# Time evolution of the state vector

[ "article:topic", "Author tag:Tuckerman", "showtoc:no" ]

The time evolution of the state vector is prescribed by the Schrödinger equation

$i\hbar {\partial \over \partial t} \vert\Psi(t)\rangle = H\vert\Psi(t)\rangle$

where $$H$$ is the Hamiltonian operator. This equation can be solved, in principle, yielding

$\vert\Psi(t)\rangle = e^{-iHt/\hbar}\vert\Psi(0)\rangle$

where $$\vert\Psi(0)\rangle$$ is the initial state vector. The operator

$U(t) = e^{-iHt\hbar}$

is the time evolution operator or quantum propagator. Let us introduce the eigenvalues and eigenvectors of the Hamiltonian $$H$$ that satisfy

$H\vert E_i\rangle = E_i \vert E_i\rangle$

The eigenvectors for an orthonormal basis on the Hilbert space and therefore, the state vector can be expanded in them according to

$\vert\Psi(t)\rangle = \sum_i c_i(t) \vert E_i\rangle$

where, of course, $$c_i(t) = \langle E_i\vert\Psi(t)\rangle$$, which is the amplitude for obtaining the value $$E_i$$ at time $$t$$ if a measurement of $$H$$ is performed. Using this expansion, it is straightforward to show that the time evolution of the state vector can be written as an expansion:

$\displaystyle \vert\Psi(t)\rangle = \displaystyle e^{-iHt\hbar}\vert\Psi(0)\rangle$

$= e^{-iHt/\hbar}\sum_i\vert E_i\rangle \langle E_i\vert\Psi(0)\rangle$

$=\sum_i e^{-iE_i t/\hbar}\vert E_i\rangle \langle E_i\vert\Psi(0)\rangle$

Thus, we need to compute all the initial amplitudes for obtaining the different eigenvalues $$E_i$$ of $$H$$, apply to each the factor $$\exp(-iE_it/\hbar)\vert E_i\rangle$$ and then sum over all the eigenstates to obtain the state vector at time $$t$$.

If the Hamiltonian is obtained from a classical Hamiltonian $$H (x, p)$$, then, using the formula from the previous section for the action of an arbitrary operator $$A (X, P)$$ on the state vector in the coordinate basis, we can recast the Schrödiner equation as a partial differential equation. By multiplying both sides of the Schrödinger equation by $$\langle x |$$, we obtain

 $\langle x\vert H(X,P)\vert\Psi(t)\rangle$ $i\hbar {\partial \over \partial t}\langle x\vert\Psi(t)\rangle$ $H\left(x,{\hbar \over i}{\partial \over \partial x}\right)\Psi(x,t)$ $i\hbar {\partial \over \partial t}\Psi(x,t)$

If the classical Hamiltonian takes the form

$H(x,p) = {p^2 \over 2m} + U(x)$

then the Schrödinger equation becomes

$\left[-{\hbar^2 \over 2m}{\partial^2 \over \partial x^2} + U(x)\right]\Psi(x,t)= i\hbar {\partial \over \partial t}\Psi(x,t)$

which is known as the Schrödinger wave equation or the time-dependent Schrödinger equation. In a similar manner, the eigenvalue equation for $$H$$ can be expressed as a differential equation by projecting it into the $$X$$ basis:

 $\langle x\vert H\vert E_i\rangle$ $E_i \langle x\vert E_i\rangle$ $H\left(x,{\hbar \over i}{\partial \over \partial x}\right)\psi_i(x)$ $E_i \psi_i(x)$ $\left[-{\hbar^2 \over 2m}{\partial^2 \over \partial x^2} + U(x)\right]\psi_i(x)$ $E_i \psi_i(x)$

where $$\psi_i(x) = \langle x\vert E_i\rangle$$ is an eigenfunction of the Hamiltonian.