spectral theorem say about the eigenvectors?

 

[paranormal]

Yes, your reasoning is correct. Since M is a Hermitian matrix, it will have real eigenvalues. Additionally, since it describes a rotating system of mass, it must be defined in the real plane and have 3 axes (x, y, and z). This means that the matrix will have 3 distinct eigenvalues, corresponding to each of the axes. Therefore, there are 3 eigenvalues of M, and as you mentioned, they can be found by considering the diagonal elements of the diagonalized form of M.