What happens when you send a Bell state through depolarizing channel?

Question

For noise parameter $Q$ and a density matrix $\rho$, we know that the depolarization channel $\mathcal{E}$ would act like:

$$ \mathcal{E}(\rho) = (1 - Q)\rho +Q\frac{I}{2}, $$ where $I$ is the identity matrix in 2 dimensions. Now, an interesting question is, what happens to a Bell state $|\Phi^{+}\rangle = \frac{1}{2}(|00\rangle + |11\rangle)$, when we apply $\mathcal{E}$ to each qubits of the Bell state separately. What would be the resulting state?

Meaning, what is:

$$ (\mathcal{E} \otimes \mathcal{E})(|\Phi^{+}\rangle \langle\Phi^{+}| ) $$ Remembering that, operation on the first qubit would collapse the state. Is it physically possible?

Answer 1

From my understanding of what you are asking, you may take the product of two depolarization operations, using the reduced density matrix of each qubit in the Bell state in the expression.

Let's denote our two qubits as $\mathrm{A}$ and $\mathrm{B}$. The Bell state of these two qubits is then: $$ |\beta_{00} \rangle =\frac{|0 \rangle_\mathrm{A} \otimes |0 \rangle_\mathrm{B} + |1 \rangle_\mathrm{A} \otimes |1 \rangle_\mathrm{B}}{\sqrt{2}} = \frac{|00 \rangle + |11 \rangle}{\sqrt{2}} $$ With a density matrix: $$ \rho = | \beta_{00} \rangle \langle \beta_{00} | = \frac{ |00 \rangle \langle 00| + |00 \rangle \langle 11 | + |11 \rangle \langle 00| + |11 \rangle \langle 11 |}{2} $$

Which makes the two reduced density matrices: $$ \rho_\mathrm{A} = \text{tr}_\mathrm{B}(\rho) = \frac{|0 \rangle \langle 0| + |1 \rangle \langle 1|}{2} = \frac{I}{2} $$

$$ \rho_\mathrm{B} = \text{tr}_\mathrm{A}(\rho) = \frac{|0 \rangle \langle 0| + |1 \rangle \langle 1|}{2} = \frac{I}{2} $$

The operation can then be defined as (assuming the same noise parameter $Q$ on both applications of the operation): $$ \mathcal{E} \otimes \mathcal{E} = (Q \; \frac{I}{2} + (1-Q) \rho_\mathrm{A}) \otimes (Q \; \frac{I}{2} + (1-Q) \rho_\mathrm{B}) $$

Which ultimately simplifies to: $$ \mathcal{E} \otimes \mathcal{E} = Q(2-Q) \; \frac{I}{2} \otimes \frac{I}{2} + (1-Q)^2 \; \rho $$

Because both reduced density matrices for the Bell state given are equivalent to the density matrix for a completely mixed state, i.e. $\frac{I}{2}$.

Hope this helps!

Answer 2

As also discussed here you can write your channel as $$\mathcal E = p\mathcal E_{dp} + (1-p) \operatorname{Id}, \quad \mathcal E_{dp}(\rho)\equiv\operatorname{Tr}(\rho)I/d.$$ where $p\in[0,1]$ and, in your case, $d=2$.

You thus have $$\mathcal E\otimes\mathcal E=p^2 \mathcal E_{dp}\otimes \mathcal E_{dp} + p(1-p) [\mathcal E_{dp} \otimes \operatorname{Id} + \operatorname{Id}\otimes \mathcal E_{dp}] + (1-p)^2 \underbrace{\operatorname{Id}\otimes \operatorname{Id}}_{\equiv \operatorname{Id}}.$$

When the input is a maximally entangled state, $\sqrt d|\Phi\rangle=\sum_k |u_k,v_k\rangle$ for some $u_k,v_k$ with $\langle u_k,u_\ell\rangle=\langle v_k,v_\ell\rangle=\delta_{k\ell}$, we get $$ (\mathcal E_{dp}\otimes \mathcal E_{dp})\Phi = \frac{1}{d}\sum_{jk} \mathcal E_{dp}(|u_j\rangle\!\langle u_k|)\otimes \mathcal E_{dp}(|v_j\rangle\!\langle v_k|) = \frac{1}{d^2}I\otimes I\equiv \frac{1}{d^2}I, \\ (\mathcal E_{dp}\otimes \operatorname{Id})\Phi = \frac{1}{d}\sum_j \mathcal E_{dp}(|u_j\rangle\!\langle u_k|)\otimes |v_j\rangle\!\langle v_k| = \frac{1}{d^2}I\otimes I\equiv \frac{1}{d^2}I. $$ where we are using the notation $\Phi\equiv |\Phi\rangle\!\langle \Phi|$. In conclusion, $$(\mathcal E\otimes\mathcal E)\Phi= \underbrace{[p^2+2p(1-p)]}_{=p(2-p)} \frac{I}{d^2} + (1-p)^2\Phi.$$