Here is something that I do not understand.
According the concept of entanglement, the quantum state
$$| \psi \rangle=\frac{1}{\sqrt{2}}(| 0 \rangle+| 3 \rangle)=\frac{1}{\sqrt{2}}(| 00 \rangle+| 11 \rangle)$$
which is the quantum state of two qubits (in a two atom system), cannot be written as a tensor product of quantum states of two separate qubits. More specifically
$$| \psi \rangle\neq (\alpha_0|0\rangle +\alpha_1|1\rangle)\otimes (\beta_0|0\rangle +\beta_1|1\rangle)$$
for all possible $\alpha_0,\alpha_1,\beta_0,\beta_1 \in \mathbb{R}$.
On the other hand, I have noticed that
$$\mathbb{R}^2 \otimes \mathbb{R}^2 \cong \mathbb{R}^4$$
which means that there should be a one-to-one mapping between them. So, I expect $(\alpha_0|0\rangle +\alpha_1|1\rangle)\otimes (\beta_0|0\rangle +\beta_1|1\rangle)$ to produce $\frac{1}{\sqrt{2}}(| 0 \rangle+| 3 \rangle)$, as a member of $\mathbb{R}^4$, for some coefficients $\alpha_0,\alpha_1,\beta_0,\beta_1 \in \mathbb{R}$. But it is not possible.
What am I missing?
Thanks in advance for any hints or answers.
You forget that the tensor product of spaces consists of linear combinations of pure tensors.
$$|00 \rangle + |11 \rangle = |0\rangle \otimes |0\rangle + |1\rangle \otimes |1\rangle$$