# NOT Universal Operator and Computational basis

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This is the relationship between density operator and Bloch vector: $$\rho= \frac{1}{2}({\bf{\hat{1}}}+{\bf{b}}.\boldsymbol{\hat{\sigma}})$$

We define the NOT Universal Operator in the following way:

$$U: {\bf{b}}\to -{\bf{b}}/3$$

My question is - How does NOT Universal Operator act on the elements of the computational basis: $|0\rangle, |1\rangle$?

This post imported from StackExchange Physics at 2014-06-11 15:05 (UCT), posted by SE-user user15940
retagged Jun 11, 2014
Related physics.stackexchange.com/questions/26864/…

This post imported from StackExchange Physics at 2014-06-11 15:05 (UCT), posted by SE-user twistor59
I put some bold symbols in your question because the quantities involved are vectors. Can I ask you to confirm 1) the bold is OK, and that 2) you really meant to have the /3? I thought UNOT just mapped antipodally, so there should be no -3 there.

This post imported from StackExchange Physics at 2014-06-11 15:05 (UCT), posted by SE-user twistor59
This definition of the Universal NOT seems to differ from the one given in the related question: the OP's (current) definition maps states antipodally but introduces a "damping" of the Bloch vector.

This post imported from StackExchange Physics at 2014-06-11 15:05 (UCT), posted by SE-user Juan Bermejo Vega

Homework, downvoted.

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Note As it has been said in the comments, this definition of Universal-NOT gate seems to differ from others discussed in other posts [1]. This answer uses the definition proposed by the OP, i.e.

$$\rho=\frac{1}{2}(I+\vec{b}\cdot\vec{\sigma}) \quad \longrightarrow \quad U(\rho)=\frac{1}{2}\left(I-\frac{1}{3}\vec{b}\cdot\vec{\sigma}\right)$$ Where I use the symbol $I$ for the identity matrix to avoid confusion with 1 and $\vec{b}\in \mathbb{R}^3$ denotes the Bloch vector.

We write the density matrices of the computational basis states explicitly: $$\rho_a=|a\rangle\langle a |= \frac{1}{2}(I+\vec{b}_a\cdot\vec{\sigma}),$$ where $a\in\{0,1\}$. Expanding this expression readily yields the vectors $b_a$: $$\vec{b}_a=(0,0,\pm1).$$ Applying your definition of $U$ to these density operators, the action of the operator on basis states can be obtained directly: $$U(|0\rangle\langle 0|)= \frac{1}{2}(I- \frac{1}{3}\sigma_z)=\frac{1}{3}|0\rangle\langle 0| +\frac{2}{3}|1\rangle\langle 1 |$$ $$U(|1\rangle\langle 1|)= \frac{1}{2}(I+ \frac{1}{3}\sigma_z)=\frac{2}{3}|0\rangle\langle 0| +\frac{1}{3}|1\rangle\langle 1 |$$ We can observe that, because the factor $1/3$ that "damps" the Bloch vector, pure basis states evolve into mixed states; notice that, intuitively, states get closer to the totally mixed state $I/2$ if you make the Bloch vector $\vec{b}$ go to zero.

This post imported from StackExchange Physics at 2014-06-11 15:05 (UCT), posted by SE-user Juan Bermejo Vega
answered Jan 1, 2013 by (285 points)
+1 Interested to know if the /3 was intentional..

This post imported from StackExchange Physics at 2014-06-11 15:05 (UCT), posted by SE-user twistor59
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