Pure states and irreducible *-representations If $A$ is a $C^*$-algebra, then a $*$-representations of $A$ is irreducible if and only if the corresponding state is a pure state.
What happens if don't insist on $A$ being a $C^*$-algebra? Is one of the two directions still true? Are there examples in which this Theorem fails?
To give a detailed version of my question: Let $A$ be an unital algebra over $\mathbb{C}$ with involution $*$. Let $\varphi: A \to \mathbb{C}$ be a state, i.e. a $\mathbb{C}$-linear map which satisfies $\varphi(a^* a) \geq 0$ for all $a \in A$ and $\varphi(1)=1$. The set of states on $A$ is convex and we call $\varphi$ a pure state if it is an extreme point of this set. The set $N_{\varphi}=\{a \in A: \varphi(a^*a)=0\}$ is a left ideal of $A$. When does one of the following hold:
$\varphi$ is a pure state $\Rightarrow$ $N_{\varphi}$ is a maximal left ideal
$\varphi$ is a pure state $\Leftarrow$ $N_{\varphi}$ is a maximal left ideal
?
Edit: Why is $N_{\varphi}$ a left ideal? To see this, we first have to prove the CSI: $|\varphi(b^* a)|^2 \leq \varphi(a^*a) \varphi(b^* b)$. We have for all $v=(v_1,v_2) \in \mathbb{C}^2$ that $\varphi((v_1 a + v_2 b)^* (v_1 a + v_2 b))\geq 0$. This means that the matrix $\begin{pmatrix} \varphi(a^*a) & \varphi(b^* a) \\ \varphi(a^* b) & \varphi(b^*b) \end{pmatrix}$ is positive semidefinite. Taking the determinant implies the CSI.
Now using the CSI, the claim is easy: $\varphi((ua)^*ua)=\varphi(a^*(u^*ua))$, thus $\varphi(a^*a)=0$ implies $\varphi((ua)^*ua)=0$. One checks similarly the additive property.
 A: There is a counterexample for: $N_{\phi}$ is a left maximal ideal $\implies \phi $ is a pure state.
[Edit:let $B$ be the free non-commutative algebra spanned by $U$ and $V$. We have a antilinear map $^*:B \rightarrow B$ that maps every product $\alpha \times E_1 E_2... E_n$ to $\overline{\alpha}\times  E_n E_{n-1}...E_1$ where $E_i \in \{U,V\}$ for all $i \in \{1,...,n\}$ and $\alpha \in \mathbb{C}$. If $p,q \in B$, $(pq)^*=q^*p^*$, $(p+q)^*=p^*+q^*$ and $(\alpha p)^*= \overline{\alpha} p^*$. Let $I$ be the ideal spanned by $(UV-VU+i)$. Let $A=B/I$. If $p \in I, p^* \in I$ because $(UV-VU+i)^*=-(UV-VU+i)$, so if $p-q \in I$, $p^*-q^*\in I$. So $^*$ is well defined on $A$]
Let $X=\overline{U}$ and $Y=\overline{V}$.
$A$ operates on the set of the functions $C^{\infty}$ from $\mathbb{R}$ to $\mathbb{C}$.
$(X.f)(x):=x\times f(x)$
$(Y.f)(x):=i \times f'(x)$
We have $X^*:=X$ and $Y^*:=Y$
We have $XY-YX=-i$
Let $E:=x \mapsto e^{-x^2}$
For all $a\in A$, we define $\phi(a)=\int_{-\infty}^{\infty}E(x)\times(a.E)(x).dx$
Then $\phi(a^*a)=\int_{-\infty}^{\infty}E(x)\times(a^*a.E)(x).dx=\int_{-\infty}^{\infty}\overline{(a.E)(x)}\times(a.E)(x).dx\geq 0$ 
We can normalize $\phi$ to obtain $\phi(1)=1$.
$N_{\phi}$ is the set of elements in $A$ that annihilate $E$.
Then $N_{\phi}$ is a maximal left ideal because:
if $b \notin N_{\phi}$, $(b.E)(x)=P(x)e^{-x^2}$ for some $P \in \mathbb{C}[X]-\{0\}$, so $(b -P(X))E=0$ 
so $b\equiv P(X) $ modulo $N_{\phi}$. If $\deg(P)=n$, $Yb \equiv YP(X)$.
$YP(X)=P(X)Y+Q(X)$ with $\deg(Q) =\deg(P)-1$, because $Q=iP'$.
$Y+2iX\in N_{\phi}$.
So $YP(X)\equiv P(X)(-2iX)+Q(X)$.
$YP(X)\equiv -2iXP(X)+Q(X)$.
So $(Y+2iX)P(X)=Q(X)$.
So $\mathbb{C} \subset (Y+2iX)^nb+N_{\phi}$ and $N_{\phi}$ is a left maximal ideal.
But $\phi$ is not a pure state.
