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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)

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up vote 3 down vote accepted

You haven't really asked a precise question, or given an example, but I think I know what you're getting at.

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, $G|\phi_0\rangle = |\phi_0\rangle~,~~ G|\psi_0\rangle = -|\psi_0\rangle$.

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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
Yes, in the case of continuous groups (Lie groups), you can do the analysis in terms of the infinitesimal generators (the Lie algebra). I think it's nice to also consider just working with the symmetry operators themselves, i.e. the unitary operators, because discrete symmetries like the reflection in my example (or parity in 3D) cannot be given as the exponential of some anti-Hermitean operator. – Rhys Nov 5 '12 at 10:40
@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

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