Understanding the Relationship between Hamilton and Momentum Operators

  • Context: Graduate 
  • Thread starter Thread starter dream_chaser
  • Start date Start date
  • Tags Tags
    Hamilton Operators
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
2 replies · 2K views
dream_chaser
Messages
5
Reaction score
0
why i[tex]\hbar[/tex]([tex]\partial[/tex]/[tex]\partial[/tex]t+i[tex]\Omega[/tex])=i[tex]\hbar[/tex]exp(-i[tex]\Omega[/tex]t)[tex]\partial[/tex]/[tex]\partial[/tex]texp(i[tex]\Omega[/tex]t)
 
Physics news on Phys.org
[tex]i\hbar(\frac{d}{dt}+ i\Omega) = i\hbar(exp(-i \Omega t) \frac{d}{dt} exp(i \Omega t)[/tex]

Well if exp(iOt) is your wavefunction, the RHS is just [tex]i\hbar(i \Omega )[/tex]

are you sure this equation is right? Looks like momentum operator, not hamilton.