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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?

Yes. I just wanted to show that the energy eigenkets are also eigenkets to the operator ##i\hbar \frac{\partial}{\partial t}##.

Can't ##H:=-\frac{\hbar ^2}{2m} \frac{\partial ^2}{\partial x^2} + V(x) ## act on ##\Psi (x,t)## as well?

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?