Density matrix of an ammonia molecule

[Lebnm]

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?

 

[DrClaude]

You forgot a couple of $i$'s.

 

[Lebnm]

How so?

 

[DrClaude]

How did you get the sines?

 

[Lebnm]

I will put the calculations: being $| 1 \rangle$ and $| 2 \rangle$ the eigenstates of $\hat{H}$ with eigenvalues $E_{0} - A$ and $E_{0} + A$, respectively. So $$| 1 \rangle = \frac{1}{\sqrt{2}}( |+\rangle + |-\rangle ),$$ $$| 2 \rangle = \frac{1}{\sqrt{2}}( |+\rangle - |- \rangle).$$ Writing $\rho(0)$ using these kets and applying the time evolutions operators to them, we get $$\rho (t) = \frac{1}{2}\left ( |1 \rangle \langle 1 | + e^{-i2At}|1 \rangle \langle 2 | + e^{i2At}|2 \rangle \langle 1 | + |2 \rangle \langle 2 |\right ).$$ Rewriting this in the $\left\{ |+\rangle, |-\rangle \right\}$ basis, we have $$\rho (t) = \frac{1}{4} \left [ (|+ \rangle \langle + | - |+ \rangle \langle - | - |- \rangle \langle + | + |- \rangle \langle - |) \\ + e^{-i2At}(|+ \rangle \langle + | + |+ \rangle \langle - | - |- \rangle \langle + | - |- \rangle \langle - |) \\ + e^{i2At}(|+ \rangle \langle + | - |+ \rangle \langle - | + |- \rangle \langle + | - |- \rangle \langle - |)\\ + (|+ \rangle \langle + | + |+ \rangle \langle - | + |- \rangle \langle + | + |- \rangle \langle - |) \right ].$$ From this follow my answer. I get the sines combining the exponentials.

 

[DrClaude]

$$
e^{-i 2 At} - e^{i 2 At} = -2 i \sin(2 A t)
$$

 

[Lebnm]

Oh, it's true... I don't believe I have forgotten this haha. But my density matrix is incorret, isn't it? it couldn't have a imaginary factor multiplying an external product...

 

[Truecrimson]

I didn't check the calculation, but having $i$ in the off-diagonal elements is fine as long as your density matrix is Hermitian.