Gell-Mann and Low theorem

In

Let ${\displaystyle |\Psi _{0}\rangle }$ be an eigenstate of ${\displaystyle H_{0}}$ with energy ${\displaystyle E_{0}}$ and let the 'interacting' Hamiltonian be ${\displaystyle H=H_{0}+gV}$, where ${\displaystyle g}$ is a coupling constant and ${\displaystyle V}$ the interaction term. We define a Hamiltonian ${\displaystyle H_{\epsilon }=H_{0}+e^{-\epsilon |t|}gV}$ which effectively interpolates between ${\displaystyle H}$ and ${\displaystyle H_{0}}$ in the limit ${\displaystyle \epsilon \rightarrow 0^{+}}$ and ${\displaystyle |t|\rightarrow \infty }$. Let ${\displaystyle U_{\epsilon I}}$ denote the evolution operator in the interaction picture. The Gell-Mann and Low theorem asserts that if the limit as ${\displaystyle \epsilon \rightarrow 0^{+}}$ of

${\displaystyle |\Psi _{\epsilon }^{(\pm )}\rangle ={\frac {U_{\epsilon I}(0,\pm \infty )|\Psi _{0}\rangle }{\langle \Psi _{0}|U_{\epsilon I}(0,\pm \infty )|\Psi _{0}\rangle }}}$

exists, then ${\displaystyle |\Psi _{\epsilon }^{(\pm )}\rangle }$ are eigenstates of ${\displaystyle H}$.

Note that when applied to, say, the ground-state, the theorem does not guarantee that the evolved state will be a ground state. In other words, level crossing is not excluded.

Proof

As in the original paper, the theorem is typically proved making use of Dyson's expansion of the evolution operator. Its validity however extends beyond the scope of perturbation theory as has been demonstrated by Molinari. We follow Molinari's method here. Focus on ${\displaystyle H_{\epsilon }}$ and let ${\displaystyle g=e^{\epsilon \theta }}$. From Schrödinger's equation for the time-evolution operator

${\displaystyle i\hbar \partial _{t_{1}}U_{\epsilon }(t_{1},t_{2})=H_{\epsilon }(t_{1})U_{\epsilon }(t_{1},t_{2})}$

and the boundary condition ${\displaystyle U_{\epsilon }(t_{2},t_{2})=1}$ we can formally write

${\displaystyle U_{\epsilon }(t_{1},t_{2})=1+{\frac {1}{i\hbar }}\int _{t_{2}}^{t_{1}}dt'(H_{0}+e^{\epsilon (\theta -|t'|)}V)U_{\epsilon }(t',t_{2}).}$

Focus for the moment on the case ${\displaystyle 0\geq t_{1}\geq t_{2}}$. Through a change of variables ${\displaystyle \tau =t'+\theta }$ we can write

${\displaystyle U_{\epsilon }(t_{1},t_{2})=1+{\frac {1}{i\hbar }}\int _{\theta +t_{2}}^{\theta +t_{1}}d\tau (H_{0}+e^{\epsilon \tau }V)U_{\epsilon }(\tau -\theta ,t_{2}).}$

We therefore have that

${\displaystyle \partial _{\theta }U_{\epsilon }(t_{1},t_{2})=\epsilon g\partial _{g}U_{\epsilon }(t_{1},t_{2})=\partial _{t_{1}}U_{\epsilon }(t_{1},t_{2})+\partial _{t_{2}}U_{\epsilon }(t_{1},t_{2}).}$

This result can be combined with the Schrödinger equation and its adjoint

${\displaystyle -i\hbar \partial _{t_{1}}U_{\epsilon }(t_{2},t_{1})=U_{\epsilon }(t_{2},t_{1})H_{\epsilon }(t_{1})}$

to obtain

${\displaystyle i\hbar \epsilon g\partial _{g}U_{\epsilon }(t_{1},t_{2})=H_{\epsilon }(t_{1})U_{\epsilon }(t_{1},t_{2})-U_{\epsilon }(t_{1},t_{2})H_{\epsilon }(t_{2}).}$

The corresponding equation between ${\displaystyle H_{\epsilon I},U_{\epsilon I}}$ is the same. It can be obtained by pre-multiplying both sides with ${\displaystyle e^{iH_{0}t_{1}/\hbar }}$, post-multiplying with ${\displaystyle e^{iH_{0}t_{2}/\hbar }}$ and making use of

${\displaystyle U_{\epsilon I}(t_{1},t_{2})=e^{iH_{0}t_{1}/\hbar }U_{\epsilon }(t_{1},t_{2})e^{-iH_{0}t_{2}/\hbar }.}$

The other case we are interested in, namely ${\displaystyle t_{2}\geq t_{1}\geq 0}$ can be treated in an analogous fashion and yields an additional minus sign in front of the commutator (we are not concerned here with the case where ${\displaystyle t_{1,2}}$ have mixed signs). In summary, we obtain

${\displaystyle \left(H_{\epsilon ,t=0}-E_{0}\pm i\hbar \epsilon g\partial _{g}\right)U_{\epsilon I}(0,\pm \infty )|\Psi _{0}\rangle =0.}$

We proceed for the negative-times case. Abbreviating the various operators for clarity

${\displaystyle i\hbar \epsilon g\partial _{g}\left(U|\Psi _{0}\rangle \right)=(H_{\epsilon }-E_{0})U|\Psi _{0}\rangle .}$

Now using the definition of ${\displaystyle \Psi _{\epsilon }}$ we differentiate and eliminate derivatives ${\displaystyle \partial _{g}(U|\Psi _{0}\rangle )}$ using the above expression, finding

${\displaystyle {\begin{aligned}i\hbar \epsilon g\partial _{g}|\Psi _{\epsilon }\rangle &={\frac {1}{\langle \Psi _{0}|U|\Psi _{0}\rangle }}(H_{\epsilon }-E_{0})U|\Psi _{0}\rangle -{\frac {U|\Psi _{0}\rangle }{{\langle \Psi _{0}|U|\Psi _{0}\rangle }^{2}}}\langle \Psi _{0}|(H_{\epsilon }-E_{0})U|\Psi _{0}\rangle \\&=(H_{\epsilon }-E_{0})|\Psi _{\epsilon }\rangle -|\Psi _{\epsilon }\rangle \langle \Psi _{0}|H_{\epsilon }-E_{0}|\Psi _{\epsilon }\rangle \\&=\left[H_{\epsilon }-E\right]|\Psi _{\epsilon }\rangle .\end{aligned}}}$

where ${\displaystyle E=E_{0}+\langle \Psi _{0}|H_{\epsilon }-H_{0}|\Psi _{\epsilon }\rangle }$. We can now let ${\displaystyle \epsilon \rightarrow 0^{+}}$ as by assumption the ${\displaystyle g\partial _{g}|\Psi _{\epsilon }\rangle }$ in left hand side is finite. We then clearly see that ${\displaystyle |\Psi _{\epsilon }\rangle }$ is an eigenstate of ${\displaystyle H}$ and the proof is complete.

References