# Feynman, Hibbs Perturbations and Energy

I am currently self-studying from Feynman & Hibbs’ Quantum Mecahnics and Path Integrals, but having an issue understanding a step in their development of first-order perturbations. They define $$V_{mn}(t)=\int_{-\infty}^\infty \phi_m^*(x)V(x,t)\phi_n(x)\,dx,$$ where $\phi_m(x)$ is an unperturbed eigenfunction with energy $E_m$, and $V(x,t)$ is a perturbing potential. This gives (eventually) the probability of transition from state $n$ to $m$ within time $T$ as $$P(n\rightarrow m)=\frac{|V_{mn}|^2}{(E_m-E_n)^2}\left[4\sin^2\frac{(E_m-E_n)T}{2\hbar}\right]$$ This all makes sense to me. But on p150, they specialise to the case where values of the final energy, $E_m$, lie in a continuum, defining $\rho(E)\,dE$ as the number of states with energy lying between $E$ and $E+dE$. We now have, as the probability of any transition: $$\int|V_{mn}|^2\frac{4\sin^2[(E_m-E_n)T/2\hbar]}{(E_m-E_n)^2}\rho(E_m)\,dE_m$$ At this point, they spend some time demonstrating that the final states that contribute most to this integral will be those with energy $E_m$ that is in the neighbourhood of, nearly, or approximately the same as $E_n$ (all three descriptions are used). They then use the approximation that $|V_{mn}|$ and $\rho(E_m)$ vary slowly enough in this neighbourhood that we can treat them as constant, taking them outside the integral. Using $\int_{-\infty}^\infty[(sin^2 x)/x^2]\,dx = \pi$, they write (p151) “we obtain the result that the probability for a transition to some state in the continuum is $$P(n\rightarrow m)=\frac{2\pi}{\hbar}|V_{mn}|^2\rho(E_n)T,$$ and that the energy in the final state is the same as the energy in the initial state.” (emphasis mine).

I have no issue following the mathematics up to this final result, but I cannot see why they have moved from describing the final energy as lying in the continuum, approximately equal to the initial energy, to claiming the result that it is the same. A few paragraphs later, they explicitly replace $\rho(E_n)$ with $\delta(E_m-E_n)$, but again, I can’t see why.

They have not indicated that this result arises from $T\rightarrow\infty$; could that be what I am missing?

If the contribution is from state close to $E_n$, they can assume that the level density doesn't vary much in that within that range in energy. Replace $\rho(E_m)$ with its value at $E_n$. The delta comes from
$\frac{\sin^2(xT)}{Tx^2} \rightarrow \pi \delta(x), T \rightarrow \infty.$