# 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.

• Have you tried sketching $y=x^2+\epsilon x^3$? – jacob1729 Feb 23 at 17:11

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.
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.
• @DvijMankad Also see en.wikipedia.org/wiki/Ghost_(physics). Ghosts typically appear in higher derivative field theories. They lead to unbounded Hamiltonians, and violate unitarity. Srednicki's example is not that of a ghost, but having a $\phi^3$ potential has the same result of an unbounded Hamiltonian. Also check the second half of pg 124, Peskin. – Avantgarde Feb 23 at 21:37