Specific Solutions of the Time-Independent Schrödinger Equation #

1. 1D TISE #

$$ \begin{align} \left[-\frac{1}{2}\frac{d^2}{dx^2}+V(x)\right]\psi(x)=E\psi(x) \end{align} $$

You can experiment with these solutions using the following web tool: 1D Time-Independent Schrödinger Equation Solver via Finite Difference Method.

1.1. Infinite Square Well Potential #

• Potential

$$ \begin{eqnarray} V(x)=\left\{ \begin{aligned} &0 &&( 0\lt x \lt L)\\ &\infty &&( x\le 0\hspace{0.5em} \text{or}\hspace{0.5em} L\le x) \end{aligned} \right. \end{eqnarray} $$

• Eigenvalues

$$ \begin{align} E_n=\frac{(n+1)^2\pi^2}{2 L^2},\hspace{2em}(n=0,1,2,\cdots) \end{align} $$

• Eigenfunctions

$$ \begin{eqnarray} \psi_n(x)=\left\{ \begin{aligned} &\sqrt{\frac{2}{L}}\sin\left[\frac{(n+1)\pi}{L}\left(x+\frac{L}{2}\right)\right] && (0\lt x \lt L)\\ & 0 && ( x\le 0 \hspace{0.5em}\text{or}\hspace{0.5em} L\le x) \end{aligned} \right. \end{eqnarray} $$

• Specific Energy Eigenvalues

• Solution Method

  For the interval $0 < x < L$, the Schrödinger equation with $V(x) = 0$ has solutions in the form of sine and cosine functions. Considering the boundary conditions $\psi(x = 0) = 0$ and $\psi(x = L) = 0$, the wavefunctions must take the form of sine functions with zeroes at $x = 0$ and $x = L$.

1.2. Harmonic Potential #

• Potential

$$ \begin{align} V(x)=Ax^2 \end{align} $$

• Eigenvalues

$$ \begin{align} E_n=\sqrt{2A}\left(n+\frac{1}{2}\right),~~(n=0,1,2,\cdots) \end{align} $$

• Eigenfunctions

$$ \begin{gather} \psi_n(x)=\frac{1}{\sqrt{\alpha}}\frac{\pi^{-1/4}}{\sqrt{2^n n!}}H_n(x/\alpha)\exp\left(-\sqrt{\frac{A}{2}} x^2\right)\\ \alpha=\left(2A\right)^{-1/4} \end{gather} $$

Here, $H_n(x)$ represents Hermite polynomials, with specific values as follows:

$$ \begin{align} H_0(z)&= 1 \nonumber\\ H_1(z)&= 2z \nonumber\\ H_2(z)&= 4z^2-2 \nonumber\\ H_3(z)&= 8z^3-12z \nonumber\\ H_4(z)&= 16z^4-48z^2+12 \nonumber\\ \vdots \nonumber \end{align} $$

For $A=1/2$:

$$ \begin{align} \psi_0(x)&=\pi^{-1/4}\exp(-x^2/2) \\ \psi_1(x)&=\pi^{-1/4}\sqrt{2}x\exp(-x^2/2) \\ \psi_2(x)&=\pi^{-1/4}\frac{1}{\sqrt{2}}(2x^2-1)\exp(-x^2/2) \\ \psi_3(x)&=\pi^{-1/4}\frac{1}{\sqrt{3}}(2x^3-3x)\exp(-x^2/2) \\ \psi_4(x)&=\pi^{-1/4}\frac{1}{2\sqrt{6}}(4x^4-12x^2+3)\exp(-x^2/2) \end{align} $$

• Specific Energy Eigenvalues

  Quantum Number $n$	Analytical Solution
  0	0.5
  1	1.5
  2	2.5
  3	3.5
  4	4.5
  …
  10	10.5
  20	20.5
  30	30.5
  40	40.5
  50	50.5

• Solution Method

  Solutions can be derived using creation and annihilation operators or by solving the differential equation via asymptotic behavior at $x \to \pm\infty$. These methods are well-known and not detailed here.

1.3. Triangular Quantum Well #

• Potential

$$ \begin{eqnarray} V(x)=\left\{ \begin{aligned} &\infty && (x\lt 0) \\ &\alpha x && (0 \lt x) \end{aligned} \right. \end{eqnarray} $$

• Eigenvalues

$$ \begin{align} E_n=-\left(\frac{\alpha^2}{2}\right)^{1/3}a_n \end{align} $$

Here, $a_n$ denotes the zeros of the Airy function¹.

• Eigenfunctions

$$ \begin{align} \psi_n(x)=C\cdot \text{Ai}\left(\left(\frac{2}{\alpha^2}\right)^{1/3}(\alpha x-E_n)\right) \end{align} $$

Here, $\text{Ai}(x)$ is the Airy function¹,².

• Specific Energy Eigenvalues

  For $\alpha = 1$, some specific energy eigenvalues are:

  Quantum Number $n$	Analytical Solution
  0	1.8557570814892386
  1	3.2446076240031596
  2	4.3816712392861303
  3	5.3866137807905003
  4	6.3052630065857747
  …
  10	10.8669420487522821
  20	16.8461586901918032
  30	21.8969179157570224
  40	26.4182304793452509
  50	30.5803354172938597

• Solution Method

  For the interval $0 < x < \infty$, the Schrödinger equation with $V(x) = \alpha x$ has solutions in the form of Airy functions. The boundary conditions $\psi(x = 0) = 0$ and $\psi(x \to \infty) = 0$ select the solution as the Airy function $\text{Ai}(x)$, which decays asymptotically. The solution is chosen so that its zero coincides with $x = 0$.

1.4. F. Morse Potential³ #

• Potential

$$ \begin{align} V(x)=A\left(e^{-2\alpha x}-2e^{-\alpha x}\right) \end{align} $$

• Eigenvalues

$$ \begin{align} E_n=-A\left[1-\frac{\alpha}{\sqrt{2A}}\left(n+\frac{1}{2}\right)\right]^2 \end{align} $$

Here, $n$ is a positive integer starting from zero and ranges up to the maximum value $n_\text{max}$, satisfying:

$$ \begin{align} \frac{\sqrt{2A}}{\alpha}\gt n+\frac{1}{2} \end{align} $$

• Eigenfunctions

$$ \begin{align} \psi(x)=e^{-\xi/2} \xi^s w(\xi) \end{align} $$

Here, $w(\xi)$, $\xi$, and $s$ are defined as follows:

$$ \begin{gather} w(\xi)=M(-n,2s+1,\xi) \\ \xi=\frac{2\sqrt{2A}}{\alpha}e^{-\alpha x} \\ s=\frac{\sqrt{-2E}}{\alpha} \end{gather} $$

where $M(a, b, x)$ is the confluent hypergeometric function⁴.

This problem has a finite number of discrete spectrum levels. If:

$$ \begin{align} \frac{\sqrt{2A}}{\alpha}\gt n+\frac{1}{2} \end{align} $$

then the discrete spectrum generally does not exist.

• Specific Energy Eigenvalues

  For $A = 5$ and $\alpha = 1$, there are only three solutions, with the following eigenvalues:

  Quantum Number $n$	Analytical Solution
  0	-3.5438611699158100
  1	-1.3815835097474309
  2	-0.2193058495790518
  3	-

• Solution Method

See reference ³.

1.5. $1/(\cosh^2)$ Potential #

• Potential

$$ \begin{align} V(x)=-\frac{V_0}{\cosh^2 \alpha x} \end{align} $$

• Eigenvalues

$$ \begin{align} E_n=-\frac{\alpha^2}{8}\left[-(1+2n)+\sqrt{1+\frac{8V_0}{\alpha^2}}\right]^2 \end{align} $$

where $n$ is determined by the condition:

$$ \begin{align} n\lt s = \frac{1}{2}\left(-1+\sqrt{1+\frac{8V_0}{\alpha^2}}\right) \end{align} $$

resulting in a finite number of discrete levels.

• Eigenfunctions

$$ \begin{gather} \psi(x)=(1-\xi^2)^{\xi/2}F\left(\varepsilon-s, ~\varepsilon+s+1; ~\varepsilon+1; ~\frac{1-\xi}{2}\right)\\ \varepsilon=\frac{\sqrt{-2E}}{\alpha},\hspace{1em}\xi=\tanh \alpha x \end{gather} $$

where $F(a, b; c; x)$ is the hypergeometric function⁵.

• Specific Energy Eigenvalues

  Quantum Number $n$	Analytical Solution
  0	-3.885009803959261
  1	-1.9750294118777854
  2	-0.7050490197963089
  3	-0.0750686277148326
  4	–

• Solution Method

  See reference ³.

1.6. Coulomb Potential (Radial Direction) #

• Potential

$$ \begin{align} V(x)=\frac{l(l+1)}{2x^2}-\frac{Z}{x} \end{align} $$

$l=0,1,\cdots,\hspace{2em} Z>0$

• Eigenvalues

$$ \begin{align} E_n=-\frac{1}{2n^2},\hspace{2em}(n=1,2,\cdots,) \end{align} $$

• Eigenfunctions

$$ \begin{align} \psi(x)=x\cdot \left(\frac{2}{n}\right)^{3/2}\sqrt{\frac{n-l-1}{2n[(n+l)!]}}\cdot e^{-Zx/n}\cdot \left(\frac{2Zx}{n}\right)^l \cdot L_{n-l-1}^{(2l+1)}\left(\frac{2Zx}{n}\right) \end{align} $$

$$ \begin{align} \psi(x=0)=0,\hspace{1em} \psi(x\to\infty)=0 \end{align} $$

$$ \begin{align} \int_{0}^{\infty} |\psi(x)|^2 dx = 1 \end{align} $$

Here, $C$ is the normalization constant, and $L_n^{(k)}(x)$ is the associated Laguerre polynomial⁶, satisfying:

$$ \begin{align} L_0^{(k)}(x)&=1 \nonumber\\ L_1^{(k)}(x)&=-x+k+1 \nonumber\\ L_2^{(k)}(x)&=\frac{1}{2}[x^2-2(k+2)x+(k+1)(k+2)] \nonumber\\ L_3^{(k)}(x)&=\frac{1}{6}[-x^3+3(k+3)x^2-3(k+2)(k+3)x+(k+1)(k+2)(k+3)] \nonumber\\ \vdots\nonumber \end{align} $$

を満たします。

• Solution Method

  Consider the asymptotic behavior as $x \to 0$ and $x \to \infty$, and assume the true solution takes the form:

$$ \begin{align} \text{(solution) $=$ (asymptotics at $x\to 0$) $\times$ (asymptotics at $x\to\infty$) $\times$ (the unknown function)} \end{align} $$

Substituting this into the Schrödinger equation reveals that the unknown function is the associated Laguerre polynomial.

2. 2D TISE #

References #