,
as a function over all space, it does not have a unique momentum. (Think
of our F.T. arguments - you need to add up waves of various momenta to get a
packet that lives only inside a box)
Notice that
,
One can build
out of two waves:
The first term represents a wave moving to the right, with wavelength 2L/n, hence momentum = nh/2L. The second term is a left moving wave, with p=-nh/2L.
This means there is a superposition of two different momenta, and there is thus an uncertainty in the momentum given by
This is a wave function with well defined energy, but the momentum is not unique. The uncertainty principle is satisfied automatically by this particle. The particle is bounded in space, which is what causes an uncertainty in the momentum. (But, if it oscillates forever, it is not bounded in time, which means it can have a definite energy)
Let's calculate the probability that the particle is, say, somewhere between x1=0.4L and x2=0.6L.
(The last is the classical result)
On the other hand, what is the expectation value of x?
I leave this integral as an exercise. Units say it must be proportional to L^2. Common sense says some fraction* L^2.
Answer is <x> = (2/L) L^2/4 = L/2, the midpoint. (Good!) True for all n, too. Even for n=2, where Prob(x=L/2)=0, it is still true that <x> = L/2.
I.e., it's just as likely to be in the left or right side.
Let's think of a different kind of (classical) system, to get more of a feel for this probability distribution. Think of a swinging pendulum, and measure its position at random times. (or, take a million pendula, and measure each one once)
The pendulum is slow near the ends, so it spends more time there. This means that it is more likely to be found near the endpoints.
So, a plot of the "classical wave function", or in other words the probability distribution of x, is:
For quantum systems, we can't measure the position over and over, because measurements usually mess it up. But we can imagine many identical systems. This fits with the probabilistic interpretation.
Remember that for a single measurement, probability is a weird concept.
After the measurement, the answer to the question "was it in some small region,
,
is no longer
,
but yes, or no! It only makes sense when you think of an ensemble
of systems.
How could one physically realize our particle in a 1-D box? F+T have a neat answer:
Potential is zero everywhere in middle. But, potential rises to V0 at either end, quite quickly.
So, if L'-L << L, and V0 -> infinity, this becomes close to an ideal box.
It's interesting to ask what the wave function would look like for this real system, rather than the ideal box.
Further, one could imagine all sorts of other potentials to trap electrons in.
The next part of this course will deal with the qualitative features of the
S.E., and how to make reasonable guesses, and sketches, of the wave function
.
This is useful for lots of reasons. It often gives insight into the physics,
real (3-D) problems can often be mapped into 1-D ones, and it allows you to
check, after you've calculated a solution, that it's reasonable. We will begin
with a (still ideal) case of a potential like the above, with L'=L, but V0
finite. The is, we want to solve the S.E. inside a potential like this:
Classically, a low energy particle rattles around in there, and a high energy particle, with E>V0, is free.
This problem can indeed be solved exactly (and we will do this, soon!) But first let's be qualitative: Imagine first that E<V0. Classically,
Why is it forbidden in regions I and III? Because E=K.E. + V. If E<V, then K.E. is negative, which is impossible!
But quantum mechanics is different. We have a wave, not a particle. Waves can exist in regions where they cannot propagate. Let's rewrite the S.E. as follows, for x<0 or x>L:
,
where
is manifestly positive. In regions I and III,
.
is not a usual wave, outside the well. It's an exponential function. In
region III, imagine what happens as x gets very large. The part of the wave
function proportional to "a", which is a growing exponential, blows up. This
would mean there was a large probability the particle is outside
the box. So, we reject this term, i.e. we require a=0. Similarly, in region I,
we discover that we can't have the "b" term, hence there, b=0. How about the
middle region? There,
Inside, we'll have sin's and cos's. Outside, we'll have damped exponentials.
Of course,
and
must still be continuous at the boundary (since
is finite everywhere, now) We expect a typical solution to look something
like
(Excuse my crude sketches! The curves throughout this section should of course be smooth: a pure sine wave inside, and a pure damped exponential outside)
Recall that
,
which means that the bigger
,
the bigger is E (if V=0).
Put another way, the more the wave function wiggles, the bigger is
,
hence the bigger is E. I can find lower energy solutions by making them wiggle
less. The lowest I can possibly draw would be
You can't wiggle any less than this!
Some features to notice: recall the infinite square well ground state looked like this:
This wave function has slightly smaller wavelength (it hits zero a bit
earlier), i.e. higher energy. So, the finite well must lower the ground
state energy a bit. (We can be slightly less wiggly, since the boundary is in
some sense a bit sloppier). As the potential V0 gets larger, our exponential
cutoff
gets bigger, and the "matching point" will drop down towards the ground,
i.e. it will smoothly go to an infinite box solution.
The top curve has a small V0, and so is weakly exponential. The bottom curve has a big V0, and hence is strongly exponential outside the box (i.e. it's very close to 0). These wave functions have not been normalized, if they had, they would all have to have unit area. (What do you think that would mean for the height of curve with small V0 compared to the one with big V0, compared to how they've been drawn?)
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