I am currently learning about quantum channels and am sadly stuck at a rudimentary problem, where I don't understand how to find the Kraus matrices of a quantum channel.

The amplitude damping channel is described by the following action on the computational basis states: \begin{aligned} \mathcal E: |0\rangle \otimes |0\rangle &\mapsto |0\rangle \otimes |0\rangle, \\ \mathcal E: |1\rangle \otimes |0\rangle &\mapsto \sqrt{1-p} |1\rangle \otimes |0\rangle + \ \sqrt{p} |0\rangle \otimes |1\rangle. \end{aligned}

Now unfortunately I don’t know where to start to find the Kraus operators. Also, I was wondering how I could check if this evolution takes place unitarily?


2 Answers 2


The question is essentially about how to relate the unitary/isometric representation of the dephasing channel with its representation in terms of Kraus operators.

In general, an isometric representation of a channel $\Phi$ is a writing of the form $\Phi(\rho)=\operatorname{tr}_2[V\rho V^\dagger]$ for some isometry $V$. This is the representation you're starting with: you're describing/defining the amplitude damping channel as the isometry defined in the computational basis as $$V|0\rangle= |0\rangle\otimes|0\rangle, \qquad V|1\rangle = \sqrt{1-p}|1\rangle\otimes|0\rangle+\sqrt p |0\rangle\otimes|1\rangle.$$ Note that this is an isometry because it's a linear operator of the form $V:\mathbb{C}^2\to(\mathbb{C}^2)^{\otimes2}$ such that $V^\dagger V=I_2$. Equivalently, it's a matrix with orthonormal columns. As a matrix you can represent this as $$V = \begin{pmatrix} 1 & 0 \\ 0 & \sqrt p \\ 0 & \sqrt{1-p} \\ 0&0 \end{pmatrix}.$$

Given any such isometric representation, you can find a Kraus representation for $\Phi$ using the operators $A_a\equiv (I\otimes \langle a|)V$, with $|a\rangle$ a(ny) orthonormal basis for the environmental space (the second space used in the definition of $V$). See also this and this for more details about the general proceedure. In the case at hand, this gives $$A_0 = (I\otimes \langle0|)V = \begin{pmatrix}1&0\\0&\sqrt{1-p}\end{pmatrix}, \qquad A_1 = (I\otimes \langle1|)V = \begin{pmatrix}0&\sqrt p\\0&0\end{pmatrix}.$$

Finding a different set of Kraus operators

For the sake of completeness, note that you can also find different Kraus operators from this isometric representation, using different bases in the formula above. For example, using the eigenbasis of the Pauli $X$ matrix, you get $$A_0' \equiv (I\otimes \langle +|)V = \frac1{\sqrt2}\begin{pmatrix}1 & \sqrt p \\ 0 & \sqrt{1-p}\end{pmatrix}, \qquad A_1' \equiv (I\otimes\langle -|) V = \frac{1}{\sqrt2}\begin{pmatrix}1 & -\sqrt p\\ 0 & \sqrt{1-p}\end{pmatrix}.$$ Note that these are related to the previous Kraus operators via $\sqrt2 A_0'=A_0 + A_1$ and $\sqrt2 A_1'=A_0-A_1$.

Prove equivalence of Kraus decompositions

Though they look different, these Kraus operators describe the same identical channel as the ones derived above. And the same applies to any set of Kraus operators derived this way. You can, for example, verify this by showing that both Kraus decompositions correspond to the same Choi matrix, which you obtain by computing $J(\Phi)=\sum_a \operatorname{vec}(A_a)\operatorname{vec}(A_a)^\dagger$ with $\operatorname{vec}(A_a)$ vectorisation of $A_a$. For more details on this process see e.g. this and this. You'll find that regardless of which Kraus decomposition you use you'll get $$J(\Phi) = \begin{pmatrix}1 & 0 & 0 &\sqrt{1-p} \\ 0&p&0&0\\0&0&0&0\\\sqrt{1-p}&0&0&1-p\end{pmatrix}.$$

It's easy to verify that this Choi has eigendecomposition

$$J(\Phi) = (2-p) \mathbb{P}_u + p \mathbb{P}_v, \qquad \mathbb{P}_u\equiv |u\rangle\!\langle u|, \\ |u\rangle \equiv \frac{1}{\sqrt{2-p}}\begin{pmatrix}1\\0\\0\\\sqrt{1-p}\end{pmatrix}, \qquad |v\rangle\equiv\begin{pmatrix}0\\1\\0\\0\end{pmatrix}.$$ Remember that rank-1 decompositions of the Choi corresponding bijectively to Kraus decompositions of the channel. The first Kraus decompositions above corresponds to the eigendecomposition of the Choi. It is however interesting to note that the second decomposition does not. It instead corresponds to a different rank-1 decomposition, namely, $$J(\Phi) = \mathbb{P}_{w_1} + \mathbb{P}_{w_2}, \\ |w_1\rangle \equiv \frac{1}{\sqrt2}(\sqrt{2-p} |u\rangle + \sqrt p |v\rangle), \qquad |w_2\rangle \equiv \frac{1}{\sqrt2}(\sqrt{2-p} |u\rangle - \sqrt p |v\rangle).$$ You can verify directly that if you unvectorise these two (unnormalised) vectors you get precisely $A_0',A_1'$. You can more characterise all possible rank-1 decompositions of a positive semidefinite operators (and thus all possible Kraus decompositions of a channel) as discussed e.g. here and here.


Regarding your first point, I recommend you read the follwoing reference: Approximating Decoherence Processes for the Design and Simulation of Quantum Error Correction Codes on Classical Computers.

The evolution is not unitary. Note that you are asking for Kraus operators of a quantum channel. Generally quantum channels are completely-positive trace-preserving (CPTP) maps given by $\mathcal{N}(\rho)=\sum_k E_k\rho E_k^\dagger$, with the set of $E_k$ being the Kraus operators. These maps are not unitary evolutions in general. The amplitude damping channel is not a unitary evolution.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.