# Why is the partial trace of this subsystem equal to this? [closed]

I am doing my bachelors dissertation based on an article by David Deutsch. He defines the action of a quantum gate as: $$U = \sum_{x, y \in \mathcal{Z}_{2}} |x \dot{+}y\rangle|y\rangle\langle x|\langle y|$$ where $\dot{+}$ is the OR operator from boolean algebra. Then, according to me: He then takes the partial trace: $$\text{Tr}_{2}[U(|\psi\rangle\langle\psi|\otimes\hat{\rho})U^{\dagger}]$$

and it is equal to: $$\frac{1}{2}\hat{I} + \text{Re}[(\langle 0|\psi\rangle\langle\psi|1\rangle)(|0\rangle\langle 1|+|1\rangle\langle 0|)]$$ EDIT: where does the $\frac{1}{2}\hat{I}$ comes from?

• Let's index your particles kets. $U = \sum_{x,y\in Z_2}|x\dot + y\rangle_1 |y\rangle_2 \langle x |_1 \langle y|_2$. Then what would $U\dagger$ be? – Ali Jul 12 '18 at 10:06
• Would it be $$U^{\dagger} = \sum_{x, y \in \mathbb{Z}_{2}} \langle y| \langle x \dot{+} y| y\rangle |x\rangle$$ – Marco Luna Jul 12 '18 at 17:01
• No, you would get an operator. The point of my last comment was to not mix the states. – Ali Jul 12 '18 at 17:33
• After about two hours from now, I will write an answer and explain what you are missing. :) – Ali Jul 12 '18 at 17:37
• Can you link the article? – Ali Jul 12 '18 at 19:52

Let's rewrite $U$ as, so it's more clear what it does:

\begin{aligned} U &= \sum_{x, y \in \mathbb{Z}_{2}} |x \dot{+}y\rangle|y\rangle\langle x|\langle y| \\ &= \sum_{x, y \in \mathbb{Z}_{2}} |x \dot{+}y\rangle_1|y\rangle_2\langle x|_1\langle y|_2 \\ &= \sum_{x, y \in \mathbb{Z}_{2}} \left(|x \dot{+}y\rangle_1\langle x|_1\right)\left(|y\rangle_2\langle y|_2\right) \ .\\ \end{aligned}

This operator leaves the first qubit untouched and maps the second qubit to the logical OR of the qubits. I have to say this is a peculiar operator, that's not even unitary. Normally people use the XOR gate(which is unitary) in the quantum information field. I guess the paper predates most of quantum information algorithms.

Anyway, $U^\dagger$ is straightforward to calculate:

$$U^\dagger = \sum_{x, y \in \mathbb{Z}_{2}} \left(|x\rangle_1\langle x \dot{+}y|_1\right)\left(|y\rangle_2\langle y|_2\right) \ .$$

\begin{aligned} \text{Tr}_2\left[ U \left(|\psi\rangle\langle\psi|\otimes\rho\right)U^\dagger\right] &= \text{Tr}_2 \left[\sum_{w,x, y,z \in \mathbb{Z}_{2}} \left(|x \dot{+}y\rangle_1\langle x|_1\right)\left(|y\rangle_2\langle y|_2\right) \left(|\psi\rangle\langle\psi|_1\otimes\rho_2\right) \left(|w\rangle_1\langle w \dot{+}z|_1\right)\left(|z\rangle_2\langle z|_2\right)\right] \\ &= \text{Tr}_2\left[\sum_{w,x, y,z \in \mathbb{Z}_{2}} \left(|x \dot{+}y\rangle\langle x|\psi\rangle\langle\psi|w\rangle\langle w \dot{+}z| \right)_1\otimes \left(|y\rangle\langle y|\rho |z\rangle\langle z|\right)_2 \right] \\ &=\sum_{u\in \mathbb{Z}_2} \sum_{w,x, y,z \in \mathbb{Z}_{2}} \left(|x \dot{+}y\rangle\langle x|\psi\rangle\langle\psi|w\rangle\langle w \dot{+}z| \right) \left(\langle u|y\rangle\langle y|\rho |z\rangle\langle z|u\rangle \right) \\ &= \sum_{u\in \mathbb{Z}_2} \sum_{w,x \in \mathbb{Z}_{2}} \left(|x \dot{+}u\rangle\langle x|\psi\rangle\langle\psi|w\rangle\langle w \dot{+}u| \right)\left(\langle u|\rho |u\rangle \right) \\ &= \sum_{w,x \in \mathbb{Z}_{2}} \langle x|\psi\rangle\langle\psi|w\rangle\left(\langle 0|\rho |0\rangle|x \rangle\langle w | +\langle 1|\rho |1\rangle|1 \rangle\langle 1 |\right) \end{aligned}

Now, the paper says $\rho = \frac{1}{2}\hat{I} + \text{Re}[(\langle 0|\psi\rangle\langle\psi|1\rangle)(|0\rangle\langle 1|+|1\rangle\langle 0|)]$, is a solution to the equation $\text{Tr}_2\left[ U \left(|\psi\rangle\langle\psi|\otimes\rho\right)U^\dagger\right] = \rho$, which I think you can check yourself by substituting it in the above equation.