Schrodinger equation

The Schrödinger equation is a fundamental equation in Quantum Mechanics that describes how the quantum state of a system changes over time.

See the derivation of Schrodinger equation from the postulates of QM.

Time-dependent Schrödinger equation (TDSE):

$$ i\hbar \frac{\partial}{\partial t} |\psi(t)\rangle = \hat{H} |\psi(t)\rangle \tag{1} $$

Where:

$i$ is the imaginary unit.

$\hbar$ is the reduced Planck constant, equal to the Planck constant $h$ divided by $2\pi$.

$\frac{\partial}{\partial t}$ is the partial derivative with respect to time.

$|\psi(t)\rangle$ is the state vector (or wave function) of the system, which is a function of time.

$\hat{H}$ is the Hamiltonian operator, which represents the total energy of the system.

Time-independent Schrödinger equation (TISE):

When the Hamiltonian operator does not explicitly depend on time, we can consider the time-independent Schrödinger equation:

$$ \hat{H} \psi(x) = E \psi(x)\tag{2} $$

Where:

$\psi(x)$ is the spatial part of the wave function (state vector), which does not depend on time.

$E$ are the energy eigenvalues of the system.

The function $\psi(x)$ is called a stationary state or an eigenstate of the Hamiltonian $\hat{H}$. A stationary state is a state in which the probability distribution of the particle's position (or other observables) does not change with time. This does not mean that the wavefunction, which is indeed

$$ |\psi(t)\rangle = e^{-i E t/\hbar} \psi(x), $$

doesn't change with time, but rather than its absolute square, $|\psi(x)|^2$, which gives the probability density of finding the particle at position $x$, remains unchanged.

In quantum mechanics, any state $|\psi(t)\rangle$ can be expressed as a superposition of these stationary states (eigenstates of the Hamiltonian), which is a direct consequence of the linearity of the Schrödinger equation.

Relation between TDSE and TISE:

Suppose we have a solution $\psi_E(x)$ for equation (2) for every $E \in \mathbb R$. Suppose also that the set $\{\psi_E\}_{E\in \mathbb R}$ spans the space of function with which we are dealing. Then, for every $t$:

$$ |\psi(t)\rangle=\int c^E(t) \psi_E(x) dE $$

Substituting into (1):

$$ \int c_E'(t) \psi_E(x) dE=\int c^E(t) E\psi_E(x) dE $$

and therefore it should be:

$$ (c^E)'(t)=c^E(t) E. $$

So $c_E(t)=K_E e^{Et}$, with $K_E$ constants to be determined by the initial data. For example, suppose we are given

$$ |\psi(0)\rangle=f(x)=\int c^E_0 \psi_E(x) dE. $$

Then $c_0^E=c_E(0)=K_E e^{E0}=K_E$, and finally

$$ |\psi(t)\rangle=\int c^E(t) \psi_E(x) dE=\int c_0^E e^{Et} \psi_E(x) dE $$

Related: Fourier transform

________________________________________

Author of the notes: Antonio J. Pan-Collantes

antonio.pan@uca.es

INDEX: