Why every state evolving infinite time becomes the ground state in QFT?

Question

For any state $|\phi \rangle $ evolving infinite time $$\lim\limits_{t\rightarrow \infty} e^{-iHt}|\phi\rangle=\lim\limits_{t\rightarrow \infty} e^{-iHt}|n\rangle\langle n|\phi\rangle$$ Let $t\rightarrow t(1-i\eta)$ , then only the lowest energy eigenstate in $|\phi\rangle$ survives. Therefore $$\lim\limits_{t\rightarrow \infty} e^{-iHt}|\phi\rangle=\lim\limits_{t\rightarrow \infty} e^{-iHt}|n\rangle\langle n|\phi\rangle\propto |m\rangle$$ where $|m\rangle$ is the lowest lowest energy eigenstate in $|\phi\rangle$. For general states which have ground state evolving infinite time, they all become the ground state.

What's the physical and mathematical meaning of this operation of artificially introducing this small decay?

This result is so weird and always used in qft. How could it happen?

For example, let $|\phi\rangle =\alpha |E_0\rangle+\beta|E_1\rangle$ and $|\alpha|^2+|\beta|^2=1$, then for any time the probability of the state $|\phi\rangle$ being $|E_1\rangle$ is $|\beta|^2$. It contradicts with the result above. So how to resolve this puzzle?

Answer 1

I think, from the way you formulated the question, you lost the context of this trick, and then it indeed doesn't make a lot of sense. The point is that in QFT, you want to compute quantities corresponding to the full interacting Hamiltonian, $H$. In practice, however, we only know the eigenstates of the free Hamiltonian $H_{0}$: the plane waves $|k\rangle=exp(-ikx)$ (disregarding spin).

The nice thing is that for most calculations (in the end, we want to know cross sections of certain processes) it is enough to know the Green's functions of the theory $G(x-y)=\langle\Omega|T\phi(x)\phi(y)|\Omega\rangle$. These functions are defined as the field operators 'sandwished' between the ground state of the full Hamiltonian. And we can in fact write them in function of our plane wave spectrum!

Indeed, let's try to evolve the ground state of the free Hamiltonian $|0\rangle$ in time, using the full Hamiltonian $H$. Then we have: $$e^{-iHT}|0\rangle=?$$ We can now fill in the energy spectrum (that we don't know!) of the full Hamiltonian: $H|n\rangle=E_{n}|n\rangle$: $$e^{-iHT}|0\rangle=e^{-iHT}\Sigma_{n}|n\rangle\langle n|0\rangle=\Sigma_{n}e^{-iE_{n}T}|n\rangle \langle n|0\rangle.$$ The ground state of $H$, denoted as $|\Omega\rangle$, -- which we want to obtain -- can now be extracted using the trick you described: $$e^{-iHT}|0\rangle=e^{-iE_{0}T}|\Omega\rangle \langle \Omega|0\rangle+\Sigma_{n\neq0}e^{-iE_{n}T}|n\rangle \langle n|0\rangle$$ $$|\Omega\rangle=\mathrm{lim}_{T\rightarrow\infty(1-i\epsilon)}(e^{-iE_{0}T}\langle\Omega|0\rangle)^{-1}e^{-iHT}|0\rangle$$

There is nothing esoterical about this, no-one said time is imaginary, the only statement that is made is that this relation between the vacuum state of $H$: $|\Omega\rangle$ and the vacuum state of $H_{0}$: $|0\rangle$, is correct and can be subsequently exploited. Indeed, if the interaction is small, the Dirac or interaction picture can be used, and we find an expression for the Green function only in terms of things we can calculate (the Feynman diagrams!) (observe that the unknown factors $(e^{-iE_{0}T}\langle\Omega|0\rangle)^{-1}$) have disappeared: $$\langle\Omega|T\phi(x)\phi(y)|\Omega\rangle=\mathrm{lim}_{T\rightarrow\infty(1-i\epsilon)}\frac{\langle 0 |T\phi_{I}(x)\phi_{I}(y)e^{-i\int dt H_{I}(t)}|0\rangle}{\langle 0 |Te^{-i\int dt H_{I}(t)}|0\rangle}.$$

I learned this from Peskin & Schroeder, so for a more complete answer, see their book "An introduction to quantum field theory", 1995, Westview Press, pp.86.

Answer 2

In QFT it is not to get down to the ground state, but to choose a correct propagator (amongst variety of Green's function). In other words, it is applying or taking into account the boundary conditions.

However, for "incomplete" systems, decaying may really mean getting down to the ground state due to interaction of some sort like irreversibly absorbing excitations by the environment or so.

Answer 3

Your argument is valid for unitary evolution. However turning the time into the complex plane you make it non-unitary.

You may say that you introduce small decay for every state $$e^{-iHt}=e^{-iHt-\eta Ht}$$ with the ground energy state the slowest to decay (stable if you set $E_0=0$)