# About hamilton operators

1. Aug 31, 2007

### dream_chaser

why i$$\hbar$$($$\partial$$/$$\partial$$t+i$$\Omega$$)=i$$\hbar$$exp(-i$$\Omega$$t)$$\partial$$/$$\partial$$texp(i$$\Omega$$t)

2. Sep 1, 2007

### dextercioby

In the RHS you have something that resembles the unitary trasformation of an operator. Where did you get the equation from ?

3. Sep 1, 2007

### K.J.Healey

$$i\hbar(\frac{d}{dt}+ i\Omega) = i\hbar(exp(-i \Omega t) \frac{d}{dt} exp(i \Omega t)$$

Well if exp(iOt) is your wavefunction, the RHS is just $$i\hbar(i \Omega )$$

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

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