# Symmetry and overlapping of ground states

In a quantum mechanics, there is the following formula to derive the zero energy $E_0$ of a perturbed Hamiltonian $$H = H_0 + V$$ knowing the zero energy $W_0$ of the free Hamiltonian $H_0$: $$E_0 = W_0 + i\frac{d}{dt}\text{ln}R(t)|_{t\rightarrow\infty(1-i\eta)}$$ The exponential killing the excited states faster than the lowest energy one. However, one needs to suppose a non-vanishing overlap of the two ground states $|\phi_0\rangle$ for $H_0$ and $|\psi_0\rangle$ for $H$. I read that if two have different symmetry they must be orthogonal but I didn't manage to derive why.

Let's suppose that $G$ is a symmetry of $|\phi_0\rangle$ then $G|\phi_0\rangle=0$ and $\langle\psi_0|G|\phi_0\rangle=0$ and if $|\psi_0\rangle$ is not invariant under $G$ I still need to have $$G|\psi_0\rangle \propto |\psi_0\rangle$$ to derive $\langle\psi_0|\phi_0\rangle=0$. ($|\psi_0\rangle$ needs to be an eigenvector of $G$ with non-vanishing eigenvalue)

-

You have misunderstood what it means for the two states to 'have different symmetry'. Suppose, as you say, that $G$ is some operator representing a symmetry of the system. This means that $G$ is unitary, and $[G, H_0] = [G,V] = 0$ (we could also consider $[G,V] \neq 0$, but I don't think this is what you need).
Since $G$ is unitary and commutes with $H_0$, the ground state of $H_0$ will also be an eigenstate of $G$ (here we assume non-degeneracy of the ground state): $G|\phi_0\rangle = \lambda|\phi_0\rangle$ for some complex number $\lambda$. The same argument applies for $H$, so $G|\psi_0\rangle = \lambda'|\psi_0\rangle$ for some (possibly different) complex number $\lambda'$. Now consider $\langle\psi_0|G|\phi_0\rangle$. We can let $G$ act either 'forwards' on $|\phi_0\rangle$, or 'backwards' on $\langle\psi_0|$ (Exercise: show that $\langle\psi_0|G = \lambda'\langle\psi_0|$ as a consequence of unitarity of $G$), to get $$\lambda\langle\psi_0|\phi_0\rangle = \lambda'\langle\psi_0|\phi_0\rangle ~,$$ and therefore $$(\lambda-\lambda')\langle\psi_0|\phi_0\rangle = 0 ~.$$ So if $\lambda' \neq \lambda$, we find $\langle\psi_0|\phi_0\rangle = 0$.
Example: A good example would be a particle moving in one dimension, with $H_0 = \frac{p^2}{2m} + \lambda x^4$, and $V = -\mu^2 x^2$, where $\lambda$ and $\mu$ are real constants. There is a symmetry $G: x\to -x$; the ground state of $H_0$ is even under this symmetry, whereas the ground state of $H = H_0 + V$ is odd. In symbols, $G|\phi_0\rangle = |\phi_0\rangle~,~~ G|\psi_0\rangle = -|\psi_0\rangle$.
Thank you Rhys for this clear explanation. As you said I was seeing in a more radical way: the definition of "state invariant under a specific symmetry" was for me that the eigenvalue of the state under this symmetry needed to be zero so, for non-infinitesimal symmetry transformation:$$U(\alpha)|\phi_0\rangle = \text{e}^{i\alpha G}|\phi_0\rangle = |\phi_0\rangle$$ equivalent to $G|\phi_0\rangle>=0$ once you develop the exponential in series. Then I get that if you ask them to the two ground states to have different eigenvalues and the generator to be Hermitean you derive the right result. – Learning is a mess Nov 5 '12 at 9:50
@Qmechanic: Yes, and if you consider $\langle\psi_0|G|\psi_0\rangle$, you will find $\frac{1}{\overline\lambda'} = \lambda'$ (this was the exercise I left in my answer). In other words, $|\lambda'| = 1$, and this is true for all eigenvalues of a unitary operator. – Rhys Nov 5 '12 at 20:04