Time evolution of the state vector

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:

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

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

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

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 $\langle x\vert$, 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.