Spontaneous breaking of supersymmetry I am a beginner in SUSY Quantum mechanics. I had read that the symmetry is spontaneously broken if $A \left|\psi \right>_n^{(1)}\neq 0 $ and symmetric if $A \left|\psi \right>_n^{(1)}= 0. $ But analyzing the spectrum of broken symmetry I can see a symmetry in the energy values, and in case of unbroken symmetry ground state (zero energy) $H_1$ does not have corresponding energy state in $H_2$. This shows an asymmetry in spectrum of partner potentials in unbroken case.
I think I am wrong somewhere.
What do we mean by symmetry and the spontaneous breaking of symmetry here? Which symmetry is broken here?
Edit:
$A$  is the Annihilation operator. $\left|\psi \right>_n^{(1)}$ is the eigen vector for $H_1$. $H_1$ and $H_2$ are the Hamiltonian corresponding to the partner potentials $V_1$ and $V_2$.
 A: This is a perfect storm of notational dissonance between QM and QFT. Your statement

I had read that the symmetry is spontaneously broken if $A \left|\psi \right>_n^{(1)}\neq 0 $ and symmetric if $A \left|\psi \right>_n^{(1)}= 0. $

is inapposite and misconstrued—justly paradoxical. I suspect labels on symmetric phase versus SSB might not be productive here.


*

*Review of SSQM
$$
Q=\begin{pmatrix} 0& 0\\ A& 0 \end{pmatrix}, \qquad Q^\dagger=\begin{pmatrix} 0& A^\dagger\\ 0& 0 \end{pmatrix},
$$
so 
$$
H= Q Q^\dagger+ Q^\dagger Q=  \begin{pmatrix} A^\dagger A& 0\\ 0& A A^\dagger \end{pmatrix}\equiv \begin{pmatrix} H_1& 0\\ 0& H_2 \end{pmatrix}.
$$
Now 
$$
A|\psi_n^{(1)}\rangle = |\psi_{n-1}^{(2)}\rangle, \qquad A^\dagger |\psi_n^{(2)}\rangle = |\psi_{n+1}^{(1)}\rangle, \qquad E^{(1)}_{n+1}= E_n^{(2)}  \qquad E_0^{(1)}=0.
$$
So $\psi^{(2)}_n$ and $\psi^{(1)}_{n+1}$ have the same eigenvalue under H, and they are degenerate pairs-- (unbroken symmetry),
$$
Q \begin{pmatrix}  \psi^{(1)}_{n+1}\\ 0 \end{pmatrix} = \begin{pmatrix}  0\\ \psi^{(2)}_{n} \end{pmatrix} .
$$
But the ground state is unique, an unpaired susy-singlet,  and non-degenerate:
$$
Q \begin{pmatrix}  \psi^{(1)}_{0}\\ 0 \end{pmatrix} = 0, \qquad A|\psi^{(1)}_0\rangle = 0 ,
$$
so, if you chose, you might call it SSBroken (but Witten calls it unbroken). A Susy rotation does not couple it with other states. So the bottom of the spectrum, the ground state, lacks the symmetry of the rest of the spectrum. 
This is a narrow answer to your question, but does not alleviate the perfect storm of dissonance.



*

*The QFT paradigm The above stands in sharp contrast to QFT with its infinite d.o.f. 


To avoid confusion, I'll call the infinitesimal Susy charge here ${\cal Q}$, so that the real correspondent to the finite Q above is, instead, the full group 
super-transformation, $\exp i\bar\theta {\cal Q}$, for a Grassmann angle θ. In QFT, the symmetric (unbroken) phase is characterized by 
$$
{\cal Q}|0\rangle= 0, \Longrightarrow \exp (i\bar\theta {\cal Q})|0\rangle=|0\rangle .
$$
For $[H,{\cal Q}]=0$, eigenstates of the hamiltonian,
$$
H \phi |0\rangle= E \phi |0\rangle 
$$
are susy-rotated to degenerate ones,
$$
H(e^{i\bar\theta {\cal Q}} \phi e^{-i\bar \theta {\cal Q}} )|0\rangle = 
e^{i\bar\theta {\cal Q}} H\phi  |0\rangle = E (e^{i\bar\theta {\cal Q}} \phi e^{-i\bar \theta {\cal Q}} )|0\rangle,
$$
evocative of "what you had read". So, it is here that this is the hallmark of degeneracy of the spectrum, while the vacuum is unique, similarly to above.
In the SSB phase, we have the opposite: most states are non-degenerate, by failure of the above argument, but the vacuum is now degenerate. It is not unique:
$$
{\cal Q}|0\rangle\neq 0  \Longrightarrow \qquad |\Omega\rangle= \exp (i\bar\theta {\cal Q})|0\rangle\neq |0\rangle .
$$
This state is degenerate with the vacuum and roils and bubbles with goldstinos. 
So, at the end of the day, your question dramatizes the stark difference between QM and QFT. The takeaway may well be that sticking to facts rather than labels might be the sanest option. 
A: I found this very confusing when I first studied it as well. It becomes much clearer when you approach the subject from the quantum field theory view and are already familiar with symmetry breaking from the Higgs mechanism.
What we mean when we say that a state of a quantum system exhibits some symmetry is that it is invariant under this symmetry. Since continuous symmetries are represented by (some representation) of a Lie group, this statement is equivalent to
$$ e^{i \theta_j J_j} |\psi\rangle = |\psi\rangle,\tag{1}$$
where the $J_j$ are the generators of the group. For a state to be invariant under arbitrary elements of the symmetry group we therefore need
$$ J_j|\psi\rangle = 0. $$
In that sense, only the ground state of supersymmetric Hamiltonians is itself supersymmetric.
But this already calls on another important notion of symmetry: the symmetry of the Hamiltonian. The Hamiltonian is said to possess some symmetry, if it commutes with the symmetry generators; this implies that both the previous state as well as the state after some symmetry transformation have the same energy. It does not imply that the eigenstates of the Hamiltonian are themselves invariant under the symmetry. It does imply, however, that the eigenstates form (depending on the symmetry group possibly infinitely dimensional) multiplets of eigenstates that only transform among each other. Thus, symmetry transformations only rotate you through those degenerate multiplets but don't move you to other multiplets.
A thing that might confuse you in the particular case of supersymmetry is that one rarely talks about susy group transformations but only looks at the algebra. Supersymmetry is not a discrete symmetry, as you might think when you look at statements like
$$ Q|\text{boson}\rangle = |\text{fermion}\rangle \tag{2}$$
but rather a continuous symmetry, acting on states in the same manner as in (1). Thus, if you have some non-trovial action of the susy generator as in (2), then the state will not be invariant under finite group transformations (but rather covariant). 
Also, note the spontaneous symmetry breaking refers to the symmetry of the ground state only. The Hamiltonian keeps the symmetry independently of ssb (i.e. it commutes with the symmetry generators). 
Thus, a symmetry is never really gone in the case of spontaneously broken symmetry, it is merely hidden in the precise sense that there is no unique vacuum state but a (possibly infinitely) degenerate multiplet of vacua.
