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# Re: [Phys-L] electron location & wave function

• From: John Denker <jsd@av8n.com>
• Date: Wed, 10 Apr 2013 22:44:47 -0700

On 04/10/2013 08:41 PM, Paul Lulai wrote:
Hi. I am somewhat comfortable with electrons being described by a
wave equation.

Good.

It has been too long since my coursework involving wave equation
items. When an atom is excited and the electron transitions to
different levels, does the wave function adjust, allowing the
electron to exist in spots inbetween what was alowable prior to the
excitation, or does the electron disappear / reappear? I thought the
transitions energy levels. Others have told me the electron just
zaps to the other level. However I don't know this well enough to
figure out what I am making up and what they are making up.

Let me start by answering a slightly different question. This serves
as a "warm-up exercise" in preparation for the main event.

Consider photon polarization. We set up a polarizing beam splitter
with two photon counters.

\ X-polarized
==================\------------------> counter
|\
|
| Y-polarized
|
V
counter

We use a polarizer to prepare a huge number of photons, all polarized
at 45° relative to the X direction. Ideally we observe that 50% of
the photons are counted as being X-polarized while the other 50% get
counted as being Y-polarized.

More generally, if the incoming beam is polarized at an angle θ
relative to the X axis, then a fraction cos^2(θ) get counted as X
while a fraction sin^2(θ) gets counted as Y.

-- We ask a very specific question. We describe the setup and we
describe rather precisely how the measurement gets done.
-- For some values of θ, quantum mechanics makes a categorical
prediction, i.e. 100% probability or 0% probability. For other
values of θ, quantum mechanics cannot predict the outcome of any
particular measurement; it only predicts the probability i.e.
the ensemble average.
-- The photons in the X branch of the analyzer really are X-polarized.
In the X-branch, you can stick in as many additional polarized filters
as you want all aligned in the X directions, and all the photons get
through. (This is the operational definition of eigenstate.)

QM is rather picky about the questions it will answer. There are lots
of seemingly-reasonable questions for which there is no answer. Actually
goes back to Galileo. In this department (like many others), Newton
went to school on Galileo. he summed up the idea by saying "Hypotheses
non fingo" in response to questions that were outside the scope of the
theory.

This is directly relevant to the question Paul asks. For any particular
atom in the ensemble, you can measure the energy or you can measure
the position of the electron, but you can't do both. The Heisenberg
uncertainty principle forbids.

There are ways of getting /some/ traction on this kind of question.
For example, you can prepare an ensemble of atoms, and measure the
energy in some of them and measure the position in others. In particular,
you can prepare atoms in an energy eigenstate and then measure the
position. The answer will be probabilistic, but that's OK. To see
exactly what this looks like, there is an animation here:
http://www.av8n.com/physics/wavefunctions.htm#sec-animation

Note the parallel between the atomic physics and the warm-up exercise:
We use one set of coordinates to prepare the state, and then rotate
to another set of coordinates to analyze the state.

Preparation Analysis
polarization: angle θ X, Y
atom: energy position

As a somewhat separate point: In quantum mechanics, there are *some*
things that are quantized, as the name suggests. However, non-experts
tend to wildly overestimate how much stuff is quantized.

In this context is is super-important to specify what is being measured
and how. If you want to talk about a particular way of measuring the
energy, that's fine. In contrast, if you just ask about "the" energy,
the question is very likely to be unanswerable.

For example, photon polarization is quantized. If you analyze any
particular photon using the (X,Y) basis, then *every* photon comes
out of the analyzer along the X-branch or the Y-branch as diagrammed
above. There are no other possibilities. That is to say, there are
no other possibilities _for this analyzer_.

Meanwhile, however, there are lots of other analyzers. X and Y are
not the only states. They are not even the only basis states! You
could equally well choose the circular polarization basis (RCP,LCP).
Every photon is either RCP or LCP. There are no other possibilities.
That is to say, there are no other possibilities _if_ you decide to
measure the photon by projecting it onto this particular basis set
(RCP,LCP).

To cut to the chase: Non-experts like to say that energy is quantized.
They take this as the defining property of energy and/or the defining
property of quantum mechanics. The problem is, it's just not true.
Note the contrast:
-- If you decide to measure the atom by projecting it onto the energy
eigenstate basis, then sure, the energy will be quantized.
++ However, the energy states are not the only states. They are not
even the only basis states!

If you want to see a movie of a state that is not even remotely an energy
eigenstate, but is nevertheless a perfectly fine state, indeed a quite
important state, see
http://www.av8n.com/physics/coherent-states.htm#fig-glauber-movie

This makes contact with the original question as follows: Starting
from an energy eigenstate, such as the ground state of an atom, it
is as easy as π/2 to create a state that is _halfway_ between that
state and the next higher energy state. This is analogous to creating
a photon that is polarized halfway between X and Y. People do it all
the time. In the language of pulsed NMR, people talk about a π/2
tipping pulse. The resulting state is not an energy eigenstate, but
it is a perfectly fine state.

At the risk of slightly overstating things, one could reasonably say
that it is the /analyzers/ that cause things to be quantized ... and
in between analyzers, there is no reason to imagine that anything is
quantized.

In particular, imagine sending an X-polarized beam into an optically
active medium. As the photon moves along, its polarization vector
will rotate, like an airplane doing a long series of aileron rolls.
above measures quantized polarization, but if there's no analyzer
there's no quantization.

Others have told me the electron just zaps to the other level.

That seems to say that an atom necessarily exists in a state of definite
energy and necessarily "zaps" from one energy state to another ... which
is just completely wrong.

I thought the energy added would adjust the wave function while the
electron transitions energy levels.

Certainly the wavefunction adjusts in response to a perturbation, such
as a π/2 tipping pulse. The part about transitioning from one energy
level to another is not helpful. Energy states are not the only states.
They are not even the only basis states!

does the wave function adjust, allowing the electron to exist in
spots inbetween what was alowable prior to the excitation,

Certainly the wavefunction adjusts in response to a perturbation. The
adjustment usually involves some new "spots" but that's not the whole
story. Remember, we are talking about nonstationary states here, so
the electron is busy moving from spot to spot. One way I can visualize
this kind of "adjustment" is by reference to this:
http://www.av8n.com/physics/coherent-states.htm#fig-glauber-movie
Imagine scaling up both X and P by 3 percent. That increases the
energy of the oscillator by 6 percent. The larger orbit visits some
different "spots" but that's not the whole story.

or does the electron disappear / reappear?

The disappearance or reappearance of an electron would violate the laws
of physics several times over, including
-- conservation of charge
-- conservation of lepton number
-- conservation of angular momentum
-- conservation of energy
++ et cetera.

Seems kinda unlikely.