State Vector and Density matrix

Question

I have a slight confusion with the two.

From what I have understood, state vector describes pure states, which means that with a probability $1$, our state is going to be in that state. However, if we have a superposition state, how do we actually determine that the probability is classical and not the one due to quantum superposition?
Can we actually use the density matrix to write the state vector? I am confused because say we consider a two level system, where I define the density matrix as the following in $Sz$ basis: \begin{align} \rho = \frac{1}{2}(|0><0| + |1><1|) \end{align} This means that when we use $Tr(P_n\rho)$ to find the probabilities, we get a $\frac{1}{2}$ for both |0>, |1> state. So why can we not write the following state vector: \begin{align} |\psi> = \frac{1}{\sqrt2}(|0> +|1>) \end{align}
Lastly, can we say that a superposition state is a pure state, if the classical probability for it is $1$ or do we have to ensure the condition $Tr(\rho^2)=1$; are they equivalent?

Does this answer your question? How is a quantum superposition different from a mixed state? — Tobias Fünke, Jul 29 '22 at 22:12
Yes, that answers the 2nd question @JasonFunderberker, thank you! I am still unsure about the first and last question. — Megan mars, Jul 29 '22 at 22:35
I don't understand the first question, to be honest. Regarding your third question: A density matrix is pure if and only if $\rho^2=\rho \Longleftrightarrow \rho=|\psi\rangle\langle \psi|$ with $|\psi\rangle$ normalized $\Longleftrightarrow \mathrm{Tr}\rho^2=1$. — Tobias Fünke, Jul 29 '22 at 22:40
Do you understand that what you call superposition is a statement on the basis you chose? — Cosmas Zachos, Jul 29 '22 at 22:45
Yes, that does. Also, I understood the first question. Thank you so much! — Megan mars, Jul 29 '22 at 22:49

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