# In exercise 8.23 of Nielsen and Chuang why is the quantum operation no longer trace-preserving?

For context, $$\mathcal{E}_{AD}$$ is the quantum operation for amplitude damping on a single physical qubit, with operation elements given by:

$$\mathcal{E}_{AD} = E_0 \rho E_0^{\dagger} + E_1 \rho E_1^{\dagger}$$ , where:

$$E_0 = \begin{pmatrix}1 & 0\\\ 0 & \sqrt{1-\gamma}\end{pmatrix}$$

$$E_1 = \begin{pmatrix}0 & \sqrt{\gamma}\\\ 0 & 0\end{pmatrix}$$.

These operation elements for this given operator-sum representation was given earlier in the book and is quoted above without proof. This operation is clearly trace preserving, as:

$$E_0^{\dagger} E_0 + E_1^{\dagger} E_1 = \begin{pmatrix}1 & 0\\\ 0 & 1-\gamma\end{pmatrix}+\begin{pmatrix}0 & 0\\\ 0 & \gamma\end{pmatrix} = I$$

This is expected because we are not measuring the system or environment in any part of the process.

In the question, however, we deal with two physical qubits to construct the logical qubits $$|0_L\rangle = |01\rangle$$ and $$|1_L\rangle = |10\rangle$$ in the dual-rail representation. For the operation elements we attempt to obtain in the question, I find that the resulting quantum operation of having amplitude damping occur on both of the physical qubits is no longer trace-preserving:

$$E_0^{dr \dagger} E_0^{dr} + E_1^{dr \dagger} E_1^{dr} = (1-\gamma)I + \gamma ( |01\rangle \langle 00| + |10\rangle \langle 00|)(|00\rangle \langle 01| + |00\rangle \langle 10|)$$

$$= (1-\gamma)I + \gamma (|01\rangle \langle 01| + |10\rangle \langle 10| + |01\rangle \langle 10| + |10\rangle \langle 01|)$$

$$= I + \gamma (|01\rangle \langle 10| + |10\rangle \langle 01|)$$,

And as such we get these off-diagonal terms. We can obtain a trace-preserving quantum operation by splitting up the second operation element into two, such that:

$$E_1^{dr} = \sqrt{\gamma} ( |00\rangle \langle 01|)$$,

$$E_2^{dr} = \sqrt{\gamma} ( |00\rangle \langle 10|)$$.

But this will be a different quantum operation to what the question wants us to get. So I conclude that either the show-that is incorrect, and we are supposed to get 3 separate operation elements for the operation, rather than the two in the question, or alternatively somehow the act of applying two amplitude damping quantum operations on two physical qubits is no longer trace-preserving. If it's the latter case could someone please explain why this is?

For simplicity, let us call $$\rho$$ the quantum state before entering the channel $$\mathcal{E}_{AD}\otimes \mathcal{E}_{AD}$$. We can write the whole operation as: $$\begin{eqnarray} (\mathcal{E}_{AD}\otimes \mathcal{E}_{AD})\circ\rho=(\mathcal{E}_{AD}\otimes I)(I\otimes \mathcal{E}_{AD}) \circ\rho. \end{eqnarray}$$ We perform the operations sequentially, because the order is indifferent given that both act on different subspaces and commute. The resulting state reads: $$\begin{eqnarray} \tilde{\rho}=(E_0\otimes E_0)\rho (E_0^\dagger\otimes E_0^\dagger)+ (E_0\otimes E_1)\rho (E_0^\dagger\otimes E_1^\dagger) + (E_1\otimes E_0)\rho (E_1^\dagger\otimes E_0^\dagger) +(E_1\otimes E_1)\rho (E_1^\dagger\otimes E_1^\dagger). \end{eqnarray}$$ You have now a set of 4 Kraus operators given by $$K_{j=1,...,4}=\{E_0\otimes E_0, E_0\otimes E_1,E_1\otimes E_0,E_1\otimes E_1$$}.
One can construct these operators now and verify that $$\sum_{j=1}^4 K_j^\dagger K_j = I$$ for arbitrary $$\gamma$$.
You are correct that there is a mistake in the problem and one needs three summands in the Kraus form instead of two. @Zaratuthustra in another answer explained that the tensor square $$\mathcal{E}_{AD} \otimes \mathcal{E}_{AD}$$ can be expressed with $$4$$ Kraus operators, and when we restrict the input to the subspace spanned by $$\left|01\right>$$ and $$\left|10\right>$$, there remain three summands with Kraus operators $$\sqrt{1-\gamma}(\left|01\rangle\langle 01\right| + \left|10\rangle\langle 10\right|)$$, $$\sqrt{\gamma}\left|00\rangle\langle 10\right|$$ and $$\sqrt{\gamma}\left|00\rangle\langle 01\right|$$.