.
But, you might ask, how can you ever tell its there? Since all you ever measure
is transitions between levels, how can you tell if the bottom level has zero
energy or not? There are in fact
many ways to "observe" the zero point energy.
Comment on zero point energy. Last time, I pointed out that the ground state energy is
.
But, you might ask, how can you ever tell its there? Since all you ever measure
is transitions between levels, how can you tell if the bottom level has zero
energy or not? There are in fact
many ways to "observe" the zero point energy.
A beautiful, but tricky way was discovered by Mulliken in 1924 (!) He looked at
the spectra of BO (Boron monoxides), where the Boron's were different isotopes,
in one case with mass 10, in another mass 11. These molecules will have
different reduced mass (slightly) By comparing the vibrational spectra we've
just discussed, and also the rotational spectra (which we haven't
discussed yet), he discovered that the data fit better if his energy levels
starts at
rather than 0. So even though that zero point energy cannot be radiated away,
it still has observable consequences. Also, it is consistent with the
uncertainty principle, which says you cannot be at rest completely, and have
some known separation. I haven't seen Mulliken's paper, but here's my guess at
how the analysis might have gone... Molecules have some equilibrium separation
r_0. If E_0=0, then the radius r=r_0 is fixed. But if E_0>0, then you expect
some sort of residual "vibration", so r spreads out. Put mathematically,
<r> = r_0
Now look at rotations. The energy of rotation is given classically by
Now, if
was just some constant number, this would simply be a modification of r_0, and
there's really no other way to know or find out r_0, so we're stuck. But,
That is, the zero point energy tells you the energy of the small residual
oscillation, and thus in the end
itself depends on the reduced mass! Two different isotopes have the same spring
constant, but different reduced mass, which means
varies with the 2 isotopic molecules, so the rotational energies shift just a
bit, in a precisely known way with
.
(If E_0 was 0, then
would also be 0, and the rotational spectra should look the same)
We've been looking at bound states so far, where E<V, so the wave
function goes to 0 at infinity. Next, we're going to consider the case where
E>V, i.e. the particle is UNBOUND. So, the second derivative will always be
"concave towards the axis", i.e. always "sin"-like. And, it will not
have to go to zero at infinity! This makes normalization a pain, but we'll just
basically ignore the problem. So, you can expect a typical wave function to
look something like
.
We continue using the time independent S.E., i.e. we have infinitely long wave
trains with fixed momentum (k), and imagine we're watching a steady state
situation.
In fact, that's why our normalization is tricky - we now have a continuous flow of particles from - infinity to + infinity. (Normalization in such a case is usually done in terms of "# of particles/unit length", or something like that)
In the first case, consider a potential with an "up-step" in it:
Imagine sending in a continuous "flow" of particles from the left, i.e. the
wave function starts off like
(remember, "-ikx" in the exponent represents particles moving to the
left, +ikx => moving to the right)
Classically: If E = E(lo)<V0, the particles will bounce, and you'll have complete reflection.
If E = E(hi)>V0, the particles will slow down, but keep going, and you'll have complete transmission.
Quantum: If E=E(lo)<V0, you expect a wave function like:
You will get a small bit of "leakage", but it exponentially dies away.
In the steady state, you still won't get any transmission, because the wave can't propagate. So you still get complete reflection. (However, this sets up standing waves to the left, because of interference)
This is a sort of analog of total internal reflection of light from a boundary. There's actually a tiny (exponential) "leakage" of E and B fields outside the glass, but in the steady state, all the light bounces.
But what about E=E(hi)>V0? Now something radically new occurs!
Continuity of the wave function at the boundary, and continuity of the first derivative at the boundary, give:
Take the first equation, multiply it by k2, and subtract the 2 eqns:
Probability is always proportional to the square of the wave function:
The fraction transmitted must be 1 minus the above (because the fraction transmitted + fraction reflected is everything!) which gives
So, this is not classical - I'm getting partial reflection! Since k1 > k2,
we see that 0 < R < 1. (Kind of like light hitting a plate of glass - some reflects. It's really very similar. Waves hitting a sudden change of wavelength means some waves transmit, and some reflect.)
If you think a bit about the above formulas, you'll notice that something is weird here. Let's go back and solve for B/A0, the quantity which should tell us about how much gets transmitted. (T=|B/A0|^2)
Now wait! This is not T!? This is, in fact, T (k1 /k2). Why? Shouldn't |B/A0|^2=T?
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