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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.

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

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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.

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  • $\begingroup$ Can you suggest me some source where I can read up about the link between the existence of the ground state and unitarity? Thanks! $\endgroup$ – Feynmans Out for Grumpy Cat Feb 23 at 17:21
  • $\begingroup$ A Hamiltonian with a linear potential is selfadjoint. $\endgroup$ – Keith McClary Feb 23 at 20:10
  • $\begingroup$ @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. $\endgroup$ – Avantgarde Feb 23 at 21:37

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