Let $H$ be a Hilbert space.
Let $p,s$ be two density operators.
How do I show that the trace of their product lies between $0$ and $1$, i.e. $0\leq$tr$(ps)\leq1$?
I know that density operators are operators $A$ where $A\geq 0$ and tr$(A)=1$.
Furthermore I know that a density operator can be written as $A=\sum_{i=1}^ne_i|f_i\rangle\langle f_i|$, where $e_i\geq 0$, $\sum_i e_i=1$, and $|f_i\rangle$ form an orthonormal basis for $H$.
My attempt was this: we write $p=\sum_i e_i|f_i\rangle\langle f_i|$ and $s=\sum_i a_i|f_i\rangle\langle f_i|$. Then
$$ps=(\sum_i e_i|f_i\rangle\langle f_i|)(\sum_i a_i|f_i\rangle\langle f_i|)=\sum_ie_ia_i|f_i\rangle\langle f_i|.$$
Then we find tr$(ps)=\sum_ie_ia_i$ tr ($|f_i\rangle\langle f_i|)=\sum_i e_ia_i\langle f_i|f_i\rangle=\sum_i e_ia_i$.
Is this a correct way of proving the question? Why must $\sum_i e_ia_i$ be between $0$ and $1$?
Best Answer
$\newcommand{\bra}[1]{\langle{#1}\rvert} \newcommand{\ket}[1]{\lvert{#1}\rangle} \newcommand{\braket}[2]{\langle{#1}|{#2}\rangle} \newcommand{\N}{\mathbb{N}} \DeclareMathOperator{\tr}{tr}$For the sake of brevity I'll write the basis as $\{\ket{i}\}_{i\in\N}$. If $p=\sum_{i\in\N}a_i\ket{i}\bra{i}$ and $s=\sum_{i\in\N}b_i\ket{i}\bra{i}$ then their product is (you have to use two different indices!) \begin{equation} ps=\sum_{i\in\N}a_i\ket{i}\bra{i}\sum_{j\in\N}b_j\ket{j}\bra{j}= \sum_{i\in\N}\sum_{j\in\N}a_ib_j\ket{i}\braket{i}{j}\bra{j}= \sum_{i\in\N}\sum_{j\in\N}a_ib_j\delta_{ij}\ket{i}\bra{j}= \sum_{i\in\N}a_ib_i\ket{i}\bra{i} \end{equation} so its trace is \begin{equation} \tr(ps)=\sum_{k\in\N}\bra{k}ps\ket{k}= \sum_{k\in\N}\sum_{i\in\N}a_ib_i\braket{k}{i}\braket{i}{k}= \sum_{k\in\N}\sum_{i\in\N}a_ib_i\delta_{ki}\delta_{ik}= \sum_{k\in\N}a_kb_k \end{equation} then, since $a_i\le 1$ and $b_i\le 1$ you have $a_ib_i\le a_i$ $\forall i\in\N$, therefore \begin{equation} \tr(ps)=\sum_{k\in\N}a_kb_k\le\sum_{k\in\N}a_k=\tr(p)=1. \end{equation}