# Exercise 5

## from Revision class solutions

### Exercise 5.

Consider the decay of a two-level atom from its excited state to its ground state. Let the probability of this decay be $$p$$. The spontaneous emission of a photon accompanies this decay.

1. Name the quantum channel that can be used to model this process and write its Kraus operators. What process does each of these Kraus operators correspond to? Give reasons for your answer.
2. What is a unital channel? Is the channel in (1) unital?
3. Suppose the atom is originally in a state $$\rho := \sum_{\alpha,\beta=0}^1 \rho_{\alpha\beta}| \alpha \rangle\langle \beta |$$, where $$| \alpha \rangle,| \beta \rangle$$, $$\alpha,\beta\in\{0,1\}$$, denote orthonormal basis states of the Hilbert space of the atom. By considering the action of a unitary operator $$U$$ on the atom and its environment, deduce how the state of the atom changes under the action of one use of the channel in (1).
4. Let $$AB$$ denote a bipartite system which is in state $$\rho_{AB}$$. Show that the mutual information of the system cannot increase under the action of a completely positive trace-preserving map on the subsystem $$B$$ alone.
1. This is called the amplitude damping channel. It has two Kraus operators, $A_1 = \begin{pmatrix} 1 & 0 \\ 0 & \sqrt{1-p} \end{pmatrix}, \qquad A_2 = \begin{pmatrix} 0 & \sqrt{p} \\ 0 & 0 \end{pmatrix}$ in the basis where $$| 0 \rangle = \begin{pmatrix} 1\\0 \end{pmatrix}$$ corresponds to the ground state of the atom, and $$| 1 \rangle = \begin{pmatrix} 0 \\ 1 \end{pmatrix}$$ corresponds to the excited state.
The Kraus operator $$A_1$$ corresponds to the process that the ground state of the system remains invariant while the excited state is invariant only with probability $$1-p$$, since $$A_1 | 0 \rangle = | 0 \rangle$$ while $$A_1 | 1 \rangle = \sqrt{1-p}| 1 \rangle$$. The operator $$A_2$$ corresponds to the decay of the excited state to the ground state with probability $$p$$, since $$A_2 | 0 \rangle = 0$$ while $$A_2 | 1 \rangle = \sqrt{p} | 0 \rangle$$.
2. A unital channel is a channel $$\Lambda$$ such that $$\Lambda(I)=I$$, i.e. it preserves the identity matrix. The amplitude damping channel is not unital, since $A_1 I A_1^\dagger + A_2 I A_2^\dagger = \begin{pmatrix} 1 & 0 \\ 0 & 1-p \end{pmatrix} + \begin{pmatrix} p & 0 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 1+p & 0 \\ 0 & 1-p \end{pmatrix}$ which is not the identity matrix for $$p\neq 0$$.
3. We can write $\rho_A = \begin{pmatrix} \rho_{00} & \rho_{01} \\ \rho_{10} & \rho_{11} \end{pmatrix}$ where we write $$A$$ to label the Hilbert space of the atom. Let us write $$E$$ for the environment and choose a bipartite unitary operator $$U_{AE}$$ such that $$U_{AE} | 00 \rangle_{AE} =| 00 \rangle_{AE}$$ and $$U_{AE}| 10 \rangle_{AE} = \sqrt{1-p}| 10 \rangle_{AE} + \sqrt{p} | 01 \rangle_{AE}$$, where $$| 0 \rangle_E$$ is the vacuum state of the photon, and $$| 1 \rangle_E$$ the excited state corresponding to the emission of the photon.

As an aside, it was pointed out in the revision class that this definition does not completely specify the unitary; we only say what happens when the environment is in state $$| 0 \rangle_E$$, since that’s all we need to model the interaction between the atom (in any state) and the environment when the environment starts in the state $$| 0 \rangle_E$$. To truly construct a unitary, however, one needs to specify what happens when the photon is in excited state. I believe there may be more than one choice that gives a unitary evolution; one choice1 is $U | 11 \rangle = | 11 \rangle, \qquad U | 01 \rangle = \sqrt{1-p} | 01 \rangle - \sqrt{p} | 10 \rangle.$

Getting back to the problem, we consider the joint final state, $\sigma_{AE} := U_{AE} \rho_A \otimes | 0 \rangle\langle 0 |_E U^\dagger_{AE}$ which models the state of the atom and the photon after the interaction. We \begin{aligned} \sigma_{AE} &= \sum_{\alpha,\beta=0}^1 U_{AE} \rho_{\alpha\beta} | \alpha \rangle\langle \beta |_A \otimes | 0 \rangle\langle 0 |_E U^\dagger_{AE}\\ &= \rho_{00} U_{AE}| 00 \rangle\langle 00 | U^\dagger_{AE} +\rho_{10} U_{AE} | 10 \rangle\langle 00 | U^\dagger_{AE} + \rho_{01} U_{AE} | 00 \rangle\langle 10 | U^\dagger_{AE} + \rho_{11} U_{AE} | 10 \rangle\langle 10 | U^\dagger_{AE}\\ &= \rho_{00} | 00 \rangle\langle 00 | + \rho_{10} (\sqrt{1-p}| 10 \rangle + \sqrt{p} | 01 \rangle)\langle 00 | \\ &\qquad + \rho_{01} | 00 \rangle ( \sqrt{1-p}\langle 10 | +\sqrt{p} \langle 01 |) +\rho_{11} (\sqrt{1-p}| 10 \rangle + \sqrt{p} | 01 \rangle)( \sqrt{1-p}\langle 10 | +\sqrt{p} \langle 01 |) \end{aligned} where we have dropped the labels $$A$$ and $$E$$ for ease of notation. Now, we can trace out $$E$$ term by term to find \begin{aligned} \operatorname{tr}_E \sigma_{AE} &= \rho_{00} | 0 \rangle\langle 0 | + \rho_{10} \sqrt{1-p} | 1 \rangle\langle 0 | +\rho_{01} \sqrt{1-p} | 0 \rangle\langle 1 | + \rho_{11} [ p| 0 \rangle\langle 0 | + (1-p) | 1 \rangle\langle 1 | ]\\ &= \begin{pmatrix} \rho_{00} + p\rho_{11} & \sqrt{1-p} \rho_{10} \\ \sqrt{1-p}\rho_{01} & (1-p) \rho_{11} \end{pmatrix} \end{aligned} which describes the result of applying the channel once to the input state $$\rho$$.
4. We can write the mutual information of a state $$\rho_{AB}$$ in terms of the relative entropy, as $I(A:B)_\rho = D(\rho_{AB} \| \rho_A \otimes \rho_B).$ Likewise, if we apply a quantum channel $$\Lambda$$ to the be system alone, we apply $$\operatorname{id}\otimes \Lambda$$ to the bipartite state $$\rho_{AB}$$, yielding $$\sigma_{AB} := \operatorname{id}\otimes \Lambda (\rho_{AB})$$. So the mutual information can be written $I(A:B)_{\sigma} = D( \sigma_{AB}\| \sigma_A\otimes \sigma_B ).$ But $$\sigma_A = \rho_A$$, since only the identity channel was applied on the $$A$$ part, while $$\sigma_B = \Lambda(\rho_B)$$. Thus, \begin{aligned} I(A:B)_\sigma &= D( \sigma_{AB} \| \rho_A \otimes \Lambda(\rho_B) ) \\ &= D(\operatorname{id}\otimes \Lambda (\rho_{AB}) \| \operatorname{id}\otimes \Lambda (\rho_A\otimes \rho_B) ). \end{aligned} By the data processing inequality for the relative entropy, since $$\operatorname{id}\otimes \Lambda$$ is a CPTP map, we have $D(\operatorname{id}\otimes \Lambda (\rho_{AB}) \| \operatorname{id}\otimes \Lambda (\rho_A\otimes \rho_B) ) \leq D(\rho_{AB} \| \rho_A \otimes \rho_B)$ which proves $$I(A:B)_\sigma \leq I(A:B)_\rho$$ as desired.

1. This choice seems fairly obvious in retrospect, but I found it by guessing and checking with my current favorite scientific programming language, Julia.