I'm still a beginner in the quantum aspect of computer science (self-studying). So please bear with me if this may sound like a stupid question.
I know Hadamard gate transforms qubit into a superposition state with $\frac{1}{2}$ chance that the outcome would be 0 or 1 when the wave function collapses:
$$\frac{\left|0\right\rangle + \left|1\right\rangle}{\sqrt2} \text{ for }\left|0\right\rangle \\ \frac{\left|0\right\rangle - \left|1\right\rangle}{\sqrt2} \text{ for }\left|1\right\rangle$$
For already-in-a-superposition qubit, it would be just some linear transformation.
$$\alpha\left|0\right\rangle + \beta\left|1\right\rangle \rightarrow \alpha \frac{\left|0\right\rangle + \left|1\right\rangle}{\sqrt2} + \beta \frac{\left|0\right\rangle - \left|1\right\rangle}{\sqrt2} = \frac{\alpha+\beta}{\sqrt2}\left|0\right\rangle + \frac{\alpha-\beta}{\sqrt2}\left|1\right\rangle $$
But I got really confused when I see the $\Sigma \left|x\right\rangle$ in the equation on the Wikipedia article about Deutsch–Jozsa algorithm.
A Hadamard transformation is applied to each bit to obtain the state
Why is it doing a summation with a ever-increasing value of $x$? My understanding of a qubit is that it can only be in either $\left|0\right\rangle$, $\left|1\right\rangle$ or a superposition state.
What the equation is doing above appears to indicate that if $n$ is 2, we would get
$$\frac{1}{\sqrt{2^{3}}} \left|0\right\rangle (\left|0\right\rangle - \left|1\right\rangle) + \frac{1}{\sqrt{2^{3}}}\left|1\right\rangle (\left|0\right\rangle - \left|1\right\rangle) + \\\frac{1}{\sqrt{2^{3}}} \left|2\right\rangle (\left|0\right\rangle - \left|1\right\rangle) + \frac{1}{\sqrt{2^{3}}} \left|3\right\rangle (\left|0\right\rangle - \left|1\right\rangle)$$
How do I make sense of the $\left|2\right\rangle$ and $\left|3\right\rangle$?
[Update]
As I was writing the question, I remembered that qubits in the state of $\left|1\right\rangle$$\left|0\right\rangle$ are often expressed as $\left|10\right\rangle$. Since we are working with two-state system here I figured out that the 2 and 3 should be converted into binary in this context. And therefore the expression should be interpreted as
$$\frac{1}{\sqrt{2^{3}}} (\left|0\right\rangle (\left|0\right\rangle - \left|1\right\rangle) + \frac{1}{\sqrt{2^{3}}}(\left|1\right\rangle (\left|0\right\rangle - \left|1\right\rangle) + \\\frac{1}{\sqrt{2^{3}}} (\left|10\right\rangle (\left|0\right\rangle - \left|1\right\rangle) + \frac{1}{\sqrt{2^{3}}} (\left|11\right\rangle (\left|0\right\rangle - \left|1\right\rangle)$$
It appears that things are making a lot more sense now. But as I take a deeper look into the expression above, two questions come to me:
Why do I see so many $\left|0\right\rangle - \left|1\right\rangle$ but not a single $\left|0\right\rangle + \left|1\right\rangle$ ? (Intuitively I expect to see just one $\left|0\right\rangle - \left|1\right\rangle$ since we can only get $\frac{\left|0\right\rangle - \left|1\right\rangle}{\sqrt2}$ when it is in the state of $ \left|1\right\rangle$ and there is only one qubit in $ \left|1\right\rangle$).
Why are there only $2^{n}$ possible states with each having the chance of $(\sqrt{2^{3}})^2$ when the input is $2^n+1$ qubits? Adding four $(\sqrt{2^{3}})^2$ together doesn't give us 1.