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I would like to know how to prove that $$tr(ST) = tr(TS)$$ for two linear operators $S$ and $T$ on a finite-dimensional Hilbert space $\mathcal{H}$, given the following definition of the trace operator: $$tr(T)=\sum_{i=1}^n{\langle T e_i, e_i \rangle}$$

for an orthonormal basis $\{ e_i \}_{i=1}^n$ in $\mathcal{H}$. Is there a way to prove this without representing the operators as matrices?

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  • $\begingroup$ You may find worthwhile the related blog by T. Tao: The cyclic property of the trace is derived, and explicit choices of orthonormal bases are not made. Along the way you encounter a central relation connecting trace and determinant: $\,\det(1 +\epsilon A) = 1+\epsilon\operatorname{tr}A+\text{O}(\epsilon^2)\,$. $\endgroup$ – Hanno Oct 5 '18 at 5:19
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I am not sure that this is different from writing an operator as a matrix, but you can argue as follows: for any operator $U$ you can write $$U(\cdot) = \sum_j U(e_j) \langle \cdot , e_j \rangle$$ and thus for the trace of $ST$ you can find: $$Tr(ST) = \sum_{i,j} \langle S(e_j), e_i \rangle \langle T(e_i), e_j \rangle $$ Now in the last formula you can commute the terms in the product. Hence the result.

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It sort of boils down to looking at matrices but you may decompose the identity as: $$ 1 = \sum_j |e_j \rangle \langle e_j| $$ which inserted in the trace yields $$ {\rm tr} (ST) = \sum_i \langle e_i |ST e_i\rangle = \sum_i \sum_j \langle e_i |S e_j\rangle \langle e_j | T e_i \rangle $$ and you may interchange the sums and the two factors in the product.

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