Blog > The Equivalences of the Choi-Jamiolkowski Isomorphism (Part I)

## The Equivalences of the Choi-Jamiolkowski Isomorphism (Part I)

October 16th, 2009

The Choi-Jamiolkowski isomorphism is an isomorphism between linear maps from Mn to Mm and operators living in the tensor product space Mn ⊗ Mm. Given any linear map Φ : Mn → Mm, we can define the Choi matrix of Φ to be

It turns out that this association between Φ and CΦ defines an isomorphism, which has become known as the Choi-Jamiolkowski isomorphism. Because much is already known about linear operators, the Choi-Jamiolkowski isomorphism provides a simple way of studying linear maps on operators — just study the associated linear operators instead. Thus, since there does not seem to be a list compiled anywhere of all of the known associations through this isomorphism, I figure I might as well start one here. I’m planning on this being a two-parter post because there’s a lot to be said.

### 1. All Linear Maps / All Operators

By the very fact that we’re talking about an isomorphism, it follows that the set of all linear maps from Mn to Mm corresponds to the set of all linear operators in Mn ⊗ Mm. One can then use the singular value decomposition on the Choi matrix of the linear map Φ to see that we can find sets of operators {Ai} and {Bi} such that

To construct the operators Ai and Bi, simply reshape the left singular vectors and right singular vectors of the Choi matrix and multiply the Ai operators by the corresponding singular values. An alternative (and much more mathematically-heavy) method of proving this representation of Φ is to use the Generalized Stinespring Dilation Theorem [1, Theorem 8.4].

### 2. Hermicity-Preserving Maps / Hermitian Operators

The set of Hermicity-Preserving linear maps (that is, maps Φ such that Φ(X) is Hermitian whenever X is Hermitian) corresponds to the set of Hermitian operators. By using the spectral decomposition theorem on CΦ and recalling that Hermitian operators have real eigenvalues, it follows that there are real constants {λi} such that

Again, the trick is to construct each Ai so that the vectorization of Ai is the ith eigenvector of CΦ and λi is the corresponding eigenvalue. Because every Hermitian operator can be written as the difference of two positive semidefinite operators, it is a simple corollary that every Hermicity-Preserving Map can be written as the difference of two completely positive linear maps — this will become more clear after Section 4. It is also clear that we can absorb the magnitude of the constant λi into the operator Ai, so we can write any Hermicity-preserving linear map in the form above, where each λi = ±1.

### 3. Positive Maps / Block Positive Operators

A linear map Φ is said to be positive if Φ(X) is positive semidefinite whenever X is positive semidefinite. A useful characterization of these maps is still out of reach and is currently a very active area of research in quantum information science and operator theory. The associated operators CΦ are those that satisfy

In terms of quantum information, these operators are positive on separable states. In the world of operator theory, these operators are usually referred to as block positive operators. As of yet we do not have a quick deterministic method of testing whether or not an operator is block positive (and thus we do not have a quick deterministic way of testing whether or not a linear map is positive).

### 4. Completely Positive Maps / Positive Semidefinite Operators

The most famous class of linear maps in quantum information science, completely positive maps are maps Φ such that (idk ⊗ Φ) is a positive map for any natural number k. That is, even if there is an ancillary system of arbitrary dimension, the map still preserves positivity. These maps were characterized in terms of their Choi matrix in the early ’70s [2], and it turns out that Φ is completely positive if and only if CΦ is positive semidefinite. It follows from the spectral decomposition theorem (much like in Section 2) that Φ can be written as

Again, the Ai operators (which are known as Kraus operators) are obtained by reshaping the eigenvectors of CΦ. It also follows (and was proved by Choi) that Φ is completely positive if and only if (idn ⊗ Φ) is positive. Also note that, as there exists an orthonormal basis of eigenvectors of CΦ, the Ai operators can be constructed so that Tr(Ai*Aj) = δij, the Kronecker delta. An alternative method of deriving the representation of Φ(X) is to use the Stinespring Dilation Theorem [1, Theorem 4.1] of operator theory.

### 5. k-Positive Maps / k-Block Positive Operators

Interpolating between the situations of Section 3 and Section 4 are k-positive maps. A map is said to be k-positive if (idk ⊗ Φ) is a positive map. Thus, complete positivity of a map Φ is equivalent to Φ being k-positive for all natural numbers k, which is equivalent to Φ being n-positive. Positivity of Φ is the same as 1-positivity of Φ. Since we don’t even have effective methods for determining positivity of linear maps, it makes sense that we don’t have effective methods for determining k-positivity of linear maps, so they are still a fairly active area of research. It is known that Φ is k-positive if and only if

Operators of this type are referred to as k-block positive operators, and SR(x) denotes the Schmidt rank of the vector x. Because a vector has Schmidt rank 1 if and only if it is separable, it follows that this condition reduces to the condition that we saw in Section 3 for positive maps in the k = 1 case. Similarly, since all vectors have Schmidt rank less than or equal to n, it follows that Φ is n-positive if and only if CΦ is positive semidefinite, which we saw in Section 4.

Update [October 23, 2009]: Part II of this post is now online.

References:

1. V. I. Paulsen, Completely Bounded Maps and Operator Algebras, Cambridge Studies in Advanced Mathematics 78, Cambridge University Press, Cambridge, 2003.
2. M.-D. Choi, Completely Positive Linear Maps on Complex Matrices, Lin. Alg. Appl, 285-290 (1975).
1. April 29th, 2010 at 01:19 | #1

Maybe I am wrong, but it seems to me that in:

“it follows that the set of all linear maps from Mn to Mm corresponds to the set of all linear operators on Mn ⊗ Mm”

it should say at the end Cn ⊗ Cm (where by C I mean the Complex numbers)

2. April 29th, 2010 at 10:33 | #2

Thanks for the correction, Abel. I meant to say “in” Mn ⊗ Mm instead of “on” Mn ⊗ Mm there — it is fixed now.

3. November 10th, 2010 at 12:54 | #3

Dude. This is awesome. Very deep.

Mn and Mm are any two manifolds?

What are applications of Hermiticity-preserving maps? I assume it’s from quantum…

4. January 17th, 2016 at 19:04 | #4

Hi, there might be something wrong in section 4. As to choi(or Kraus) representation for completely positive maps, you can only take A_i mutually orthogonal. If you want to make A_i normalized, you have to write like this psi(X) = sum_i lambda_i A_i X A_i ^\dagger, right?

5. January 17th, 2016 at 21:36 | #5

@Kun Fang – Eep, you’re absolutely right. They can be taken mutually orthogonal, but of course there has to be a scaling factor somewhere. Either inside the A_i’s (in which case they aren’t orthonormal as I claimed), or you can put a lambda_i in front of them.

Good catch, thanks!

6. February 2nd, 2016 at 16:36 | #6

In 2, How can one ensure that $I \otimes \Psi$ is Hermicity preserving when $\Psi$ is Hermiticity preserving? It seems that, if you want to insist that $C_{\Psi}$ is a Hermitian operator, we need not only Hermiticity-preservity of $\Psi$ but ‘completely Hermiticity preservity’ of $\Psi$.

7. February 3rd, 2016 at 13:47 | #7

@Guldam Kwak – It follows from the fact that every Hermitian operator $X \in M_n \otimes M_n$ can be written as a sum of the form $\sum_i A_i \otimes B_i$, where each $A_i$ and $B_i$ is Hermitian. Then $(I \otimes \Psi)(X) = \sum_i A_i \otimes \Psi(B_i)$, which is Hermitian.

In other words, $\Psi$ being Hermiticity-preserving is indeed equivalent to it being “completely Hermiticity-preservity”.

1. October 23rd, 2009 at 08:09 | #1
2. November 20th, 2009 at 08:15 | #2
3. July 11th, 2011 at 20:04 | #3