1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.