Let $A,B$ be postitive semi-definite operators on a finite dimensional Hilbert space. Is the following true? \begin{equation} A \geq B \ \Rightarrow A^{\otimes n} \geq B^{\otimes n} \quad n=1,2,3,... \end{equation} i.e. tensor power is an operator monotone function?
(If $A$ commutes with $B$ it seems to be true, but what about in the non-commutative case?)
Denote the inner product space by $V$. Since $A\ge B$, \begin{aligned} \langle\otimes_{i=1}^nu_i, A^{\otimes n}(\otimes_{i=1}^nu_i)\rangle =\prod_i\langle u_i, Au_i\rangle \ge\prod_i\langle u_i, Bu_i\rangle =\langle\otimes_{i=1}^nu_i, B^{\otimes n}(\otimes_{i=1}^nu_i)\rangle \end{aligned} for any $n$ vectors $u_1,u_2,\ldots,u_n\in V$. It follows that $\langle v, A^{\otimes n}v\rangle\ge\langle v, B^{\otimes n}v\rangle$ for every $v\in V^{\otimes n}$, because $\{\otimes_{i=1}^nu_i: u_1,\ldots,u_n\in V\}$ contains an orthogonal basis of $V^{\otimes n}$. Hence $A^{\otimes n}\ge B^{\otimes n}$.
Alternatively, by a continuity argument, you may assume that $A\ge B>0$. The condition $A\ge B>0$ then implies that $I\ge A^{-1/2}BA^{-1/2}$. Hence $I^{\otimes n}\ge(A^{-1/2}BA^{-1/2})^{\otimes n}=(A^{-1/2})^{\otimes n}B^{\otimes n}(A^{-1/2})^{\otimes n}$. Consequently, $A^{\otimes n}\ge B^{\otimes n}$.
However, I have reservations to call the tensor power "operator monotone", because we usually speak of the monotonicity of a matrix function or the lack of it only when the function is one that stems from a scalar-valued function and maps a matrix to another matrix of the same size.