# The hamiltonian of single trapped ion model: ion trap quantum computer

This question is related to the textbook, "Quantum Computation and Quantum Information written by M. A. Nielsen and I. L. Chuang.

I tried to derive a relation when we deal with a model system where a single ion is trapped. (page 317-318)

The free particle hamiltonian is given as

$$H_{0} = \hbar \omega_{0}S_{z}+\hbar \omega_{z}a^{\dagger}a,$$ which stands for two electron spin states and vibrational states from the harmonic potential generated from an ion trap.

And, the authors considers exerting electromagnetic field which gives the perturbation:

$$H_{I} = -\vec{\mu} \cdot \vec{B},$$ and they say that it can be approximated to

$$H_{I} \approx \left[\frac{\hbar \Omega}{2} \left(S_{+}e^{i(\varphi-\omega t)}+ S_{-}e^{-i(\varphi-\omega t)} \right) \right] + \left[ i \frac{\eta \hbar \Omega}{2} \left\{S_{+}a+S_{-}a^{\dagger}+S_{+}a^{\dagger}+S_{-}a \right\} \left(e^{i(\varphi-\omega t)}-e^{-i(\varphi-\omega t)} \right)\right]$$

I'm wondering if there's some missing terms of the above relation.

The rest stuff is my derivation:

The given magnetic field owing to the EM field is $\vec{B} = B_{1} \hat{x} \cos(kz- \omega t + \varphi),$ and the magnetic dipole moment of atom is $\vec{\mu} = \mu_{m} \vec{S}$. Thus, the perturbation is the following: \begin{align*} H_{I} &= -\mu_{m}S_{x}B_{1} \cos(kz-\omega t + \varphi) \\ &=-\mu_{m} B_{1} \frac{S_{+}+S_{-}}{2} \cdot \frac{1}{2} \left(e^{i(kz-\omega t + \varphi)} + e^{-i(kz-\omega t + \varphi)} \right) \\ &= \frac{\hbar \Omega}{2} \left(S_{+}+S_{-} \right) \left(e^{i(kz-\omega t + \varphi)} + e^{-i(kz-\omega t + \varphi)} \right), \end{align*} where $\Omega = -\mu_{m}B_{1}/2 \hbar$.

Now, I put $z = z_{0}(a^{\dagger}+a)$ because the space is quantized according to the harmonic potential, and assumed the Lamb-Dicke parameter, $\eta = k z_{0} \rightarrow 0$. Then $$e^{ikz} = e^{i \eta (a^{\dagger}+a)} \approx 1 + i \eta (a^{\dagger}+a).$$

Put this to the original equation, then

\begin{align*} H_{I} & \approx \frac{\hbar \Omega}{2} \left(S_{+}+S_{-} \right) \left\{ \left[ 1 +i \eta(a^{\dagger}+a) \right] e^{i(\varphi - \omega t)}+ \left[ 1 -i \eta(a^{\dagger}+a) \right] e^{-i(\varphi - \omega t)} \right\} \\ &= \left[\frac{\hbar \Omega}{2} \left(S_{+}e^{i(\varphi-\omega t)}+ \color{blue}{ S_{-}e^{i(\varphi-\omega t)} + S_{+}e^{-i(\varphi-\omega t)} }+ S_{-}e^{-i(\varphi-\omega t)} \right) \right] + \left[ i \frac{\eta \hbar \Omega}{2} \left\{S_{+}a+S_{-}a^{\dagger}+S_{+}a^{\dagger}+S_{-}a \right\} \left(e^{i(\varphi-\omega t)}-e^{-i(\varphi-\omega t)} \right)\right] \end{align*}

So, the above blue terms are missing in the equation from the book.

Did I make some mistakes?

I guess I found the justification for the approximation.

If we see the hamiltonian in interaction picture, then

$$H_{I}' = \frac{\hbar \Omega}{2} \left(S_{+}e^{i[\varphi-(\omega-\omega_{0}) t]}+ \color{blue}{ S_{-}e^{i[\varphi-(\omega+\omega_{0}) t]} + S_{+}e^{-i[\varphi-(\omega+\omega_{0}) t]} }+ S_{-}e^{-i[\varphi-(\omega-\omega_{0}) t]} \right)$$

When the EM field is resonant $\omega = \omega_{0}$, only two terms are dominant. $$H_{I}' \approx \frac{\hbar \Omega}{2} \left(S_{+}e^{i\varphi}+ S_{-}e^{-i\varphi} \right)$$

So, the blue terms may be insignificant if the authors tried to explain the qubit control in resonant region.

Also, it seems to be customary to omit these high frequency term. I find it's one of examples for rotating wave approximation.

• Nice job answering your own question. This is one of the best ways to learn :-) – DanielSank Jul 7 '18 at 5:21