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given a multi particle state I have to calculate the reduced density matrix where I trace out the third particle

for this I first calculate the corresponding 2D density matrix with the bra vector of the state; enter image description here

the partial trace is then given by the expression below where I sum over the Eigenbasis of the third particle enter image description here

but I don't get how they get from the first line to the second one in the partial trace calculation; how does all the terms cancel? I think they cancel somehow due to orthogonality reasons but I don't see how the third bra/ket vector acts on the density matrix.. can someone explain in detail how this calculation works?

thanks in advance

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    $\begingroup$ Please use MathJax to format your equations rather than upload pictures $\endgroup$ – Aaron Stevens Feb 5 at 17:57
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They insert unity by way of the completeness relation of the states

$$ \sum_{|s_1, s_2, s_3 \rangle\ \in\ \mathrm{Basis(\mathcal{H})} \mathrm{}} |s_1, s_2, s_3\rangle \langle s_1, s_2, s_3 | = \hat{1}. \tag{1} $$

where the sum is taken over all spin states in your basis. You insert this resolution of unity to each term in your first equation.

For example, if you had just a single spin 1/2 particle (1) would become

$$ | \uparrow\rangle\langle \uparrow| + |\downarrow\rangle\langle\downarrow| = \hat{1}. $$

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  • $\begingroup$ thank you for your answer. i see want you mean but nontheless i dont get how the unity acts on the density operator such that from a triple state remains a state with only two particles; can you please explain this step by step for $$_3\langle\uparrow|\psi\rangle\langle\psi|\uparrow\rangle_3$$ $\endgroup$ – jeffs Feb 5 at 18:59

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