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So I think I got the main idea of finding the possible energies given some simple hamiltonians (like particle in an infinite well, harmonic oscillator and so on). In class, we were given a challenge problem to think about:

something along the lines of what are the energies of a particle confined to some long wiggly-noodle of length say $a$. I'm 99% sure this was all the information I was given, and I'm stumped.

Normally, I would write out the hamiltonian and then solve Schrödinger's equation and applying the appropriate boundary conditions but I don't even know how to start here. How do I go about tackling this problem?

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TLDR: The problem is exactly equivalent to one of an infinite potential well of length $a$.

Setting up

First let's try and set it up classically, then we can move on to the quantum case.

So let's say you have a squiggly wire of length $a$, and your particle is some bead that rolls moves around constrained on this squiggly wire. The wire has a shape given by the vector $\vec{X}(s)$, where $s$ is some number that labels positions on the wire. For every $s$ you get a position vector giving you a point on the wire.

To write a lagrangian we need to some generalised coordinates, so we'll go with $s$, which we used to label a position on the wire. A configuration of this system (a particle on a squiggly noodle) is given by the position of the particle on the noodle. If at time $t$ the particle is at position label $s$ on the noodle, the position of the particle will be $\vec{X}(s(t))$.

Writing the Lagrangian

To write the lagrangian: $L = T - V$, we need to write the kinetic energy, and potential energy. There's no explicit potential energy in this (One could argue that you have infinite potential at the ends of the noodles but that can be taken care of using boundary conditions). For the kinetic energy we simply write $\frac{1}{2}m \vec{V}\cdot \vec{V} $.

$$\vec{V} = \frac{d\vec{X}}{dt} = \frac{\partial\vec{X}}{\partial s}\frac{\partial s}{\partial t} = \vec{X}' \dot{s}$$

And hence:

$$ L = T - V = \frac{1}{2}m \vec{V}\cdot \vec{V} - 0 = \frac{1}{2}m \dot{s}^2 (\vec{X}'\cdot \vec{X}')$$

Notice that $(\vec{X}'\cdot \vec{X}')$ is just some function of $s$ that depends ONLY on the shape of the wire and how we chose to label it with $s$. So let's just call it $(\vec{X}'\cdot \vec{X}') = f(s)$ for now.

Finding the conjugate momentum

We know our generalised coordinate is $s$. We can find its conjugate momentum, by $P = \frac{\partial L}{\partial \dot{s}}$

We find:

$$P = \frac{\partial L}{\partial \dot{s}} = m \dot{s} (f(s))$$

So:

$$ L = \frac{1}{2}m \dot{s}^2 (f(s)) = \frac{P^2}{2 m f(s)} $$

That's your lagrangian. Now that looks awfully similar to a free particle or particle in a well hamiltonian, but with that weird $f(s)$ term at the bottom.

Quantizing this awful thing

To quantise it, you'd write the hamiltonian:

$$ H = \frac{\hat{P}^2}{2 m f(\hat{s})} $$

and put the commutation relation $[\hat{s},\hat{P}] = i\hbar$

Given $f(s)$ for your curve how would you solve it? It'd probably be really difficult.

However, there's a trick.

Now for the trick

So far we've chosen an arbitrary parameter for labelling points on our squiggly noodle. But what if we chose a specific way of labelling them. We now say that $s$ is actually the distance along the wiggly-noodle from a starting point.

What does this change? Let's look at the the condition on $\vec{X}$. So if we move along the noodle by very small $\Delta s$ , we should get a resultant displacement vector: $\vec{\Delta X} = \vec{X(s + \Delta s)} - \vec{X(s)} = \vec{X}'\Delta s$. The length of the displacement vector should be $\Delta s$, by definition of $s$. So:

$$ \vec{\Delta X}\cdot\vec{\Delta X} = (\Delta s)^2 $$

and

$$ \vec{\Delta X}\cdot\vec{\Delta X} = (\vec{X}'\Delta s)\cdot (\vec{X}'\Delta s) = \vec{X}'\cdot\vec{X}' (\Delta s)^2 $$

$$ \implies \vec{X}'\cdot\vec{X}' = f(s) = 1 $$

This means that if we choose $s$ to be the distance along the noodle, $f(s) = 1$.

Solving it now

Hence we have that $L = \frac{1}{2}m \dot{s}^2 $ and hence: $P = m \dot{s}$, as well as $H = \frac{P^2}{2m}$

So we can now quantize it:

$$H = \frac{\hat{P}^2}{2m} $$ $$ [\hat{s},\hat{P}] = i\hbar $$

and condition that $\psi(s) = 0$ on the endpoints of your squiggly-noodle. i.e. $s = [0,a]$ and $\psi(0) = 0$ and $\psi(a) = 0$.

This is MATHEMATICALLY EQUIVALENT to the problem of a particle in an infinite well of length $a$.

It's the same system!

Hence, they are the same systems with exactly the same energy spectrums.

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  • $\begingroup$ Wow! That was surprisingly interesting and elegant! Thanks a bunch. $\endgroup$ Jan 19 '18 at 18:32

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