Are the Eigenvalues of the Zero Ket Always Zero?

[bjnartowt]

Homework Statement

I am wondering if I can make the sweeping generalization that the eigenvalues of the zero ket are zero. I further generalize that the zero ket is not of interest, as far as physical observables occur.

Homework Equations

the eight axioms of vector spaces.
http://en.wikipedia.org/wiki/Vector_space

(didn't feel like typing all 8 out. i already put those into my notes).

The Attempt at a Solution

$$\exists \left| 0 \right\rangle :{\rm{ }}\left| u \right\rangle + \left| 0 \right\rangle \equiv \left| u \right\rangle & & & & \left| 0 \right\rangle \equiv 0 \cdot \left| \alpha \right\rangle$$

also, scalar multiplication is commutitive and compatible with operator "multiplication" (i.e., front-multiplying).

$$A\left| 0 \right\rangle = A(0 \cdot \left| \alpha \right\rangle ) = 0 \cdot A\left| \alpha \right\rangle = 0 \cdot a \cdot \left| \alpha \right\rangle = 0 \cdot \left| \alpha \right\rangle = \left| 0 \right\rangle$$

and therefore,
$$A\left| \alpha \right\rangle = 0 \cdot \left| 0 \right\rangle$$

 

[vela]

Observables, not kets, have eigenvalues, so it doesn't make sense to talk about the eigenvalues of the zero ket. Also, as far as eigenvectors go, they are by definition non-zero, so again it doesn't make sense to consider the zero ket an eigenvector of an observable.