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I met a Hamiltonian containing the derivative of the Dirac delta potential:

In order to do it we use a method described in [9]. We define a formal Hamiltonian $$ \tag{2}\tilde{H}_{abcd}=-\frac{{\rm d}^2}{{\rm d}x^2}+a\delta\left(x\right)+b\delta'\left(x\right)+c\delta\left(x\right)\frac{{\rm d}}{{\rm d}x}+d\delta'\left(x\right)\frac{{\rm d}}{{\rm d}x} $$

It is surprising to see terms like $b \delta'(x)$, how should one interpret $ \delta'(x)$?

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  • $\begingroup$ I suggest that you need to act the derivative of delta function in another functions ($\int \delta '(x)f(x)$) and uses integration by parts. $\endgroup$ – Nogueira Jan 24 '15 at 20:31
  • $\begingroup$ But how to understand the Hamiltonian? $\endgroup$ – Jiang-min Zhang Jan 25 '15 at 8:36
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    $\begingroup$ I do not know what the reference [9] in your question is. Anyway the book "Solvable Models in Quantum Mechanics" by Albeverio et al. treats the mathematical aspects of this type of problem. $\endgroup$ – Urgje Jan 25 '15 at 13:21
  • $\begingroup$ Sorry I cannot find it now. It is some mathematical physics paper. $\endgroup$ – Jiang-min Zhang Aug 23 '15 at 20:15
  • $\begingroup$ @Jiang-minZhang Google leads to link.springer.com/article/10.1007/BF01597402, with Ref. [9] being Grossmann A., Wu T. T.: J. Math. Phys.25 (1984) 1742. $\endgroup$ – Norbert Schuch Aug 23 '15 at 22:02
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Take this $\delta '(x)$ and apply in an arbitrary function $f(x)$.

$$ \int_{a}^{b} \delta'(x) f(x)\ \mathrm{d}x = f(x) \delta(x) |_{a}^{b} - \int_{a}^{b} \delta(x) f'(x)\ \mathrm{d}x = -f'(0) $$

Then $ \delta '(x) \rightarrow -\delta (x) \frac{\mathrm{d}}{\mathrm{d}x}$.

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    $\begingroup$ Another perspective on the final result shows that the derivative operator and the delta function anti-commute. $\endgroup$ – Mozibur Ullah Aug 24 '15 at 4:06
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$\delta'$ is the charge density that generates a dipole. That is, the charge density of two nearby point charges of equal and opposite magnitude in the limit as they get closer and closer to each other.

Imagine approximating the delta function with a smooth bump function, and it becomes clear what is going on. enter image description here

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  • $\begingroup$ am improvement is desirable in your graphs. The barrier is not finite, it's infinite, and so the dip before it. To show that one can leave an opening on the peak and on the bottom of the dip. That hints the barrier continues to $\infty $ and the dip to $-\infty $ . $\endgroup$ – Sofia Jan 25 '15 at 16:11
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    $\begingroup$ @Sofia In the limit as the bump becomes a delta function, the dip/peak go to infinity. This is certainly true. However, the point of the graph is to illustrate what happens before the limit, when the delta function is approximated by a smooth bump function. In this case the dip and peak are large but finite. $\endgroup$ – Nick Alger Jan 26 '15 at 4:07
  • $\begingroup$ I take it you are aware that there is a whole introductory textbook on distribution theory "An Introduction to Fourier Analysis and Generalised Functions" by M. J. Lighthill that defines distributions as sequences like this. $\endgroup$ – WetSavannaAnimal Aug 24 '15 at 10:37
  • $\begingroup$ @Sofia There is an approach to distribution theory that defines a distribution as a sequence of functions (see my comment above): Nick is simply using this approach, which can be made every bit as rigorous as other approaches. $\endgroup$ – WetSavannaAnimal Aug 24 '15 at 10:39
  • $\begingroup$ @WetSavannaAnimalakaRodVance Hi Savanna! How are you? If you mean passing a series of functions to limit, i.e. narrower and narrower, taller and taller, yes I agree with you. Best wishes! $\endgroup$ – Sofia Aug 24 '15 at 16:35
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The physical meaning of a potential looking like the Delta function is a potential acting only near the origin $x = 0$, and along the rest of the axis the particle is free. I saw such potential-barriers used in nuclear physics.

A potential like the derivative of the Delta function, $\delta '(x)$ is an approximation of a potential that along all the $x$ axis is zero, and only near the origin it displays a very thin, though infinitely high, potential barrier, followed by a very deep potential-well. More than that your book should explain why this form was convenient to them. About the treatment of this potential, integration by parts, or Fourier transforms, will help you to get rid of these unpleasant functions. One can work for instance in the linear momentum representation.

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For (most) physical purpuses, you can safely think of the Dirac delta function $\delta(x-x_0)$ as some function not vanishing only around $x_0$, and with the property that its integral is normalized to one: $$ \tag{1} \int dx \,\delta(x) = 1.$$ With this I mean that you can think of the delta as being a proper function, satisfying (1) and being not vanishing only in a very narrow$^\dagger$ interval around $x_0$.

So given this view, what is $\delta'(x)$? Nothing but the "usual" derivative of whatever function $\delta(x)$ is. And this is the crucial point: we don't know what $\delta$ really looks like apart from the localization and the integral property, so while there is no problem in defining its derivative, we don't know what it looks like.

So how can we use it? Well it turns out (as shown in the other answer) that when $\delta'(x)$ appears in an integral with another function we can "transfer" (integrating by parts) the derivative to the other function, and safely carry on with the calculation.


$^\dagger$meaning smaller that all the other physical quantities involved in the given calculation

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$\delta'(x)$ is a scale invariant barrier, where the S-matrix and phase shifts do not depend of the momentum.

A recent divulgative article is Point interactions: boundary conditions or potentials with the Dirac delta function (De Vincenzo - Sánchez) Canadian Journal of Physics 10/2010; 88(11):809-815. DOI: 10.1139/P10-060 Another interesting reference can be http://arxiv.org/abs/quant-ph/0406158 where it is argued that the parametrisation has some gauge freedom.

But if you want an interpretation, so I argue to look at it as some scale-invariant object. Already the point of being supported in a single point implies some amusing property under scaling, as it must be mapped towards another interaction having support in a point, so you can guess all the families will do for nice fixed points and renormalisation lines in the space of potentials with compact support. Besides, the $\delta'$ potential -alone- can be argued to have dimensions of inverse length squared, the same that the kinetic term, and so some invariance of the whole hamiltonian under scaling $x \to \lambda x$ can be expected.

Indeed if you apply the formulae of the first reference to $V(x) = g_2 \ \delta'(x)$ you get conditions

$$ u(0^+) =\mu \ u(0^-), \ \ \mu\, u'(0^+) = u'(0^-)$$

that allow to solve for the $S$-matrix, or if you prefer the Transmision and Reflection coefficient. Now for instance for the left wave we will have in $0^-$ the sum of incident and reflected: $$u_k(0^-)= e^{ikx}+ e^{-ikx} R^l = (1 + R^l) $$ and its derivative $$u'_k(0^-)= ik (e^{ikx} - e^{-ikx} R^l) = ik (1 - R^l)$$ and similarly in $0^+$ the transmited wave $$ u_k(0^+)= e^{ikt} T^l = T^l, \ u'_k(0^+)= ik e^{ikt} T^l = ik T^l$$

so you see the magic of this particular boundary condition: the $ik$ factors can cancel and the transmission and reflection coefficients do not depend of $k$

$$ \mu T^l = (1-R^l), T^l = \mu (1+R^l)$$

A modern reference relating delta derivatives to scattering is http://iopscience.iop.org/0305-4470/36/27/311 "On the existence of resonances in the transmission probability for interactions arising from derivatives of Dirac's delta function"

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