The following paragraph is from Schwartz Sec 4.2.1

Using OFPT we would calculate the energy shift using

$$\Delta E_n = \langle\psi_n\rvert H_{int} \rvert \psi_n\rangle +\sum_{m,m \ne n} \frac{\rvert \langle\psi_n\rvert H_{int} \rvert \psi_m\rangle \rvert ^2}{E_n-E_m}$$

This is the standard formula from time-independent perturbation theory. The basic problem is that we have to sum over all possible intermediate states $\rvert \psi_m \rangle$, including ones that have nothing much to do with the system of interest (for example, free plane waves). It is still true in field theory that there are only a finite number of states below any given energy level $E$, so that as $E \to \infty$, $\frac{1}{E-E_n}\to0$. The catch is that there are an infinite number of states, and their phase space density goes as $\int d^3k \sim E^3$, so that you get $\frac{E^3}{E-E_n}\to \infty$ and perturbation theory breaks down. This is exactly what Oppenheimer found.

I have 3 doubts in the above text

$1$.What does Schwartz mean with the states which has nothing to do with our system of interest? Why does it cause us some uneasiness to sum over states that have nothing to do with our system of interest, where he gives an example of plane waves but perturbation does not regard whether the state has little or large impact in the summation. We just add all the terms mechanically to get the change in energy. For example, say we have $\psi_1,\psi_2,\psi_3$ when we do calculation of $\Delta E_1$ we have two terms in second-order $\frac{\rvert \langle\psi_2\rvert H_{int} \rvert \psi_1\rangle \rvert ^2}{E_2-E_1}$,$\frac{\rvert \langle\psi_3\rvert H_{int} \rvert \psi_1\rangle \rvert ^2}{E_3-E_1}$ now suppose latter term comes from the wavefunction($\psi_3$) which has nothing to do with our system but still we have thrown it in our summation to get right result because that's how we derived the formulae for $\Delta E_n$ the set of all wavefunction used in summation must be complete.

$2.$ It doesn't matter whether there are finite number of states below any given energy $E$ since we are summing over all states we have to even add those states which have much higher energy than $E$ so what does he want to convey when he says as $E \to \infty$, $\frac{1}{E-E_n}\to0$.

$3.$ What is phase space density? And how $\int d^3k \sim E^3$? If I assume $H$ to be of form $H_{int}=\int {\frac{d^3k}{{(2 \pi)}^3} \omega_k(a_k^{\dagger}a_k + \frac{1}{2})}$ then ${\langle\psi_n\rvert H_{int} \rvert \psi_m\rangle} \sim (const)\int d^3k \omega_k \sim E$ since $E=\hbar \omega_k$ and we are summing over all possible $k$ which will lead to an exponent of 2 rather than 3 in $\frac{E^3}{E-E_n}$ though the sum will diverge in both the case.

  • $\begingroup$ Concerning 3. Is this a relativistic theory ? By phase space density, he might means here the number of states at a given energy (see also density of state). Here it could be $\rho(E)=\int d^d k \delta(E-\epsilon_k)$ with $\epsilon_k$ the energy of a state with momentum $k$. Then $\int d^d k=\int dE \rho(E)$. $\endgroup$ – Adam Sep 20 '18 at 7:07
  • $\begingroup$ Concerning 2. It seems to me that $E$ is the energy that is summed over (i.e. $E\equiv E_m$, while $E_n$ is the fixed energy of the unperturbed state we are interested in. $\endgroup$ – Adam Sep 20 '18 at 7:08

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