# Partial trace question: how does this term become zero?

I have recently been introduced to the partial trace but I have a question about how my lecturer recognises 'zero' and 'non-zero' terms. Consider the state:

$\psi_{AB} = c_{00}|00\rangle + c_{01}|01\rangle + c_{10}|10\rangle + c_{11}|11\rangle$

where the entangled pair of qubit in full notation is

$|00\rangle = |0\rangle \otimes |0\rangle$, etc...

Upon trying the density matrix for system B (the second one of the pair in the tensor product), my lecturer takes the partial trace over system A (the first one of the pair in the tensor product). However, he makes a step from one line to the next stating that he has from experience recognised and not included the zero terms but I don't understand how some of the terms are zero:

I.e. for example looking at the $c_{00}c_{10}^{*} |00 \rangle \langle10|$ term within the expansion of the expression within the partial trace $Tr_{A}(...+ c_{00}c_{10}^{*} |00 \rangle \langle10| + ... )$, how does this become zero in the final expression?

I.e.

Am I correct in saying:

$Tr_{A} [ c_{00}c_{10}^{*} |00\rangle\langle 10|] = |0\rangle\langle0| c_{00}c_{10}^{*}$

Thank you, I'm just a bit confused about the ways to find the partial trace quickly and correctly

The statement:

$Tr_{A} [ c_{00}c_{10}^{*} |00\rangle\langle 10|] = |0\rangle\langle0| c_{00}c_{10}^{*}$

is incorrect.

In fact, using the summing over a basis idea of the partial trace:

$Tr_{A} [ c_{00}c_{10}^{*} |00\rangle\langle 10|] = (\langle0| \otimes I)( c_{00}c_{10}^{*} |00\rangle\langle 10|)(|0\rangle \otimes I) + (\langle1| \otimes I)( c_{00}c_{10}^{*} |00\rangle\langle 10|)(|1\rangle \otimes I)$

$= [\langle1| \otimes (|0\rangle\langle0|] (|0\rangle \otimes I)$

$= 0$

as $\langle1|0\rangle=0$