A checkable Lindbladian condition

Summary

Physicists often define a Lindbladian superoperator as one whose action on an operator $B$ can be written as

(1) $\begin{align*} \mathcal{L}[B] = -i [H,B] + \sum_i \left[ L_i B L_i^\dagger - \frac{1}{2}\left(L_i^\dagger L_i B + B L_i^\dagger L_i\right)\right], \end{align*}$

for some operator $H$ with positive anti-Hermitian part, $H-H^\dagger \ge 0$ , and some set of operators $\{L^{(i)}\}$ . But how does one efficiently check if a given superoperator is Lindbladian? In this post I give an “elementary” proof of a less well-known characterization of Lindbladians:

A superoperator $\mathcal{L}$ generates completely positive dynamics $e^{t\mathcal{L}}$ , and hence is Lindbladian, if and only if $\mathcal{P} \mathcal{L}^{\mathrm{PT}} \mathcal{P} \ge 0$ , i.e.,
$\mathrm{Tr}\left[B^\dagger (\mathcal{P} \mathcal{L}^{\mathrm{PT}} \mathcal{P})[B]\right] \ge 0$

for all $B$ . Here “ $\mathrm{PT}$ ” denotes a partial transpose, $\mathcal{P} = \mathcal{I} - \mathcal{I}^{\mathrm{PT}}/N = \mathcal{P}^2$ is the “superprojector” that removes an operator’s trace, $\mathcal{I}$ is the identity superoperator, and $N$ is the dimension of the space upon which the operators act.

Thus, we can efficiently check if an arbitrary superoperator $\mathcal{L}$ is Lindbladian by diagonalizing $\mathcal{P}\mathcal{L}^{\mathrm{PT}} \mathcal{P}$ and seeing if all the eigenvalues are positive.

Contents

1 Summary
2 A quick note on terminology
3 Intro
4 Main result
5 Comments
6 Appendix A: Definitions
7 Appendix B: Lemmas
8 Footnotes

A quick note on terminology

The terms superoperator, completely positive (CP), trace preserving (TP), and Lindbladian are defined below in Appendix A in case you aren’t already familiar with them.

Confusingly, the standard practice is to say a superoperator $\mathcal{S}$ is “positive” when it is positivity preserving: $B \ge 0 \Rightarrow \mathcal{S}[B]\ge 0$ . This condition is logically independent from the property of a superoperator being “positive” in the traditional sense of being a positive operator, i.e., $\langle B,\mathcal{S}[B]\rangle \ge 0$ for all operators (matrices) $B$ , where

$\langle B,C\rangle \equiv \mathrm{Tr}[B^\dagger C] = \sum_{n=1}^N \sum_{n'=1}^N B^\dagger_{nn'} C_{n'n}$

is the Hilbert-Schmidt inner product on the space of $N\times N$ matrices. We will refer frequently to this latter condition, so for clarity we call it op-positivity, and denote it with the traditional notation $\mathcal{S}\ge 0$ .

Intro

It is reasonably well known by physicists that Lindbladian superoperators, Eq. (1), generate CP time evolution of density matrices, i.e., $e^{t\mathcal{L}}[\rho] = \sum_{b=0}^\infty t^b \mathcal{L}^b[\rho]/b!$ is completely positive when $t\ge 0$ and $\mathcal{L}$ satisfies Eq. (1). This evolution is furthermore trace-preserving when $H$ is Hermitian.^a

Indeed, for any one parameter family of CP maps $\mathcal{S}_t$ obeying the semigroup property $\mathcal{S}_{t_2}\mathcal{S}_{t_1} = \mathcal{S}_{t_1+t_2}$ that is differentiable about $\mathcal{S}_{t=0}=\mathcal{I}$ , the family is necessarily generated by some Lindbladian superoperator: $\mathcal{S}_t = e^{t\mathcal{L}$ . The Hamiltonian and Lindblad operators defining the Lindbladian superoperator in Eq. (1) can be extracted from the eigendecomposition of $\mathcal{S}_t$ for small $t$ . Although this procedure is highly enlightening, it does not yield an easily “checkable” criterion for when a given superoperators can be put in the Lindbladian form. How could we easily see whether $\mathcal{L}$ satisfies Eq. (1) without searching exhaustively through all choices of $H$ and $L_i$ ?

It has long been known, but is not always widely appreciated by physicists^b, that CP maps are exactly those superoperators^c $\mathcal{S}$ that are op-positive under the partial transpose operation^d: $(\mathcal{S}^{\mathrm{PT}})_{(nm)(n'm')} = \mathcal{S}_{(nn')(mm')}$ , where $\mathcal{C}_{\mathcal{S}} = \mathcal{S}^{\mathrm{PT}}$ is called the Choi matrix.^e (We use the index convention $(\mathcal{S}[B])_{nn'} = \sum_{m=1}^N \sum_{m'=1}^N \mathcal{S}_{(nn')(mm')} B_{mm'}$ .) This is just one of several elegant relationships between the most important properties of a superoperator, considered as a map on density matrices, and its corresponding Choi matrix $\mathcal{C}_{\mathcal{S}} = \mathcal{S}^{\mathrm{PT}}$ :

Map property		Choi property
Map preserves Hermiticity: $(\rho=\rho^\dagger) \Rightarrow (\mathcal{S}[\rho]=\mathcal{S}[\rho]^\dagger$ )	⇔	Choi is Hermitian: $\mathcal{S}^{\mathrm{PT}} = (\mathcal{S}^{\mathrm{PT}})^\dagger$
Map is CP: $\mathcal{S} \otimes \mathcal{I}^{(\tilde{N})} \ge 0$	⇔	Choi is op-positive: $\mathcal{S}^{\mathrm{PT}} \ge 0$
Map is trace-preserving: $\mathrm{Tr} [\mathcal{S}[\rho]] =\mathrm{Tr} [\rho]$	⇔	Unit “outer” trace of Choi: $\mathrm{Tr}_{\mathrm{out}} \mathcal{S}^{\mathrm{PT}} = I$
Map is unital: $\mathcal{S}[I] = I$	⇔	Unit “inner” trace of Choi: $\mathrm{Tr}_{\mathrm{in}} \mathcal{S}^{\mathrm{PT}} = I$

Here, we define the “outer” and “inner” partial traces^f as, respectively,

$(\mathrm{Tr}_{\mathrm{out}} \mathcal{S})_{nm} = \sum_{p=1}^N \mathcal{S}_{(pn)(pm)}, \quad(\mathrm{Tr}_{\mathrm{in}} \mathcal{S})_{nm} = \sum_{n'=1}^N \mathcal{S}_{(np)(mp)} .$

The equivalences in the above table can all be checked explicitly with index manipulation.

We can use the first two equivalences in the table to show the following.

Main result

We define $\mathcal{P}$ to be the superprojector that removes an operator’s trace, $\mathcal{P} \equiv \mathcal{I} - \mathcal{I}^{\mathrm{PT}}/N = \mathcal{P}^2$ , so that $\mathcal{P}[B] = B - (\mathrm{Tr}[B]/N)I$ .

Totally true fact: These two statements are equivalent:

$e^{t\mathcal{L}}$ is completely positive for any $t\ge 0$ .
$\mathcal{P} \mathcal{L}^{\mathrm{PT}} \mathcal{P}$ is an op-positive superoperator.

Proof: First, we’ll show that (1) implies (2).

If $e^{t\mathcal{L}}$ is CP, then $(e^{t\mathcal{L}})^{\mathrm{PT}} = \mathcal{I}^{\mathrm{PT}} + t \mathcal{L}^{\mathrm{PT}} + O(t^2) \ge 0$ . For this to hold for arbitrarily small $t$ , it must also be true when dropping the $O(t^2)$ terms for all sufficiently small $t$ , i.e., for all positive $t$ below some threshold. Then we use our first lemma, which is proved in Appendix B.

Lemma 1: The superoperator $\mathcal{I}^{\mathrm{PT}} + t \mathcal{L}^{\mathrm{PT}}$ is an op-positive superoperator for all sufficiently small $t$ if and only if $\mathcal{P} \mathcal{L}^{\mathrm{PT}} \mathcal{P}$ is an op-positive superoperator.

Applying Lemma 1, we conclude that (1) implies (2).

Now we’ll show that (2) implies (1). If $\mathcal{P} \mathcal{L}^{\mathrm{PT}} \mathcal{P}$ is op-positive, then by Lemma 1 we know $(\mathcal{I} + t \mathcal{L})^{\mathrm{PT}} \ge 0$ for sufficiently small $t$ . Now we make use of our second lemma, also proved in Appendix B.

Lemma 2: If the partial trace of a superoperator is positive, then the partial traces of all positive powers of that superoperator are also positive, i.e., $\mathcal{S}^{\mathrm{PT}}\ge 0 \Rightarrow (\mathcal{S}^a)^{\mathrm{PT}}\ge 0$ for all positive integers $a$ .

If $(\mathcal{I} + t \mathcal{L})^{\mathrm{PT}} \ge 0$ for sufficiently small $t$ , then by Lemma 2 we know, for the same values of $t$ , that the object

$[(\mathcal{I} + t \mathcal{L})^a]^{\mathrm{PT}} = \sum_{b=0}^a \binom{a}{b} t^b (\mathcal{L}^b)^{\mathrm{PT}}$

is an op-positive superoperator (for any positive integer $a$ ). If we define

$v_c(t\mathcal{L}) \equiv \sum_{a=0}^c \mu_a^{(c)} (\mathcal{I} + t \mathcal{L})^a, \qquad \mu_a^{(c)} \equiv \frac{1}{a!} \sum_{d=0}^{c-a} \frac{(-1)^d}{d!}$

then one can then check that $\mu_a^{(c)} \ge 0$ and $v_c(t\mathcal{L}) = \sum_{b=0}^c t^b\mathcal{L}^b/b!$ . Since $v_c(t\mathcal{L})^{\mathrm{PT}}$ is a mixture (convex combination) of op-positive superoperators $[(\mathcal{I} + t \mathcal{L})^a]^{\mathrm{PT}}$ , it itself is an op-positive superoperator for all $c$ , and hence its limit

$\lim_{c\to\infty} v_c(t\mathcal{L}) ^{\mathrm{PT}} = \lim_{c\to\infty} \sum_{b=0}^c \frac{t^b}{b!} (\mathcal{L}^b)^{\mathrm{PT}} = (e^{t \mathcal{L}})^{\mathrm{PT}}$

is also an op-positive superoperator. This makes $e^{t \mathcal{L}}$ completely positive for sufficiently small $t$ , but since complete positivity is preserved under composition we conclude that $e^{t \mathcal{L}}$ is CP for all $t\ge 0$ . ☐

Therefore, we can take

(2) $\begin{align*} \mathcal{P} \mathcal{L}^{\mathrm{PT}} \mathcal{P} \ge 0 \end{align*}$

to be an equivalent definition to say that $\mathcal{L}$ is a Lindbladian superoperator. It can be supplemented with the trace-preserving condition $\mathrm{Tr}_{\mathrm{out}} \mathcal{L}^{\mathrm{PT}} = 0$ (implying $\mathrm{Tr}[ \mathcal{L}[B]]=0$ for all $B$ ) to define the subset of Lindbladians generating CPTP evolution. Lindblad called this subset “completely dissipative”, and it is equivalent to Eq. (2) with Hermitian $H$ .

Comments

Although Eq. (2) is not as useful as Eq. (1) for understanding the action of a Lindbladian, it is much easier to use Eq. (2) to check whether a given superoperator is Lindbladian.

With a bit of manipulation, we can re-write Eq. (2) in a more quantum-information-y (and less linear-algebraic) way:

$\overline{P}_\Psi[ (\mathcal{L}\otimes \mathcal{I})(|\Psi \rangle\langle \Psi|)] \overline{P}_\Psi \ge 0,$

where $|\Psi \rangle = N^{-1} \sum_{n=1}^N|n\rangle|n\rangle$ is some maximally entangled state and $\overline{P}_\Psi=I - |\Psi \rangle\langle \Psi|$ projects onto the orthogonal subspace.^g (This condition is independent of the choice of basis $\{|n\rangle\}$ and hence the choice of maximally entangled state.)

If you’ve interested in learning more, Tarasov’s “Quantum Mechanics of Non-Hamiltonian and Dissipative Systems” is the most thorough yet readable monograph I’ve found.

Appendix A: Definitions

For our purposes, superoperators are just linear operators on the vector space of linear operators.^h If we represent finite-dimensional linear operators as $N\times N$ matrices, then superoperators are $N^2\times N^2$ matrices. A superoperator $\mathcal{S}$ can be indexed as $\mathcal{S}_{(nn')(mm')}$ and its action $\mathcal{S}[B]$ on an operator $B$ is given by the matrix elements

$(\mathcal{S}[B])_{nn'} = \sum_{m=1}^N \sum_{m'=1}^N \mathcal{S}_{(nn')(mm')} B_{mm'}.$

Here, $B_{mm'}$ are the matrix elements of $B$ and the parentheses on $(mm')$ just emphasize that we treat this as a joint index (taking $N^2$ values) of $\mathcal{S}$ . (It does not denote antisymmeterization.)

Lindbladians are the subset of superoperators that can be put in the form

$\begin{align*} \nonumber \mathcal{L}_{(nn')(mm')} &= -i H_{nm} \delta_{n'm'} +i \delta_{nm} H_{n'm'} \\ &+ \sum_j \left[ L^{(j)}_{nm} L^{(j)\dagger}_{m'n'} - \frac{1}{2}\sum_\ell \left(L^{(j)\dagger}_{n\ell} L^{(j)}_{\ell m} \delta_{n'm'} + \delta_{nm} L^{(j)\dagger}_{m'\ell} L_^{(j)}{\ell n'}\right)\right] \end{align*}$

for some Hermitian operator $H=H^\dagger$ (the Hamiltonian) and some set of operator $\{L^{(j)}\}$ (the Lindblad operators), where of course $B^\dagger_{mn} = (B_{nm})^*$ . You can express this more elegantly as

$\mathcal{L} = -i (H \odot I - I \odot H) + \sum_i L_i\odot L_i^\dagger - (L_i^\dagger L_i \odot I + I \odot L_i^\dagger L_i),$

where “ $\odot$ ” is just a tensor product with a bit of syntactic sugar: Given any two operators $B$ and $C$ , the superoperator $B\odot C$ is defined to have action $(B\odot C)[E] = BEC$ .

As explained near the beginning, we distinguish two notions of superoperator “positivity”:

$\mathcal{S}$ is positivity preserving when $B\ge 0 \Rightarrow \mathcal{S}[B] \ge 0$ .
$\mathcal{S}$ is op-positive when $\langle B,\mathcal{S}[B]\rangle \ge 0$ for all $B$ .

When the first condition holds, people usually just say that $\mathcal{S}$ is a “positive map”, but this can be confusing because it is logically independent of $\mathcal{S}$ being a “positive operator” when thought of, naturally enough, as an operator acting on the space of matrices (the second condition).

A superoperator $\mathcal{S}$ is said to be completely positive (CP) when $\mathcal{S}\otimes \mathcal{I}^{\tilde{N}}$ is positive for all positive integers $\tilde{N}$ , where $\mathcal{I}^{\tilde{N}}$ is the indentity superoperator on a separate space of $\tilde{N}\times \tilde{N}$ matrices. The tensor product on superoperators is naturally defined as $(\mathcal{S}\otimes \mathcal{T})(B\otimes C) = (\mathcal{S}[B])\otimes(\mathcal{T}[C])$ , extended by linearity. (Note that we do not necessarily require CP maps to preserve operator trace.) Complete positivity is a strengthening of positivity preservation (not op-positivity).

Appendix B: Lemmas

Proof. For a fixed vector $|1\rangle$ and Hermitian operator $B$ , consider the family of Hermitian operators $G_\epsilon = |1\rangle\langle 1| + \epsilon B$ for all $\epsilon \ge 0$ . If another vector $|v\rangle$ has non-zero overlap with $|1\rangle$ , then

is positive for sufficiently small $\epsilon$ . Therefore, if there is a $|v\rangle$ such that $\langle v|G_\epsilon|v\rangle <0$ for arbitrarily small $\epsilon>0$ , we know that vector is orthogonal to $|1\rangle$ . Such a vector exists if and only if $P B \bar P$ is not a positive operator, where $P = I - |1\rangle\langle 1|$ .☐

Proof. If $\mathcal{S}^{\mathrm{PT}}$ is positive then it has the eigendecomposition $\mathcal{S}^{\mathrm{PT}}[B] = \sum_i \mu_i E^{(i)} \mathrm{Tr}[ E^{(i)\dagger} B]$ or, with indices, $\mathcal{S}^{\mathrm{PT}}_{(ab)(cd)} = \sum_i \mu_i E^{(i)}_{ab} E^{(i)*}_{cd}$ . The eigenvalues $\mu_i$ are positive and the eigenoperators $E^{(i)}$ are orthonormal under the Hilbert-Schmidt norm. The previous expression also gives us the matrix elements $\mathcal{S}_{(ac)(bd)}$ for the original superoperator, allowing us to use simple index manipulation to show that

$(\mathcal{S}^n)^{\mathrm{PT}}_{(ab)(cd)} = \sum_{i_1, \ldots i_n} \lambda_{i_1}\cdots \lambda_{i_n} \left(E^{(i_1)}\cdots E^{(i_n)}\right)_{ab} \left(E^{(i_1)}\cdots E^{(i_n)}\right)^*_{cd}.$

This expression is manifestly positive since it’s a mixture (convex combination) of superprojectors, so we conclude $(\mathcal{S}^n)^{\mathrm{PT}} \ge 0$ for positive integers $n$ .☐

Remark. This is equivalent to the statement that complete positivity is preserved by composition.

Footnotes

(↵ returns to text)

Non-hermitian $H$ can be useful to account for the system decaying into a mode outside the space of density matrices under consideration. In this case, the trace should be non-increasing, so we require that $(H-H^\dagger)/2 \ge 0$ , i.e., the anti-Hermitian part of $H$ is a positive operator.↵
An ever-present barrier in navigating the mathematical literature on quantum information is disentangling (1) the basic finite-dimentional linear algebra ideas from (2) the mathematical machinery for handling infinite dimensions (e.g., the distinction between W*-algebras and the more general C*-algebras, ultraweak continuity, etc.). This is really too bad, because finite-dimensional ideas are almost always the physical ones. I propose the following maxim of quantum mechanical pedagogy: All ideas should be presented first in the finite-dimensional context.↵
Note, the op-positivity of a superoperator and the op-positivity of the Choi matrix are logically independent. This is true even though the op-positivity of the Choi matrix implies (but is not implied by) the superoperator preserving positivity of the operators it acts on: $\mathcal{P}^{\mathrm{PT}} \ge 0 \Rightarrow (B \ge 0 \Rightarrow \mathcal{P}[B]\ge 0)$ .↵
This is a property that is invariant under local unitary operations applied to the input or output density matrices, but not under general unitaries. In that sense, it “knows” about the super index structure $(nn')$ , but does not care the bases in which you do the sum over $n$ and $n'$ .↵
Indeed, the eigenoperators of an op-positive Choi matrix (which are orthonormal with respect to the Hilbert-Schmidt inner product on operators) are just the Krauss operators of the corresponding CP map.↵
These adjectives refer to whether we are summing over the indices associated with the input matrix or output matrix of the corresponding superoperator.↵
This version is presented in Sec. 7.1.2 of Michael Wolf’s “Quantum Channels & Operations” [PDF]. The proof given in this blog post differs in that it is “elementary”, making use only of basic linear algebra facts.↵
We assume the range and domain of our superoperators are the same space since we will be considering semigroups generated by them.↵

2 Comments

ouyang
August 29, 2022 at 5:26 am

Sir, thank you for your detailed explanation!
For my research project, I have recently struggled to understand the condition named Conditional Completely Positivity of Lindbladian you mentioned, but due to the lack of knowledge of algebra, I’ve been very confused about it. But luckily now I find your blog, so I just need to figure out the relationship between Choi matrix and superoperator. Thanks so much to your instructive article!
- Jess Riedel
  August 29, 2022 at 12:45 pm
  
  Thanks for taking the time to let me know you found this useful! Keeps me motivated.

A checkable Lindbladian condition

Summary

A quick note on terminology

Intro

Main result

Comments

Appendix A: Definitions

Appendix B: Lemmas

Footnotes

2 Comments

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