"Harmonic oscillator in quantum mechanics"의 두 판 사이의 차이

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*  해밀토니안<br><math>H(P,X)=\frac{P^2}{2m}+\frac{m}{2}\omega^{2}X^2</math><br>
 
*  해밀토니안<br><math>H(P,X)=\frac{P^2}{2m}+\frac{m}{2}\omega^{2}X^2</math><br>
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<h5>Schrodinger equation</h5>
 
<h5>Schrodinger equation</h5>
  
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* [[Schrodinger equation]]
  
 
 
 
 

2011년 2월 8일 (화) 03:30 판

introduction

 

 

 

harmonic oscillator in classical mechanics
  • 질량 m, frequency \(\omega\) 인 조화진동자
  • 해밀토니안
    \(H(p,q)=\frac{p^2}{2m}+\frac{m}{2}\omega^{2}q^2\)
  • 해밀턴 방정식
    \(\dot{q}=\partial H/\partial p=\frac{p}{m}\)
    \(\dot{p}=-\partial H/\partial q=-m\omega^{2}q\)
  • 운동방정식
    \(\ddot{x}=-\omega^{2} x\) 즉 \(\ddot{x}+\omega^{2} x=0\)

 

 

quantum harmonic oscillator
  • 해밀토니안
    \(H(P,X)=\frac{P^2}{2m}+\frac{m}{2}\omega^{2}X^2\)

 

 

creation and annhilation operators
  • the position operators and momentum operators satisfy the relation
    \([X,P] = X P - P X = i \hbar\)Heisenberg group and Heisenberg algebra
  • define operators as follows
    \(a =\sqrt{m\omega \over 2\hbar} \left(x + {i \over m \omega} p \right)\)
    \(a^{\dagger} =\sqrt{m \omega \over 2\hbar} \left( x - {i \over m \omega} p \right)\)
  • Hamiltonian
    \(H = \hbar \omega \left(a^{\dagger}a + 1/2\right)\)
  • Commutation relation
    \(\left[a , a^{\dagger} \right] = 1\)
    \(\left[ H, a \right]= - \hbar \omega a\)
    \(\left[ H, a^\dagger \right] = \hbar \omega a^\dagger\)

 

 

energy  eigenstates
  • Assume that Planck’s constant equals 1
  • a harmonic oscillator that vibrates with frequency \(\omega\) can have energy \(\frac{\omega}{2}, (1 +\frac{1}{2})\omega, (2 +\frac{1}{2})\omega,(3 +\frac{1}{2})\omega,\cdots\) in units where
  • The lowest energy is not zero! It’s \(\omega/2\). This is called the ground state energy of the oscillator.

 

 

Schrodinger equation

 

 

path integral formulation

 

 

 

Groundstate correlation functions

 

 

 

Green's funtion

 

 

history

 

 

related items

 

 

encyclopedia

 

 

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expositions

 

 

articles

 

 

 

question and answers(Math Overflow)

 

 

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experts on the field

 

 

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