24 Schrodinger PDE

 24.1 Schrodinger PDE with zero potential (Logan p. 30)
 24.2 Schrodinger PDE with initial and boundary conditions. Zero potential
 24.3 Initial value problem with Dirichlet boundary conditions. Zero potential
 24.4 Solve a Schrodinger equation with potential over the whole real line
 24.5 Schrodinger equation, with initial conditions. Zero potential (Griffiths p. 47)
 24.6 Schrodinger equation, with initial conditions. Infinite square well potential (Griffiths p. 47)
 24.7 In 2 space dimensions

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24.1 Schrodinger PDE with zero potential (Logan p. 30)

problem number 152

From page 30, David J Logan textbook, applied PDE textbook.

Solve \[ I h \frac{\partial f}{\partial t} = - \frac{h^2}{2 m} \frac{\partial ^2 f}{\partial x^2} \]

With boundary conditions

\begin{align*} f(0,t) &= 0\\ f(L,0) &=0 \end{align*}

Mathematica

\[ \left \{\left \{f(x,t)\to \underset{n=1}{\overset{\infty }{\sum }}e^{-\frac{i h n^2 \pi ^2 t}{2 L^2 m}} c_n \sin \left (\frac{n \pi x}{L}\right )\right \}\right \} \]

Maple

\[ f \left ( x,t \right ) =\sum _{n=1}^{\infty }{\it \_C1} \left ( n \right ) \sin \left ({\frac{n\pi \,x}{L}} \right ){{\rm e}^{{\frac{-i/2h{\pi }^{2}{n}^{2}t}{m{L}^{2}}}}} \]

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24.2 Schrodinger PDE with initial and boundary conditions. Zero potential

problem number 153

Solve for \(f(x,y,t)\) \[ I \frac{\partial f}{\partial t} = - \frac{\hbar ^2}{2 m} \left ( \frac{\partial ^2 f}{\partial x^2} + \frac{\partial ^2 f}{\partial y^2} \right ) \]

With boundary conditions

\begin{align*} f(0,y,t) &= 0\\ f(1,y,t) &=0 \\ f(x,1,t) &=0 \\ f(x,0,t) &=0 \end{align*}

And initial conditions \(f(x,y,0)=\sqrt{2} \left ( \sin (2\pi x) \sin (\pi y) + \sin (\pi x) \sin (2 \pi y) \right )\)

Mathematica

\[ \left \{\left \{f(x,y,t)\to \sqrt{2} e^{-\frac{5 i \pi ^2 \text{hBar}^2 t}{2 m}} (\sin (2 \pi x) \sin (\pi y)+\sin (\pi x) \sin (2 \pi y))\right \}\right \} \]

Maple

\[ f \left ( x,y,t \right ) =\sin \left ( \pi \,x \right ) \sqrt{2}{{\rm e}^{{\frac{-5/2\,i{{\it hBar}}^{2}t{\pi }^{2}}{m}}}} \left ( 2\,\sin \left ( y\pi \right ) \cos \left ( \pi \,x \right ) +\sin \left ( 2\,y\pi \right ) \right ) \]

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24.3 Initial value problem with Dirichlet boundary conditions. Zero potential

problem number 154

Taken from Mathematica DSolve help pages

Solve for \(f(x,t)\) \[ I \frac{\partial f}{\partial t} = - 2 \frac{\partial ^2 f}{\partial x^2} \]

With boundary conditions

\begin{align*} f(5,t) &= 0\\ f(10,t) &=0 \\ \end{align*}

And initial conditions \(f(x,2)=f(x)\) where \(f(x)=-350 + 155 x - 22 x^2 + x^3\)

Mathematica

\[ \left \{\left \{f(x,t)\to \underset{n=1}{\overset{\infty }{\sum }}\frac{100 \left (7+8 (-1)^n\right ) e^{-\frac{2}{25} i n^2 \pi ^2 (t-2)} \sin \left (\frac{1}{5} n \pi (x-5)\right )}{n^3 \pi ^3}\right \}\right \} \]

Maple

\[ f \left ( x,t \right ) =\sum _{n=1}^{\infty }{\frac{ \left ( 800+700\, \left ( -1 \right ) ^{n} \right ) \sin \left ( 1/5\,n\pi \,x \right ){{\rm e}^{-{\frac{2}{25}}\,i{\pi }^{2}{n}^{2} \left ( t-2 \right ) }}}{{n}^{3}{\pi }^{3}}} \]

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24.4 Solve a Schrodinger equation with potential over the whole real line

problem number 155

Taken from Mathematica DSolve help pages

Solve for \(f(x,t)\) \[ I \frac{\partial f}{\partial t} = - \frac{\partial ^2 f}{\partial x^2} + 2 x^2 f(x,t) \]

With boundary conditions

\begin{align*} f(-\infty ,t) &= 0\\ f(\infty ,t) &=0 \end{align*}

Mathematica

\[ \left \{\left \{f(x,t)\to \underset{n=0}{\overset{\infty }{\sum }}e^{-\frac{x^2}{\sqrt{2}}-2 i \sqrt{2} \left (n+\frac{1}{2}\right ) t} c_n \text{HermiteH}\left (n,\sqrt [4]{2} x\right )\right \}\right \} \]

Maple

\[ \text{ sol=() } \] Maple does not support \(\infty \) in boundary conditions

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24.5 Schrodinger equation, with initial conditions. Zero potential (Griffiths p. 47)

problem number 156

Taken from Introduction to Quantum mechanics, second edition, by David Griffiths, page 47.

Solve for \(f(x,t)\) \[ I h \frac{\partial f}{\partial t} = - \frac{h^2}{2 m} \frac{\partial ^2 f}{\partial x^2} \]

With initial conditions \(f(x,0) = A x (a-x)\) for \(0\leq x \leq a\) and zero otherwise.

Mathematica

\[ \left \{\left \{f(x,t)\to \frac{\left (\frac{1}{2}+\frac{i}{2}\right ) (-1)^{3/4} A \sqrt{h} \sqrt{t} \left (-i \sqrt{\pi } m x^2 \text{Erfi}\left (\frac{\left (\frac{1}{2}+\frac{i}{2}\right ) \sqrt{m} (a-x)}{\sqrt{h} \sqrt{t}}\right )+i \sqrt{\pi } a m x \text{Erfi}\left (\frac{\left (\frac{1}{2}+\frac{i}{2}\right ) \sqrt{m} (a-x)}{\sqrt{h} \sqrt{t}}\right )+i \sqrt{\pi } a m x \text{Erfi}\left (\frac{\left (\frac{1}{2}+\frac{i}{2}\right ) \sqrt{m} x}{\sqrt{h} \sqrt{t}}\right )+\sqrt{\pi } h t \text{Erfi}\left (\frac{\left (\frac{1}{2}+\frac{i}{2}\right ) \sqrt{m} (a-x)}{\sqrt{h} \sqrt{t}}\right )-(1+i) a \sqrt{h} \sqrt{m} \sqrt{t} e^{\frac{i m x^2}{2 h t}}-(1+i) \sqrt{h} \sqrt{m} \sqrt{t} x e^{\frac{i m (a-x)^2}{2 h t}}-i \sqrt{\pi } m x^2 \text{Erfi}\left (\frac{\left (\frac{1}{2}+\frac{i}{2}\right ) \sqrt{m} x}{\sqrt{h} \sqrt{t}}\right )+\sqrt{\pi } h t \text{Erfi}\left (\frac{\left (\frac{1}{2}+\frac{i}{2}\right ) \sqrt{m} x}{\sqrt{h} \sqrt{t}}\right )+(1+i) \sqrt{h} \sqrt{m} \sqrt{t} x e^{\frac{i m x^2}{2 h t}}\right )}{\sqrt{2 \pi } m^{3/2} \sqrt{\frac{h t}{m}}}\right \}\right \} \]

Maple

\[ f \left ( x,t \right ) =-2\,A \left ({\frac{i/2}{\pi }\int _{-\infty }^{\infty }\!{\frac{1}{{s}^{3}}{{\rm e}^{{\frac{-i/2h{s}^{2}t}{m}}+isx}}}\,{\rm d}s}-{\frac{i/2}{\pi }\int _{-\infty }^{\infty }\!{\frac{1}{{s}^{3}}{{\rm e}^{{\frac{-i/2s \left ( hst+2\,am \right ) }{m}}+isx}}}\,{\rm d}s}+1/4\,{\frac{a}{\pi }\int _{-\infty }^{\infty }\!{\frac{1}{{s}^{2}}{{\rm e}^{{\frac{-i/2h{s}^{2}t}{m}}+isx}}}\,{\rm d}s}+1/4\,{\frac{a}{\pi }\int _{-\infty }^{\infty }\!{\frac{1}{{s}^{2}}{{\rm e}^{{\frac{-i/2s \left ( hst+2\,am \right ) }{m}}+isx}}}\,{\rm d}s} \right ) \]

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24.6 Schrodinger equation, with initial conditions. Infinite square well potential (Griffiths p. 47)

problem number 157

Taken from Introduction to Quantum mechanics, second edition, by David Griffiths, page 47. This is the same as the above problem but has an extra \(V(x) f(x,t)\) terms where \(V(x)\) is the infinite square well potential defined by \(V(x)=0\) if \(0\leq x \leq a\) and \(V(x)=\infty \) otherwise.

Solve for \(f(x,t)\) \[ I \hslash \frac{\partial f}{\partial t} = - \frac{\hslash ^2}{2 m} \frac{\partial ^2 f}{\partial x^2} + V(x) f(x,t) \]

With initial conditions \(f(x,0) = A x (a-x)\) for \(0\leq x \leq a\) and zero otherwise.

Mathematica

\[ \left \{\left \{f(x,y,t)\to \sqrt{2} e^{-\frac{5 i \pi ^2 h t}{m}} \left (\sin (\pi x) \sin (3 \pi y)+\sin (2 \pi x) \sin (\pi y) e^{\frac{5 i \pi ^2 h t}{2 m}}\right )\right \}\right \} \]

Maple

\[ f \left ( x,t \right ) =\cases{Ax \left ( a-x \right ) &$0\leq x$\ and \ $x\leq a$\cr 0&otherwise\cr }+\sum _{n=1}^{\infty }{\frac{{t}^{n} \left ( U\mapsto{\frac{-i\cases{0&$0\leq x$\ and \ $x\leq a$\cr \infty &otherwise\cr }U}{h}}^{ \left ( n \right ) } \right ) \left ( \cases{Ax \left ( a-x \right ) &$0\leq x$\ and \ $x\leq a$\cr 0&otherwise\cr } \right ) }{n!}} \]

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24.7 In 2 space dimensions

problem number 158

Added December 20, 2018.

Example 28, taken from https://www.mapleprimes.com/posts/209970-Exact-Solutions-For-PDE-And-Boundary--Initial-Conditions-2018

Solve for \(f(x,y,t)\) \[ I \hslash \frac{\partial f}{\partial t} = - \frac{\hslash ^2}{2 m} \left ( \frac{\partial ^2 f}{\partial x^2} + \frac{\partial ^2 f}{\partial y^2} \right ) \]

With initial conditions \(f(x,y,0) = \sqrt{2} \left ( \sin (2 \pi x) \sin (\pi y)+ \sin (\pi x) \sin (3 \pi y) \right )\) and boundary conditions \begin{align*} f(0,y,t) &= 0 \\ f(1,y,t) &= 0 \\ f(x,1,t) &= 0 \\ f(x,0,t) &= 0 \end{align*}

Mathematica

\[ \text{DSolve}\left [\left \{i h f^{(0,1)}(x,t)=f(x,t) \left (\begin{array}{cc} \{ & \begin{array}{cc} 0 & 0\leq x\leq a \\ \infty & \text{True} \\\end{array} \\\end{array}\right )-\frac{h^2 f^{(2,0)}(x,t)}{2 m},f(x,0)=\left (\begin{array}{cc} \{ & \begin{array}{cc} A (a-x) x & 0\leq x\leq a \\ 0 & \text{True} \\\end{array} \\\end{array}\right )\right \},f(x,t),\{x,t\},\text{Assumptions}\to a>0\right ] \]

Maple

\[ f \left ( x,y,t \right ) =\sin \left ( \pi \,x \right ) \sqrt{2} \left ( 2\,{{\rm e}^{{\frac{-5/2\,i{\it hbar}\,t{\pi }^{2}}{m}}}}\sin \left ( y\pi \right ) \cos \left ( \pi \,x \right ) +{{\rm e}^{{\frac{-5\,i{\it hbar}\,t{\pi }^{2}}{m}}}}\sin \left ( 3\,y\pi \right ) \right ) \]