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Assume a potential of the form \begin{eqnarray}V(x) &=& \frac{1}{2}m\omega^2x^2,~-x_0\leq x\leq x_0,\\&=& 0,\hspace{2.5cm}{\rm otherwise} \end{eqnarray} where $x_0$ is a finite positive number. Surely tunneling is possible in this case and the particle can escape to $x\to\pm\infty$. How to qualitatively draw the stationary states of this problem (of course, without solving it in detail). In particular, how different will the stationary state in this case be from those of the actual harmonic oscillator potential $V(x)=\frac{1}{2}m\omega^2x^2, ~\forall x$? If someone could draw the stationary states that will be helpful.

Can it be solved on a computer to obtain the stationary states?

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  • $\begingroup$ You might find these interesting: asc.ohio-state.edu/physics/ntg/H133/handouts/wavefunctions.pdf and falstad.com/qm1d $\endgroup$ – robphy Oct 18 '19 at 7:51
  • $\begingroup$ @robphy Thanks. they are useful $\endgroup$ – mithusengupta123 Oct 18 '19 at 17:26
  • $\begingroup$ Just to be clear, are you looking for a plot of $\psi\left(x\right)$ (or $\psi^2\left(x\right)$)? For various energies? FWIW, your situation is very close to that of a resonant tunneling diode (RTD), and you can find plots of $\psi\left(x\right)$ or $\psi^2\left(x\right)$ for that siutation pretty easily online. E.g. tinyurl.com/rltvsdy . If you want to see something similar for your problem, I'll try to cook up a plot this evening or tomorrow. $\endgroup$ – lnmaurer Dec 5 '19 at 16:29
  • $\begingroup$ Well, I am actually looking for $|\psi|^2(x)$. $\endgroup$ – mithusengupta123 Dec 5 '19 at 17:04
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How to qualitatively draw the stationary states of this problem (of course, without solving it in detail)

Start with the obvious facts:

  1. The potential is symmetric: $V(-x)=V(x)$.
  2. The potential is bounded and is a potential barrier of finite width.
  3. The solution for $|x|>x_0$ is a plane wave.

The first fact lets you classify the solutions in two kinds: even and odd ones. This reduces the problem on $x\in\mathbb R$ to two problems on $x\in[0,\infty)$. The even solutions will have homogeneous Neumann boundary condition at $x=0$, while the odd ones will have the Dirichlet boundary there.

The second fact lets you immediately say that there'll be continuous spectrum. Moreover, the third fact tells you that the problem is a free particle scattering problem, where an incident particle can have any energy, so there are no gaps in the spectrum.

From this follow several ways to proceed.

Purely qualitative approach

If you really don't want to solve the problem quantitatively, and only want to qualitatively sketch the eigenstates, without much regard to exact energies and other numeric details, then you can approximate the finite parabolic potential with a square barrier of some width:

suggested simplification of parabolic potential

If you are able to sketch eigenstates for this potential, you're done.

Quantitative approach

This is actually quite easy if you can program a solver for ODE Cauchy problems. A method even as simple as Euler's should be enough to get started here.

In particular, you use the insight given by the above mentioned obvious facts and start with a Cauchy problem for each parity of the solution you're looking for. For the odd solution you can impose e.g. the condition

$$\psi_{\text{odd}}(0)=0,$$ $$\left.\frac{d\psi_{\text{odd}}(x)}{dx}\right|_{x=0}=1,$$

and for the even solution you can impose

$$\psi_{\text{even}}(0)=1,$$ $$\left.\frac{d\psi_{\text{even}}(x)}{dx}\right|_{x=0}=0.$$

Solving the Cauchy problems with these initial conditions won't give you normalized (in any sense) solutions, but I guess you don't require them. If you do, then note that these states can't be normalized to probability of 1, since they correspond to unbounded motion, so a different normalization scheme is required.

After you do solve the Cauchy problems and extend the odd and even solutions calculated e.g. for $x>0$ to the space of $x<0$, you'll get something like the following solution for the case of $m=1$, $\omega=1$, $x_0=2$.

solutions for different energies

In particular, how different will the stationary state in this case be from those of the actual harmonic oscillator potential $V(x)=\frac{1}{2}m\omega^2x^2, ~\forall x$?

For this, I guess, the best answer would be to simply do the numerical calculation and plot the solutions side-by-side. It's very fortunate that one of the problems has continuous spectrum, since we can now take any eigenstate of a SHO and find the solution to the modified problem with the same energy. Below is a set of comparisons of the first several eigenstates of SHO (blue) and of the modified equation (orange) with the same values of $m$, $\omega$ and $x_0$ as taken above.

Comparison of solutions for SHO and the modified equation

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Because tunneling can happen at all accessible energies, the spectrum will be purely continuous, i.e., there will be no stationary states in the Hilbert space of square integrable wave functions. The solutions of $H\psi=E\psi$ corresponding to the continuous spectrum have real $E$ and are concentrated near $x=0$, with sinusoidal tails outside $[-x_0,x_0]$. To construct them numerically, pick an energy $E$ and solve a 2-point boundary-value problem in $[-x_0,x_0]$, with boundary conditions at $x=\pm x_0$ matching those of a free solution with energy $E$.

With outgoing boundary conditions at infinity, the Schrödinger equation will have Gamov states (= Siegert states) with complex eigenvalues, corresponding to resonances, unstable states decaying into the continuum. Gamov states satisfy the equation $H\psi=E\psi$ with complex $E$; the imaginary part is the inverse lifetime of the state. These states form a discrete set and are the analogues of the harmonic oscillator wave functions.

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  • $\begingroup$ By stationary states I meant solution of $H\psi=E\psi$. Doesn't matter if they are not normalizable. I wanted some sketch of a typical $\psi$. $\endgroup$ – mithusengupta123 Dec 5 '19 at 15:23
  • $\begingroup$ No need to solve a boundary-value problem. This can be solved even as a Cauchy problem, starting with arbitrary nonzero initial conditions. This works thanks to the spectrum having no gaps, so the Cauchy problem solution will always remain bounded. $\endgroup$ – Ruslan Dec 5 '19 at 18:09
  • $\begingroup$ @Ruslan: Then one must match instead the boundary to the free trailing parts. Fortunately, this is always possible since the continuum spectrum is doubly degenerate. $\endgroup$ – Arnold Neumaier Dec 5 '19 at 19:31
  • $\begingroup$ Well, I was thinking of numerical solution, where you just march forward using an explicit solver. Then you don't have to match, nor do you have to handle the second solution of the Hermite-polynomial equation (that solution is non-elementary). $\endgroup$ – Ruslan Dec 5 '19 at 19:47
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We can numerically determine the first n eigenfunctions in a box $-L\le x\le L$ with such a potential. We put $$H=-\frac{1}{2}\nabla ^2+\frac{1}{2}x^2$$ for $-4\le x\le 4$, and $$H=-\frac{1}{2}\nabla ^2$$ for $|x|>4$. Figure 1 shows the modulus of the wave function for n = 30 and L = 20. Eigenvalues are shown above. It can be seen that there are solutions with high and low probability located in the region $-4\le x\le 4$.

Figure 1

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