# Why can the partial trace be written as $\text{Tr}_B(\rho)= \sum_k (1 \otimes \langle k|) \rho (1 \otimes |k \rangle)$?

I don't really understand a notation that I stumbled upon regarding a partial trace. According to the definition I have, partial trace is

$$\rho_A=\text{Tr}_B(\rho_{AB}):= \sum_k (1_A \otimes \langle k|_B) (\rho_{AB}) (1_A \otimes |k \rangle _B)$$

I am not sure how to actually use it. I am given an example for $$\rho_{AB}=|\psi^+\rangle \langle \psi^+|$$ so that $$\rho_{AB}=\frac{1}{2}(|00\rangle \langle 00|+|00\rangle \langle 00| +|11 \rangle \langle 00| + |11\rangle \langle 11|)$$

and

\begin{align} \rho_A:= &1_A \otimes \langle 0| (\rho_{AB}) 1_A \otimes |0 \rangle+1_A \otimes \langle 1| (\rho_{AB}) 1_A \otimes |1 \rangle \\ = &1_A \otimes \langle 0| (\frac{1}{2} |00\rangle \langle 00|) 1_A \otimes |0 \rangle+1_A \otimes \langle 1| (\frac{1}{2}|11\rangle \langle 11|) 1_A \otimes |1 \rangle .\end{align}

But how do we choose only these specific terms of $$\rho_{AB}$$? How can I then calculate partial trace using this method for $$\rho_{AB}=\frac{1}{3}|00 \rangle \langle 00|+\frac{2}{3}|++\rangle \langle ++|$$?

## 1 Answer

Your two-quantum-number basis is defined as $$|\phi_1\phi_2\rangle \equiv |\phi_1\rangle_A \otimes |\phi_2\rangle_B \equiv |\phi_1\rangle_A|\phi_2\rangle_B$$. When you act on such a state with another one, the substates act together. For example, $$\left[\ _A\langle 0| _B\langle0| \right] \left[ |00\rangle\langle11| \right] \left[ |0\rangle_A|0\rangle_B\right] = \left[\langle 0 | 0\rangle\langle 1 | 0 \rangle \right]_A \otimes \left[\langle 0 | 0\rangle\langle 1 | 0 \rangle \right]_B = 0.$$

That's why in your first case, you only have two terms that are non-zero. For your last case, you need to rewrite your $$|+\rangle$$ states in terms of $$|0\rangle$$ and $$|1\rangle$$ or at least know the scalar product between $$|+\rangle$$ and $$|0,1\rangle$$.