$\mathrm{Tr}(XY) = \mathrm{Tr}(YX)$? I'm trying to understand the Dirac notation to understand quantum mechanics better. I'm trying to show the above relation using the Dirac notation.
Given
$$\mathrm{Tr}(X)~=~\sum_j\langle j|X|j\rangle$$
and
$$\mathrm{Tr}(Y)~=~\sum_j\langle j|Y|j\rangle.$$
Thus,$$\mathrm{Tr}(XY)~=~\sum_{ij}\langle j|X|i\rangle \langle i|Y|j\rangle.$$
And isn't that just equal to the following 
$$\mathrm{Tr}(XY)~=~\sum_j \langle j|XY|j\rangle?$$
But that shouldn't equal the following
$$\mathrm{Tr}(XY)~=~\sum_j\langle j|YX|j\rangle,$$
since $XY \neq YX$, right?
 A: Remember that $\Sigma_i |i \rangle \langle i |$ is the identity operator.
Then $Tr(XY) = \Sigma_j \langle j| XY |j  \rangle 
= \Sigma_j  \langle j| X \left(\Sigma_i |i \rangle \langle i |\right)Y |j  \rangle 
= \Sigma_j  \Sigma_i \langle j| X  |i \rangle \langle i |Y |j  \rangle 
=\Sigma_j  \Sigma_i \langle i |Y |j  \rangle \langle j| X  |i \rangle 
= \Sigma_i  \Sigma_j \langle i |Y |j  \rangle \langle j| X  |i \rangle 
=  \Sigma_i  \langle i |Y \left( \Sigma_j |j  \rangle \langle j | \right) X  |i \rangle 
=\Sigma_i  \langle i |Y  X  |i \rangle 
= Tr(YX)$.
So the two traces are equal.
A: Throwing a wrench in the works, we mention that 
$${\rm Tr}[X,Y]~=~0$$ 
does not always hold for two linear operators 
$$X:D(X)\subseteq H~\to~ H\qquad\text{and}\qquad Y:D(Y)\subseteq H~\to~ H$$ in an infinite-dimensional vector space $H$. This is not just an academic remark. Take for instance the trace on both sides of the canonical commutation relations 
$$ [\hat{x},\hat{p}]~=~i\hbar{\bf 1}, $$
cf. Example 1 in Ref. 1 and this Phys.SE post.
References:


*

*F. Gieres, Mathematical surprises and Dirac’s formalism in quantum mechanics, arXiv:quant-ph/9907069. (Hat tip: David Bar Moshe.)

A: To extend @NowIGetToLearnWhatAHe's answer, it should be noted that it's a general theorem of linear algebra that traces are cyclical, so it's always true that:
$$Tr(ABCDE...XYZ) = Tr(ZABCDE...XY)$$
From this, your result follows as a special case.
