Why does the $\phi$-cubed theory have no ground state? In the book of Sredinicki's, he claimed that the $\phi^3$ theory has no ground state, hence this is not a physical theory.
My question is that I can't see why this system has no ground state. And I don't understand either the explaination he gave. For example, what does "roll down the hill" really mean? What's the case for a harmonic oscillator pertuibed by a $q^3$ term? Maybe it's better if someone can explain it using the quantum harmonic case.

 A: In quantum theory we usually require that the Hamiltonian $H$ is bounded from below and that the system has a ground state. This is intimately related to unitarity. The $\phi^3$ theory violates this. 
A: Work on a spatial lattice of finite extent so that the field operators and the Hamiltonian are well-defined as (unbounded) operators on a Hilbert space. Consider any Hamiltonian of the form
$$
H = \epsilon^D \sum_x \Big(\Pi^2(x)+V\big(\phi(x)\big)\Big)
\tag{1}
$$
where $\epsilon$ is the lattice spacing, the sum is over all lattice sites, $D$ is the number of spatial dimensions, and $V(\phi)$ is an arbitrary polynomial. The commutation relation is
$$
\big[\phi(x),\Pi(y)\big]=i\frac{\delta_{x,y}}{\epsilon^D}.
\tag{2}
$$
Suppose that a ground state $|0\rangle$ exists. By definition, this is a state that satisfies
$$
 \psi_\text{diff}\equiv
 \frac{\langle \psi|H|\psi\rangle}{
 \langle \psi|\psi\rangle}
 - 
 \frac{\langle 0|H|0\rangle}{
 \langle 0|0\rangle}
 \geq 0
\tag{3}
$$
for all states $|\psi\rangle$. For any real number $a$, the unitary operator
$$
U(a)\equiv\exp\left(-ia\epsilon^D\sum_x\Pi(x)\right)
\tag{4}
$$
satisfies
$$
U^\dagger(a)\phi(x) U(a)=\phi(x)+a,
\tag{5}
$$
so $U^\dagger(a)H U(a)$ is the same as $H$ but with $\phi$ replaced by $\phi+a$ inside $V(\phi)$. Now consider the state 
$$
|\psi\rangle\equiv U(a)|0\rangle
\tag{6}
$$
where $|0\rangle$ is the alleged ground state. Then the quantity (3) is
$$
 \psi_\text{diff} =
 \frac{\langle 0|V_a-V|0\rangle}{
 \langle 0|0\rangle}
\tag{7}
$$
with $V_a(\phi)\equiv V(\phi+a)$. Now suppose that $V(\phi)$ is a cubic polynomial with non-zero cubic term. Then the quantity (7) is a cubic polynomial in the real variable $a$ with non-zero cubic term. Since $a$ is an arbitrary real number, this polynomial attains negative values for values of $a$ of the appropriate sign and with sufficiently large magnitude. This contradicts the assumption that $|0\rangle$ was a ground state, so this completes the proof.
