# How does a $2 \pi$ pulse in Cirac Zoller give a -1 sign to the state?

I understand the first step in the Cirac-Zoller controlled-phase gate; about how to move the state from the electronic state to the vibrational mode state. However, I am unable to understand how a $$2\pi$$ pulse gives the -1 sign to the state and how it is applied conditioned on the vibrational mode. Any help is appreciated.

• Can you include a reference to the "Cirac-Zoller controlled-phase gate"?
– glS
Commented Feb 4, 2019 at 18:53
• Can you elaborate on what you do (and do not) understand about it? Commented Feb 5, 2019 at 2:24
• It is okay. He is asking about the implementation of C-NOT gate using the scheme of Cirac-Zoller (en.wikipedia.org/wiki/Cirac-Zoller_Controlled-NOT_Gate) where the second step is a $2\pi$ "pulse". I can answer the question in detail. Commented Feb 5, 2019 at 14:08
• I think I can close this question. I have posted another question on the Hamiltonian of the system. If I can get answer to that, I can figure out the CIrac-Zoller gate myself. Commented Feb 6, 2019 at 3:09
• @SiddhāntSingh I see, thanks for the reference. Ideally, that should have been in the original question itself. Anyway, I've reopened the post now. Feel free to write up your answer. Commented Feb 6, 2019 at 10:44

CNOT in ion traps is not implemented in one go. It is decomposed in terms of SWAP gate. The general decomposition they use is:

$$$$\operatorname{CNOT}_{jk}=H_k \operatorname{SWAP}^{-1}_k C_j(Z) \operatorname{SWAP}_kH_k$$$$ where $$j,k$$ are the labels for the control and target ion (qubit) respectively, in the array. $$H$$ is Hadamard gate and $$C(Z)$$ is the control-Z operation.

Hadamard is just rotations, so we can ignore that for a while and focus on the middle three gates, they are the three crucial steps of the algorithm of Cirac-Zoller implementation in the right order.

The circuit is as shown (Source: Nielsen and Chuang, pg.323)

The Control-Z is what the $$2\pi$$ pulse is in the second step.

If you follow this scheme of Nielsen and Chuang (sec-7.6.4), the $$2\pi$$ pulse implements this operation as follows:

We need to switch $$|01\rangle \leftrightarrow |10\rangle$$ for the control-Z. This is done as follows. We need a way to swap qubits between the atom’s internal spin state and the phonon state (little perturbation over the spin states). This can be done by tuning a laser to the frequency $$\omega_0-\omega_z$$, and arranging for the phase (which can always be done by the laser) to be such that we perform the rotation $$R_y(\pi)$$ on the subspace spanned by $$|01\rangle$$ and $$|10\rangle$$, which is just the unitary transform

$$$$C_{SWAP}=|00\rangle \langle 00|+|01\rangle \langle 10|-|10\rangle \langle 01|+|11\rangle \langle 11|$$$$ on the $$|00\rangle, |01\rangle, |10\rangle, |11\rangle$$ space. If the initial state is $$a|00\rangle + b|10\rangle$$ (that is, the phonon is initially $$|0\rangle$$), then the state after the swap is $$a|00\rangle + b|01\rangle$$, so this accomplishes the desired swap operation, which is the second step in Cirac-Zoller. Note here that $$2\pi$$ pulse is just the $$R_y(\pi)$$ here which is equivalent by just a convention.

The frequency notation is explained in the last image.

I would suggest to read thoroughly the section 7.6 of Nielsen and Chuang to make a complete understanding.