possible values of quantity in matrix mechanics

Question

Suppose that there is a quantity described by the matrix as the following:

$M = \begin{bmatrix} 3 & 0 & -i \\ 0 & 1 & 0 \\ i & 0 &3\end{bmatrix}$

How we determine possible values of this quantity? (According to the book, it says that some possible values are 1,2 and 4, but it does not say how it determined this.)

Answer 1

The allowed values are eigenvalues of $M$ i.e. numbers $\lambda$ such that there exists a nonzero vector $\vec v$ (called the eigenvector) such that $$ M\cdot \vec v = \lambda \vec v $$ One may shift both terms to the left hand side to rewrite it as $$ (M - \lambda\cdot {\bf 1}) \cdot \vec v = 0 $$ Because the rows may be combined by coefficients in $\vec v$ to get zeroes, they must be linearly dependent. Consequently, it must be true that $$ \det (M - \lambda\cdot {\bf 1}) = 0.$$ This equation is known as the characteristic equation. If you know how to compute the determinant, you may easily see that the characteristic equation (with the "characteristic polynomial" on one side) is $$ -\lambda^3 +7 \lambda^2-14\lambda+8 = 0.$$ Cubic equations may be solved by Cardano's formula. However, it's easier to guess. Indeed, $1,2,4$ are the three solutions as you should verify.

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Answer 2

Hints:

1. The possible values for an observable in quantum mechanics is the eigenvalues of the corresponding self-adjoint operator or matrix.

2. More concretely, the entries of the matrix can be reordered into a block form $$\begin{bmatrix} 3 & -i & 0 \\ i & 3 &0 \\ 0 & 0 & 1\end{bmatrix}.$$ One can find the eigenvalues of each block independently.