# Calculate probability of a state after depolarization

Let's say I have a particle in the quantum state $$|+\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$$, represented as a density operator (1st matrix) that went through a depolarizing chanel (2nd matrix). Let's call the depolarized matrix $$D_p$$.

$$\begin{bmatrix} .5 & .5 \\ .5 & .5 \end{bmatrix} \rightarrow \begin{bmatrix} .5 & .43 \\ .43 & .5 \end{bmatrix}$$ Now, I have two of these $$D_p$$, and their resulting product state is: $$D_p^{\otimes 2} = \begin{bmatrix} .5 & .43 \\ .43 & .5 \end{bmatrix} \otimes \begin{bmatrix} .5 & .43 \\ .43 & .5 \end{bmatrix}.$$ Now, if I want want to calculate the probability of finding some state $$|\psi\rangle = |0\rangle \otimes |0\rangle$$ in the above mentioned product system, then this is what I do: $$p(|\psi\rangle | D_p^{\otimes 2}) = trace(\psi\rangle\langle \psi | D_p^{\otimes 2}).$$ As you can see, calculating this trace is a $$O(N^3)$$ complexity operation and becomes very slow for even a small number of particles, i.e. for $$D_p^{\otimes 10}$$ or higher. Is there a principled way to calculate these probabilities? Without using any matrix multiplication?

Cross-posted on Physics.SE

But even if this is not the case, you can compute $$\langle \psi\vert D_p^{\otimes k}\vert\psi\rangle$$ rather than the trace, then it only scales as $$O(N^2)$$, $$N=2^k$$.
Finally, you could use that $$D_p^{\otimes k}$$ is a tensor product and multiply $$\vert\psi\rangle$$ with one $$D_p$$ at a time. Then, it is easy to see that each of these operations only sums over one index with $$2$$ settings, which takes $$2N$$ operations, so the total number of operations is $$2kN = O(kN) = O(N\log(N))$$.
Norbert's answer is correct, but just for the sake of being explicit: $$\langle 0|^{\otimes N} D_P^{\otimes N}|0\rangle^{\otimes N}=\left(\langle 0|D_p|0\rangle\right)^{\otimes N}=\langle 0|D_p|0\rangle^{N}=\frac{1}{2^N}.$$