The Qubit

Let $x_{1}$ be the eigenstate corresponding to the 1 state, and let $x_{0}$ be the eigenstate corresponding to the 0 state. Let $X$ be the total state of our state vector, and let $w_{1}$ and $w_{0}$ be the complex numbers that weight the contribution of the base states to our total state, then in general:

$$\left\vert X \right\rangle = w_{0} * \left\vert x_{0} \right\... ...e + w_{1} * \left\vert x_{1} \right\rangle \equiv (w_{0},w_{1})$$

At this point it should be remembered that $w_{0}$ and $w_{1}$, the weighting factors of the base states are complex numbers, and that when the state of $X$ is measured, we are guaranteed to find it to be in either the state:

$$0 * \left\vert x_{0} \right\rangle + w_{1} * \left\vert x_{1} \right\rangle \equiv (0,w_{1})$$

or the state

$$w_{0} * \left\vert x_{0} \right\rangle + 0 * \left\vert x_{1} \right\rangle \equiv (w_{0},0)$$

This is analogous to a system you should be more familiar with, a vector with real weighting factors in the a two dimensional plane. Let the base states for this two dimensional plane be the unit vectors $x$, and $y$. In this case we know that the state of any vector $V$ can be described in the following manner:

$$V = x_{0} * x + y_{0} * y \equiv (x_{0},y_{0})$$

We can further restrict our state vector to be a unit vector in a Hilbert space. It is not necessary from a physics perspective for the state vector to be a unit vector (by which I mean it has a length of 1), but it makes for easier calculations further on, so I will assume from here on out that the state vector has length 1. This assumption does not invalidate any claims about the behavior of the state vector. To see how to convert a state vector of any length to length 1 see appendix A.