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Schrodinger's equation


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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.


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.

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