System without ground state is not real in nature? We know that Coulomb force is real phenomena in nature and with Coulomb potential $V(x) \thicksim -\frac{1}{|x|}$ lowest energy is bounded in hydrogen atom. But it's mathematically clear that if Coulomb potential is replaced with strange potential $V(x) \thicksim -\frac{1}{|x|^ {10}}$ the ground state energy cannot be bounded from below and system is not stable/possible in quantum mechanical sense. So,  is the following implication true? $$\text{Force is not possible in QM   }\rightarrow \text{   Force is not real in nature}$$
 A: 
the ground state energy cannot be bounded from below and system is not stable/possible in quantum mechanical sense. 

in fact, the system could radiate an infinite amount of energy (photons) which for example would make a fundamental law such as the conservation of energy useless.

So, is the following implication true?
  Force is not possible in QM → Force is not real in nature

no, that is not necessarily true. 
Keep in mind that the theory of Quantum Mechanics is our description of nature, it's not that nature has to follow the description we come up with. If at some point we observe an effect which can't be described by Quantum Mechanics (and we are sure that there are no mistakes made in the observation) then we will have to extend / generalize it.
For example, the fact that in Newtonian mechanics all inertial frames share the same universal time does not imply that this is actually what is happening in nature (which it isn't in fact).
A: There is another reason why such a potential needs further attention. The point is that the system must have a unitary time evolution which is generated by a self-adjoint Hamiltonian. Thus one has to verify that $H={-{(∂/∂_{\bf{x}})}^2}/(2m)+V(x)$ defined on a suitable dense domain (for instance smooth square integrable functions vanishing in a neighbourhood of $0$) has a self-adjoint extension. There can be more than one in which case the physics should lead to the right choice. In case $V(x)$ is non-negative one can start from the associated quadratic form (Friedrichs extension) to obtain a self-adjoint operator.
http://en.wikipedia.org/wiki/Friedrichs_extension
http://en.wikipedia.org/wiki/Extensions_of_symmetric_operators
