- #1
Lebnm
- 31
- 1
- Homework Statement
- Consider a ammonia molecule ##NH_{3}##. The ##N## atom can be above (state ##| + \rangle## or below (state ##| - \rangle##) the plane formed by the ##H## atoms. The hamiltonian of the system ##\hat{H}## is showed below, writed in the ##\left \{ |+ \rangle, |- \rangle \right \}## basis. Knowing that in ##t = 0## the system is in the state ##| + \rangle##, the objective is determine the density matrix ##\rho## at any time ##t > 0##.
- Relevant Equations
- H = \begin{pmatrix}
E_{0} & -A\\
-A & E_{0}
\end{pmatrix}
In ##t = 0##, we have ##\rho (0) = | + \rangle \langle + |##. The time evolution of the density matrix is given by ##\rho(t) = e^{-i\hat{H}t} \rho (0) e^{i\hat{H}t}## (I am considering ##\hbar = 1##). I can write the state ##| + \rangle ## as a linear combination of the eigenstates of the hamiltonian. Making this, it's easy to compute the actions of the time evolutions operators. So I get the ##\rho (t)## writed white the eigenkets of ##\hat{H}##. Writing these eigenkets in the basis ##\left \{ |+ \rangle, |- \rangle \right \}##, I get the following result: $$\rho (t) = \frac{1}{2}\left \{ \left [ 1 + cos(2At) ] | + \rangle \langle + | + [1 - cos(2At)] | - \rangle \langle - | + sin(2At)| - \rangle \langle + | - sin(2At)| + \rangle \langle - | \right ] \right \}.$$It look likes wrong to me, because of the two last terms. If they were canceled, this density matrix would make sense, but the sum of them is not zero, is it? So it's more likely that I missed the calculations?