0.8 The trace

The trace is an operation which turns outer products into inner products, \operatorname{tr}\colon |b\rangle\langle a| \longmapsto \langle a|b\rangle. We have just seen that any linear operator can be written as a sum of outer products, and so we can extend the definition of trace (by linearity) to any operator. Alternatively, for any square matrix A, the trace of A is defined to be the sum of its diagonal elements: \operatorname{tr}A = \sum_k \langle e_k|A|e_k\rangle = \sum_k A_{kk}. You can show, using this definition or otherwise, that the trace is cyclic (i.e. \operatorname{tr}(AB) = \operatorname{tr}(BA)) and linear (i.e. \operatorname{tr}(\alpha A+\beta B) = \alpha\operatorname{tr}(A)+\beta\operatorname{tr}(B), where A and B are square matrices and \alpha and \beta complex numbers). Moreover, \begin{aligned} \operatorname{tr}|b\rangle\langle a| &= \sum_k \langle e_k|b\rangle\langle a|e_k\rangle \\&= \sum_k \langle a|e_k\rangle\langle e_k|b\rangle \\&= \langle a|\mathbf{1}\rangle|b\rangle \\&= \langle a|b\rangle. \end{aligned} Here, the second term can be viewed both as the sum of the diagonal elements of |b\rangle\langle a| in the |e_k\rangle basis, and as the sum of the products of two complex numbers \langle e_k|b\rangle and \langle a|e_k\rangle. We have used the decomposition of the identity, \sum_k|e_k\rangle\langle e_k|=\mathbf{1}. Given that we can decompose the identity by choosing any orthonormal basis, it is clear that the trace does not depend on the choice of the basis.

!!to-do: mention what this whole package of data all bundled up looks like from the categorical pov!!