# Pure state density-operator can not be expressed as non trivial linear combination

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

1. $$\rho$$ is self-adjoint
2. $$\rho$$ is positive
3. $$\mathrm{tr} \rho=1$$.

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: $$$$\label{eq:trsqkonvex} \begin{split} \mathrm{tr} \rho^{2}&\leq \lambda^{2} \mathrm{tr} \rho_1^{2} +2 \lambda (1-\lambda) \sqrt{\mathrm{tr} \rho_1^{2}\mathrm{tr} \rho_2^{2}} + (1-\lambda)^{2} \mathrm{tr} \rho_2^{2} \\ &= \left( \lambda \sqrt{\mathrm{tr} \rho_1^{2}} + (1-\lambda) \sqrt{\mathrm{tr} \rho_2^{2}} \right)^{2} \\ &\leq \lambda \mathrm{tr} \rho_1^{2} + (1-\lambda) \mathrm{tr} \rho_2^{2} \leq 1. \end{split}$$ \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).

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.