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I am studying quantum computing a little bit by myself, and I have simple questions.

I didn't find a clear definition of what is a completely positive and trace-preserving (CPTP) map. The best I've found was here .

To summarize it -
Let $\,T:H\to H $ be a mapping with $H$ the space of operators (that means). In order for $T(\cdot)$ to be CPTP, it needs to verify:

  1. $T(\cdot)$ need to be a linear mapping
  2. $\forall A \in H$ we have $ \text{Tr}(A)=\text{Tr}(T(A))$. It means that $T(\cdot)$ is trace-preserving.
  3. $T(\cdot)$ need to be a positive map

I have a question about this last point.

  • What is a "completely" positive map?
  • More precisely, what does it mean concerning the matrix representation of $T(\cdot)$? Does it mean that all its elements are positive? Or (completely) positive map for an application $T(\cdot)$ is equivalent to say that the matrix representation of $T(\cdot)$ is a positive semi-definite matrix?
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    $\begingroup$ To study quantum computing, these lectures by Artur Ekert are a good start. CPTP maps are explained in lecture 7. $\endgroup$ Sep 21, 2023 at 5:39

2 Answers 2

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[A]

States lie in Hilbert space $\mathcal{H_S}$.

  • $|\psi\rangle \in \mathcal{H_S}\,.$

Operators, density operators lie in the bounded operator space of $\mathcal{H}_S$.

  • $\rho \in \mathcal{B}(\mathcal{H}_S)\,. $

Maps (super-operators) acting on these operators lie in the operator space of operators/density matrices.

  • $\Phi \in \mathcal{B}\big[\mathcal{B}(\mathcal{H}_S)\big]\,.$
  • $\Phi: \mathcal{B}(\mathcal{H}_S) \mapsto \mathcal{B}(\mathcal{H}_S) \,.$

Let $X \in \mathcal{B}(\mathcal{H}_S)\,.$

$\therefore \Phi(X) \in \mathcal{B}(\mathcal{H}_S)\,.$


Let $\Phi$ be a map. $\Phi$ is said to be a CPTP map if

  1. $\Phi$ is trace-preserving
  2. $\Phi$ is linear
  3. $\Phi$ is positive
  4. $\Phi$ is completely positive.

1. Trace-preserving

$\Phi$ is trace-preserving if $\text{Tr}(X) = \text{Tr}(\Phi(X))\,.$


2. Linear

$\Phi$ is linear if $\Phi(aX_1 + bX_2) = a\Phi(X_1) + b\Phi(X_2)\,.$


3. Positive

$\Phi$ is positive if for a given $X\geq0$, then $\Phi(X)\geq0 \,. $

  • $\geq0$ in this context means it is positive-semidefinite, i.e., has non-negative eigenvalues.

This condition is more like positivity-preserving.


4. Completely Positive

$\Phi$ is completely positive if $(\Phi \otimes \mathcal{I}^{(d)}_R)(A) \geq 0$ is true for all dimensions of $\mathcal{H}_R \geq 1$, given that $A\geq0\,.$

where

  • $A \in \mathcal{B}(\mathcal{H}_S \otimes \mathcal{H}_R)\,.$
  • $\mathcal{H}_R$ is some auxiliary Hilbert space with dimension $d$.
  • $\mathcal{I}^{(d)}_R \in \mathcal{B}\big[\mathcal{B}(\mathcal{H}_R)\big]$ is an identity map acting on the auxiliary Hilbert space of dimension $d$.

This condition is also more like complete-positivity-preserving.

Condition 4 of complete positivity supersedes the positivity condition, and it turns out that conditions 1,2 & 4 are necessary & sufficient for a map to be CPTP. (Since for $d=1$, complete positivity reduces to ordinary positivity)


If you want an intuition as to why this complete positivity condition is needed, and positivity is not sufficient:

  • Imagine you have two qubits. You apply some positive map on them; hence, the resultant density matrix will also be positive since your map was positive, hence a valid quantum state.

  • Now, imagine you have 10 qubits which are all entangled with each other. You apply some operation, i.e. some quantum map, to your first two qubits, and you are doing nothing to the remaining eight qubits. This is the same as applying identity to the rest of the eight qubits. But now, due to entanglement, even if you only apply a positive map on the first two qubits and do nothing to the remaining eight qubits, the resultant density matrix may have negative eigenvalues which are not allowed/do not correspond to any valid physical quantum state. Hence, a positive map does not correspond to a physical process.

  • However, if your map was a CPTP map on the first two qubits, it will ensure that even if your system is entangled with the remaining eight qubits, this operation on the first two-qubit will not cause any negative eigenvalues in the whole density matrix of 10 qubits.

In short -

  • CPTP map on a subsystem will always ensure that the whole system will result in a valid quantum state after applying the map, and not just the subsystem.

  • A positive map will always give you a valid state of the subsystem, but if there is the presence of entanglement with other parts of the system, then the resultant state of your whole system may not be a valid quantum state after the application of a positive map on the subsystem.


As for the matrix representation of $\Phi$, matrix $\Phi$ can be anything, but for every positive semi-definite operator $X$, $\Phi(X)$ must also be a positive semi-definite, i.e., it has non-negative eigenvalues.

As I have said above, when we say a map is positive, what we really mean is that it is positivity-preserving. It maps every positive operator to another positive operator.


[B]

As for your second question, every CPTP map can be represented in a Kraus operator-sum representation format and for each physical process, there are infinite choices of these Kraus OSRs, and hence it is a surjective mapping.

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  • $\begingroup$ Thank a lot for you clear answer. Just to avoid confusion in my head. In order for an quantum gate (susceptible to modify the state of the qubit) to be feasible this quantum gate must accept a matrix representation that is unitary. And in order for quantum channel (whose purpose is to transmit the qbit without altering it between quantum gate) to be feasible is to admit a matrix representation that is CPTC? Am i right? $\endgroup$
    – X0-user-0X
    Sep 21, 2023 at 14:40
  • $\begingroup$ Note: By quantum gate I meant quantum logic gate. $\endgroup$
    – X0-user-0X
    Sep 21, 2023 at 14:53
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    $\begingroup$ Yes, you are correct that quantum gates must be unitary matrices which govern how your qubits will evolve in the case where there are no outside interactions, your qubits are perfectly isolated. As for the CPTP maps, they are much more general than that. They represent any valid quantum operation on your system in the presence of outside interactions, imperfect isolations as well as in the absence of outside interactions. They can represent not just quantum channels, but also perfect/imperfect gate operations, evolution of system in presence/absence of outside interactions, even measurements. $\endgroup$
    – FDGod
    Sep 21, 2023 at 18:13
  • $\begingroup$ So can I conclude from what you wrote that all CPTP are Unitary matrices but that not all unitary matrices are CTPT? $\endgroup$
    – X0-user-0X
    Sep 22, 2023 at 14:29
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    $\begingroup$ You can conclude that you can write operation of any unitary acting on your state as a CPTP map but you cannot write every CPTP map as a unitary acting on your state. CPTP maps are not “unitary matrices” themselves. You can represent every $U \rho U^{\dag}$ operation as a CPTP map. $\endgroup$
    – FDGod
    Sep 22, 2023 at 15:15
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Definitions

Let $\mathcal{H}$ be a complex Hilbert space. It turns out that the set $L(\mathcal{H})$ of all linear operators on $\mathcal{H}$ is also a Hilbert space. Let $I_\mathcal{H}$ denote the identity map on $\mathcal{H}$.

Definition 1. A linear map $\Phi:L(\mathcal{H})\to L(\mathcal{H})$ is said to be trace-preserving if $$ \mathrm{tr}(\Phi(A))=\mathrm{tr}(A)\tag1 $$ for every linear operator $A\in L(\mathcal{H})$.

Definition 2. A linear map $\Phi:L(\mathcal{H})\to L(\mathcal{H})$ is said to be positive if it sends every positive semidefinite operator to a positive semidefinite operator.

Definition 3. A linear map $\Phi:L(\mathcal{H})\to L(\mathcal{H})$ is said to be completely positive if the linear map $\Phi\otimes I_\mathcal{H'}$ is positive for every Hilbert space $\mathcal{H'}$.

Finally,

Definition 4. A linear map $\Phi:L(\mathcal{H})\to L(\mathcal{H})$ is said to be CPTP if it is completely positive and trace-preserving.

Intuition

The concept of the CPTP map arises in the analysis of the constraints that a mathematical function on $L(\mathcal{H})$ must satisfy in order for it to correspond to a transformation realizable as physical dynamics of an open quantum system$^1$. First and foremost, any such function must be linear. However, this is not sufficient.

The key fact motivating the above definitions is that in density matrix formalism of quantum mechanics the spectrum of a linear operator encodes a classical probability distribution. Therefore, if a linear function on $L(\mathcal{H})$ is to represent a physical dynamical transformation, then it must send an operator whose spectrum is a valid probability distribution to another operator with such a spectrum.

More precisely, if the input eigenvalues sum up to one, then the output eigenvalues must sum up to one. This is what definition 1 says. Further, if the input eigenvalues are non-negative, then the output eigenvalues must be non-negative. This is what definition 2 says. Somewhat unexpectedly, it turns out that if a positive map acts on a subsystem, then a map$^2$ acting on a larger system may fail to be positive. This is why we need definition 3. Finally, definition 4 collects all these constraints into one concept: the CPTP map.

Matrix representation

In general, the matrix of $\Phi:L(\mathcal{H})\to L(\mathcal{H})$ in an arbitrary operator basis does not immediately divulge its complete positivity. However, we can think of $\Phi$ as an order $4$ tensor with four indices: input row, input column, output row, output column rather than an order $2$ tensor (aka a matrix). Looking at $\Phi$ from this perspective, we see that there is more than one way to represent it as a matrix. Indeed, we can combine the four indices into a pair of composite indices in a few different ways.

If input row and output row indices are combined to give the matrix row index and input column and output column indices are combined to give the matrix column index, then the resulting matrix $C(\Phi)$ is called the Choi matrix.

This matrix is different than the matrix $K(\Phi)$ obtained by following the standard linear algebra procedure for a given choice of operator basis in $L(\mathcal{H})$. However, $C(\Phi)$ makes it easier to determine complete positivity of $\Phi$ than $K(\Phi)$, because of the following

Theorem. $\Phi$ is completely positive if and only if $C(\Phi)$ is positive semidefinite.


$^1$ The formalism of quantum channels makes an important hidden assumption that the input to the channel and its environment are independent. Crucially, if the environment contains systems that participated in input preparation then the dynamics may fail to be described by a linear map. See closing remarks in Chapter 8 of Nielsen & Chuang.
$^2$ Transpose $T(A)=A^T$ is positive since it preserves the eigenvalues. However, $T\otimes I$ sends rank one projector onto $\mathrm{span}(|00\rangle,|11\rangle)$ to a matrix with eigenvalue $-1$.

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    $\begingroup$ I used to think $\mapsto$ is the proper notation for showing the mapping. Just looked up the meaning of $\mapsto$ and $\to$ in the context of algebra maps. Sorry about the edit. $\endgroup$
    – FDGod
    Sep 21, 2023 at 8:44
  • $\begingroup$ Thank a lot for you clear answer. Just to avoid confusion in my head. In order for an quantum gate (susceptible to modify the state of the qubit) to be feasible this quantum gate must accept a matrix representation that is unitary. And in order for quantum channel (whose purpose is to transmit the qbit without altering it between quantum gate) to be feasible is to admit a matrix representation that is CPTC? Am i right? $\endgroup$
    – X0-user-0X
    Sep 21, 2023 at 14:40
  • $\begingroup$ Note: By quantum gate I meant quantum logic gate. $\endgroup$
    – X0-user-0X
    Sep 21, 2023 at 14:53

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