Basic Definitions:
Let $\mathcal{H}$ be a Hilbert space. Define a density operator $\rho \in \mathcal{L}(H)$ (continous linear operators from $\mathcal{H}$ into itself) by
- $\rho$ is self-adjoint
- $\rho$ is positive
- $\mathrm{tr} \rho=1$.
A density operator is called a pure state, if it is the projection onto a one dimensional subspace of $\mathcal{H}$.
This is equivalent to $\mathrm{tr} \rho^2=\mathrm{tr}\rho$.
The Problem:
I want to prove that $\rho$ is a pure state $\Leftrightarrow$ $\rho$ can not be expressed as a nontrivial linear combination of density operators.
What i have done:
I have allready proven $\Leftarrow$ by showing that every mixed (non-pure) state is the non trivial convex linear combination of other density operators.
I am stuck on $\Rightarrow$. I have proven, that a pure state cannot be expressed as a nontrivial convex linear combination of density operators. I did this by showing the convexity of $\mathrm{tr} \rho^2$.
Let $ ̣\rho= \lambda \rho_1 +(1- \lambda) \rho_2$ with $\lambda \in [0,1]$ and $\rho_1,\rho_2$ density operators. Then:
$$\begin{equation}
\label{eq:trsqkonvex}
\begin{split}
\mathrm{tr} \rho^{2}&\leq \lambda^{2} \mathrm{tr} \rho_1^{2} +2 \lambda (1-\lambda) \sqrt{\mathrm{tr} \rho_1^{2}\mathrm{tr} \rho_2^{2}} + (1-\lambda)^{2} \mathrm{tr} \rho_2^{2} \\
&= \left( \lambda \sqrt{\mathrm{tr} \rho_1^{2}} + (1-\lambda) \sqrt{\mathrm{tr} \rho_2^{2}} \right)^{2} \\
&\leq \lambda \mathrm{tr} \rho_1^{2} + (1-\lambda) \mathrm{tr} \rho_2^{2} \leq 1.
\end{split}
\end{equation}
\tag{1}
$$
Where the Cauchy-Schwarz inequality was used at the first $\leq$ and the convexity of $t \to t^2$ at the second.
Where im stuck:
Now i want to show $\Rightarrow$ for a general linear combination.
Its not hard to show that if $\rho = \alpha \rho_1 + \beta \rho_2$, with $\alpha, \beta \in \mathbb{C}$, $\rho_1,\rho_2$ density operators,
then $\alpha,\beta$ are of the form $\alpha =\lambda,\beta = (1-\lambda)$ with $\lambda \in \mathbb{R}$.
If $\lambda \in \mathbb{R}\setminus [0,1]$, then the argument i have used in Eq. 1 will break down, since $\lambda$ or $(1-\lambda)$ will have to be replaced with their absolute value.
My idea is to show that the general linear combination must be a convex one to begin with. I would be grateful for any hints or complete proofs (even alternative proofs).
Best Answer
I figured it out i think. The statement is simply untrue. Consider the example $$\rho_1 =\frac{1}{2} ( | 1 \rangle \langle1| +|2 \rangle \langle 2 | ),$$ $$\rho_2 = |2 \rangle \langle 2 |. $$ Where $| i \rangle \in \mathcal{H} , i=1, 2$ are an orthonormal system. Then $$2 \rho_1 + (1-2)\rho_2 = | 1 \rangle \langle1|.$$ A pure state. So convexity of the lin. combination is necessary on the RHS of the problem.