sriram Posted June 11, 2006 Share Posted June 11, 2006 Guy's i have a doubt in Schrodinger's equation! Im a beginner, so, anyone plz help me to understand the full equation clearly. Thanks in advance. Link to comment Share on other sites More sharing options...

Klaynos Posted June 11, 2006 Share Posted June 11, 2006 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.... Link to comment Share on other sites More sharing options...

Perturbation Posted June 11, 2006 Share Posted June 11, 2006 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. Start with your state-vector [imath]|\psi (0)\rangle[/imath] defined at initial time t=0 and perform a unitary evolution [math]|\psi (t)\rangle =e^{-iHt/\hbar}|\psi (0)\rangle[/math] Differentiate with respect to time [math]\frac{\partial}{\partial t}|\psi (t)\rangle =\frac{-iH}{\hbar}e^{-iHt/\hbar}|\psi (0)\rangle =\frac{-iH}{\hbar}|\psi (t)\rangle[/math] Putting this in the usual form gives [math]H|\psi (t)\rangle =i\hbar\frac{\partial}{\partial t}|\psi (t)\rangle[/math] 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 [math]\psi (\vec{x}, t)=\langle\vec{x} |\psi\rangle[/math] obeys [math]\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)[/math] 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 [math]\left(i\gamma^{\mu}\partial_{\mu}-m\right)\psi (\vec{x}, t)=0[/math] 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. Link to comment Share on other sites More sharing options...

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