While studying thermoionic emission from metals I wanted to get a feeling for the problem with classical mechanics before delving into quantum mechanics. The potential used to model the situation is this one:

$$V(x)=V_0 \Theta(x)$$

Where $\Theta (x)$ is the Heaviside step function. If we want the classical force for this potential we differentiate:

$$F_x = - \frac{dV}{dx}= - V_0 \delta(x)$$

Where $\delta(x)$ is the Dirac delta function. This gives an equation of motion of the type:

$$m \ddot{x} = -V_0 \delta(x)$$

With $m$ and $V_0$ positive parameters and the dots denote differentiation with respect to time. My question is: how to treat this equation? It turns out that the problem is much simpler in quantum mechanics if we try to solve the time independent Schroedinger equation.

Thanks in advance.

  • $\begingroup$ I know the question is somewhat "physicsy" but the core of it is mathematical, so I thought this should go here instead of physics stack exchange. $\endgroup$ – Gennaro Marco Devincenzis Jan 3 '17 at 15:01
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    $\begingroup$ We've: $$\text{V}\space'\left(x\right)=\frac{\partial}{\partial x}\left(\text{V}_0\cdot\Theta\left(x\right)\right)=\text{V}_0\cdot\Theta'\left(x\right)=\text{V}_0\cdot\delta\left(x\right)=-\text{m}\cdot\color{red}{\text{x}''\left(t\right)}$$ But that $\text{x}''\left(t\right)$ gives displacement?! $\endgroup$ – Jan Eerland Jan 3 '17 at 15:08
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    $\begingroup$ what do you mean with no time dependence in QM. Do we consider the time independent Schrödinger equation $(\frac{p^2}{2m}+V(x))\psi(x)=E\psi(x)$.? $\endgroup$ – tired Jan 3 '17 at 15:08
  • $\begingroup$ @tired yes, I was considering the time independent Schroedinger equation. There is time dependence in quantum mechanics obviously. I should probably phrase that better. EDIT: Original post edited. $\endgroup$ – Gennaro Marco Devincenzis Jan 3 '17 at 15:09
  • $\begingroup$ @JanEerland $\ddot{x}$ is the acceleration, $x(t)$ position as a function of time. $\endgroup$ – Gennaro Marco Devincenzis Jan 3 '17 at 15:11

As in the QM case, the usual way to solve differential equations involving delta-functions is to solve them piecewise on each domain. We first re-cast the equation to solve for the speed $v = dx/dt$ as a function of $x$: $$ \frac{d^2 x}{dt^2} = \frac{dv}{dt} = \frac{dv}{dx} \frac{dx}{dt} = v \frac{dv}{dx} = \frac{1}{2} \frac{d}{dx} \left(v^2 \right). $$ Our equation is now \begin{equation} \frac{m}{2} \frac{d}{dx} \left(v^2 \right) = - V_0 \delta(x). \qquad \qquad (1) \end{equation}

We can now note that for the regions $x < 0$ and $x > 0$, we have $$ \frac{m}{2} \frac{d}{dx} \left(v^2 \right) = 0, $$ which implies that the solution is $$ v(x) = \begin{cases} v_- & x < 0 \\ v_+ & x>0 \end{cases} $$ To find the relationship between $v_-$ and $v_+$, we integrate equation (1) in a small interval $[-\epsilon, \epsilon]$ around 0: \begin{align*} \frac{m}{2} \int_{-\epsilon}^{\epsilon} \frac{d}{dx} \left(v^2 \right) \, dx &= - V_0 \int_{-\epsilon}^{\epsilon} \delta(x) \, dx \\ \frac{m}{2} \left[ v^2 \right]_{-\epsilon}^{\epsilon} &= - V_0 \\ \frac{m}{2} \left(v_+^2 - v_-^2 \right) &= - V_0. \end{align*} This latter equation can be recognized as energy conservation across the boundary $x = 0$: $\Delta KE = - \Delta PE$.

The solution for $v(x)$ is then $$ v(x) = \begin{cases} v_0 & x < 0 \\ \sqrt{v_0 - 2V_0/m} & x > 0 \end{cases} $$ If you want the solution for $x(t)$, you can then integrate this with respect to time.

Alternately, if you want to skip the step of finding $v(x)$, you can instead use the identity $$ \delta(x(t)) = \sum_i \frac{1}{|\dot{x}(t_i)|} \delta(t - t_i) $$ where the sum runs over the zeroes of the function $x(t)$. This then allows us to recast this equation solely in terms of $x$ as a function of $t$. One can oncesagain solve this piecewise between successive zeroes of the function $x(t)$, and integrate over small intervals of $t$ surrounding these zeros to "patch" the piecewise solutions together. In this case, the solutions for $x(t)$ "between" the zeroes will be simply linear functions of $t$, which means that you will only have one zero for $x(t)$, and applying the above techniques will yield the same sort of solution.

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There are two approaches. One is just to think about energy conservation. To the left of the step the energy is $\frac 12mv_0^2$, so as long as that is greater than the step, the velocity will reduce to keep the energy constant and the new velocity will be $\sqrt{v_0^2-\frac 2mV}$. If you want to solve the equation, you integrate across the time the particle crosses the step. The delta integrates to a step function, the acceleration integrates to the velocity, and you get a step function in velocity just as before. The second approach is like what you would do in quantum mechanics.

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How to treat this equation?

Well, in the usual manner: as the equation of motion is time independent we write down the first integral - energy conservation

$$E = \frac{m v^2}{2}+V_0 \theta (x)=\frac{m v_0^2}{2}$$

Here $v_0$ is the velocity of the particle for $x<0$

For definiteness let us assume $V_0 > 0$.

We have to distinguish two cases of initial conditions $(1)\; v_0 < 0$, $(2)\; v_0 > 0$.

In the first case the particle moves indefinitely along the negative x-axis with velocity $v_0$.

In the second case the particle is either reflected at the potential barrier, if

$$(2a)\;0 < v_0 < v_c $$

where the critical velocity is defined as

$$v_c = \sqrt{\frac{2 V_0}{m}}$$

and moves further on as in case (1), or, for suffiently high velocity

$$(2b) \; v_0 > v_c $$

the particle is decelerated to

$$v_1 = \sqrt{v_0^2 - v_c^2}$$

and continues the motion into the region $x>0$ with verlocity $v_1$.

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