Why do smaller objects become harder to break? When grabbing a typical tree branch of at least two feet, it's so easy to snap with a less than one inch circumference that even a toddler can do it.
However, after breaking it, the smaller halves with multiply from the center, going like this:
1 - BB (before break)
2 - AB (after break)
4 - AB (half of second break equals four)
8 - AB (continuing multiplication factors of two)
And so on....
But what I want to know is why this is ... in quantum physics.
I want to know why the more you break it, the harder it is to keep breaking it from the center of each half.
This applies to tearing as well ... if you tear a cardboard box in half down the middle it's easy, but if you turn it over and continue on it keeps getting harder as the pieces get smaller.
 A: This has to do with deflection
When one grabs a twig and tries to break it, the ends are pulled together to bow the stick till it breaks.
The deflection ("bending") of a stick is proportional to the cube of its length. So, with the same force, a stick of length $L$ will be deflected 8 times less than a stick of length $2L$ with the same force.
Now, one side of a bent twig is stretched and the other side is contracted. This stretching/contraction is proportional to the deflection, and when it reaches a limit (Breaking stress), the twig breaks. 
So what's happening here is that as you shorten the twig, you lose the capacity to generate deflection, and thus it becomes harder to break.
A: A slightly different take on Manishearth's answer is to think in detail of the forces and torques involved it independently from deflexions (which follow from the Young's modulus and the second moment of area of the twig's cross section about the appropriate axis). 
Here is a drawing of my twig. Not very pretty, but it shows all the forces on the twig (neglecting gravity)

The vertical forces must sum to nought, so as you heave down on the end of the twig with force $F$, there is a vertical balancing force $-F$ exerted on the twig from its "cantilever bearing" i.e. the join to the tree. This appears as shear forces directed upwards along the plane of the cross section. Note that this shear is independent of the twig's length $L$, so, although shear does contribute the failure, it is a constant and so not the deciding factor here.
The other equilibrium condition is the rotational one. If we reckon torques about the origin in the centre of the twig's cross-section at the "bearing", then there is a torque of magnitude $F\, L$ from the force $F$ you put on the end of twig that must be balanced by the torques of forces normal to the cross-section at the twig's "bearing". I imagine these as two concentrated forces $T$ and $-T$ (horizontal force components sum to nought) spread by distance $w / 2$ as shown, where $w$ is the twig cross section's width vertical. Of course the forces are distributed and you need to look at the Euler-Bernoulli theory for a fuller description, but the concentrated forces will do for our argument. Summing all torques to nought yields:
$$w\,T = F\, L$$
or 
$$T = \frac{F\, L}{w}$$
so this torque is proportional to the twig's length. This is proportional to the forces normal to the cross-section that pull on the molecular bonds fastening the branch to the tree. For the same force downwards, as the twig's length increases, so do these forces; in the end the forces in the upper half of the twig reach the tensile strength of the wood and failure begins at the very top of the "bearing". This is an unstable situation as now the same torque must be derived from less cross-section, stresses increase even further, the crack swiftly propagates through the cross-section of twig breaks off the tree.
The full description of static problem in the small deflexion limit is given by Euler-Bernoulli beam theory (see the Wikipedia page for this). Euler-Bernoulli theory also works very well for oscillating beams as long as the spatial frequency of the beam's bends is small compared to the acoustic wavelength at the temporal frequency in question. A fuller theory that overcomes this approximation is afforded by Timoshenko beam theory (again, see the Wiki page with this name) but even this assumes that beam cross section stay planar (i.e. the points in the beam on a cross sectional plane are mapped by the stress and strain to other, rotated planes whereas in practice these surfaces are curved). The only full solution is numerical.
