My quantum mechanics textbook says that when a particle (in the classical case) comes across a potential-step barrier of finite height, if it has sufficient energy to surmount the barrier, it will continue on with reduced kinetic energy.

I'm finding this hard to understand since force is given by $$F=-\frac{dV(x)}{dx}$$

For a step barrier, this should give an infinite force acting (in the opposite direction) on the particle when it comes in contact with the barrier.

The only other thing I can think of is to model the force on the particle by a dirac delta function, so we effectively see it getting an impulse in the opposite direction, which could lower its kinetic energy. Is this reasoning right?


3 Answers 3


You can actually do this using $\delta$ functions as well if you'd like (provided you're careful). Let $V(x) = V_0\theta(x-x_0)$, then $$ F(x) = -V'(x) = -V_0\delta(x-x_0) $$ For simplicity, let's take the potential barrier to be located at $x_0 = 0$ so that by Newton's second law, the equation of motion becomes $$ -V_0\delta(x(t)) = m \ddot x(t) $$ Let's solve this equation subject to the initial data $$ x(0) = -x_a, \qquad \dot x(0) = v_a, \qquad x_a>0, \qquad v_a>0 $$ In other words, the particle is approaching the potential barrier from the left with speed $v_a$. If we integrate both sides with respect to time from $t=0$ to some $t>t_*=x_a/v_a$ ($t_*$ is simply the time at which it gets to the origin), then we have $$ v(t) = v_a - \frac{V_0}{m}\int_0^t \delta(x(t')) dt' $$ now here's the tricky part, to perform the integral, we use the distributional identity $$ \delta(f(x)) = \sum_{\{x_i:f(x_i) = 0\}} \frac{\delta(x-x_i)}{|f'(x_i)|} $$ and we obtain $$ \delta(x(t)) = \frac{\delta(t-t_*)}{v(t_*)} $$ so we get $$ v(t) = v_a - \frac{V_0}{m}\int_0^t\frac{\delta(t'-t_*)}{v(t_*)} dt' = v_a - \frac{V_0}{m v(t_*)} $$ The trouble is, what should we pick for $v(t_*)$, the velocity when the particle hits the barrier? It turns out that we need to pick it to equal the average of the velocities before and after it crosses the barrier in order to satisfy energy conservation (and to therefore be consistent with the smoothing procedure outlined in Michael Brown's response) $$ v(t_*) = \frac{1}{2}(v(t) + v_a) $$ and this gives $$ v(t) = v_a - \frac{2 V_0}{m(v(t) + v_a)} $$ Which is the same as the conservation of energy equation that determines the velocity after encountering the barrier; $$ \frac{1}{2} m v(t)^2 = \frac{1}{2} mv_a^2 - V_0. $$

  • $\begingroup$ Dear Josh, I have an inquiry. So I'm totally cool with your reasoning up to the moment when you evaluate the integral and obtain $v(t)$. I'm also fine with how you stipulated energy conservation from which you obtained $v(t_{*}$. My problem is this: it's true that energy is conserved always and that $$\frac{1}{2} m v(t)^2 = \frac{1}{2} mv_a^2 - V_0.$$ However, one should consider the physical situation at hand. For example, if there's a ball falling from a certain height $z=h$ from the ground($z=0$) at rest then it's true that its energy will always be constant and given by $1/2mv^2 +mgz=E$ $\endgroup$
    – Omar Nagib
    Feb 24, 2017 at 18:50
  • $\begingroup$ However this formula should be used with reservation, It's invalid for $z<0$ because we know from the physical situation we have that the ball will hit the ground and bounce back, and in fact it will never cross the ground $z=0$. If we applied the formula for $z<0$ without taking this into account, then it's true that it will give us a certain kinetic energy for the the ball, but the result will be meaningless. By the same token in your answer, the ball should experience an infinite repulsive force at $x=0$ and should be reflected back towards $x=- \infty$. $\endgroup$
    – Omar Nagib
    Feb 24, 2017 at 18:55
  • $\begingroup$ Using the equation $\frac{1}{2} m v(t)^2 = \frac{1}{2} mv_a^2 - V_0$ for $x>0$(since you assume that the ball will cross the barrier and will have a velocity $v(t)$) ignores(implicitly) that the ball will be reflected at $x=0$ and won't travel in the positive x direction in the first place(just like the example I set above in the last comment). $\endgroup$
    – Omar Nagib
    Feb 24, 2017 at 19:09
  • $\begingroup$ @OmarNagib It's not true that the ball will be reflected in this case if the initial momentum is large enough. The delta function yields an "infinite" force for only an "infinitely small" time in such a way that the total impulse in the end is finite. $\endgroup$ Feb 25, 2017 at 17:07
  • $\begingroup$ Right. I think you should append this comment in your answer as it will clear up a great deal of confusion that OP(and people like me) were having, namely, how the ball got pass the infinite repulsive force. $\endgroup$
    – Omar Nagib
    Feb 25, 2017 at 17:24

Yes, the derivative of a step function is a Dirac delta. You can see this by integrating the delta function: $$ \Theta(x)=\int_{-\infty}^x \delta(x') \mathrm{d}x'$$ where $$\Theta\left(x\right)=\begin{cases} 1 & x>0\\ 0 & x<0 \end{cases}$$ (note that $\Theta(0)$ is not defined by this prescription. If you use a symmetric representation of the $\delta$ function you'll get $\Theta(0)=\frac{1}{2}$, but that's not important right now.)

The more physical way of thinking about it is to smooth the potential out to some function which goes from zero to the maximum in some finite distance. For example:

$$V(x) = \frac{V_0}{2} \left[1 + \tanh\left(\frac{x}{\ell}\right)\right]$$

which looks like this:

graphed of smoothed step potential

In this case the width of the potential step is of the order $\ell$. The force is the derivative of this:

$$ F = - \frac{V_0}{2\ell} \mathrm{sech}^2 \left( \frac{x}{\ell} \right) $$

which is everywhere finite, but becomes very large near $x=0$ in the limit $\ell\rightarrow0$.

This makes the classical mechanics problem nice and well defined. You can integrate the equations of motion for a classical particle in this potential and see what it does. You will find that if its kinetic energy is less than $V_0$ it will have a turning point and reflect back to $x=-\infty$. On the other hand, if its energy is greater than the barrier it will continue on with a final velocity determined by energy conservation:

$$ \frac{1}{2} m v_f^2 = \frac{1}{2} m v_i^2 - V_0 $$

This holds even in the limit $\ell\rightarrow 0$. The interesting differences in the quantum theory are barrier penetration and finite probability of reflection even above the barrier.

Just for fun I've plotted some streamlines, picking some simple numbers for $m,V_0,\ell$. If you haven't seen enough classical mechanics yet this is called a phase plane. The two axes are the two variables $x$ and $v=\dot{x}$, and the curves show how they change with time. You can see that particles which come in from the left reflect if they don't have enough energy, but pass through to the right with a velocity decrease if they go ever the barrier. All particles incident from the right pass through the step (and in fact gain velocity in the negative direction). Plots like this are a great way of gaining insight into classical mechanical systems.

phase plane for a single particle in a smooth step potential

  • $\begingroup$ Nice phase plot! $\endgroup$ Feb 11, 2013 at 17:27
  • 1
    $\begingroup$ Mathematica baby! $\endgroup$
    – Michael
    Feb 12, 2013 at 0:15

One way to define force is as dp/dt - rate of change in momentum. Or, over s finite time interval, $\Delta p = \int F dt$ F, as you say, is a dirac function, infinitely strong but experienced by the particle (imagining it as classical) for zero time. Zero times infinity - or rather, the limit of a very tiny duration times a very huge force - is some finite number.

  • $\begingroup$ "the limit of a very tiny duration times a very huge force - is some finite number." - Not necessarily. It depends on how you take the limits. You do make the point that $0\times\infty$ can be anything and potentially finite, however. $\endgroup$
    – Michael
    Feb 11, 2013 at 8:37

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