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?

  • $\begingroup$ 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? $\endgroup$ – Gautam Shenoy May 4 at 11:50
  • $\begingroup$ @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. $\endgroup$ – 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.

  • $\begingroup$ In the calculations on the norm of $\sigma$, shouldn't the third equals sign from the left be a "$\leq$"? $\endgroup$ – CS1994 May 7 at 13:55
  • $\begingroup$ No, they are all equalities. $\endgroup$ – Martin Argerami May 7 at 13:59
  • $\begingroup$ Are we not applying the triangle inequality there? $\endgroup$ – CS1994 May 7 at 14:35
  • $\begingroup$ No. It's an equality, and it is by definition. If you want to put a name to it (although that wouldn't be entirely correct) it would "Pythagorean Theorem". $\endgroup$ – Martin Argerami May 7 at 14:49
  • $\begingroup$ Sorry, yes I now agree after expanding the norms into scalar products and calculating explicitly. Thanks for the answer and patience :-) $\endgroup$ – CS1994 May 7 at 14:56

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