# Why can $|\Psi (t=0)\rangle$ be written as a coherent superposition of some eigenkets?

Why can $$|\Psi (t=0)\rangle$$ be written as a coherent superposition of some eigenkets?

One of the approaches to solve time dependent Schrodinger equation $$i\hbar \frac{\partial |\Psi(t)\rangle}{\partial t} = \hat{H} |\Psi(t)\rangle$$ is to solve the time-independent part $$\hat{H} |\Psi\rangle = E |\Psi\rangle$$ and using the eigen vectors & eigen values, find the $$|\Psi(t)\rangle$$. In this we make use of the time propagator $$\mathcal{U}(t,t_o)$$ that acts on $$|\Psi (t=0)\rangle$$.

So, when we write $$|\Psi (t)\rangle = \mathcal{U}(t,t_o) \ |\Psi(0)\rangle$$ whys is it that we can write the wavefunction at t=0 as coherent superposition of eigen states of the time-independent schrodinger equation.

• the propagator $U(t,t_0)$ acts on $| \psi(t_0) \rangle$ and not $| \psi(0) \rangle$ – lakehal May 17 at 13:00

• @EverydayFoolish I'm not sure I understand. Are you asking why $t=0$ is used instead of $t=1$ for example? This is true for all times – Aaron Stevens May 17 at 3:27
The Schrodinger equation is a first order equation in time, hence we need an initial condition for the wave function $$\psi(\textbf{r},t=0)$$ to solve for later time. The wave function at time $$t=0$$ is a superposition of whatever you want since it is arbitrary, like the initial position and velocity in classical mechanics. However, it is more practical to express it in the basis of the eigenstates of the Hamiltonian.