A constant electric field

• The potential for a constant electric field: \(V(x) = e \tilde{E} \, x\)
• The time-independent Schrodinger equation:

\[ -\frac{\hbar^2}{2m} \frac{d^2 \psi(x)}{dx^2} + e\tilde{E}x \psi(x) = E \psi(x) \]

A constant electric field

\[ -\frac{\hbar^2}{2m} \frac{d^2 \psi(x)}{dx^2} + e\tilde{E}x \psi(x) = E \psi(x) \]

Rewrite the equation to clarify its form:

\[ \begin{aligned} \frac{d^2 \psi(x)}{dx^2} &= \frac{2m e\tilde{E}}{\hbar^2} (x-\frac{E}{e\tilde{E}}) \psi(x)\\ &= c\,(x - d) \psi(x) \end{aligned} \]

Where \(c = \frac{2m e\tilde{E}}{\hbar^2}\) and \(d = \frac{E}{e\tilde{E}}\).

This looks very much like the (solvable) Airy equation:

\[ \frac{d^2 f(z)}{dz^2} = z f(z) \]

\(\longrightarrow\) we need to find a suitable substitution for \(z\)

Suitable substitution for z(x)

Assume a linear form for \(z = \alpha x + \beta\) and rewrite the Airy equation

\[ \frac{d^2 \psi(x)}{dx^2} = \frac{d^2 f(z)}{dz^2} \left( \frac{dz}{dx} \right)^2 = \alpha^2 z f(z) = (\alpha^3 x + \beta \alpha^2) f(z) \]

Airy equation solutions

\[ \frac{d^2 f(z)}{dz^2} = z \, f(z), \qquad \quad \psi(x) = f(z) = C \, \mathrm{Ai}(z) + D \, \mathrm{Bi}(z) \]

\[ \begin{aligned} \mathrm{Ai}(z) &= \frac{1}{\pi} \int_0^\infty \cos\left(\frac{t^3}{3} + zt\right) dt\\ \mathrm{Bi}(z) &= \frac{1}{\pi} \int_0^\infty \left[\exp\left(-\frac{t^3}{3} + zt\right) + \sin\left(\frac{t^3}{3} + zt\right)\right] dt\\ \end{aligned} \]

Boundary conditions

\[ \frac{d^2 f(z)}{dz^2} = z \, f(z), \qquad \quad \psi(x) = f(z) = C \, \mathrm{Ai}(z) + D \, \mathrm{Bi}(z) \]

\[ \begin{aligned} &x = 0 \longrightarrow z_0 = -c^{1/3} d &\qquad C \, \mathrm{Ai}(z_0) + D \, \mathrm{Bi}(z_0) = 0\\ &x = L \longrightarrow z_L = c^{1/3}(L-d) &\qquad C \, \mathrm{Ai}(z_L) + D \, \mathrm{Bi}(z_L) = 0\\ \end{aligned} \]

\[ \begin{pmatrix} \mathrm{Ai}(z_0) & \mathrm{Bi}(z_0)\\ \mathrm{Ai}(z_L) & \mathrm{Bi}(z_L)\\ \end{pmatrix} \pmatrix{C \\ D} = \pmatrix{0 \\ 0} \]

Inverse cannot exist \(\longrightarrow\) determinant is zero:

\[ \det \begin{pmatrix} \mathrm{Ai}(z_0) & \mathrm{Bi}(z_0)\\ \mathrm{Ai}(z_L) & \mathrm{Bi}(z_L)\\ \end{pmatrix} = \mathrm{Ai}(z_0) \, \mathrm{Bi}(z_L) - \mathrm{Ai}(z_L) \, \mathrm{Bi}(z_0) = 0\\ \]

\(\longrightarrow\) Numerically solutions \(z_0(E)\), \(z_L(E)\)

Dimensionless units

Choice of units

Eigenenergies & eigenstates

1. Numerically solve determinant equation for eigenenergies \(E_n\),
2. Use \(E_n\) in boundary conditions to obtain \(\psi_n(x)\)
3. Normalize by \(\int_0^L |\psi(x)|^2 dx = 1\)

\[ \begin{aligned} &\textrm{Eigenenergies} \, \,E_n: \qquad \qquad\qquad\mathrm{Ai}(z_0) \, \mathrm{Bi}(z_L) - \mathrm{Ai}(z_L) \, \mathrm{Bi}(z_0) = 0\\ &\\ &\textrm{Eigenstates} \, \,\psi_n(x) = C \,\mathrm{Ai}(z) + D \,\mathrm{Bi}(z): \qquad\frac{D}{C} = -\frac{\mathrm{Bi(z_0)}}{\mathrm{Ai(z_0)}} \end{aligned} \]

Remember that \(z\) scales with energy

\[ z = c^{1/3}(x - d) = \left( \frac{2 m e\tilde{E}}{\hbar^2} \right)^{1/3} \left(x - \frac{E}{e\tilde{E}}\right) = \left(\frac{\pi}{\nu_L}\right)^{\frac{2}{3}} \left(\frac{x}{L}\nu_L - \varepsilon\right) \]

Final solution

• Eigenenergies \(\varepsilon_n\) have zero determinant: find roots
• Fill in \(\varepsilon_n\) to get eigenstates \(\psi_n(x)\)

Finite basis approximation

Steps to reach to the solutions:

1. Expand the solution in a known basis
2. Limit the amount of energy levels \(\longrightarrow\) matrix algebra
3. Solve the eigenvalue equations for eigenenergies & eigenstates

Dimensionless Hamiltonian

The dimensionless Hamiltonian for infinite well is obtained by:

• \(z = x/L\), \(E \longrightarrow E/E^\infty_1\),
• electric field \(\nu_L\) in units of \(V_L/E^\infty_1\)

\[ \hat{H} = -\frac{1}{\pi^2}\frac{d^2}{dz^2} + \nu_L \, (z - 1/2) \quad \textrm{ with } \quad \hat{H}_0 = -\frac{1}{\pi^2}\frac{d^2}{dz^2} \]

Compute the elements of the Hamiltonian (matrix)

\[ H_{mn} = -\frac{1}{\pi^2} \int \psi_m(z) \frac{d^2}{dz^2} \psi_n(z) dz \,\, + \,\,\int \nu_L \, (z - 1/2) \psi_m(z) \psi_n(z) dz, \]

With \(\,\psi_n(z) = \sqrt{2} \sin(n \pi z)\)

Hamiltonian: overlap integrals

\[ H_{mn} = \langle \psi_m | \hat{H} | \psi_n \rangle = -\frac{1}{\pi^2} \int \psi_m(z) \frac{d^2}{dz^2} \psi_n(z) dz \,\, + \,\,\int \nu_L \, (z - \frac{1}{2}) \psi_m(z) \psi_n(z) dz, \]

With the eigenstates \(\psi_n(z) = \sqrt{2}\sin(n\pi z)\)

\[ \begin{aligned} H_{mn} &= -\frac{1}{\pi^2} \int_0^1 \psi_m(z) \frac{d^2}{dz^2} \psi_n(z) dz \,\, + \,\,\int_0^1 \nu_L \, (z - 1/2) \psi_m(z) \psi_n(z) dz,\\ &= n^2\delta_{mn} \,\, + \,\,\nu_L \int_0^1 \, (z-1/2) \sin(m\pi z) \sin(n\pi z) dz,\\ \end{aligned} \]

The second integral can be calculated:

\[ \int_0^1 \, (z-1/2) \sin(m\pi z) \sin(n\pi z) dz = \left\{\,\,\begin{aligned} \frac{4nm}{\pi^2(m^2-n^2)^2} & \qquad \textrm{if }\,\, m+n \,\,\textrm{ is odd} \\ & \\ 0 & \qquad \textrm{if } \,\,m+n\,\, \textrm{ is even} \end{aligned}\right. \]

Hamiltonian: overlap integrals

We see that the integral has two different contributions:

• Diagonal elements \(\,H_{nn}\,\) are defined by \(\,\hat{H}_0\,\) with \(\,\,E_n = n^2\, E_1^\infty\)
• Other elements \(\,H_{n, m \neq n}\,\) determined by perturbing potential \(\,\,\hat{H}_p = \hat{H} - \hat{H}_0\)

\[ H_{nn} = n^2 \qquad \left\{\,\, \begin{aligned} H_{mn} &= - \nu_L \, \frac{4nm}{\pi^2(m^2-n^2)^2} & \quad \textrm{if }\,\,n = m \pm 1, m \pm 3, \dots \\ & & \\ H_{mn} &= 0 & \quad \textrm{if }\,\, n = m \pm 2, m \pm 4, \dots\\ \end{aligned} \right. \]

• The Hamiltonian gives contributions of eigenstates \(\,\,\psi_n(z) = \sqrt{2}\sin(n\pi z)\)
• Eigenenergies \(\,E_n\) and potential \(\,\nu_L\) are in units of \(\,\,E_1^\infty = \frac{\hbar^2\pi^2}{2mL^2}\)

Solving the eigenvalue equation

The eigenvalue equation is:

\[ \hat{H} \psi_n(z) = E_n \psi_n(z) \]

If we numerically calculate the overlap integrals:

\[ H_{mn} = \pmatrix{ 1 & -0.54 & 0\\ -0.54 & 4 & -0.584\\ 0 & -0.584 & 9\\ } \]

When comparing resulting eigenvalues:

Comparing Eigenstates

Finite difference approximation

Steps to reach to the solutions:

1. Discretize the wave function (finite basis)
2. Finite Difference Method to discretize the Schr"odinger equation
3. Limited discrete points \(\longrightarrow\) matrix algebra (Requires finite box)
4. Solve the eigenvalue equations for eigenenergies & eigenstates

Discretizing a function

• A function on a computer as a vector:

\[ f(x) = \pmatrix{f(x_1) \\ f(x_2) \\ \vdots \\ f(x_N)} \longrightarrow \pmatrix{f_1 \\ f_2 \\ \vdots \\ f_N} \]

A normalized state

• A normalized state requires: \(\langle f |f \rangle = 1\)
• If we define the normalized state as ket and bra:

\[ |f \rangle = |f(x_j) \sqrt{\delta x}\rangle = \pmatrix{f(x_1) \\ f(x_2) \\ \vdots \\ f(x_N)} \longrightarrow \pmatrix{f_1 \\ f_2 \\ \vdots \\ f_N} \]

\[ \langle f | = \langle f^*(x_j) \sqrt{\delta x}| \longrightarrow \pmatrix{f_1^* \sqrt{\delta x} & f_2^* \sqrt{\delta x} & \cdots & f_N^* \sqrt{\delta x}} \]

And the inner product is:

\[ \langle f | f \rangle = \langle f(x) | f(x)\rangle \delta x = \sum_{j=1}^N |f_j|^2 \delta x \quad\longleftrightarrow\quad \int_a^b |f(x)|^2 dx \]

Finite differences

• Hamiltonian contains second derivative
• Derivative of a discrete function?

\[ \textrm{Central difference scheme: } \quad \frac{df(x)}{dx} \longrightarrow \frac{\delta f(x)}{\delta x} = \frac{f_{i+1} - f(i-1)}{2\delta x} \]

Second derivative

• Hamiltonian contains second derivative
• Derivative of a discrete function?

\[ \textrm{Central difference scheme: } \quad \frac{df(x)}{dx} \longrightarrow \frac{\delta f(x)}{\delta x} = \frac{f_{i+1} - f(i-1)}{2\delta x} \]

• Second derivative by using central difference scheme with \(\delta/2\)

\[ \frac{d^2 f(x)}{dx^2} \longrightarrow \frac{\frac{f_{i+1} - f_{i}}{\delta x} - \frac{f_{i} - f_{i-1}}{\delta x}}{\delta x} = \frac{f_{i+1} + f_{i-1} - 2 f_i}{\delta x^2} \]

Matrix form

\[ \hat{H} = -\frac{\hbar^2}{2m} \frac{d^2}{dx^2} + V(x) \quad \longrightarrow \quad -\frac{\hbar^2}{2m} \frac{f_{i+1} + f_{i-1} - 2 f_i}{\delta x^2} + V_i \]

• The Hamiltonian can be written as a matrix operator

\[ \hat{H} = -\frac{\hbar^2}{2m\,\delta x^2} \pmatrix{ 1 & 0 & & & & & \\ 1 & -2 & 1 & & & & \\ & 1 & -2 & 1 & & & \\ & & \ddots & \ddots & \ddots & & \\ & & & 1 & -2 & 1 & \\ & & & & 1 & -2 & 1 \\ & & & & & 0 & 1 \\ } + V(x) \mathbb{1} \]

Matrix form (dimensionless)

\[ \hat{H} = -\frac{1}{\pi^2} \frac{d^2}{dx^2} + V(x) \quad \longrightarrow \quad -\frac{1}{\pi^2} \frac{f_{i+1} + f_{i-1} - 2 f_i}{\delta x^2} \,+\, \nu_L \left(\frac{i}{N}-\frac{1}{2}\right) \]

• The Hamiltonian can be written as a matrix operator (\(E\), \(\nu_L\) in units of \(E_1^\infty\))

\[ -\frac{1}{\pi^2\,\delta x^2} \pmatrix{ 1 & 0 & & & \\ 1 & -2 & 1 & & \\ & \ddots & \ddots & \ddots & \\ & & 1 & -2 & 1 \\ & & & 0 & 1 \\ } + \pmatrix{ V_1 & & & & \\ & V_2 & & & \\ & & \ddots & & \\ & & & V_{N-1} & \\ & & & & V_{N} \\ } \]

Matrix form (dimensionless)

\[ \hat{H} = -\frac{1}{\pi^2} \frac{d^2}{dx^2} + V(x) \quad \longrightarrow \quad -\frac{1}{\pi^2} \frac{f_{i+1} + f_{i-1} - 2 f_i}{\delta x^2} \,+\, \nu_L \left(\frac{i}{N}-\frac{1}{2}\right) \]

• The Hamiltonian can be written as a matrix operator (\(E\), \(\nu_L\) in units of \(E_1^\infty\))

\[ -\frac{1}{\pi^2\,\delta x^2} \pmatrix{ 1 & 0 & & & \\ 1 & -2 & 1 & & \\ & \ddots & \ddots & \ddots & \\ & & 1 & -2 & 1 \\ & & & 0 & 1 \\ } + \frac{\nu_L}{2N} \pmatrix{ -N & & & & \\ & -N+2 & & & \\ & & \ddots & & \\ & & & N-2 & \\ & & & & N \\ } \]

\(\longrightarrow\) Solve the eigenvalue equation: \(\,\,\hat{H} |\psi_n\rangle = E_n |\psi_n\)

Solving Large matrix equations

import numpy as np
from scipy.sparse import diags_array
from scipy.sparse.linalg import eigsh, LaplacianNd

# Parameters
n = 500; L = 1.0; dx = L/(n-1)
x = np.linspace(0, L, n)
V = 5 * (x - L/2)

# Calculate E_n, Psi_n
lap = LaplacianNd(
  grid_shape=(n, ), boundary_conditions='dirichlet'
).tosparse().astype(np.float64)
Vmat = diags_array([V], offsets=[0]).toarray()
E_n, Psi_n = eigsh(-lap/np.pi**2/dx**2 + Vmat, k=3, which="SM")
print("Eigenvalues E1, E2, E3:\n\t\t" + str(np.round(E_n, 3))) 

Eigenvalues E1, E2, E3:
        [0.729 4.038 8.976]

Infinite well with electric field

\[ \hat{H} = -\frac{1}{\pi^2}\frac{d^2}{dz^2} + \nu_L \, (z - 1/2) \quad \textrm{ with } \quad \hat{H}_0 = -\frac{1}{\pi^2}\frac{d^2}{dz^2} \]

• Choosing initial basis functions not necessary
• Accurate potential \(V(x)\) (\(\delta x\) dependent)
• BAD eigenenergies accuracy? Wave function accuracy GOOD?

Wave function comparison

Perturbation theory

Steps to reach to the solutions:

1. \(\hat{H} = \hat{H}_0 + \gamma\hat{H}_p\) Assume the perturbation to be small
2. Expand both the eigenenergies & eigenstates into power series in \(\gamma\)
3. Recursive relations for eigenenergies & eigenstates

Example of standard perturbation

• Putting on a “small” electric field
• Approximating the effect of \(V(x) = e \tilde{E} (x - L/2) \equiv \varepsilon z\) by power series:

\[ E_m = E^{(0)}_m + \alpha_1 \varepsilon + \alpha_2 \varepsilon^2 + \dots \]

Perturbation part of Hamiltonian

• Independent of the actual perturbation form
• Known solutions for the unperturbed part \(\hat{H}_0\)

\[ \hat{H}_0 |\psi_n\rangle = E_n |\psi_n\rangle \]

• Perturbation of \(\hat{H}_0\) by “small” perturbing part \(\hat{H}_p\):

\[ \hat{H} = \hat{H}_0 + \gamma\hat{H}_p \]

Power series expansion

\[ \hat{H}|\phi\rangle = (\hat{H}_0 + \gamma\hat{H}_p) |\phi\rangle = E |\phi\rangle \]

Expand \(E\) and \(|\phi\rangle\) as power series in \(\gamma\):

\[ E = E^{(0)} + \gamma E^{(1)} + \gamma^2 E^{(2)} + \gamma^3 E^{(3)} + \gamma^4 E^{(4)} + \dots \]

\[ |\phi\rangle = |\phi^{(0)}\rangle + \gamma |\phi^{(1)}\rangle + \gamma^2 |\phi^{(2)}\rangle + \gamma^3 |\phi^{(3)}\rangle + \gamma^4 |\phi^{(4)}\rangle + \dots \]

Zeroth order perturbation

\[ \begin{aligned} &(\hat{H}_0 + \gamma\hat{H}_p) \left( |\phi^{(0)}\rangle + \gamma |\phi^{(1)}\rangle + \gamma^2 |\phi^{(2)}\rangle + \dots \right) = \\ & \qquad\left(E^{(0)} + \gamma E^{(1)} + \gamma^2 E^{(2)} + \dots \right) \left( |\phi^{(0)}\rangle + \gamma |\phi^{(1)}\rangle + \gamma^2 |\phi^{(2)}\rangle + \dots \right) \end{aligned} \]

Matching orders

\[ \begin{aligned} &(\hat{H}_0 + \gamma\hat{H}_p) \left( |\psi_m \rangle + \gamma |\phi^{(1)} \rangle + \gamma^2 |\phi^{(2)}\rangle + \dots \right) = \\ & \qquad\left(E_m + \gamma E^{(1)} + \gamma^2 E^{(2)} + \dots \right) \left( |\psi_m \rangle + \gamma |\phi^{(1)} \rangle + \gamma^2 |\phi^{(2)}\rangle + \dots \right) \end{aligned} \]

Matching orders in \(\gamma\)

\[ \begin{aligned} \hat{H}_0 |\psi_m \rangle &= E_m |\psi_m \rangle\\ &\\ \hat{H}_0 |\phi^{(1)}\rangle + \hat{H}_p|\psi_m \rangle &= E_m |\phi^{(1)}\rangle + E^{(1)} |\psi_m \rangle\\ &\\ \hat{H}_0 |\phi^{(2)}\rangle + \hat{H}_p|\phi^{(1)}\rangle &= E_m |\phi^{(2)}\rangle + E^{(1)} |\phi^{(1)}\rangle + E^{(2)} |\psi_m\rangle\\ \end{aligned} \]

Matching orders CTU’d

\[ \begin{aligned} \hat{H}_0 |\psi_m \rangle &= E_m |\psi_m \rangle\\ \hat{H}_0 |\phi^{(1)}\rangle + \hat{H}_p|\psi_m \rangle &= E_m |\phi^{(1)}\rangle + E^{(1)} |\psi_m \rangle\\ \hat{H}_0 |\phi^{(2)}\rangle + \hat{H}_p|\phi^{(1)}\rangle &= E_m |\phi^{(2)}\rangle + E^{(1)} |\phi^{(1)}\rangle + E^{(2)} |\psi_m\rangle\\ \end{aligned} \]

Rewrite highest order state to the left-hand-side:

\[ \begin{aligned} (\hat{H}_0 - E_m) |\psi_m \rangle &= 0\\ (\hat{H}_0 - E_m)|\phi^{(1)}\rangle &= (E^{(1)} - \hat{H}_p)|\psi_m \rangle\\ (\hat{H}_0 - E_m)|\phi^{(2)}\rangle &= (E^{(1)} - \hat{H}_p) |\phi^{(1)}\rangle + E^{(2)} |\psi_m\rangle\\ \end{aligned} \]

First order perturbation theory

\[ \begin{aligned} (\hat{H}_0 - E_m) |\psi_m \rangle &= 0\\ (\hat{H}_0 - E_m)|\phi^{(1)}\rangle &= (E^{(1)} - \hat{H}_p)|\psi_m \rangle\\ (\hat{H}_0 - E_m)|\phi^{(2)}\rangle &= (E^{(1)} - \hat{H}_p) |\phi^{(1)}\rangle + E^{(2)} |\psi_m\rangle\\ \end{aligned} \]

Left-multiply by the bra \(\langle \psi_m |\)

\[ \begin{aligned} \textrm{left-hand-side:}\quad \langle \psi_m |(\hat{H}_0 - E_m)|\phi^{(1)}\rangle &= \langle \psi_m |(E_m - E_m)|\phi^{(1)}\rangle = 0\\ \textrm{right-hand-side:}\quad \langle \psi_m |(E^{(1)} - \hat{H}_p)|\psi_m \rangle &= E^{(1)} - \langle\psi_m | \hat{H}_p|\psi_m\rangle\\ &\\ \Longrightarrow E^{(1)} = \langle\psi_m | \hat{H}_p|\psi_m\rangle \end{aligned} \]

• First order correction to the eigenenergy: \(E_m + E^{(1)}\)

First order eigenstate

• Expand first order correction eigenstate: \(|\phi^{(1)}\rangle = \sum_n a_n^{(1)} \, | \psi_n \rangle\)

• Filling in:

\[ \begin{aligned} (\hat{H}_0 - E_m) |\phi^{(1)}\rangle &= (E^{(1)} - \hat{H}_p)|\psi_m \rangle\\ \end{aligned} \]

\[ \begin{aligned} \textrm{left-hand-side:}\quad \langle \psi_i |(\hat{H}_0 - E_m)|\phi^{(1)}\rangle &= (E_i - E_m)\langle \psi_i | \phi^{(1)}\rangle = (E_i - E_m) a_i^{(1)}\\ \textrm{right-hand-side:}\quad \langle \psi_i |(E^{(1)} - \hat{H}_p)|\psi_m \rangle &= E^{(1)} \langle\psi_i | \psi_m \rangle - \langle \psi_i | \hat{H}_p |\psi_m \rangle\\ &\\ \Longrightarrow \qquad \,\,\,\,\, a_i^{(1)} &= \frac{\langle \psi_i | \hat{H}_p |\psi_m \rangle}{E_m - E_i}\\ &\\ \Longrightarrow \qquad |\phi^{(1)}\rangle &= \sum_{n \neq m} \frac{\langle \psi_n | \hat{H}_p |\psi_m \rangle}{E_m - E_n} |\psi_n \rangle \end{aligned} \]

Electric field as perturbation

Up to first order we have:

\[ \begin{aligned} |\psi_m \rangle \longrightarrow |\psi_m \rangle + |\phi^{(1)}_m\rangle &= |\psi_m \rangle + \sum_{n \neq m} \frac{\langle \psi_n | \hat{H}_p |\psi_m \rangle}{E_m - E_n} |\psi_n \rangle\\ E_m \longrightarrow \quad\,\,\, E_m + E^{(1)} &= E_m + \langle\psi_m | \hat{H}_p|\psi_m\rangle \end{aligned} \]

The Hamiltonian in dimensionless units:

\[ \hat{H} = -\frac{1}{\pi^2}\frac{d^2}{dz^2} + \nu_L \, (z - 1/2) \quad \textrm{ with } \quad \hat{H}_0 = -\frac{1}{\pi^2}\frac{d^2}{dz^2} \]

• The Hamiltonian gives contributions of eigenstates \(\,\,\psi_n(z) = \sqrt{2}\sin(n\pi z)\)
• Eigenenergies \(\,E_n\) and potential \(\,\nu_L\) are in units of \(\,\,E_1^\infty = \frac{\hbar^2\pi^2}{2mL^2}\)

First order correction

The correction in the energies is zero (no change):

\[ \begin{aligned} E_m + E^{(1)}_m &= E_m + \langle\psi_m | \hat{H}_p|\psi_m\rangle\\ &= m^2 \,\, + \,\,\nu_L \int_0^1 \, (z-1/2) \sin^2(m\pi z) dz,\\ &= m^2 + 0 = m^2 \end{aligned} \]

\[ \begin{aligned} |\psi_m \rangle + |\phi^{(1)}_m \rangle &= |\psi_m \rangle + \sum_{n \neq m} \frac{\langle \psi_n | \hat{H}_p |\psi_m \rangle}{E_m - E_n} |\psi_n \rangle\\ &= |\psi_m \rangle \,\, - \sum_{n \neq m} \frac{2\nu_L}{m^2 - n^2} \,\, \int_0^1 \, (z-1/2) \sin(m\pi z) \sin(n\pi z) dz \,\, |\psi_n \rangle\\ &= |\psi_m \rangle \,\, - \sum_{n = m \pm 1, \pm 3, \dots} \frac{2\nu_L}{m^2 - n^2} \,\, \frac{4nm}{\pi^2(m^2-n^2)^2} \,\, |\psi_n \rangle\\ \end{aligned} \]

First order correction CTU’d

Where the second integral was obtained from:

\[ \int_0^1 \, (z-1/2) \sin(m\pi z) \sin(n\pi z) dz = \left\{\,\,\begin{aligned} \frac{4nm}{\pi^2(m^2-n^2)^2} & \qquad \textrm{if }\,\, m+n \,\,\textrm{ is odd} \\ & \\ 0 & \qquad \textrm{if } \,\,m+n\,\, \textrm{ is even} \end{aligned}\right. \]

So we have for eigenenergies and eigenstates:

\[ \begin{aligned} E_m + E^{(1)}_m &= m^2\\ | \psi_m \rangle + | \psi^{(1)}_m \rangle &= \sqrt{2}\sin(m\pi z) \,\, + \sum_{n = m \pm 1, \pm 3, \dots} \frac{8nm \nu_L\sqrt{2}}{\pi^2(m^2-n^2)^3} \,\, \sin(n\pi z)\\ \end{aligned} \]

Wave function comparison

For 2nd order perturbation theory: see Chapter 6 of David Miller’s book