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 Feb 4 at 18:53
• Can you elaborate on what you do (and do not) understand about it? – Norbert Schuch Feb 5 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. – Siddhant Singh Feb 5 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. – user2508039 Feb 6 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. – Sanchayan Dutta Feb 6 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.