eigenvalueprobabilityquantum mechanicswavefunction
in quantum mechanics we always see the eigenvalue equation $\hat A\psi=\lambda\psi$ and $\lambda$ is the probability amplitude meaning $\lambda^2$ is the actual probability of finding the system in the state $\psi$. We are also told that $\lambda$ must be real, but what would be wrong with having $\lambda$ imaginary because it's square is real as a physical quantity must be?
So why can we never have a complex $\lambda$?
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.
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:
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.
If you want to model a measurement as a hermitian operator, you must assign distinct real numbers to the possible outcomes, whether they naturally correspond to real numbers or not. This is a limitation of the operator formalism and not a fundamental property of measurement in quantum mechanics.
If your position measurement amounts to noting which grain of a photographic plate darkened, you could number the grains 1, 2, ... in an arbitrary order, and that would be a valid representation of the measurement, though perhaps not a very useful one.
If you want the outcome of the measurement to be represented by a point in $\mathbb R^d$, you have to split the measurement into $d$ measurements of a single coordinate and imagine that you are performing all of them in sequence (which creates no problems since they commute). This is just to make the formalism happy, and has no meaning beyond that.
Best Answer
$\lambda$ is not the probability amplitude. And your equation $\hat A\psi=\lambda\psi$ means that $\psi$ is an eigenstate of $A$ with eigenvalue $\lambda$ where $\lambda$ is a physical quantity associated with $\hat A$ upon measurement of $\hat A$. Hence $\lambda$ itself must be real.
For example, if $\hat A$ was the momentum operator, then $\lambda$ is a value of momentum. And in the ubiquitous equation, $$\hat H\Psi=E\Psi$$ where $\hat H$ is the Hamiltonian, $E$ is energy.
You have a few misconceptions. Say we have a system described by a state $$\Psi=\sqrt{\frac{1}{3}}\psi_1 +\sqrt{\frac{2}{3}}\psi_2$$ Then the coefficients $\sqrt{\frac{1}{3}}$ and $\sqrt{\frac{2}{3}}$ are probability amplitudes and the corresponding square of these coefficients i.e., $\frac 13$ and $\frac 23$ are probabilities (corresponding to the likelihood of finding the system in states $\psi_1$ and $\psi_2$ respectively).
When you look at the equation $\hat A\psi=\lambda\psi$ (with Hermitian or self-adjoint $\hat A$) by definition we need a coefficient $\lambda$ that will change the length of the state $\psi$ and not a complex $\lambda$ since multiplication by imaginary numbers will rotate a state.
Also, if the eigenvalues were imaginary, this would mean the operator $\hat A$ will not be Hermitian. This is because $\hat A$ will have imaginary elements on the diagonal. This means it cannot possibly be Hermitian since the rule $$A^\dagger=(A^T)^*=A$$ cannot possibly hold.
Remember that the rule is that $\hat A$ must be Hermitian since it is also a requirement that $\hat A$ has a complete set of eigenvalues.