This is a good question (or at least I think so because I struggled with the same question when I first studied quantum mechanics).
Notational Clarity
First of all, let's establish some notation. I will call the vector-space $\mathbb{V}$ and I will assume that it is finite-dimensional. Nothing of relevance hinges upon its finity and it'll save me some time. I will denote vectors in the vector-space as $\vert v\rangle$. Now, there are three different $0$s that you are playing with.
- The scalar $0\in\mathbb{R}\subset\mathbb{C}$. In particular, this is the null element of the field over which the vector space is defined.
- The null vector $\vert\Phi\rangle\in\mathbb{V}$ which is the null element of the vector-space itself, i.e., $\vert{\Phi}\rangle+\vert{v}\rangle=\vert{v}\rangle,\forall\vert{v}\rangle\in\mathbb{V}$.
- The ground-state $\vert0\rangle$ of the Hamiltonian such that $\hat{H}\vert 0\rangle=E_{0}\vert0\rangle$ where $E_0\in\mathbb{R}$ is the ground-state energy of the Hamiltonian. If $E_0 = 0$ then $\hat{H}\vert0\rangle=0\vert0\rangle=\vert\Phi\rangle\neq 0$.
- Notice here that multiplying a vector by a scalar gives you a vector, not a scalar. In particular, if you multiply a vector with the null-emement of the field over which the vector-space is defined then you get the null-element of the vector-space, not the null element of the field (i.e., not the scalar zero).
- I don't think this is something you are particularly confused about but in the interest of a broader audience, it should also be noted that $\vert 0\rangle\neq\vert\Phi\rangle$. The quickest way of seeing this is to note that $\langle 0\vert 0\rangle=1 $ and $\langle \Phi\vert\Phi\rangle=0$.
Onto Your Questions
We must get the same state on the other side of the equation, otherwise, it's not even an eigenvalue equation!
Mathematically, there is nothing too confusing here. As already pointed out in the comments, $0\vert v\rangle=\vert\Phi\rangle,\forall\vert v\rangle\in\mathbb{V}$. And thus, clearly, if $A\vert\psi\rangle=\vert \Phi\rangle$ the eigenvalue equation is satisfied by $\vert\psi\rangle$ with the eigenvalue being $0$.
Look at it this way if it helps: The eigenvalue equation says that if the two quantities $A\vert\psi\rangle$ and $e\vert\psi\rangle$ are equal then $\vert\psi\rangle$ is an eigenstate of $A$ with the eigenvalue $e$. OK, so, let's check if this criterion is satisfied for an eigenstate with zero eigenvalue: The first quantity is $A\vert\psi\rangle=\vert\Phi\rangle$ and the second quantity is $0\vert\psi\rangle = \vert \Phi\rangle$. Voila! The criterion is obviously satisfied.
Now, it is completely irrelevant that $0\vert \lambda\rangle=\vert \Phi\rangle$ for $\vert\lambda\rangle\neq\vert\psi\rangle$. The eigenvalue equation does not say that there cannot exist a $\vert\lambda\rangle\in\mathbb{V}$ such that $A\vert\psi\rangle=e\vert\lambda\rangle$ for $\vert\psi\rangle$ to be an eigenstate of $A$ with the eigenvalue $e$.
If I do a physical measurement on a state and I get zero "eigenvalue" (actually, measurement value) corresponding to some operator (which is doing the measurement) with the state collapsing to a different state then it's not actually the "measurement". I know it's confusing but perhaps someone can understand.
No, you're misunderstanding one of the two things here:
- If $\vert\psi\rangle$ is an eigenstate of $A$ with the eigenvalue $0$ then it is true that $A\vert\psi\rangle=0\vert\lambda\rangle$ for $\vert\lambda\rangle\neq\vert\psi\rangle$. BUT, this does not make $\vert\lambda\rangle$ an eigenstate of $A$. For $\vert\lambda\rangle$ to be an eigenstate of $A$, it would have to satisfy the eigenvalue equation $A\vert\lambda\rangle=0\vert\lambda\rangle$, not $A\vert\psi\rangle=0\vert\lambda\rangle$. So, there is no reason the post-measurement state of a measurement that yielded an eigenvalue $0$ for $A$ would be anything other than $\vert\psi\rangle$ because $\vert\psi\rangle$ is the eigenstate.
- Another thing to keep in mind is that the eigenvalue equation is not to be interpreted in the following way in quantum mechanics: you measure $A$ on a state $\vert\psi\rangle$ and that corresponds to the mathematical action $A\vert\psi\rangle$. And then, the output of what you get is whatever comes out on the righthand side of the mathematical expression when you write down $A\vert\psi\rangle=...$. This is not what quantum mechanics says. I can imagine that if one thinks this way then they might be confused as to what would happen in the case of zero eigenvalue because you can write down the right-hand side in many different ways.
Consider a scaled harmonic oscillator such that the ground state energy is 0. Now, \begin{align} H &= \hat{n}\ \hbar \omega \\ H |0\rangle &= \hat{n}\ \hbar \omega |0\rangle = \vec{0} \ne 0\\
a |0\rangle &= 0 \ne \vec{0}
\end{align}
This, to me, clearly says that $ A |\psi\rangle = 0\ |\phi\rangle$ is not an eigenvalue equation.
I am not sure how this tells you what you think it tells you but you are making a mistake of confusing the scalar and the vector zeroes. In particular, the correct version is $\hat{H}\vert 0\rangle = 0 \vert 0\rangle = \vert\Phi\rangle $ and $a\vert 0\rangle = \vert \Phi\rangle$. In your notation, my $\vert\Phi\rangle$ is $\vec{0}$. Hope this helps.
