0.9 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. Equivalently, for any square matrix $A$ , the trace of $A$ can be defined to be the sum of its diagonal elements: $\operatorname{tr}A = \sum_k \langle e_k|A|e_k\rangle = \sum_k A_{kk}.$ In fact, the trace of $A$ is equal to the sum of the eigenvalues of $A$ , even in the case where $A$ is not diagonalisable.

You can show, using this definition or otherwise, that the trace is cyclic¹⁸ ( $\operatorname{tr}(AB) = \operatorname{tr}(BA)$ ) and linear ( $\operatorname{tr}(\alpha A+\beta B) = \alpha\operatorname{tr}(A)+\beta\operatorname{tr}(B)$ , where $A$ and $B$ are square matrices and $\alpha$ and $\beta$ complex numbers).

To recover the first definition from the second, we argue as follows: $\begin{aligned} \operatorname{tr}|b\rangle\langle a| &= \sum_k \langle e_k|b\rangle\langle a|e_k\rangle \\&= \sum_k \langle a|e_k\rangle\langle e_k|b\rangle \\&= \langle a|\mathbf{1}|b\rangle \\&= \langle a|b\rangle. \end{aligned}$ 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.

Note that “cyclic” does not mean the same thing as “permutation invariant”! It is not true in general that $\operatorname{tr}(ABC)=\operatorname{tr}(CBA)$ , but only that $\operatorname{tr}(ABC)=\operatorname{tr}(BCA)=\operatorname{tr}(CAB)$ , i.e. we can only cyclically permute the operators.↩︎