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Let $\rho = \begin{bmatrix}\ 1&0 \\ 0&0 \end{bmatrix}$, $\rho' = \begin{bmatrix}\ 0&0 \\ 0&1 \end{bmatrix}$, $\rho'' = \dfrac{1}{2}\begin{bmatrix}\ 1&1 \\ 1&1 \end{bmatrix}$ (all density operators).

Consider a physical operation $\phi$ such that $\phi(\rho) = \rho$, $\phi(\rho') = \rho'$, $\phi(\rho'') = \dfrac{1}{5}\begin{bmatrix}\ 4&2 \\ 2&1 \end{bmatrix}$.

Why is $\phi$ not a realisable physical operation? It certainly preserves trace and positivity...

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What is your definition of "physical operation?" – Chris White May 20 '13 at 22:01
Do we have any particular knowledge about $\phi$? is $\phi$ a linear operator? – Alex A May 20 '13 at 22:10
@AlexA: Yes, $\phi$ is a linear operator. – wemblem May 20 '13 at 22:31
@wemblem Can you please edit your question to make clear what you are trying to ask? In particular what is an "affine map" ? As per your definition a "physical operation" should be an affine map; then why don't you simply check if the map is "affine" or not? (whatever that means) – user10001 May 20 '13 at 23:34
@wemblem: As Norbert implies in his answer below, "physical operations" are completely positive, trace preserving maps (CPTP maps). Anything you can do to a state is always a CPTP map, conversely any CPTP map can be physically realized. The map in the question is not CPTP. I don't think I'd use the phrase "affine" here, since such maps are linear superoperators. – Dan Stahlke May 21 '13 at 1:06
up vote 2 down vote accepted

Your map fails to be completely positive. If you apply it to half of a maximally entangled state $(|0\rangle|0\rangle+|1\rangle|1\rangle)/\sqrt{2}$, you can easily see that $\phi(\rho)=\rho$ and $\phi(\rho')=\rho'$ imply that $\phi(|0\rangle\langle1|) = \alpha |0\rangle\langle1|$ and $\phi(|1\rangle\langle0|) = \alpha^* |1\rangle\langle0|$ for the resulting state to be positive (with $|\alpha|\le1$). However, this is incompatible with the last condition.

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How does one represent the maximally entangled state in matrix form? – wemblem May 21 '13 at 6:50
Would it be $\frac{1}{\sqrt 2}\begin{bmatrix}\ 1&0 \\ 0&1 \end{bmatrix}$? – wemblem May 21 '13 at 7:53
You should read up on some basics of quantum information ... it is $\left[\begin{matrix}1&0&0&1\\0&0&0&0\\0&0&0&0\\1&0&0&1\end{matrix}\right]/2$, and you want to apply $\phi\otimes I$, i.e., you apply $\phi$ to every $2\times 2$ block individually. For $\phi$ to be a physical operation, the resulting state must be positive. – Norbert Schuch May 21 '13 at 8:21
So it is trivial that $\dfrac{1}{\sqrt 2}\begin{bmatrix}\ 1&0 \\ 0&0 \end{bmatrix} \otimes \dfrac{1}{\sqrt 2}\begin{bmatrix}\ 1&0 \\ 0&0 \end{bmatrix}$ + $\dfrac{1}{\sqrt 2}\begin{bmatrix}\ 0&0 \\ 0&1 \end{bmatrix} \otimes \dfrac{1}{\sqrt 2}\begin{bmatrix}\ 0&0 \\ 0&1 \end{bmatrix}$ = $\dfrac{1}{2}\begin{bmatrix}\ 1&0&0&0 \\ 0&0&0&0 \\ 0&0&0&0 \\ 0&0&0&1 \end{bmatrix}$, but how does the $\dfrac{1}{2}\begin{bmatrix}\ 1&1 \\ 1&1 \end{bmatrix}$ term come into play? – wemblem May 21 '13 at 18:56
Sorry, I've only ever dealt with completely positive maps in the sense that transposition fails to be completely positive. – wemblem May 22 '13 at 8:28

A physical quantum operation ${\cal E}$ can be described as a map between the set of density operators of the form

$${\cal E}(\rho) =\sum_k E_k \rho E_k^{\dagger}, \qquad \sum_kE_k^{\dagger}E_k \leq {\bf 1},$$

cf. Ref. 1. As Norbert Schuch correctly notes this implies that a physical quantum operation ${\cal E}$ must be a completely positive map. In this answer we note that OP's example fails to be quantum operation for an even more elementary reason: Its image (of the Bloch ball $B^3$) doesn't lie inside the Bloch ball

$$ \rho~=~\frac{1}{2}({\bf 1} + \sigma), \qquad \sigma = \sum_{i=1}^3x^i\sigma_i, \qquad \sum_{i=1}^3(x^i)^2\leq 1.$$

In detail, let

$$\rho^{+} ~=~ \begin{bmatrix}\ 1&0 \\ 0&0 \end{bmatrix}, \quad \rho^{-} ~=~ \begin{bmatrix}\ 0&0 \\ 0&1 \end{bmatrix}, \quad \rho_{1} ~=~ \frac{1}{2}\begin{bmatrix}\ 1&1 \\ 1&1 \end{bmatrix}, \quad \rho_{2} ~=~ \frac{1}{5}\begin{bmatrix}\ 4&2 \\ 2&1 \end{bmatrix},$$


$$ {\cal E}(\rho^{\pm})~=~\rho^{\pm}, \qquad {\cal E}(\rho_1)~=~\rho_2. $$

In other words, the North and the South pole of the Bloch sphere $S^2$ are fixed points, and the pure state $(1,0,0)$ is mapped to the pure state $(\frac{4}{5},0,\frac{3}{5})$ in the $xz$ plane.

It follow from linearity that

$$ {\cal E}({\bf 1})~=~{\bf 1}, \qquad {\cal E}(\sigma_3)~=~\sigma_3, \qquad {\cal E}(\sigma_1)~=~\frac{4}{5}\sigma_1+\frac{3}{5}\sigma_3. $$

In other words, the great circle in the $xz$ plane is mapped to an ellipse in the $xz$ plane, which spends half the time outside (and half the time inside) the great circle. So the image of $ {\cal E}$ doesn't lie inside the Bloch ball as it should.


  1. M.A. Nielsen and I.L. Chuang, Quantum Computation and Quantum Information, (2011) Section 8.2.
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