How do I trace out the second qubit to find the reduced density operator? [closed]

Question

I'm doing an exercise to trace out the second qubit to find the reduced density operator for the first qubit:

$tr_2|11\rangle\langle00| = |1\rangle\langle0|\langle0|1\rangle$

I'm just wondering if I do trace for the first qubit, should I have:

$tr_1|11\rangle\langle00| = |1\rangle\langle0|\langle0|1\rangle$ or $tr_1|11\rangle\langle00| = \langle0|1\rangle|1\rangle\langle0|$ ?

In the Nielsen-and-Chuang textbook, we have $tr(|b_1\rangle\langle b_2|)=\langle b_2|b_1\rangle$. Can I say the left and right hand side are just two ways to locate an element in a matrix? Thanks!!

Answer 1

Regarding your results: Both coincide, since $\langle 0|1\rangle \in \mathbb{C}$.

Edit: In fact, the result is zero, because both states are orthogonal, which is also used in the calculation performed in the textbook.

I hope this helped.