The trace of an operator X is defined as the sum of its diagonal elements:
d
Tr(X) = \sum_{j=1}^{d} \langle \alpha_j | X | \alpha_j \rangle
(2)
where \langle \alpha_j | X | \alpha_j \rangle represents the diagonal elements of X expressed in terms of a complete and
orthonormal basis \{ |\alpha_j\rangle, j = 1, ..., d \}, prove the following properties of trace of an operator.
(i) Tr(XY) = Tr(YX).
Hint: Expand Tr(XY) using the definition of trace and then insert the completeness
relationship of the basis \{ |\alpha_j\rangle, j = 1, ..., d \} to obtain the right-hand side of the proof.
(ii) Tr(U^\dagger XU) = Tr(X)
Hint: Let W = U^\dagger X and apply Tr(WU) = Tr(UW)
(iii) Tr(|\alpha_i\rangle \langle \alpha_j|) = \delta_{ij}
(iv) Tr(|b\rangle \langle \alpha_j|) = \langle \alpha_j | b \rangle
For parts (iii) & (iv), utilize the same procedure as provided in the hint of part (i)
