Let
$$\Psi(x,t) = A(t) \psi(x)$$
Applying Schrodinger's Time dependent equation:
$$\begin{equation}
i\hbar\frac{\partial}{\partial t}\left(A(t)\psi(x)\right) = H\left(A(t)\psi(x)\right)
\end{equation}$$
Let ##\psi(x)## is an eigenfunction of ##H## with eigenvalue ##E##. So, we get...
Isn't the Hamiltonian Operator in the Schrodinger's time dependent equation is the Hamiltonian operator defined for the particular system we are considering?
How do we experimentally apply the operator ## \exp{\left(-i\phi\frac{ S_z}{\hbar}\right)}## on a quantum mechanical system? (Here ##S_z## is the spin angular momentum operator along the z-axis)
For example, on a beam of electrons?
Suppose, the energy of a particle is measured, say ##E_1##. So now the state vector of the particle is the energy eigenket ##|E_1>##.
Then the position of the particle is measured, say ##x##. As the Hamiltonian operator and the position operator are non-commutative, the state vector is changed...
Consider the particle in a box problem. The number of energy eigenbasis is 'countable' infinity. But the number of position eigenbasis is 'uncountable' infinity. x can take any value from the interval [0,L] Whichever basis I choose, shouldn't the dimensionality of the vector space be the same?