# Adding phases of two qubits

Imaging a system of two qubits which at a given step of evolution is in the state

1. $$|q_{1}(0)\rangle = |0\rangle + e^{-i\phi_{1}}|1\rangle$$,
2. $$|q_{2}(0)\rangle = |0\rangle + e^{-i\phi_{2}}|1\rangle$$,

is there a simple two qubit quantum circuit of $$n$$ gates (CNOT, Hadamard and single-qubit rotations) that produces the output

1. $$|q_{1}(n)\rangle = |0\rangle + e^{-i\left(\phi_{1}+\phi_{2}\right)}|1\rangle$$
2. $$|q_{2}(n)\rangle = |0\rangle + e^{-i\phi_{2}}|1\rangle$$

from the input above?

• Do you know $\phi_2$ in advance? Commented Mar 15, 2023 at 17:35
• No, both phases are unknown. Commented Mar 15, 2023 at 17:40
• Then I sense the answer is going to be no, no such a unitary circuit exists per no-cloning theorem Commented Mar 15, 2023 at 17:50

You've asked for a $$2$$-qubit quantum circuit, thus we are looking for a $$2$$-qubit unitary that performs the desired transformation. Note that if we take $$\theta=0$$, then $$\frac{1}{\sqrt{2}}\left(|0\rangle+\mathrm{e}^{-\mathrm{i}\theta}|1\rangle\right)=|+\rangle$$ and if we take $$\theta=\pi$$, then $$\frac{1}{\sqrt{2}}\left(|0\rangle+\mathrm{e}^{-\mathrm{i}\theta}|1\rangle\right)=|-\rangle$$. Thus, we can describe the action of the gate we're looking for in the $$XX$$ basis: \begin{align}|++\rangle\to|++\rangle\\|+-\rangle\to|--\rangle\\|-+\rangle\to|-+\rangle\\|--\rangle\to|+-\rangle\end{align} This is perfectly doable by applying a CNOT gate on the second qubit controlled by the first one. What we've shown is that if such a gate exists, then it must be a CNOT gate, whose matrix is: $$\mathsf{C}-X=\begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&0&1\\0&0&1&0\end{pmatrix}$$ Now, we start with the state: $$\left|q_1(0)\right\rangle\otimes\left|q_2(0)\right\rangle=\begin{pmatrix}1\\\mathrm{e}^{-\mathrm{i}\varphi_1}\end{pmatrix}\otimes\begin{pmatrix}1\\\mathrm{e}^{-\mathrm{i}\varphi_2}\end{pmatrix}=\begin{pmatrix}1\\\mathrm{e}^{-\mathrm{i}\varphi_2}\\\mathrm{e}^{-\mathrm{i}\varphi_1}\\\mathrm{e}^{-\mathrm{i}\left(\varphi_1+\varphi_2\right)}\end{pmatrix}$$ We then apply the CNOT to this state, yielding: $$\begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&0&1\\0&0&1&0\end{pmatrix}\begin{pmatrix}1\\\mathrm{e}^{-\mathrm{i}\varphi_2}\\\mathrm{e}^{-\mathrm{i}\varphi_1}\\\mathrm{e}^{-\mathrm{i}\left(\varphi_1+\varphi_2\right)}\end{pmatrix}=\begin{pmatrix}1\\\mathrm{e}^{-\mathrm{i}\varphi_2}\\\mathrm{e}^{-\mathrm{i}\left(\varphi_1+\varphi_2\right)}\\\mathrm{e}^{-\mathrm{i}\varphi_1}\end{pmatrix}$$ But we wanted to end up with: $$\left|q_1(n)\right\rangle\otimes\left|q_2(n)\right\rangle=\begin{pmatrix}1\\\mathrm{e}^{-\mathrm{i}\left(\varphi_1+\varphi_2\right)}\end{pmatrix}\otimes\begin{pmatrix}1\\\mathrm{e}^{-\mathrm{i}\varphi_2}\end{pmatrix}=\begin{pmatrix}1\\\mathrm{e}^{-\mathrm{i}\varphi_2}\\\mathrm{e}^{-\mathrm{i}\left(\varphi_1+\varphi_2\right)}\\\mathrm{e}^{-\mathrm{i}\left(\varphi_1+2\varphi_2\right)}\end{pmatrix}$$ Though the first three coefficients match, the last one differs. However, note that if $$\varphi_2=0$$ or if $$\varphi_2=\pi$$, these expressions match, as we expected. But as soon as $$\varphi_2\not\equiv0\pmod{\pi}$$, these expression are not equal anymore. Thus, the CNOT gate is not a valid candidate for a $$2$$-qubit circuit for realizing this operation. Since this was the only possible candidate, such a gate cannot exist.
We can do even better and show that this is not possible even if you allow more general operations like working with more qubits or measuring some of them. As hinted by Mauricio, this operation manages to clone states that are non-orthogonal, which is forbidden by the no-cloning theorem. Take the second qubit to be the one to be cloned. Then if we start with $$\varphi_1=0$$ for the first qubit, this operation achieves the following: $$|+\rangle\otimes|\psi(\theta)\rangle\to|\psi(\theta\rangle\otimes|\psi(\theta)\rangle$$ with $$|\psi(\theta)\rangle=\frac{1}{\sqrt{2}}\left(|0\rangle+\mathrm{e}^{-\mathrm{i}\theta}|1\rangle\right)$$. Clearly, this is a cloning circuit, and the states $$|\psi(\theta)\rangle$$ are not orthogonal with each others.