Exact solution to Dirac delta perturbation for particle in a box

Question

Using diagrammatic perturbation theory the energy of a particle in a box with a Dirac delta potential can be closely approximated. The following energy correction terms to the ground state energy ($\mathscr{E}_1$) up to 5th order are found. The question is if anybody recognizes some series or easy way to extend to higher orders.

\begin{align*} E^{(0)}_1 = \langle 1|\mathscr{H}_0|1\rangle &= \frac{\pi^2}{2}& \\ E^{(1)}_1 = \langle 1|\mathscr{V}|1\rangle &= 2&\\ E^{(2)}_1 = \langle 1|\mathscr{V}|\Psi_1^{(1)}\rangle &= -\frac{2}{\pi^2}& \\ E^{(3)}_1 = \langle 1|\mathscr{V}|\Psi_1^{(2)}\rangle &= -\frac{2}{3\pi^2} + \frac{8}{\pi^4} \\ E^{(4)}_1 = \langle 1|\mathscr{V}|\Psi_1^{(3)}\rangle &= \frac{4}{\pi^4} - \frac{40}{\pi^6} \\ E^{(5)}_1 = \langle 1|\mathscr{V}|\Psi_1^{(4)}\rangle &= \frac{2}{5\pi^4} - \frac{80}{3\pi^6}+ \frac{224}{\pi^8}\\ \end{align*}

$$\mathscr{E}_1=E^{(0)}_1 + E^{(1)}_1 + E^{(2)}_1 + E^{(3)}_1 + E^{(4)}_1 + E^{(5)}_1 + ...$$

I also found a source stating $E^{(5)}_1 = \frac{8}{45\pi^4}-\frac{22}{\pi^6}+\frac{200}{\pi^8}$, the difference is probably caused by human error, I am not sure who is wrong though. (I got it from a Master's student report, which I am not sure if they want to share it publically)

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System description

For a particle in an infinite square well ("particle in a box"), the solutions to the one-dimensional Schrödinger equation are known. In a box with length $a$ (for simplicity $a=1$), the particle has a wavefunction/eigenfunction $\psi_n$ with corresponding energy/eigenvalue $E_n$, where $n$ refers to the state, with $n=1$ being the ground state.

$$ \left(-\frac{1}{2}\frac{d^2}{dx^2}\right)|\psi\rangle = E|\psi\rangle\\ \psi_n = \sqrt{\frac{2}{a}}\sin\left(\frac{n\pi}{a}x\right)\\ E_n = \frac{n^2\pi^2}{2a} $$

A perturbation can be made, specifically a Dirac potential in the middle of the well. This means that the energies and wavefunctions are no longer known. Perturbation theory can approximate the perturbed system using solutions of the unperturbed system.

Look at Physics Exchange, which discusses the same problem. However, they did not state any closed form of the different correction terms.

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Perturbation theory

The repulsive Dirac potential will change the Schrödinger Equation. $$ \left(-\frac{1}{2}\frac{d^2}{dx^2}+\delta\left(x-\frac{1}{2}\right)\right)|\psi\rangle = E|\psi\rangle. $$ Perturbation theory, says that the Hamiltonian $\mathscr{H}$ can be split in the unperturbed Hamiltonian $\mathscr{H}_0$ and the perturbation $\mathscr{V}$. Some shorthand notation can be set up. $$ V_{ij} = \langle i|\mathscr{V}|j\rangle = \langle i|\delta\left(x-\frac{a}{2}\right)|j\rangle = 2\sin\left(\frac{i\pi}{2}\right)\sin\left(\frac{j\pi}{2}\right)\\ E_i^{(0)} = E_i = \frac{i^2\pi^2}{2}. $$ This shows that $V_{ij}=V_{ji}$ and note that $a=1$ was used.

Using the unperturbed wavefunctions and energies, the ground state energy of the perturbed system ($\mathscr{E}_1$) can be approximated. For the first few terms a closed form can be found, could this be extended to arbitrary order? $$\mathscr{E}_1=E^{(0)}_1 + E^{(1)}_1 + E^{(2)}_1 + E^{(3)}_1 + E^{(4)}_1 + E^{(5)}_1 + ...$$

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Diagrammatic perturbation theory

One way of finding the energy corrections is called diagrammatic perturbation theory. "Modern Quantum Chemistry" by Szabo and Ostlund has a chapter on this [1]. It comes down to the fact that diagrams can represent parts of the energy correction terms.

The rules for drawing the diagrams are as follows:

1. Draw $n$ dots vertically, in order.
2. Connect all $n$ dots together with a continuous line, so that each dot has one line passing through it.
3. Do this in all possible distinct ways. diagrams are equivalent if each and every dot is connected to an identical pair of dots.

Then arrows can be added to the lines, lines going down are "hole lines" and lines going up are called "particle lines". The perturbation ($\mathscr{V}$) is represented by the dots.

The rules to translate the diagrams into expressions are as follows:

1. Each dot contributes a factor $\langle\psi_{\text{label line in}}|\mathscr{V}|\psi_{\text{label line out}}\rangle$, to the numerator.
2. Each pair of adjacent dots contributes to the denominator factor $\sum E^{(0)}_{\text{hole}}-E^{(0)}_{\text{particle}}$, this sum runs over all lines crossing a (imaginary) horizontal line separating two adjacent dots
3. The overall sign of the expression is $(-)^{h+l}$, $h$ is the number of hole lines and $l$ is the number of closed loops (which is 1 in this system).

Below the diagrams representing the first four corrections to the energy are shown. Below that their corresponding expressions are given.

\begin{align} E^{(1)}_1 = V_{11}&\\ E^{(2)}_1 = \sum_{n\neq1}& \frac{V_{1n}^2}{\left(E_1-E_n\right)}\\ E^{(3)}_1 = \sum_{nm\neq1}& \frac{V_{1n}V_{nm}V_{m1}}{\left(E_1-E_n\right)\left(E_1-E_m\right)}-V_{11}\frac{V_{1n}^2}{\left(E_1-E_n\right)^2}\\ E^{(4)}_1 = \sum_{nmk\neq1}& \frac{V_{1n}V_{nm}V_{mk}V_{k1}}{\left(E_1-E_n\right)\left(E_1-E_m\right)\left(E_1-E_k\right)} + V_{11}^2\frac{V_{1n}^2}{\left(E_1-E_n\right)^3} \\&- V_{11}\frac{V_{1n}V_{nm}V_{m1}}{\left(E_1-E_n\right)^2\left(E_1-E_m\right)}- V_{11}\frac{V_{1n}V_{nm}V_{m1}}{\left(E_1-E_n\right)\left(E_1-E_m\right)^2} \\&- \frac{V_{1n}^2V_{1m}^2}{\left(E_1-E_n\right)\left(2E_1-E_n-E_n\right)\left(E_1-E_m\right)} - \frac{V_{1n}^2V_{1m}^2}{\left(E_1-E_n\right)^2\left(2E_1-E_n-E_n\right)} \end{align}

One can see that there would be 24 distinct diagrams for the 5th order and 120 for the 6th order. Using Mathematica the infinite sums are then evaluated, and the energies at the top of this post are found.

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Concluding, is there a series or an easy way to go to higher orders? Possibly is there even an exact solution when going to infinite order?

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References:

[1]: Szabo, A., & Ostlund, N. S. (1996). Modern Quantum Chemistry: Introduction to Advanced Electronic Structure Theory. Courier Corporation.

Answer 1

1. OP considers the TISE for an infinite square potential with a midpoint Dirac delta potential $$\begin{align}\left(-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}+V_0\delta\left(x-\frac{L}{2}\right)\right)\psi(x) ~=~& E\psi(x),\tag{1a}\cr 0~\leq~x~~\leq~L~\equiv~&a,\tag{1b}\cr \psi(0)~=~0~=~&\psi(L), \tag{1c}\cr E~\equiv~&\frac{\hbar^2k^2}{2m}, \qquad k~\geq~0, \tag{1d} \end{align}$$ OP is seeking to determine the ground state energy $E_1$ as a (not necessarily convergent) formal power series of the coupling constant $V_0$ using Rayleigh-Schrödinger (RS) perturbation theory$^1$.

2. Instead of RS perturbation theory, it is easier to directly consider the transcendental equation for the odd state eigenvalue of the wave number $k$ $$ \cot\frac{kL}{2}~=~-\frac{\kappa}{2k}, \qquad \kappa~\equiv~\frac{2m V_0}{\hbar^2}, \tag{2}$$ derived in e.g. this Phys.SE post.

3. Define a dimensionless coupling constant $$g~\equiv~\frac{m L}{\hbar^2} \frac{V_0}{\pi}\tag{3} $$ Then eq. (2) becomes $$-\frac{kL}{2}\cot\frac{kL}{2}~=~\frac{\kappa L}{4}~\equiv~ \frac{\pi}{2}g. \tag{4} $$

4. Next replace the wave number $k$ with a dimensionless phase $\phi$: $$ \frac{kL}{2} ~\equiv~(2n+1)\frac{\pi}{2} +\phi \tag{5}$$ for the $(2n\!+\!1)$th state. We can now rewrite eq. (4) as a fixed-point equation $$ \phi~=~\arctan\frac{g}{2n+1+\frac{2}{\pi}\phi}. \tag{6} $$

5. Let us from now on focus on the ground state. Eq. (6) with $n=0$ can be solved recursively in the coupling constant $g$ $$\begin{align} \phi_N~=~&\arctan\frac{g}{1+\frac{2}{\pi}\phi_{N-1}},\qquad N~\in\mathbb{N},\cr \phi_0~=~&0. \end{align}\tag{7} $$ The $N$th iteration is exact to $N$th order in the coupling constant $g$. In this way one can in principle obtain expressions for any finite order.

   E.g. Mathematica yields that the phase $\phi$ to order $N=7$ is

                    2                        4          2
                 2 g       1     8    3   8 g  (-15 + Pi )
     Out[1]= g - ---- + (-(-) + ---) g  + ---------------- + 
                  Pi       3      2                3
                                Pi             3 Pi
   
                                 6                 2        4
          1   224   20    5   2 g  (10080 - 1120 Pi  + 23 Pi )
    >    (- + --- - ---) g  - -------------------------------- + 
          5     4     2                         5
              Pi    Pi                     15 Pi
   
            1    8448   1120    1568    7       8
    >    (-(-) + ---- - ---- + ------) g  + O[g]
            7      6      4         2
                 Pi     Pi     45 Pi
   

6. After we have determined the phase $\phi$, we can use eq. (5) to find the wave number $k$, and finally eq. (1d) to find the energy $E$.

   E.g. the ground state energy $E_1$ (in energy units of $\frac{\hbar^2}{m L^2}$) to order $N=8$ is$^2$

               2
             Pi                2    8    2 Pi   3        40    4
     Out[2]= --- + 2 g Pi - 2 g  + (-- - ----) g  + (4 - ---) g  + 
              2                     Pi    3                2
                                                         Pi
   
          224    80    2 Pi   5      46    1344    560    6
    >    (--- - ---- + ----) g  + (-(--) - ---- + -----) g  + 
            3   3 Pi    5            9       4        2
          Pi                               Pi     3 Pi
   
          8448   1344    784    2 Pi   7
    >    (---- - ---- + ----- - ----) g  + 
            5      3    15 Pi    7
          Pi     Pi
   
          88   54912   9856   2464    8       9
    >    (-- - ----- + ---- - -----) g  + O[g]
          15      6      4        2
                Pi     Pi     5 Pi
   

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$^1$For higher-order corrections to RS perturbation theory, see e.g. this Phys.SE post.

$^2$To compare with OP's normalization of the $N$th-order energy correction $E_1^{(N)}$, substitute $g\to\frac{1}{\pi}\times$(OP's coupling constant).