Eigenvalues and Eigenvectors in Quantum Mechanics I can calculate them mathematically but I don't understand the physical meaning of these concepts. My knowledge  is purely algorithmic and lacks actual insight, how do I picture eigenvalues/eigenvectors in my mind.
 A: What is the physical meaning of the length of a vector? The answer I give to that question is that it's that aspect of the vector which doesn't change when you rotate your axes. What remains of the vector when you factor out its length is a unit vector, what we colloquially call "direction" when we describe a vector as having direction and magnitude.
Now, for an operator you also have elements of it that are invariant under rotations (or unitary operators - for the purpose of this explanation they're the same thing). The invariant elements are the operator's eigenvalues. What do you have left when you factor out an operator's eigenvalues? You have a rotation matrix that describes the orientation of the operator. We frequently describe the rows/columns of that orientation matrix as the eigenvectors of the matrix.
So, what makes eigenvectors special is that all observers agree on the eigenvalues of the operator, no matter what basis they're working in. The components of the eigenvectors, on the other hand, are maximally dependent on which basis you work in, and are thus, in a sense, subjective. You can derive objective information from eigenvectors, though. Just like the angles between vectors are rotation invariant, the matrices that translate one basis into another are also rotation invariant. Put more concretely, the inner product $\langle p | x\rangle = \frac{1}{\sqrt{2\pi}} e^{-ipx/\hbar}$ is basis independent, even if the representations of $|x\rangle$ and $|p\rangle$ in other bases aren't.
You can actually go more general than rotations, of course. Eigenvalues are invariant under similarity transform for operators, but the analogy to vectors is easier to picture if we restrict our attention to rotations.
A: It's easy to understand eigenvalues and eigenvectors in the context of matrices. Any n by n matrix can transform n-D space in some form. To make it easy to understand let's work with a 2by2 matrix
\begin{equation}
A=\begin{bmatrix}
1 & 1 \\
0 & 1 
\end{bmatrix}
\end{equation}
This $A$ can transform the entire 2D space. Let's check it for one vector in 2d space:
\begin{bmatrix}
0  \\
1  
\end{bmatrix} $\quad$This is nothing but $\hat{j}$ (read the column vector as $0\hat{i}+1\hat{j}$).
When $A$ acts on 2D cartesian space, every vector in the space including $\begin{bmatrix}
  0 \\ 
  1
\end{bmatrix}$ transforms as:
\begin{equation}
\begin{bmatrix}
  1 & 1 \\ 
  0 &1
\end{bmatrix} \begin{bmatrix}
  0 \\ 
  1
\end{bmatrix}= \begin{bmatrix}
  1  \\ 
  1
\end{bmatrix}
\end{equation}
So $A$ transforms $\hat{j}$ into $\hat{i}+\hat{j}$ ($=\hat{j'}$ for the transformed space). You can also check that it transforms $\hat{i}$ to $\hat{i}$ itself.
Check for any general vector and you'll see that $A$ shears the entire x-y plane. The coordinate axes in the transformed space are at $45^\circ$.
Now, $A$ is analogous to an operator in quantum mechanics and the column vectors are analogous to the wavefunction (usually called state vector in matrix language of Quantum Mechanics).
If $A$ acting on any vector gives the same vector but scaled by some factor, then this particular vector is called the eigenvector of $A$ (analogous to eigenfunction of operator $A$):
\begin{equation}
A\begin{bmatrix}
   a  \\ 
  b
\end{bmatrix}
=\lambda \begin{bmatrix}
   a  \\ 
  b
\end{bmatrix} \quad ...(*)\end{equation}
. In other words, If a vector in the original space retains it's direction in the transformed space then it is an eigenvector. In case of $A$ you can now tell that $\hat{i}$ is an eigenvector, so is any scaled version of $\hat{i}$, e.g,  $\begin{bmatrix}
   5  \\ 
  0
\end{bmatrix}$$=5$$\begin{bmatrix}
   1  \\ 
  0
\end{bmatrix}$ $=5\hat{i}$ , which is $\hat{i}$ scaled by 5, is also an eigenvector.
The scalar that scales the length of eigenvector in the transformed space is known as the eigenvalue($\lambda$, see $(*)$) corresponding to that particular eigenvector. e.g, in case of $\hat{i}$, $A\hat{i}=\begin{bmatrix}
1 & 1 \\
0 & 1 
\end{bmatrix} \begin{bmatrix}
1  \\
0  
\end{bmatrix}=\begin{bmatrix}
1  \\
0  
\end{bmatrix}=1.\begin{bmatrix}
1  \\
0  
\end{bmatrix} $, compare this to $(*)$ and you'll see that the eigenvalue corresponding to the eigenvector $\hat{i}$ is $1$.
To give an example in Quantum Mechanics(in Heisenberg's matrix mechanics form): If you operate the Hamiltonian matrix($H$, a 3by3 matrix) on an eigenvector of $H$ ($\psi$, a 3by1 column vector), then $H\psi=E\psi$, where $E$ (=Energy) is the eigenvalue of the Hamiltonian operator corresponding to the eigenvector $\psi$.
A: An eigenvector of an operator $\hat{O}$ is physically invariant (acquires a phase at most) when subjected to a transformation of which $\hat{O}$ is a generator. For example, momentum operator $\hat{p}$ is the generator for translation operation $\hat{T_a}=\exp(-i\hat{p}a/\hbar)$, and indeed an eigen-state of momentum $\psi(x)=\exp(-i p x)$ only acquires phase $\phi(a) = pa/\hbar$ upon translation, i.e. remains invariant. Another example is time evolution $\hat{U}=\exp(i\hat{H}t/\hbar)$. An eigenstate of the Hamiltonian $\hat{H}$ (i.e. the solution of the Schodinger equation) is invariant as time goes on that is,  remains "where it was". No transitions or any other physically significant dynamics occur.
