Questions and answers from earlier this semester
A question about terminology
>A quick question:in number 3, we are asked for the expectation value
>of the energy. From what I understand that would be the arithmetic
>mean of the possible energies. Therefore the expectation value isn't
>even an allowed energy, so what is the physical content of it? The
>name is also misleading because you should NOT expect to get such
>value! What is the physics of expectation values? Is this a stupid
>question?
As you surely know by now, there is no such thing as a stupid
question!!
Yes, the expectation value is indeed the mean of the possible energies
(weighted by the relative probabilities of measuring each one) It is
usually denoted {H} or perhaps {E}. (Those braces are supposed to
represent bra's and ket's, I'm having trouble getting html to print the
right characters!) That notation also tells you an alternative
(mathematically equivalent) way of computing it - it's the bracket of
the the state you're in, with the H operator, and then your state
again, i.e.
{H} = {Psi | H | Psi}.
In some problems, that's easier. The other way (what you said) I
might write out as
{H} = Sum[ |c_n|^2 E_n ]
(If that makes sense in such an ugly Ascii form) It's not hard to prove
those two expressions are the same.
Now for your questions - first, "isn't the name misleading" Yes, I'd
agree the name is quite misleading, for the reasons you stated. In most
problems (like #3, e.g. ) you can never actually *measure* the
"expected value"! (I think "Average value" might have been a slightly
better name, perhaps, but it's an old and well-used word, so we have to
live with it) It's possible, through some good luck, that the
expectation value might just HAPPEN to match up with an eigenvalue in
some problem, but that would be fortuitous, and still doesn't mean
you're necessarily guaranteed to measure it every time. (That guarantee
*only* happens if your state is an eigenstate, in which case the
expectation value is also the ONLY possible value!)
"What is the physical content"? Well, it's the average result of a
bunch of (hypothetical ) measurements . I.e. if you have a bunch of
*identical* systems all in that same starting quantum state, and
measure energy in each one, you'll get a bunch of different results.
The results must always be eigenvalues of energy. In problem 3, I
believe there were two possible results, with probabilities 1/3 and 2/3
respectively. So, in an ensemble of identical systems, your *average*
energy measurement would be the expectation value. That's the physical
content. But if I only give you ONE box, then the expectation value is
really just some sort of mathematical fiction, because you only get one
shot at measuring E and you'll only get one of the allowed
eigenvalues!
Bear in mind that many real systems do involve a large number of
identical components, so in the lab it's often quite *reasonable* to
ask questions about the expectation value. (Think "quantum statistical
mechanics")
Probabilities for *single* events are kind of weird. If I say there's a
20% chance of snow tomorrow, what do I mean? (It's either going to snow
or it won't!!) I think it does have some meaning though, beyond just
telling you that weather forecasters stink - given an ensemble of days
where all the weather data is identical to what it looks like right
now, I predict that in 1/5 of those circumstances it will snow the next
day, and in 4/5 it will not. (After enough years of data, you could
*verify* that my 20% prediction had been correct!) With weather,
however, the probability arises because we simply don't collect enough
data, and also our models aren't accurate enough. In quantum, even with
*perfect data* (i.e., being given the wave function) and even though we
believe quantum mechanics is an exact (not approximate) theory, the
probability is still there, it's intrinsic....
Dunno if this helps, let me know if it still doesn't make sense.
An epistemological question (arising from the last answer)
>
>"quantum is an exact theory"?... really? I thought all scientific
>theories were approximations. I don't think I understand how an exact
>theory is even possible! But that may be more of an epistemological
>question anyway.
>
>But still, do you mean we found the exact truth? I thought it colud be
>*very* precise, precise to many decimal places, but in my understanding
>it can never be exact! it is very mysterious.
You got me. One can never *know* if one has an exact theory, of
course.
How about this - quantum mechanics makes predictions for
non-relativistic systems of particles (with actions smaller than or of
order Planck's constant), and in the last 75 years, wherever quantum
mechanics has been applicable, (i.e. wherever the system was simple
enough that you could actually DO the quantum calculations) I know of
NO reproducible experiments or classes of experiments inconsistent with
the theory. Technically, you must extend "quantum mechanics" to
"relativistic quantum field theory" to make my claim really correct,
but that's just a generalization, and quantum mechanics is still the
non-relativistic limit.
Here's another way to phrase it - quantum field theory predictions are
consistent with *all* relevant experiments out to the best accuracy
which has been possible to date, and as those experiments have improved
in accuracy by many MANY factors of 10's over the years, the SAME
theory, without modification, has continued to accurately reproduce the
data. I can't think of a better physical theory than quantum, but
you're right, who knows if it's exact. Maybe the next decimal place
will find a discrepancy. (I suspect that's what most particle
physicists secretly HOPE for, so we have to come up with something
NEW! :-)
Thanks for keepin' me on my toes
More hard questions about the big picture...
>I'm sorry if I'm bothering you -- tell me to stop and I will.
No bother, this is a fun part of teaching quantum II, generating
questions about "real" physics!
>So do you believe in a grand-unificaiton theory of nature?
>I thought the business of unification theories was to find a quantum
>theory of relativity.
Remember the word "relativity" in your sentence above probably means
GENERAL relativity (gravity). When I talk about "relativistic quantum
field theory", the word relativity refers to SPECIAL relativity.
>I remember reading in a newspaper that
>gravitation isn't quantitized, and that was the main goal of GUT's. How
>correct is that? Where do super-strings come in?
Your statement that "gravitation isn't quantized" is correct - string
theory would effectively do that, but there's no manifestly correct
string theory yet. The standard model of particle physics is a
relativistic quantum field theory, i.e. compatible with relativistic
causality, and incorporates three of the four known fundamental forces
(strong, weak, and EM), it's just that it's not adequate to explain
gravity in a way that's consistent with quantum mechanics.
(Fortunately, I can't think of any "quantum" physics systems studied in
labs where gravity matters a hoot!) The standard model deeply unifies
weak and EM forces, but the strong force is just sort of "tacked" on in
a consistent but not well motivated or fundamentally satisfying way.
Gravity is not even included - if you care about gravity, the "standard
model" would say just to use (classical) general relativity.
Do I "believe" in Grand Unification? There's no succesful GUT yet, so
believing in it would be asking if I think there *might* be. There
might be, I don't actually have a strong opinion about it. To me, GUT's
(at least, "old fashioned" ones, like they were working on in the 70's)
involve *deeper* unification of strong, weak, and EM forces than we
currently have. I'm not sure that the word GUT necessarily implies it's
also a theory with quantum gravity in it, though some do this. For
that reason, unification in the context of a string theory seems *more*
promising to me. (By their nature, string theories automatically
incorporate gravity along with the other forces) Also, lots of
*simple* (non-string) GUT's have been worked out that don't agrees with
data (e.g. many predict proton decay at a rate faster than observed in
nature, which must be VERY small, since it hasn't been observed yet!)
There are still different possible ("simple") string theories, some of
then haven't even been worked out yet, which would incorporate gravity,
so I'm mildly optimistic about one of them being an accurate theory of
nature. (But not optimistic enough to actually work on it!)
By the way, "simple" was in quotes there for a reason - string theory
is pretty hard to grasp - my own understanding of it is fairly basic.
Ask me sometime and I can give you *some* crude idea of what a "string
theory" means compared to a regular "field theory" (and for that
matter, the difference between "field theory" and quantum mechanics!)
but if you want real details we'll have to find a local expert to
help.
Great questions, keep them coming!
HW 4 question
>What is H' for #4, is it +e^2/4*Pi*epsilon -3e^2(R^2-r^2)/8*Pi*epsilon*R^3
?
I don't think so, I'm not sure I see where you got this formula from?
(Did you type it in a little wrong?. For instance, your first term has
no r in the denominator [units?], and in your second term I'm not sure
how you could get R^2-r^2, since in the formula in the HW the r^2 has
an extra factor of 1/3 in front of it...)
>I just don't see H', and I need it to use P.T., right?
Yup. H' is *always* H - H0, where H is the *full* perturbed Hamiltonian,
and H0 is the full *un*perturbed Hamiltonian.
Here, H is kinetic energy plus the given potential, V (which differs
from the Coulomb potential only for r < R)
H0 is kinetic plus the usual old Coulomb potential, call it V0.
As always, V0 = -e^2/(4 Pi epsilon r)
So, the kinetic energy term completely *cancels* out in H'=H-H0, and so
does the potential, if r > R! (Leaving H' = 0 for r > R)
For r < R, you have H' = V-V0, where V is the first formula I have
written down in the homework, and V0 is the usual old Coulomb
potential.
Hope this helps get you started! I'll be around this afternoon.
(Don't forget that in the radial integrals you need to do, the
integration is over r^2 dr, not just dr.)
Office hours
>Hi! Can I come by your office around noon? Are you always there
>around noon? (so that I can stop asking every week) thanks
I'll try to come up to campus somewhere around noon or "noon-thirty",
all semester on Mo, Tu, and Wed - there seem to be a few people who
like to stop by then. I can't guarantee it'll be exactly 12, but I'll
bring my lunch, so it shouldn't be much later... (Every other Tues, noon-1
is a faculty meeting, too)
Peer grading?
> I just visited your web page from last year's class and noticed that
> you used a system of peer grading. It seems like a very innovative way
> to make people take homework seriously. I'm interested to know why we
> aren't using it this year. Was it unsuccessful - or simply more
> convenient to have a full time grader? No rush on an answer, just my
> curiosity itching.
Last year's peer grading was indeed quite successful. It was a
*tremendous* amount of extra work for the student the one week they had
to grade. First, I made them generate a solution set entirely on their
own, so that they really *understood* the set pretty deeply. I
typically had 2 meetings with the student grader - first 1 day after
the set was turned in, to go over their "draft solution set", to iron
out any problems, and then again a day or so later to make sure their
solutions were 100% correct. They had a one-week deadline to finish all
the grading, and of course were responsible for keeping up with the
next week's homework at the same time. So, it was quite a labor, but I
got amazingly *no* complaints, and in fact got a lot of real clear
positive feedback about it in the end. (I think people felt they
understood at least ONE homework set backwards and forwards this way!)
I decided not to do it again this semester primarily because of our
class size. Last semester there were 12 students, so each one could
grade exactly one set. We have 16 people this semester, so either a
couple people would simply get out of grading, or we'd have had to
double up graders a couple of weeks. Either scenario would be pretty
unfair, as well as logistically very complicated. So I decided not to
do it again this time around. The fact that we got assigned a very
good and conscientious grader this semester also helped swing my
decision, I must admit!
So, I guess it's good and bad news for you - grading your peer's
homework is a pretty unique and rewarding assignment, but you'd be
surprised how much time and effort it takes. Having your homework
*graded* by peers turns out to be o.k., there are plusses and minuses -
there are almost certainly more cases of wrong grading (in both
directions - unfair points taken *off*, or not *enough* points taken
off!), but I think people compensated by writing slightly neater and
easier to follow homeworks!
Anyway, hope this "scratches" your curiosity. Of course, if you still
*really* want to grade a set, perhaps we could work something out with
the grader :-)
HW 5 comment
In problem 4, I want the shift in the *energy* of the states, not the
shift in the wave functions. Sorry for any confusion about that!
HW 5 question
>Hello Dr. Pollock,
>
> I need to choose an orthogonal basis set for the unperturbed
>system. I immediately chose:
>
> 1. (1/L)*exp[(2Pi)inx/L]
> 2. (1/L)*exp[-(2Pi)inx/L]
> for n=0,1,2,3,...
>
> These are not orthogonal. Is this still alright?
>
> Thank you,
You're right to be concerned in general about orthogonality - at least,
our degenerate perturbation theory formulas were based on the
assumption that the degenerate basis states (psi_a and psi_b) were
orthonormal. But ... why exactly do you say your basis states are
*not* orthogonal? (They look orthogonal to me)
Isn't it true that the integral of *any* member of your (function
1)(complex conjugated) times *any* member of your (function 2)
(integrated from -L/2 to +L/2) vanishes?
Let me know if you still think there's a problem, and why...
HW6 questions
>Hi, Dr. Pollock,
Hi
>In problem 2, I get 8 eigenvalues of : +A, -A, +A, -A (again), +C, -C,
>0 and 0.
Looks like trouble already. For one thing, my matrix is 9x9, so there
should be 9 eigenvalues. The +C and -C ones look wrong, to me, and you
seem to be missing one.
>For the last four values I listed, my eigenvectors consist of
>all zeros. Is this correct?
Eigen*vectors* can never be all zeros, because eigenvectors have to be
normalized! In a couple cases (the last TWO), when the eigenVALUE is
zero, the eigenVector is something simple like a column with many 0's,
but one 1 in the right place. (See below for another example)
>Also, the first two pairs of eigenvalues,
>as well as the last pair, are the same. Should they have separate
>eigenvectors,
Yes, *every* eigenvector must be orthogonal to ALL others!
>or should they be a combination i.e.( +Aeigenvector) +
>(+Aeigenvector-other), therefore looking something like {1/2^.5, 1/2^.5,
>1/2^.5, 1/2^.5, 0,0,0,0,0}in vertical form.
I'm not getting this at all (?) Just remember, if the big matrix is
block diagonal, find the eigenvector of the LITTLE matrix, and then add
zero's EVERYWHERE else in the column. So, e.g. I'd expect your very
first eigenvector to be (1/2^.5, 1/2^.5, 0,0,0,0,0,0,0) (the next one
would have a minus sign on the second term) and the THIRD eigenvector
would be (0,0, 1/2^.5, 1/2^.5, 0,0,0,0,0) Do you see why I put the
zero's where I did? This is the eigenvector of a little 2x2 submatrix
that's located "two down" in the big matrix, so all those zero's are
just like place holders. This is awfully hard to do by computer - if
it doesn't make sense, better come see me!
Hope this helped a little.
> This would mean that
>instead of 8 eigenvectors, there would only be 5. Am I right in these
>assumptions?
Nope, see above. 9 eigenvectors, gotta be!!
>Thanks,
Good luck!
Steve P.
A puzzle
>I just thought of something and I'm wondering if it is conceivable in
>quantum mechanics.
>
>I'm imagining a way to communicate at a rate faster than the speed of
>light. (Therefore it's probably wrong.)
>
>Take a decay of some particle (e.g. pi0 --> e+e- & neutrinos). Now we
>look at the electrons. The spin is undefined before we measure it, and
>we can choose to measure any component. Say we measure Sx on one of
>the electrons. Then because angular momentum must be conserved, the
>positron must be an eigenstate of Sx. If we measure Sz on the
>electron, then the positron must be an eigenstate of Sz.
>
>Therefore, by measuring the spin of the positron, we know, immediately,
>what was measured some distance away. If we measure the spin of the
>positron a time t after we measure the spin of the electron and the two
>measurements are greater than (speed of light)*t away, we have
>propagated information faster than the speed of light.
>
> What's wrong with my thinking?
Your puzzle is a clever and confusing one - it's quite well known in
the literature, and goes by the name "Einstein Podolsky Rosen paradox",
or just "EPR" for short, because of a paper they wrote in the 50's,
with almost exactly your setup. We'll talk about it later this semester
in class (it's discussed in Griffiths) - the bottom line answer is that
quantum mechanics does *not* violate special relativity. Your setup is
not actually transmitting any useful information from one person to
another faster than the speed of light. Person a and person b might be
observing some sort of "faster than c correlations",(which Einstein
referred to as "spooky action at a distance") but person "a" cannot in
ANY way use your experiment to *communicate a message* to person "b"
faster than speed of light. (Person "a" *measures" the spin of the
electron, but doesn't determine it.)
Check out Griffith's appendix A, and we'll talk more about this later
this semester! Or, swing by my office sometime - it's a lot deeper
puzzle than can be resolved in a quick email!
HW7 answer
When trying to use Mathematica to do the expansion problem (out to
order alpha^4), I was having troubles because MMA refused to simplify
obvious things like Sqrt[(j+1/2)^2]. I just discovered a neat Mma
command - PowerExpand[...], which seems to do what I wanted.
Another HW7 question
>
>Hi!
>
>So, in #5, are the transitions between levels of equal values of j?
>It's not very clear from Gritths what happens. If so, then there are
>only two values of delta E for a total of 6 lines? Then there are only
>two different photon frequencies, and two different line colors? It
>doesn't sound right.
>
Griffiths hasn't told you yet about selection rules, so don't worry
about them. In problem 6.16, he wants you to find ALL possible visible
photon frequencies. Since there are 3 distinct fine structure levels
for n=3, and 2 distinct levels for n=2, there should be a total of 6
possible "lines" (or "colors" or "frequencies"). If the fine structure
levels were all equally spaced, you might have some overlap, but
because each fine structure splitting is different from all the others,
I claim you will find 6 *distinct* possible photon colors. (not two)
Hope this helps you get started
HW 12
>Hello Dr. Pollock,
>
> I tried something on problem one and it must be wrong because it
>results in h=0. Below I use P for the wave function (instead of Psi).
>
>H|Pa> = E|Pa>
>
>H0+H'|Pa> = E|Pa>
>
> =
>
> + = E
>
> 0 + h = 0
>
>h=0 !?!
> Doesn't this mean that h is always zero with the assumptions we
>made?
No - h can certainly be nonzero - think of "H0" as a B field in the z
direction, and "H'" as a B field in the x direction, nothing prevents
you from making H' be whatever you like!
It's a little hard to see what you're doing, but I think the problem
might be in your very first step:
You said H|Pa> = E|Pa>
But I don't think that's right. Pa is (presumably by definition) the
eigenfunction of H0 with eigenvalue Ea. So, it cannot ALSO be an
eigenfunction of the entire Hamiltonian, H.
Instead, you should probably think of H0 and H' as matrices. For
example,
H0 = (Ea,0
0 Eb)
Why? Because Pa and Pb are by definition our basis states, i.e. we
write Pa in matrix notation as (1,0) and Pb as (0,1). So, H0 acting on
Pa gives Ea Pa, and H0 acting on Pb gives Eb Pb, as desired.
And, H' = (0,h
h,0)
(Again, the reason is that this is what you NEED to reproduce
Griffiths' 4 matrix elements)
So, H = H0 + H' = (Ea,h
h, Eb)
h can be anything, and the eigenvalues of H are NOT anything simple,
you just have to find them. (And the eigenvectors are NOT Pa and Pb,
but some linear combos)
Does this help you get started?
Cheers,
Steve P.
Matter-antimatter
>I was wondering... when matter and anti-matter collide an explosion
>occurs and energy is released... Is that all that happens? What are the
>consequences of this interaction? I'm not sure if I'm asking the right
>question...
I guess I'm not sure what you're asking, but here goes: when matter and
antimatter interact, they can of course just interact normally, e.g.
electromagnetically, and then nothing catastrophic happens. This is
how positronium (electron + antielectron "atom") can live for awhile,
acting very much like a hydrogen atom, only lighter. But it is also
possible for matter and antimatter to annihilate, which means that the
particles disappear, and by conservation of energy *something* must be
left behind. Typically, matter and antimatter produce photons (at least
two photons are needed to conserve both total momentum and total
energy), and the combined energy of the photons equals the combined
energy of the original matter. Since E^2=p^2c^2+m^2c^4, the original
energy comes from the kinetic energy of the original stuff, and ALSO
(2nd term) from its "rest mass" (minus any potential energy of binding
there might also have been) Typically, that 2nd term dominates, and so
we say E=mc^2 for each particle to start with, and all this energy must
still be in the system after the annihilation. Of course, there are
other ways the energy can be released besides making photons - you can
produce pions, or other mesons, or even new (lighter) matter/antimatter
pairs. (So, e.g, a proton and antiproton might annihilate and produce
an electron anti-electron pair, maybe as well as some pions and
photons...)
Nature conserves certain quantities, like energy and momentum, but also
electric charge, "strangeness", "baryon number" (which counts e.g. how
many protons and neutrons you have) and more. So this tells you a lot
about what can be emitted from matter-antimatter annihilations. If a
proton-antiproton annihilate, the TOTAL initial charge is zero, so you
CANNOT convert the energy into a single electron. You need a system of
total charge zero at the end, so you might e.g. make an electron
antielectron pair.
Matter antimatter annihilation is a very efficient energy conversion
process, where the (huge) rest mass energy of two particles gets
converted almost entirely into photons. Normally, rest mass energy is
kind of inert, you can't do any WORK with it, it's just stuck there as
mass. But, after annihilation, you get photons which certainly can do
work. Since mc^2 is so big, there's a lot of potential energy in a
small amount of stuff! The universe *appears* to be mostly matter (not
much antimatter) so we won't be able to ever get at very much of the
mass energy that's available in principle. (One of the big questions
of cosmology today, by the way, is WHY there is such an imbalance of
matter and antimatter? You'd think by symmetry that there'd be equal
amounts of each) You can also *make* antimatter (which of course costs
energy, at *least* mc^2 worth for each particle you make). This might
someday be a means of energy storage. You get it back later when you
let the antimatter annihilate. Kind of like a battery - this doesn't
*produce* any energy, but might make for efficient transportation of
energy, a la Star Trek...(?)
Dunno if this answered your question, if not, try to reformulate it!
Cheers,
Steve P.
No comment
>Hi!
>
>I was just reading the virtual office hours on the web. I know where all
>the antimatter is. I keep it in my sock drawer next to my 10^6 Tesla
>magnet. Just thought I'd share that info with you--and maybe the rest of
>the class.
An anonymous question about CH. 7
>Q1 Eq. 7.14 You indicated in class that the coulomb potential was in
>the first quadrant and the repulsive potential between the protons was
>in the fourth quadrant. I don't see this, because the coulomb
>potential goes as (-1/x), which is in the first quadrant and the
>proton repulsion goes as (1/x), which is in the first quadrant (Eq.7.14).
>
>Q.2
>Eq. 7.37 -> Eq. 5.28 only the following two cases?
>1. R -> 0 (eq. 7.5, 7.50)
>2. |r_1 - r_2| -> 0 (eq. 5.27)
I'm afraid that I don't understand either question. Since your email to
me was anonymous, I don't know who to respond to about this, so I just
put it here. (If this was your question, please come by my office or
after class, and you can help explain your questions!)
TDPT
>
> If I remember correctly, we could get an exact answer with
>time-independent PT regardless of the size of the perturbation by
>taking the series out really far. This doesn't seem true for
>time-dependent PT. We are assuming that we could use the (n-1)th order
>coefficient in the nth order solution.
>
> Could we get arbitrary accuracy with TDPT like we can with TIPT?
I think the answer is that in both TIPT and TDPT you are NOT guaranteed
to get an exact answer no matter how far out you go. These methods are
both approximations based on the convergence of an infinite series. If
the terms in the series do not satisfy the usual mathematical
requirements for absolute convergence, the series does *not* converge
to the correct answer. (and probably doesn't converge at all)
I mentioned one case of an "asymptotic" series where the series
converged at first (to the correct metastable energy) and then formally
diverged, and then argued that technically that was *also* still
correct because the system was formally unbound. But, there are other
cases (like in QED) where the series may behave like that (convergence
for awhile, then divergence) where the divergence seems INcorrect, like
say for the magnetic moment of the electron. And there are other cases
(like in nuclear physics) where the first order correction is *larger*
than the starting term, and the 2nd order correction is *larger* still,
and so the series doesn't have a chance of converging. In such a case,
TIPT and TDPT are both useless, you need some other non-perturbative
approximation.
As long as the perturbation is small, so that the series formally
converges, I think that both TIPT and TDPT generally work fine, i.e.
you can get arbitrary accuracy by going out sufficiently far. TDPT has
additional problems because there is a *small* parameter (the "lambda"
in the time dependent perturbation, say) but also *time*, which can get
large, so the convergence of TDPT is a little more difficult to prove.
I think if you limit yourself to studying rates (which are time
independent for long times and sinusoidal perturbations) then the
convergence issues for TDPT are essentially identical to those of
TIPT.
Hope this answers your question -
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