Stern Gerlach with spin in opposite directions So for the Stern-Gerlach apparatus, we assume that we either have a particle spin up or spin down.  We also have the varying field, $\partial B/\partial z$.  This initial configuration results in the particle wither going to plus $\hbar$ or minus $\hbar$.  
Suppose instead of having spin up/down in the z direction, I sent it through with an initial spin aligned in the x direction (same exact configuration)?  The Hamiltonian is given (for a linear B) as
$$H=\frac{1}{2m}(p_x^2+p_z^2)-\mu \sigma_z(B_0+B'z)$$
So my equations of motion for the z direction would just give me $p_0t/m+z_0$ and $\dot p_x=0$.  Do I need to account for the spin x now instead, or will the particle go  undeflected?
 A: $\newcommand{\ket}[1]{\left| #1 \right>}$$\newcommand{\bk}[2]{\left< #1 | #2 \right>}$Notice that the eigenvectors of the operator $S_z$ spans the whole space, which means that you can write any state as a superposition, (if you prefer as a linear combination) of these states. The situation is akin to the basis vectors of usual 3d Euclidean space. You can choose tree basis vectors there, usually the following is chosen:
$$\vec x = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}\quad \vec y = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}\quad \vec z = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}\quad$$
Then you can write any vector $\vec v$ as a linear combination of these vectors that is
$$\vec v = \begin{pmatrix} a \\ b \\ c \end{pmatrix} \implies \vec v = a \vec x + b \vec y + c \vec z$$
Notice that the choice of basis vectors is not unique. I could have chosen the following basis vectors if I wanted to:
$$\vec x' = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}\quad \vec y' = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}\quad\vec z' = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}\quad$$
for which the vector $\vec v$ can be written as
$$\vec v = a\vec x' + (b-a) \vec y' + c\vec z'$$
The situation with spin 1/2 particle is almost the same. You can choose your basis vectors to be the eigenvectors of $S_z$ operator in which case you write
$$\ket{z+} = \begin{pmatrix} 1 \\  0 \end{pmatrix} \quad \ket{z-} = \begin{pmatrix} 0 \\ 1 \end{pmatrix} $$ 
$$\implies \ket{x+}= \frac{1}{\sqrt 2} \Big( \ket{z+}+ \ket{z-} \Big)\quad \ket{x-}= \frac{1}{\sqrt 2} \Big( -\ket{z+} + \ket{z-} \Big)$$
Notice that $\ket{x\pm}$ and $\ket{z\pm}$ are the eigenvectors of $S_x$ and $S_z$ operators respectively, as we wanted.
If you send a particle with spin $+x$ in a SG machine aligned in the $z$ direction, then half the time you get $+z$ and half the time you get $-z$. You can immediately see that by taking the inner product of the state $\ket{x+}$ with the states $\ket{ z \pm}$.
$$\left| \bk{z+}{x+} \right|^2 = \left( \frac{1}{\sqrt 2} \right)^2 = \frac{1}{2} \qquad{\text{and similarly}} \quad \left| \bk{z-}{x+} \right|^2 = \frac{1}{2}$$
You can also choose your basis vectors to be the eigenvectors of $S_x$ operator but everything gets messy for this problem and you don't, in general, want to do that. 
A: If you imagine a beam entering a Stern-Gerlach device, at first the beam is behind the device and then the beam could be deflected left or deflected right after it goes through the device.
A Stern-Gerlach device sends a spin up $\vert+\rangle$ to the (say) left and a spin down $\vert-\rangle$ to the right.  So it sends $\vert+\rangle\vert behind\rangle$ to $\vert+\rangle\vert left\rangle$ and sends $\vert-\rangle\vert behind\rangle$ to $\vert-\rangle\vert right\rangle$. By linearity we then know how it works on any possible state.
For instance something spin up for the x direction is eigen to $\sigma_x$ so is proportional to $\frac{1}{\sqrt{2}}\vert+\rangle+\frac{1}{\sqrt{2}}\vert-\rangle.$ 
Thus it evolves (by linearity) from $$\left(\frac{1}{\sqrt{2}}\vert+\rangle+\frac{1}{\sqrt{2}}\vert-\rangle\right)\vert behind \rangle \rightarrow \left(\frac{1}{\sqrt{2}}\vert+\rangle\vert left \rangle\right)+\left(\frac{1}{\sqrt{2}}\vert-\rangle\vert right \rangle\right).$$ 
The particle is in a linear combination (superposition) of going left and going right, and if the left and right beams are separated (by any measuring device or by just letting a normal Stern-Gerlach device finish the job) then not only is it deflected left or deflected right, but it becomes spin up (in the z direction) or spin down (in the z direction).  Measurements are about correlating the state of the thing to be measured (up/down in the z direction) with something else in a way that separates the different results.  Making them go left and right can succeed at separating them if those separate beams can interact differently to the point where they can't interfere with each other ever again in a detectable way.
Real beams have some thickness to them before they go through the device and you can use the Schrödinger equation for the actual Hamiltonian used to measure it to track the progress of the beam and you'll see the probability current smoothly split the beam into two parts, the relative size of the fraction that goes each way is determined by the squares of the amplitudes $\alpha, \beta$ where the spin state going in is $\alpha\vert+\rangle+\beta\vert-\rangle.$
See e.g. "The pilot-wave perspective on spin" by Travis Norsen in the American Journal of Physics 82(4) 337-348 (2014); http://dx.doi.org/10.1119/1.4848217 and even if you don't read the article, figure 2 (with long caption) is available (for free) if you click on the "data & media" tab.  So you can see the streamlines determined by the probability current flow from a connected beam into two forks, this is what happens in general.  It's just that nothing goes into one of the forks when it was all spin up (or all spin down).  For spin in the x direction, the left half goes left and the right half goes right, similarly for any spin direction orthogonal to the z direction.  When the spin is not orthogonal to the z direction a fraction (different than half) goes each way.
If you want to see the dynamics of a measurement, always use the Schrödinger equation and track the probability current, doing something else might seriously mislead you.
A: There are two possible answers, one for formula-writers and another for conceptually-thinking physicists.
First: the “spin $x$” is just $\sigma_x$ and its exact configurations are merely eigenvectors, like the “spin $z$” is $\sigma_z$ and its exact configurations are simply the standard basis vectors. Don’t ask what does it mean because it doesn’t mean anything.
Second: learn what is the quantum superposition (I won’t recommend the en.wikipedia article — it is ugly) and then read the 
Why, for a spin-½ particle, are the possible outcomes of measuring spin projection along any direction the same? topic. The latter will not explain anything before the quantum superposition would be actually understood.
