# Schrodinger's equation

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Guy's i have a doubt in Schrodinger's equation!

Im a beginner, so, anyone plz help me to understand the full equation clearly.

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http://en.wikipedia.org/wiki/Schrodinger_equation

Which bit don't you get?

How good is your maths education?

(why is this in chem?)

||well that's 2 edits wonder if I can think of anything else to add....

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Schrodinger quantum mechanics has the time dependence in two things: the state-vector and consequently the state operator. There is no time dependance in the operators that represent dynamical variables. The Schrodinger equation expresses the time dependence of state-vectors as a differential equation, i.e. as a dynamical equation.

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

Differentiate with respect to time

$\frac{\partial}{\partial t}|\psi (t)\rangle =\frac{-iH}{\hbar}e^{-iHt/\hbar}|\psi (0)\rangle =\frac{-iH}{\hbar}|\psi (t)\rangle$

Putting this in the usual form gives

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

This is the Schrodinger equation.

For a non-relativistic particle the Hamiltonian H is

[imath] H=\frac{P^2}{2M}+V [/imath]

And in coordinate representation the wave-function $\psi (\vec{x}, t)=\langle\vec{x} |\psi\rangle$ obeys

$\left(\frac{-\hbar^2\nabla^2}{2M}+V(\vec{x}, t)\right) \psi (\vec{x}, t)=i\hbar\frac{\partial}{\partial t}\psi (\vec{x}, t)$

The momentum operator [imath]\vec{P}[/imath] is [imath]-i\hbar\vec{\nabla}[/imath] in coordinate representation.

For a relativistic particle we have the Dirac equation

$\left(i\gamma^{\mu}\partial_{\mu}-m\right)\psi (\vec{x}, t)=0$

Heinsenberg QM differs in that the state-vectors and state operators have no explicit time depedence; instead the time dependence is in the dynamical operators, such as the Hamiltonian or Momenta.

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