## Wednesday, April 23, 2008

### Particle in a Finite Box and the Harmonic Oscillator

When we solved the system in which a particle is confined to an infinite box (that is, an infinite square well), we saw that quantum numbers arose naturally through the enforcement of continuity conditions (that the wavefunction ψ must go to zero at x=0 and x=L). Quantization of energy and position (namely, nodes at which the particle cannot exist) are directly to these quantum numbers, whose values are n=1, 2, ..., ∞, representing an infinite number of energy levels.

A particle in a finite box, however, can tunnel into the walls, in the same fashion that we saw earlier with the two barrier problems. Solving this system is not difficult but, unfortunately, has no analytical solution and must be solved either numerically or, as was done in class, graphically. On the other hand, the wavefunctions are essentially just those from the infinite box but are allowed to bleed into the wall (with the caveat that higher energy states tunnel further than the lower energy states). To summarize the major differences between the particle in a finite box and one in an infinite box:
• only a finite number of energy levels exist [called bound states]
• tunneling into the barrier is possible
• higher energy states are less tightly bound than lower states
• a particle given enough energy can break free [in other words, unbound]
The next quantum system to investigate is the one-dimensional harmonic oscillator, whose potential [from Hooke's Law] is V=1/2kx2. Plugging this into the Schrödinger Equation leads to, after some well-chosen substitutions, a differential equation solved by Hermite in the mid-1800's, and we obtain the wavefunction: ψ(x) = NvHve-q2/2, where q = αx, v is the quantum number [v=0, 1, ...] and Hv are the Hermite polynomials. Here we see energy quantization as well, giving E = (v + 1/2)hbarω. This quantum system is the only one to exhibit constant spacing but other results mirror those seen in prior examples: tunneling into classically forbidden zones (where x represents displacement from equilibrium rather than position), a nonzero ground state energy as well as the existence of nodes.

One important distinction from the particle in a box result is that the peaks in the wavefunction are not uniform. For example, for v=2 and larger, it is clear that outside peaks (representing larger displacement from x=0) have higher probability than inside peaks. As n gets large, we see another clear example of the correspondence principle.

#### 13 comments:

Anonymous said...

Could you put up the solutions to HW 3 please.....

Anonymous said...

I was hoping you could post the exam 1 topic sheet this weekend!!
Thanks,
Amanda

Anonymous said...

the HW 3 solutions posted are actually HW 1. If you could change it that would be great...

Ian said...

For Hw 3, the question for 7a varies between the actual hw question and the solutions.

rod said...

true, but that shouldn't be a problem.

maybe i should claim that i intentionally do that to keep students on their toes?

Ian said...

That would certainly do the trick. While you're at it, care to explain mathematically to me how 0!=1?

rod said...

Yes, it does seem weird that 0! = 1 but check this out:

Clearly n! = (n -1)! x n

Rearranging, we get:

(n - 1)! = n!/n

Setting n = 1 gives you that 0! = 1

Another way to impress your friends and family.

Just don't put n=0 in there cause that will make the world blow up.

Ian said...

Using n=0 would be devastating to the mind, as can be seen by my current state. If (n - 1)! = n!/n is true, then how can the result of -1!=-1? For surely you would have to use n=0

rod said...

You're not going to like this but -1! is actually complex_infinity. Yes, not just ordinary_infinity, but complex_infinity. Have fun with that.

Crystal said...

For problem 05 on HW 03, does Xtp represent the turning point between classical physics and QM when looking at kinetic energy?

I guess I'm just confused by the math when calculating Xtp in part (a) and all the different substitutions

any help or clarification maybe?

rod said...

Yay, I love it when people use the blog...

The turning point is purely classical and is often called "the classical turning point". As mentioned in lecture, it is that point at which a harmonic oscillator (whether a pendulum or a spring) will stop and turn the other direction. From classical physics, that is the instant at which the system runs out of kinetic energy and all energy is potential.

A quantum oscillator has no deterministic turning point (and is, rather, entirely probabilistic) so we should interpret the term only as an analogy to classical mechanics.

We often refer to classical analogs, as you may have noticed in lecture, because quantum mechanics obeys physical laws that are utterly un-visualizable. Although quantum mechanics is the ultimate law of the universe, we have fleeting experience with it since we, as macroscopic entities, only experience average quantities (which are, in fact, averages over unimaginable numbers of particles).

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