Perturbation Theory, Infinite Square Well, Kronig Penney This is a two part question from a paper I did, I couldn't answer it at the time and can't find sufficient help in my notes. I've been trying to solve it since but it's a lot harder than normal to get help.
Part A)
An infinite square well of width $Na$ has allowed energies and eigenfunctions
\begin{align}
 E_m^{(0)} &= \frac{\hbar^2 \pi^2 m^2}{2m_eN^2a^2}\, \\
 φ_m^{(0)} &= \sqrt{\frac{2}{Na}}\sin\left(\frac{m\pi x}{Na}\right)
\end{align}
where the superscript (0) is a label denoting the unperturbed states. Calculate the first order perturbation to the energies for a perturbation produced by a series of Dirac delta function barriers
$$
\hat{H}'=V_o \sum_{n=1}^{N-1}\delta(x-na)
$$
EDIT ---------
I understand the process for calculating the first order perturbation up to the line:
$$
E_m^{1} = \frac{V_0}{Na} \int_{0}^{Na}\sum_{n=1}^{N-1}\delta(x-na)
$$
I get stuck understanding how to integrate this sum. I feel like given the similarities between integration and summation this should be simple but I do not understand the jump to the next step.
 A: We begin with the known Hamiltonian which is also the unperturbed $$\hat H^0\left|\ m\right>=E_m^0\left|\ m\right>$$
Now we introduce the perturbation
$$(\hat H^0+\hat H')\left|\ m\right>=E_m^0\left|\ m\right>$$
If you are familiar with first order perturbation you  must be knowing the first order correction in energy eigenvalue
$$E_m^1=  \left<\psi\right|\hat H'\left|\psi\right>$$
$$\hat{H}'=V_o \sum_{n=1}^{N-1}\delta(x-na)$$
$$E_m^1=\int_0^{Na} V_o \sum_{n=1}^{N-1}\delta(x-na) \sqrt \frac{2}{Na}\sin \frac{mx\pi}{Na}\sqrt \frac{2}{Na}\sin \frac{mx\pi}{Na}$$
$$E_m^1=V_o\int_0^{Na}   \sum_{n=1}^{N-1}\delta(x-na)   \frac{2}{Na}\sin^2 \frac{mx\pi}{Na} $$
$$E_m^1=\frac {2V_o}{Na}\int_0^{Na}   \sum_{n=1}^{N-1}\delta(x-na) \sin^2 \frac{mx\pi}{Na} $$
$$E_m^1=\frac {2V_o}{Na}\int_0^{Na}   \sum_{n=1}^{N-1}\delta(x-na)\frac{1}{2}\bigr ( 1- \cos2 \frac{mx\pi}{Na}) $$
Odd terms go to zero
$$E_m^1=\frac {2V_o}{Na}\int_0^{Na}   \sum_{n=1}^{N-1}\delta(x-na)\frac{1}{2} $$
$$E_m^1=\frac {V_o}{Na}\int_0^{Na}   \sum_{n=1}^{N-1}\delta(x-na) $$
$$E_m^1=\frac {V_o}{Na}\int_0^{Na} \delta(x-a)+\delta(x-2a)+........\delta(x-(N-1)a) $$
$$E_m^1=\frac {V_o}{Na}\Bigr( \int_0^{Na} \delta(x-a)+\int_0^{Na}\delta(x-2a)+........\int_0^{Na}\delta(x-(N-1)a) \Bigr)$$
Now,if you use matlab or wolfram it would be easy to solve using the Heaviside step function or θ(x)
$$\int\delta(x-tk)= \theta(x-tk)$$
Substitute this in the integral
$$E_m^1=\frac {V_o}{Na}\Bigr(   \theta(x-a)+\theta(x-2a)+........\theta(x-(N-1)a) \Bigr)\Bigr|_0^{Na}$$
$$E_m^1 =\frac {V_o}{Na}\sum_{n=1}^{N-1}\theta(x-an)\Big|_0^{Na}$$
Let N be any positive number like 3
then
$$\sum_{n=1}^{N-1}\theta(x-an)\Big|_0^{Na}=\sum_{n=1}^{3-1}\theta(x-an)\Big|_0^{3a}$$
$$=\theta(x-a) + \theta(x-2a)+......\theta(x-(N-1)a)\Bigr|_0^{Na}$$
$$=\theta(Na-a) + \theta(Na-2a)+......\theta(Na-(N-1)a)-\theta(0-a) -\theta(0-2a)-......\theta(0-(N-1)a)$$
$$=\theta(a(N-1)) + \theta(a(N-2))+......\theta(a(N-N+1))-\theta(-a) -\theta(-2a)-......\theta(-(N-1)a)$$
$$=\theta(a(N-1)) + \theta(a(N-2))+......\theta(a)-\theta(-a) -\theta(-2a)-......\theta(-(N-1)a)$$
Using the property of the Heaviside function
$$\theta(x) - \theta(-x) =1, x>0$$
$$=\theta(a(N-1))+......\theta(a)-\theta(-a)-......\theta(-(N-1)a)= N-1 $$
This is the proof of
$$\int_0^a \delta(x-tk)=1$$
So if there are 2=n delta functions
$$\int_0^a \delta(x-t_1k)+......\int_0^a \delta(x-t_nk)=1+1=2=n$$
$$=\frac {V_o}{Na}(N-1)=\frac {V_o}{a}-\frac{V_o}{Na}$$
$$E_m = E_m^0 +E_m^1$$
$$=\frac{\hbar^2 m^2 \pi^2}{2m_e N^2 a^2}+\frac {V_o}{Na}(N-1)$$
Given
$$E_m^1 =0 $$
$$V_o = 0, N -1 =0$$
The system won't be perturbed if Vo is 0 , hence
$$N=1$$
$$E=\frac{\hbar^2m^2\pi^2}{2m_e(1)^2a^2}$$
$$m^2=\frac{2m_eEa^2}{\hbar^2\pi^2}$$
$$m=\sqrt\frac{2m_eEa^2}{\hbar^2\pi^2}$$
$$m= \frac{a}{\hbar\pi}\sqrt{2m_eE}$$
$$m=\frac{ak}{\pi}$$
$$\frac{\pi}{a}m = k$$
This result can be easily compared to the Kronig Penney result
