# Proving a vector state is pure

Let $$\mathcal{M}$$ be a until $$*$$-algebra of 3x3 complex matrices. We have the general form of a vector state $$\omega_{\psi} : \mathcal{M} \to \mathbb{C}$$ over $$\mathbb{C}^3$$ as given by $$\omega_{\psi}(A) := \langle \psi, A\psi \rangle, \quad A\in \mathcal{M}, \psi \in \mathbb{C}^3, \|\psi \|=1.$$ I need to pick an $$\psi$$ such that the above is a pure state (i.e. it cannot be written as a convex combination $$\lambda \omega_{\sigma} + (1-\lambda)\omega_{\phi}$$ of pure states $$\omega_{\sigma}, \omega_{\phi}$$ and $$\lambda \in (0,1)$$). The proof I have formulated is to pick a basis of $$\mathbb{C}^3$$ over $$\mathbb{C}$$ and take $$\psi$$ to be one of these basis vectors, for example $$\psi = \begin{pmatrix} 1\\0\\0 \end{pmatrix}.$$ In this case a convex combination would be of the form $$\langle \sqrt{\lambda}\sigma + \sqrt{1-\lambda}\phi, A\left(\sqrt{\lambda}\sigma + \sqrt{1-\lambda}\phi \right)\rangle,$$ which implies that for $$\omega_{\psi}$$ to be mixed, $$\psi$$ must be a linear combination of the form $$\sqrt{\lambda}\sigma + \sqrt{1-\lambda}\phi$$, but since it is a basis vector, this isn't the case and we have a contradiction. I have a feeling this isn't totally sufficient as a proof - does anyone have any ideas/hints?

• The way you've defined $\omega_\psi$, it appears to be a scalar. Kindly check and clarify the definition. Do you require $A\psi$ to be a pure state? – Gautam Shenoy May 4 at 11:50
• @GautamShenoy This may be a difference in language - here I mean a state is a normalised, positive linear functional $\omega_{\psi} : \mathcal{M} \to \mathbb{C}$, rather than the quantum mechanical meaning of state being just a vector. – CS1994 May 4 at 12:18

I don't think your argument works. For instance you can write $$\begin{bmatrix} 1\\ 0\\0\end{bmatrix} = \sqrt{\frac12}\,\begin{bmatrix} 1/\sqrt2\\ 1/\sqrt2\\0\end{bmatrix} + \sqrt{1-\frac12}\,\begin{bmatrix} 1/\sqrt2\\ -1/\sqrt2\\0\end{bmatrix}$$
Now, any state of the form $$\omega_\psi$$ is pure. Suppose that $$\omega_\psi=\lambda\omega_\sigma+(1-\lambda)\omega_\phi$$. Take $$A=P_\psi$$, the projection onto $$\mathbb C\psi$$. Then $$\tag1 1=\langle \psi,P_\psi\psi\rangle=\lambda\langle \sigma,P_\psi\sigma\rangle+(1-\lambda)\langle \phi,P_\psi\phi\rangle.$$ Since $$0\leq \langle \sigma,P_\psi\sigma\rangle\leq1$$ and $$0\leq \langle \phi,P_\psi\phi\rangle\leq1$$, it follows that $$1=\langle \sigma,P_\psi\sigma\rangle=\langle \phi,P_\psi\phi\rangle.$$ Then, since $$P_\psi$$ is a projection, $$1=\langle \sigma,P_\psi\sigma\rangle=\langle P_\psi\sigma,P_\psi\sigma\rangle=\|P_\psi\sigma\|^2.$$ From $$1=\|\sigma\|^2=\|P_\psi\sigma+(1-P_\psi)\sigma\|^2=\|P_\psi\sigma\|^2+\|(1-P_\psi)\sigma\|^2=1+\|(1-P_\psi)\sigma\|^2,$$ we conclude that $$(1-P_\psi)\sigma=0$$. In other words, $$P_\psi\sigma=\sigma$$. So $$\sigma=P_\psi\sigma=\langle\psi,\sigma\rangle\,\psi.$$ That is, $$\sigma=\alpha\,\psi$$, with $$|\alpha|=1$$. Similarly, $$\phi=\beta\psi$$ with $$|\beta|=1$$. Thus $$\omega_\sigma=\omega_\phi=\omega_\psi$$.
The above reasoning can be made more general. You can consider states of the form $$A\longmapsto \operatorname{Tr}(XA)$$ for some $$X$$ with $$X\geq0$$ and $$\operatorname{Tr}(X)=1$$. Your $$\omega_\psi$$ is obtained when $$X=P_\psi$$. Even in this setting, the pure states are precisely the point states $$\omega_\psi$$ for some $$\psi$$, and the proof is basically the same as above.
• In the calculations on the norm of $\sigma$, shouldn't the third equals sign from the left be a "$\leq$"? – CS1994 May 7 at 13:55