I have to solve analitically the Schrodinger equation in one-dimension with a barrier of potential (tunnel effect):

$$ih \frac{d}{dt} U(x,t) = \left[ \left(-\hbar^2 \frac{d^2}{dx^2} \right) + q V(x) \right] U(x,t)$$

where: $i$ is the imaginary unit, ($d/dt$) is the time derivative, $\hbar$ is the Reduced Plank constant, ($d^2/dx^2$) is the second derivative in space, $V(x)$ is an external potential function of $x$, $U(x,t)$ is the wave function of time and place. The barrier of potential is:

$$V(x) = \begin{cases} 0, & \mbox{if } -d<x<-L \\ V_0, & \mbox{if } -L<x<L \\ 0, & \mbox{if } L<x<d \end{cases} $$

with $d=10L$ and $V_0>0$; Also the boundary condition are: $U(-d,t)=U(d,t)=0$; and the initial condition is

$$U(x,t_{0})=\frac{1}{\sqrt{Dx}} \exp \left(i P_0 \frac{x}{\hbar} \right)$$

if $-d<x<-d+Dx$ and $U(x,t_{0})=0$ if $-d+Dx<x<d$; where $Dx<<L$ and $P_0$ is the quantum moment at t0. Also I know that at time $t_0$, the Fourier Transform of $U(x,t_{0})$ is a sinc centered in $P_0$, the aspectated value of position is $-d+Dx/2$ and the aspectated value of velocity is $P_0/m$, where $m$ is the mass of the particle.

Then I have to compare analytical results with results from FINITE ELEMENTS and FINITE DIFFERENCE method.

I hope that someone can help me to solve this problem.

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    $\begingroup$ Welcome luca82 to physics.SE! At the moment there is not really a question in your text. Maybe you could elaborate a bit what you already tried and where you got stuck. Nobody here will solve homework/assignment problems but many people are more than willing to help. $\endgroup$ – Alexander Feb 21 '12 at 20:35
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    $\begingroup$ This appears to be copied largely from your assignment, which brings us to some questions for you. What have you done? What do you understand here, and what do you not understand? Where exactly are you stuck? In particular, I see at least three ways of attempting to solve the problem listed---surely you have been able to make some progress with some of them. $\endgroup$ – dmckee Feb 21 '12 at 21:13
  • $\begingroup$ try \hbar: $\hbar$ $\endgroup$ – Emilio Pisanty Apr 22 '12 at 8:04
  • $\begingroup$ You should be careful: the solutions will be very different depending on whether $P_0^2$ is greater than $qV_0$ or not. $\endgroup$ – Emilio Pisanty Apr 22 '12 at 8:08

Without complete solution, just a road map. In principle, there is a standard way how this kind of problems is solved.

First solve stationary problem: $$ \left[ \left(-h^2 \frac{d^2}{dx^2} \right) + q V(x) \right] U_i(x) = E_i U(x).$$ As long as the potential is symmetric, I would recommend use this and search for even and odd solutions. For $E<V_0$ $$U_{i+}(x) = \begin{cases} A\cos{k(|x|-x_0)}, & \mbox{if } L<|x|<d \\ B\mathrm{ch}{\varkappa x}, & \mbox{if } -L<x<L \\ \end{cases}, $$ $$ U_{i-}(x) = \begin{cases} A\sin{k(|x|-x_0)}, & \mbox{if } L<|x|<d \\ B\mathrm{sh}{\varkappa x}, & \mbox{if } -L<x<L \\ \end{cases} $$ with $k=\sqrt{\frac{2mE}{\hbar^2}}$, $\varkappa = \sqrt{\frac{2m(V_0-E)}{\hbar^2}}$

For energies above $V_0$ replace hyperbolic functions with $\cos{k_2x}$, $\sin{k_2x}$ with $k=\sqrt{\frac{2m(E-V_0)}{\hbar^2}}$.

If you substitute these functions into equation, you'll get equation for energy. Note that as long as this form of functions automatically satisfies equation in the regions of constant potential, you should just take care of boundary conditions. Solve this equation and get a spectrum $\left\{ E_{i\pm} \right\}_{i=1}^{\infty}$ and wavefunctions $U_i(x)$. Obviously, these solutions have very nice feature when you consider non-stationary problem: $$ U_i(x,t)=e^{-i\frac{\hbar}{E_i} t}U_i(x), $$ which means that if you decompose your initial condition into these solutions (which you may do because eigenfunctions of a Hermitian operator is a complete set) $$ U(x,t_0) = \sum_1^{\infty} C_i U_i(x) $$ then time evolution of wavefunction is given by $$ U(x,t) = \sum_1^{\infty} C_i U_i(x) e^{-i\frac{\hbar}{E_i} t} $$

  • $\begingroup$ If one needs an analytic solution to the problem, won't the lack of elemental expressions for the energies (which will be roots of a transcendental equation, I should think) severely harm this approach? Without any hope of evaluating the sum, this is arguably a semi-numerical, spectral method to solve the initial-value problem $\endgroup$ – Emilio Pisanty Apr 22 '12 at 8:10
  • $\begingroup$ @episanty No, it will not. There is an enormous difference between solving stranscendent equation and finding solutions of differential equation numerically. The first is called analytic, the second is not. Anyway, there is no better solution. $\endgroup$ – Misha Apr 22 '12 at 18:40
  • $\begingroup$ Yes, it's clear that there isn't a better solution, and that the finite-element and finite-difference methods are on a different scale altogether. My point was that I'm reluctant to call this a fully analytic solution since any attempt at visualising it (and comparing with any other method) will involve numerical work (on this side). $\endgroup$ – Emilio Pisanty Apr 22 '12 at 22:19
  • $\begingroup$ Sure, one can write down an explicit (but not elemental!) formula for the solution in a way no existence theorem can give. However, I see strong parallels between this solution and numerical spectral methods - this one simply has a particularly well chosen spectral basis set, and propagating any basis function is equivalent (at least in any reasonable numerical sense) to solving the transcendental equation. $\endgroup$ – Emilio Pisanty Apr 22 '12 at 22:19
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    $\begingroup$ @episanty, there is some misconception in your point of view. You could bring the same arguments if the solution was logriphm or bessel functions or any other non-trivial function: you should had use a computer to get the numbers in that case too. Just call the solution of that transcendent equation a new function, say, $Q_n(e)$ and it will not be much differnent from Bessel or Hankel functions which are also hard to find in calculator. $\endgroup$ – Misha Apr 23 '12 at 7:17

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