- #1
yamata1
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Hello
1. Homework Statement
The dipole moment of an ammonia molecule is ##d_0=5*10^{-30} C.m##.If we apply a static electric field of ##\mathcal { E }=1*10^{6 }V*m^{-1}## to an ammonia molecule initially in the state ## |ψG⟩## where the nitrogen molecule is considered to be on the left,we make it oscillate between states ##|ψ+⟩## and ##|ψ−⟩ ##which represent two wavefunctions.What is the angular frequency of oscillation between the two states ?
In the ##(|ψ+⟩ ,|ψ−⟩)## basis:
##\hat { H } _ { \mathrm { tot. } } = \hat { H } _ { 0 } + \hat { W } = \frac { \hbar \omega _ { 0 } } { 2 } \left( \begin{array} { c c } { - 1 } & { 0 } \\ { 0 } & { 1 } \end{array} \right) - d _ { 0 } \mathcal { E } \left( \begin{array} { l l } { 0 } & { 1 } \\ { 1 } & { 0 } \end{array} \right)##
We can put the hamiltonian in a simple form : ##\hat { H } _ { t o t } = - \frac { \hbar \Omega } { 2 } \left( \begin{array} { c c } { \cos 2 \theta } & { \sin 2 \theta } \\ { \sin 2 \theta } & { - \cos 2 \theta } \end{array} \right)## with ##\tan 2 \theta = \frac { 2 d _ { 0 } \mathcal { E } } { \hbar \omega _ { 0 } } \quad - \pi / 4 < \theta < \pi / 4## et ##\omega _ { 0 }=160*10^9 Hz##[/B]
##( \frac { \hbar \Omega } { 2 } ) ^ { 2 } = ( \frac { \hbar \omega _ { 0 } } { 2 }) ^ { 2 } + ( d _ { 0 } \mathcal { E } ) ^ { 2 }## so the angular frequency is ##
\Omega=\frac { 2 } { \hbar }\sqrt{( \frac { \hbar \omega _ { 0 } } { 2 }) ^ { 2 } + ( d _ { 0 } \mathcal { E } ) ^ { 2 }}
##
By using the given values I find that ##\Omega=188000000000 \; rad.s^{-1}##.Is it correct ?
Thank you[/B]
1. Homework Statement
The dipole moment of an ammonia molecule is ##d_0=5*10^{-30} C.m##.If we apply a static electric field of ##\mathcal { E }=1*10^{6 }V*m^{-1}## to an ammonia molecule initially in the state ## |ψG⟩## where the nitrogen molecule is considered to be on the left,we make it oscillate between states ##|ψ+⟩## and ##|ψ−⟩ ##which represent two wavefunctions.What is the angular frequency of oscillation between the two states ?
Homework Equations
In the ##(|ψ+⟩ ,|ψ−⟩)## basis:
##\hat { H } _ { \mathrm { tot. } } = \hat { H } _ { 0 } + \hat { W } = \frac { \hbar \omega _ { 0 } } { 2 } \left( \begin{array} { c c } { - 1 } & { 0 } \\ { 0 } & { 1 } \end{array} \right) - d _ { 0 } \mathcal { E } \left( \begin{array} { l l } { 0 } & { 1 } \\ { 1 } & { 0 } \end{array} \right)##
We can put the hamiltonian in a simple form : ##\hat { H } _ { t o t } = - \frac { \hbar \Omega } { 2 } \left( \begin{array} { c c } { \cos 2 \theta } & { \sin 2 \theta } \\ { \sin 2 \theta } & { - \cos 2 \theta } \end{array} \right)## with ##\tan 2 \theta = \frac { 2 d _ { 0 } \mathcal { E } } { \hbar \omega _ { 0 } } \quad - \pi / 4 < \theta < \pi / 4## et ##\omega _ { 0 }=160*10^9 Hz##[/B]
##( \frac { \hbar \Omega } { 2 } ) ^ { 2 } = ( \frac { \hbar \omega _ { 0 } } { 2 }) ^ { 2 } + ( d _ { 0 } \mathcal { E } ) ^ { 2 }## so the angular frequency is ##
\Omega=\frac { 2 } { \hbar }\sqrt{( \frac { \hbar \omega _ { 0 } } { 2 }) ^ { 2 } + ( d _ { 0 } \mathcal { E } ) ^ { 2 }}
##
The Attempt at a Solution
By using the given values I find that ##\Omega=188000000000 \; rad.s^{-1}##.Is it correct ?
Thank you[/B]
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