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May 16, 1992, 12:24:03 AM5/16/92

to

Hi. I'm a high school chem teacher. My colleague, a Physics

teacher, thought the following test question was a little

too vague to answer. What do you think?

teacher, thought the following test question was a little

too vague to answer. What do you think?

Question: A single atom of a substance known to be unstable

with a half-life of 1 s is produced. After an interval of

1 s the atom has not decayed. Is it now more likely to decay

sometime before the end of the second interval (between

1 s and 2 s) ? Explain.

Thanks for any info. Marsh

Internet: CH...@UWPG02.UWINNIPEG.CA

May 17, 1992, 1:59:35 AM5/17/92

to

ch...@uwpg02.uwinnipeg.ca writes:

>Internet: CH...@UWPG02.UWINNIPEG.CA

I don't see what problem your physics teacher friend had with the

question (though as a physicist myself I appreciate your

capitalization of the word Physics. :D ) The question seems to be

probing the students knowledge of independent assortment, which is

rather fundamental in many fields, including nuclear physics. If this

is what you were trying to test the student on, then keep the

question. If a student has a problem with the question, encourage

him to write his own, more clearly stated version, and then use it. A

little feedback never hurt anyone.

ma...@ugcs.caltech.edu

May 17, 1992, 3:59:23 AM5/17/92

to

In article <maxim.706082375@molest>, ma...@molest.ugcs.caltech.edu (Random Scurve) writes...

>ch...@uwpg02.uwinnipeg.ca writes:

>

>>Hi. I'm a high school chem teacher. My colleague, a Physics

>>teacher, thought the following test question was a little

>>too vague to answer. What do you think?

>

>>Question: A single atom of a substance known to be unstable

>> with a half-life of 1 s is produced. After an interval of

>> 1 s the atom has not decayed. Is it now more likely to decay

>> sometime before the end of the second interval (between

>> 1 s and 2 s) ? Explain.

>

>ch...@uwpg02.uwinnipeg.ca writes:

>

>>Hi. I'm a high school chem teacher. My colleague, a Physics

>>teacher, thought the following test question was a little

>>too vague to answer. What do you think?

>

>>Question: A single atom of a substance known to be unstable

>> with a half-life of 1 s is produced. After an interval of

>> 1 s the atom has not decayed. Is it now more likely to decay

>> sometime before the end of the second interval (between

>> 1 s and 2 s) ? Explain.

>

> >I don't see what problem your physics teacher friend had with the

>question (though as a physicist myself I appreciate your

>question (though as a physicist myself I appreciate your

I do. How can an atom be more or less likely to decay in the second

time interval depending on whether on not it decayed in the first time

interval? If it decayed in the first time interval, the experiment

is over. There is no answer to the question "what is the probability

of an atom decaying in the second time interval after it decayed in the

first time interval?", so there is no meaning to the question of whether

the probability is more or less if the atom did not decay.

Nevertheless, the *intended* question is a good one. If you flip

a coin N times, and get N heads, what is the probability of getting

tails on the next flip? This question addresses very common misconceptions

about probability that inexperienced students tend to have.

So I suggest that the question be recast into usable form.

-Scott

--------------------

Scott I. Chase "The question seems to be of such a character

SIC...@CSA2.LBL.GOV that if I should come to life after my death

and some mathematician were to tell me that it

Do you like my new quote? -> had been definitely settled, I think I would

immediately drop dead again." - Vandiver

May 18, 1992, 6:03:05 PM5/18/92

to

In article <15MAY92....@uwpg02.uwinnipeg.ca> ch...@uwpg02.uwinnipeg.ca writes:

>Question: A single atom of a substance known to be unstable

> with a half-life of 1 s is produced. After an interval of

> 1 s the atom has not decayed. Is it now more likely to decay

> sometime before the end of the second interval (between

> 1 s and 2 s) ? Explain.

First, let me congratulate the teacher in question for seeking

out second opinions on test question clarity.

Second, the answer to this question MIGHT be YES.

Probability (initially) that the particle decays between

1 s and 2 s, is 0.25; after observing the first second and

seeing that it has not yet decayed, the probability THEN that

the particle will decay between 1 s and 2 s is 0.5.

If, however, it is meant to find whether the probability

of decay before the end of the second second (given that it did

not decay during the first second) is greater than the probability

of the particle decaying during THE FIRST SECOND, then the answer

is NO.

As stated, the question admits both possible interpretations.

John Whitmore

May 18, 1992, 5:05:59 PM5/18/92

to

ch...@uwpg02.uwinnipeg.ca writes:

>Question: A single atom of a substance known to be unstable

> with a half-life of 1 s is produced. After an interval of

> 1 s the atom has not decayed. Is it now more likely to decay

> sometime before the end of the second interval (between

> 1 s and 2 s) ? Explain.

>Question: A single atom of a substance known to be unstable

> with a half-life of 1 s is produced. After an interval of

> 1 s the atom has not decayed. Is it now more likely to decay

> sometime before the end of the second interval (between

> 1 s and 2 s) ? Explain.

"More likely" than what? Than the probability determined at the

beginning of the experiment for it to decay in the [0s,1s] interval?

Than the probability determined at the beginning of the experiment

for it to decay in the [1s,2s] interval? Than the probability that it

will decay in the [1s,2s] interval if it *had* decayed in the [0s,1s]

interval (which was apparently Scott Chase's interpretation)? Than the

probability that a student will pass the test? You can't ask a "more"

question without specifying "more than what".

Richard M. Mathews Lietuva laisva = Free Lithuania

Internet: ric...@locus.com Brivu Latviju = Free Latvia

UUCP: ...!uunet!lcc!richard Eesti vabaks = Free Estonia

MIL/BITNET: richard%l...@UUNET.UU.NET WE DID IT!!!

May 19, 1992, 8:55:25 AM5/19/92

to

In article <1992May18.2...@u.washington.edu>, wh...@milton.u.washington.edu (John Whitmore) writes:

> In article <15MAY92....@uwpg02.uwinnipeg.ca> ch...@uwpg02.uwinnipeg.ca writes:

>

>>Question: A single atom of a substance known to be unstable

>> with a half-life of 1 s is produced. After an interval of

>> 1 s the atom has not decayed. Is it now more likely to decay

>> sometime before the end of the second interval (between

>> 1 s and 2 s) ? Explain.

>

> First, let me congratulate the teacher in question for seeking

> out second opinions on test question clarity.

>

> Second, the answer to this question MIGHT be YES.

> Probability (initially) that the particle decays between

> 1 s and 2 s, is 0.25; after observing the first second and

> seeing that it has not yet decayed, the probability THEN that

> the particle will decay between 1 s and 2 s is 0.5.

>

> [stuff deleted] > In article <15MAY92....@uwpg02.uwinnipeg.ca> ch...@uwpg02.uwinnipeg.ca writes:

>

>>Question: A single atom of a substance known to be unstable

>> with a half-life of 1 s is produced. After an interval of

>> 1 s the atom has not decayed. Is it now more likely to decay

>> sometime before the end of the second interval (between

>> 1 s and 2 s) ? Explain.

>

> First, let me congratulate the teacher in question for seeking

> out second opinions on test question clarity.

>

> Second, the answer to this question MIGHT be YES.

> Probability (initially) that the particle decays between

> 1 s and 2 s, is 0.25; after observing the first second and

> seeing that it has not yet decayed, the probability THEN that

> the particle will decay between 1 s and 2 s is 0.5.

>

> John Whitmore

Hmm. Took me a few minutes to figure out your point but I think

you are right. Your probability of .25 arises because decays in the

first second are not counted. I.e. starting off with 100 particles,

50 decay in the first second, then 25 in the second.

Talking about probabilities with one particle is not intuitive

to me. Either the particle decays or it doesn't. You can't tell when,

where, or why until after it happens. It's only with a large group

of particles that probabilities become useful. Let's face it, if

it were your child's Schroedinger cat in the box you wouldn't describe

the cat's wave function to the child. You'd look in the box and see

if it was time to buy a new cat.

As for this question being given to high school students, it's

a good question to motivate discussion in the classroom and perhaps

as homework, but it doesn't strike me as a very fair test question

simply because of the different interpretations of it's meaning that

are possible, i.e. which probability are you talking about? Versus

the whole ensemble of particles you started out with? versus only those

particles that survived the first second? (no pun intended)

After all, in physics measurements we don't sit around waiting

for one particle to decay. We always deal with multiple events (or

at least we hope to). So probabilities take on a more concrete meaning.

I'd be willing to bet that if you asked the question in the

following way, you'd get more kids understanding it and answering it correctly.

"Given 100 identical particles with a half-life of 1 second, how many

decays would occur during the 2nd second."

Sorry for the tirade.

Mark Balbes

Dept. of Physics

Ohio State University

p.s. One more thing... With my phrasing of the question, you could

even get into a discussion on the errors of the measurements, thus

teaching the students that probabilities are not certainties (hence

the name :-) )

May 19, 1992, 10:26:47 AM5/19/92

to

I sent my original comments on the question directly to the requestor. My

first thoughts were that the primary way of reading the question made it

fair, but I now see the alternative interpretation. I guess it depends a

lot on something we do not know -- which is what was actually discussed

in class on this subject. My own opinion was that it was not really a

suitable question for HS unless the issues related to probability had

been discussed adequately.

first thoughts were that the primary way of reading the question made it

fair, but I now see the alternative interpretation. I guess it depends a

lot on something we do not know -- which is what was actually discussed

in class on this subject. My own opinion was that it was not really a

suitable question for HS unless the issues related to probability had

been discussed adequately.

I also suggested that the class do some experiments on decay with coin

flips that explore correlations such as whether a coin coming up heads

is more likely to come up tails the next time.

Perhaps the most productive thing to do would be to discuss the question

in class, and have the students explain why they gave the answers they

chose -- which would make it easy to decide if the student deserves credit --

or assign a short essay on the problem to those who missed it and thought

it was unfair. Then do the suitable set of experiments to eliminate any

misconceptions identified from the discussion or essays.

--

J. A. Carr | "The New Frontier of which I

j...@gw.scri.fsu.edu | speak is not a set of promises

Florida State University B-186 | -- it is a set of challenges."

Supercomputer Computations Research Institute | John F. Kennedy (15 July 60)

May 19, 1992, 2:10:02 PM5/19/92

to

>Hi. I'm a high school chem teacher. My colleague, a Physics

>teacher, thought the following test question was a little

>too vague to answer. What do you think?

>teacher, thought the following test question was a little

>too vague to answer. What do you think?

>Question: A single atom of a substance known to be unstable

> with a half-life of 1 s is produced. After an interval of

> 1 s the atom has not decayed. Is it now more likely to decay

> sometime before the end of the second interval (between

> 1 s and 2 s) ? Explain.

I think it's a very good question. In fact it gets at a common

misconception about probability, namely the unfounded belief that

probability somehow changes with previous outcomes (the dice are

getting "hot"). The real question is, "Is it fair to require a

particular class to answer it and have their grades depend on

the answer?"

If the teacher has adequately addressed this principle, by all means

ask the question. Otherwise I would like to see the teaching

upgraded. Too many people think that if you toss 3 heads in a row

with a coin, the chance of tails on the next toss increases. (And

professional "gamblers" who know better use this to separate people

from their money.)

May 19, 1992, 4:24:01 PM5/19/92

to

In article <96...@vice.ICO.TEK.COM> ha...@vice.ICO.TEK.COM (Hal Lillywhite) writes:

>In article <15MAY92....@uwpg02.uwinnipeg.ca> ch...@uwpg02.uwinnipeg.ca writes:

>>[...is this a fair question...]>In article <15MAY92....@uwpg02.uwinnipeg.ca> ch...@uwpg02.uwinnipeg.ca writes:

>>Question: A single atom of a substance known to be unstable

>> with a half-life of 1 s is produced. After an interval of

>> 1 s the atom has not decayed. Is it now more likely to decay

>> sometime before the end of the second interval (between

>> 1 s and 2 s) ? Explain.

>

>I think it's a very good question. In fact it gets at a common

>misconception about probability, namely the unfounded belief that

>probability somehow changes with previous outcomes (the dice are

>getting "hot"). The real question is, "Is it fair to require a

>particular class to answer it and have their grades depend on

>the answer?"

Actually, it's a difficult question, since it's dependent

on whether atomic decay is or is not a stationary stochastic

process. (A stationary process is one in which the probability

does not depend on time, or whatever parameter one uses as

the metric; a non-stationary process is one in which the

probability does depend on the metric--e.g., what is the

probability that a balloon will pop at a certain diameter,

as a function of the largest observed diameter).

The difficulty is that atomic decay may or may not be a

stationary process. (Of course, this is a difficulty only

to me, since someone probably knows with smug certainty

that it is or is not. Such is the nature of physical

discovery--wondering just what it is that nature already

knows).

The asker, however, is probably testing knowledge, here,

meaning that the class has been told (probably through

example rather than mathematical definition) that atomic

decay is a stationary process. I.e., nobody's said

"atomic decay is a stationary process," it's probably

more like "it doesn't matter how long it's been, the

probability doesn't change."

My answer, given my detailed ignorance, would be "I don't

know; I don't remember that it's ever been made clear to me

whether atomic decay is or is not a stationary process;"

and I would hope for half-credit for understanding the

principle but lacking the result.

--Blair

"Government is a stationary

process, but that's appropriate

only by dint of the thesaurus."

P.S. Haha. Fooled me. "Half-life" indicates a stationary

process. One-quarter credit.

May 19, 1992, 6:02:14 PM5/19/92

to

I think you mean: "Thanks for any opinion." In this case I'll support you.

I think any student who has been exposed to "the gambler's fallacy" ought

to cough up the answer (no) to this question in reflex fashion. If scien-

tists have to be more careful with language than you have been in this

context then there are too many nitpickers around. Your question is clear

to me.

Leigh

May 19, 1992, 6:11:31 PM5/19/92

to

Now that I have read the first six responses to this request I have not

changed my opinion, initially expressed before reading anyone else's.

My overall impression is that this teacher is unlikely ever again to

seek counsel from us! With the pool of knowledge available here it is a

shame that more reasonable use is not made of it all.

changed my opinion, initially expressed before reading anyone else's.

My overall impression is that this teacher is unlikely ever again to

seek counsel from us! With the pool of knowledge available here it is a

shame that more reasonable use is not made of it all.

Leigh Palmer

May 19, 1992, 7:49:41 PM5/19/92

to

>Too many people think that if you toss 3 heads in a row

>with a coin, the chance of tails on the next toss increases. (And

>professional "gamblers" who know better use this to separate people

>from their money.)

Actually the probability of *heads* increases as the string of heads

increases in length. Without the explicit hypothesis of a fair coin

one must admit the possibility of one which is unfair. Anyone who

bets tails after *ten* consecutive heads is irretrievably idealistic!

Leigh Palmer

May 19, 1992, 8:53:42 PM5/19/92

to

In the referenced article, pal...@sfu.ca (Leigh Palmer) writes:

>>Question: A single atom of a substance known to be unstable

>> with a half-life of 1 s is produced. After an interval of

>> 1 s the atom has not decayed. Is it now more likely to decay

>> sometime before the end of the second interval (between

>> 1 s and 2 s) ? Explain.

>

>>Question: A single atom of a substance known to be unstable

>> with a half-life of 1 s is produced. After an interval of

>> 1 s the atom has not decayed. Is it now more likely to decay

>> sometime before the end of the second interval (between

>> 1 s and 2 s) ? Explain.

>

>I think any student who has been exposed to "the gambler's fallacy" ought

>to cough up the answer (no) to this question in reflex fashion. If scien-

>to cough up the answer (no) to this question in reflex fashion. If scien-

Of course, you always have to remember the "gambler's retort" to the

"gambler's fallacy", but that's probably picking nits (see below).

I'd say that the question is somewhat ambiguous. I understand

perfectly well what the question is probably getting at. However,

there is a way of thinking that says that questions are better when

they have fewer *possible* interpretations.

I prefer questions that lend themselves to that greatest of test

taking aphorisms: Read the question, and then answer it. Therefore,

I would prefer to see the question reworded something like:

Question: A single atom of an isotope known to be unstable

with a half-life of 1s is produced. What is the probability that

it will decay during the interval 0s to 1s? If it does not

decay during the interval 0s to 1s, what is the probability

that it will decay during the interval 1s to 2s?

I wouldn't hesitate to ask about the actual probabilities. If they

don't understand the term "half-life" in the first place, they aren't

going to understand the concept you are getting at either, except by

rote or guessing.

If that's too pedantic sounding, how about:

Question: A single atom of a substance known to be unstable

with a half-life of 1s is produced. What is the chance that

it will decay during the first second? If it does not

decay during the first second, what is the chance

that it will decay during the second second? (If you don't

know the answers, are the 2 chances the same or different? Explain

for partial credit.)

Aside: For those that haven't heard the "gambler's retort", it is

basically summed up by the following story:

An average person, a mathematician, and a gambler see a coin tossed 10

times. Each time, it comes up heads. The three people are then asked

to bet on the outcome of the 11th toss. The average person will

usually bet on tails, because "it's got to come up soon". The

mathematician will bet on either, because the probabilities are the

same. The gambler will bet on heads every time, because at worst the

odds are the same, and you never know, the coin might be fixed.

--

"When you're down, it's a long way up

When you're up, it's a long way down

It's all the same thing

And it's no new tale to tell" ../ray\..

May 20, 1992, 11:31:46 AM5/20/92

to

In article <11...@inews.intel.com> bhou...@hopi.intel.com (Blair P. Houghton) writes:

>In article <96...@vice.ICO.TEK.COM> ha...@vice.ICO.TEK.COM (Hal Lillywhite) writes:

>>In article <15MAY92....@uwpg02.uwinnipeg.ca> ch...@uwpg02.uwinnipeg.ca writes:

>In article <96...@vice.ICO.TEK.COM> ha...@vice.ICO.TEK.COM (Hal Lillywhite) writes:

>>In article <15MAY92....@uwpg02.uwinnipeg.ca> ch...@uwpg02.uwinnipeg.ca writes:

>>>Question: A single atom of a substance known to be unstable

>>> with a half-life of 1 s is produced. After an interval of

>>> 1 s the atom has not decayed. Is it now more likely to decay

>>> sometime before the end of the second interval (between

>>> 1 s and 2 s) ? Explain.

>Actually, it's a difficult question, since it's dependent

>on whether atomic decay is or is not a stationary stochastic

>process. (A stationary process is one in which the probability

>does not depend on time[....])

>The difficulty is that atomic decay may or may not be a

>stationary process.

Houghton has hit the nail on the head here. The question has

been hotly disputed, and I've seen lots of articles on it.

(Unfortunately, I don't know any references off the top of my head.) In

a sloppy way of speaking, the question is, "does a radioactive nucleus

know how old it is?" That is, is its chance of decaying in the next

second, assuming it hasn't decayed yet, a function of how old it is, or

just a constant? Now, I realize that "everybody knows" that radioactive decay

is a stationary process, and that the term "half-life" implicitly

assumes this. On the other hand, if you do the calculations carefully,

you will find that radioactive decay is NOT a stationary process,

although it is VERY CLOSE to being one. I don't know if anyone has

actually observed this effect - it might be very hard.

If people want to flame me for this heresy I suggest that they dig up

the literature first. Here let me just sketch why it can't be a

stationary process. Let psi be the wavefunction of a "new-born

radioactive nucleus", and let P be the projection onto the space of

states in which the nucleus has already decayed. The probability of

decaying after time t is

||P exp(-itH) psi||^2

where H is the Hamiltonian. The claim is that various natural

assumptions imply that this function cannot equal exp(-kt) for t > 0,

where k is positive constant. There are different ways to proceed and

people love to argue about them...

Argument 1: Put the nucleus in a box and assume that only states below

a certain energy matter. By physics, H is bounded below and has pure

point spectrum (i.e. no continuous spectrum). Then we can write psi as

a finite linear combination of eigenstates and calculate

||P exp(-itH) psi||^2

Go ahead, do it - you get a bunch of sines and cosines, not anything

like exp(-kt).

People who know quantum will say, "Okay, sure, you're right, all motions

are quasiperiodic in such a system, what goes around comes around, but

if the box is big the answer might be so close to exp(-kt) that it's

ridiculous to worry about the discrepancy." Fine.

Argument 2: Just assume that psi is an analytic vector for H.

(Mathematicians can explain to physicists why this is reasonable.) Then

||P exp(-itH) psi||^2

is a real-analytic function. If it equals exp(-kt) for t > 0, it has to

equal exp(-kt) for t < 0, which is impossible, because then exp(-kt) >

1!!!

This is a somewhat better argument.

Even better is to do the calculation in a particular case, making as few

approximations as possible, and see what you get. This is done in the

literature.

------

I realize this is getting technical but I think it's worth posting to

sci.math and sci.edu because radioactive decay is often used to

illustrate exponential decay, and, while this is an excellent

approximation and there's certainly nothing to be ashamed of about using

it, it's interesting to note that there are controversies about this.

May 20, 1992, 12:28:14 PM5/20/92

to

pal...@sfu.ca (Leigh Palmer) writes:

>Actually the probability of *heads* increases as the string of heads

>increases in length. Without the explicit hypothesis of a fair coin

>one must admit the possibility of one which is unfair. Anyone who

>bets tails after *ten* consecutive heads is irretrievably idealistic!

This, of course, is one of the classical problems of Bayes theory; If you

suspect the coin is two-headed with probability p, then what probability

should you ascribe to the coin being two-headed after it comes up heads?

A simple application of Bayes theorem gives the result. (The Monty Hall

problem, which seems to bemuse people periodically, is another simple

application)

May 20, 1992, 3:14:58 PM5/20/92

to

I have recently (on the three boards I posted this to)

offered to email my file explaining the basics of evolutionary

biology to anyone who wanted to read it. I've sent many copies out,

but I have had mail bounce on a few requests. If you mailed

me and did not recieve a reply, send another message and include a

valid path in the body of your message. Alternately, I recently

posted the file on the newsgroup talk.origins. It is in two parts

titled "the theory of evolution" and "the theory of evolution (II)".

Here is the outline of the file:

----------------------------------------------------------

EVOLUTIONARY BIOLOGY

Introduction

What is evolution?

What isn't evolution?

What evolution isn't

Genetic variation

How is genetic variation described?

How much genetic variation is there?

Evolution within a lineage -- anagenesis

Mechanisms that decrease genetic variation

Natural selection

Sexual selection

Genetic drift

Mechanisms that increase genetic variation

Mutation

Recombination

Gene flow

Overview of anagenesis

Evolution among lineages -- cladogenesis

Speciation -- increasing biological diversity

Modes of speciation

Observed speciations

Macroevolution vs. microevolution

Extinction -- decreasing biological diversity

"Ordinary" extinctions

Mass extinctions

Conclusion

-------------------------------------------------------------

Chris Colby --- email: co...@bu-bio.bu.edu ---

"'My boy,' he said, 'you are descended from a long line of determined,

resourceful, microscopic tadpoles--champions every one.'"

--Kurt Vonnegut from "Galapagos"

May 20, 1992, 6:48:04 PM5/20/92

to

pal...@sfu.ca (Leigh Palmer) writes:

>In article <15MAY92....@uwpg02.uwinnipeg.ca> ch...@uwpg02.uwinnipeg.ca

>writes:

>>

>>Question: A single atom of a substance known to be unstable

>> with a half-life of 1 s is produced. After an interval of

>> 1 s the atom has not decayed. Is it now more likely to decay

>> sometime before the end of the second interval (between

>> 1 s and 2 s) ? Explain.

>>

>>Thanks for any info. Marsh

>I think you mean: "Thanks for any opinion." In this case I'll support you.

>I think any student who has been exposed to "the gambler's fallacy" ought

>to cough up the answer (no) to this question in reflex fashion. If scien-

>tists have to be more careful with language than you have been in this

>context then there are too many nitpickers around. Your question is clear

>to me.

The fact that the question is clear to you may be due to the fact that you

know too much 8^). Let me point out a couple of things. The question is

misleading in bringing up 'half-life'. Half-life or mean lifetime is a

statement about a sample of identical nuclei, it has no relevance to a

single nuclei. One has absolutely NO idea when a particular nuclei will

decay so all one-second time intervals are equally likely. Unless the

question was devised to expose the gambler's fallacy let me suggest a better

one.

Given N identical nuclei with mean lifetime = 1 sec what percentage decay in

the first second. Of those remaining after one second, what percentage

decay in the next second?

The answer is surprising and enlightening.. At least it was to me the first

time I encountered it.

Mark

May 20, 1992, 7:07:51 PM5/20/92

to

pal...@sfu.ca (Leigh Palmer) writes:

No. Ten heads in a row is not *that* rare. I like Martingales up to five

some times -- you win often enough to keep it interesting. (Can't go

higher than that because of house limits).

Mark

May 20, 1992, 3:44:08 PM5/20/92

to

In article <1992May20.1...@galois.mit.edu>, jb...@zermelo.mit.edu (John C. Baez) writes:

>

> Houghton has hit the nail on the head here. The question has

> been hotly disputed, and I've seen lots of articles on it.

> (Unfortunately, I don't know any references off the top of my head.) In

> a sloppy way of speaking, the question is, "does a radioactive nucleus

> know how old it is?" That is, is its chance of decaying in the next

> second, assuming it hasn't decayed yet, a function of how old it is, or

> just a constant? Now, I realize that "everybody knows" that radioactive decay

> is a stationary process, and that the term "half-life" implicitly

> assumes this. On the other hand, if you do the calculations carefully,

> you will find that radioactive decay is NOT a stationary process,

> although it is VERY CLOSE to being one. I don't know if anyone has

> actually observed this effect - it might be very hard.

>

Yes, in principle NO DECAY IS EXPONENTIAL - there's always a power law.>

> Houghton has hit the nail on the head here. The question has

> been hotly disputed, and I've seen lots of articles on it.

> (Unfortunately, I don't know any references off the top of my head.) In

> a sloppy way of speaking, the question is, "does a radioactive nucleus

> know how old it is?" That is, is its chance of decaying in the next

> second, assuming it hasn't decayed yet, a function of how old it is, or

> just a constant? Now, I realize that "everybody knows" that radioactive decay

> is a stationary process, and that the term "half-life" implicitly

> assumes this. On the other hand, if you do the calculations carefully,

> you will find that radioactive decay is NOT a stationary process,

> although it is VERY CLOSE to being one. I don't know if anyone has

> actually observed this effect - it might be very hard.

>

at very long times. There was a summary article in _Nature_ about 2-3 years

ago, I think. According to that article the power-law tail has never veen

observed for radiative or radioactive decay, because it sets in at about

40-50 half-lives, which leads to S/N problems...

This is a classica example of "results that are very well known by a very

small group", but somehow have never diffused into the mainstream. The

theoretical arguments go back at least to the 1950's.

For those of you who think of resonances as poles of the S-Matrix - there's

always a branch cut, and exponential decay arises only when you ignore it.

Robert

May 20, 1992, 12:55:58 PM5/20/92

to

> ||P exp(-itH) psi||^2

Stated more precisely, this is the probability of observing a decayed

state at time t. Note also that the "wavefunction" must also include

any fields that are involved in the decay, e.g. electromagnetic and

neutrino in many cases.

>where H is the Hamiltonian. The claim is that various natural

>assumptions imply that this function cannot equal exp(-kt) for t > 0,

>where k is positive constant. There are different ways to proceed and

>people love to argue about them...

> [ Argument 1 deleted ]

>Argument 2: Just assume that psi is an analytic vector for H.

>(Mathematicians can explain to physicists why this is reasonable.) Then

> ||P exp(-itH) psi||^2

>is a real-analytic function. If it equals exp(-kt) for t > 0, it has to

>equal exp(-kt) for t < 0, which is impossible, because then exp(-kt) >

>1!!!

>This is a somewhat better argument.

Actually, what is meant is, if it equals 1 - exp(-kt) for t > 0, it has to

equal 1 - exp(-kt) < 0 for t < 0.

You don't have to assume analyticity, just that P psi is in the domain of H.

If f(t) = ||P exp(-itH) psi||^2, then f'(0) = i <H psi, P psi> - i <P psi, H psi>

= 0 since P psi = 0. Thus f(t) can't be 1 - exp(-kt).

This argument gives you a little more than the others, since it describes

one way in which the decay is not exponential: because of the initial condition,

i.e. the assumption that the particle has definitely not decayed at t = 0,

the decay rate must start at 0.

>Even better is to do the calculation in a particular case, making as few

>approximations as possible, and see what you get. This is done in the

>literature.

>------

>I realize this is getting technical but I think it's worth posting to

>sci.math and sci.edu because radioactive decay is often used to

>illustrate exponential decay, and, while this is an excellent

>approximation and there's certainly nothing to be ashamed of about using

>it, it's interesting to note that there are controversies about this.

--

Robert Israel isr...@math.ubc.ca

Department of Mathematics or isr...@unixg.ubc.ca

University of British Columbia

Vancouver, BC, Canada V6T 1Y4

May 21, 1992, 7:08:35 PM5/21/92

to

In article <north.706402084@watop> no...@watop.nosc.mil (Mark North) writes:

>pal...@sfu.ca (Leigh Palmer) writes:

>

>>In article <15MAY92....@uwpg02.uwinnipeg.ca> ch...@uwpg02.uwinnipeg.ca

>>writes:

>>>

>>>Question: A single atom of a substance known to be unstable

>>> with a half-life of 1 s is produced. After an interval of

>>> 1 s the atom has not decayed. Is it now more likely to decay

>>> sometime before the end of the second interval (between

>>> 1 s and 2 s) ? Explain.

>>>

>>>Thanks for any info. Marsh

>

>>I think you mean: "Thanks for any opinion." In this case I'll support you.

>>I think any student who has been exposed to "the gambler's fallacy" ought

>>to cough up the answer (no) to this question in reflex fashion. If scien-

>>tists have to be more careful with language than you have been in this

>>context then there are too many nitpickers around. Your question is clear

>>to me.

>

>The fact that the question is clear to you may be due to the fact that you

>know too much 8^). Let me point out a couple of things. The question is

>misleading in bringing up 'half-life'. Half-life or mean lifetime is a

>statement about a sample of identical nuclei, it has no relevance to a

>single nuclei. One has absolutely NO idea when a particular nuclei will

>decay so all one-second time intervals are equally likely. Unless the

>question was devised to expose the gambler's fallacy let me suggest a better

>one.

>pal...@sfu.ca (Leigh Palmer) writes:

>

>>In article <15MAY92....@uwpg02.uwinnipeg.ca> ch...@uwpg02.uwinnipeg.ca

>>writes:

>>>

>>>Question: A single atom of a substance known to be unstable

>>> with a half-life of 1 s is produced. After an interval of

>>> 1 s the atom has not decayed. Is it now more likely to decay

>>> sometime before the end of the second interval (between

>>> 1 s and 2 s) ? Explain.

>>>

>>>Thanks for any info. Marsh

>

>>I think you mean: "Thanks for any opinion." In this case I'll support you.

>>I think any student who has been exposed to "the gambler's fallacy" ought

>>to cough up the answer (no) to this question in reflex fashion. If scien-

>>tists have to be more careful with language than you have been in this

>>context then there are too many nitpickers around. Your question is clear

>>to me.

>

>The fact that the question is clear to you may be due to the fact that you

>know too much 8^). Let me point out a couple of things. The question is

>misleading in bringing up 'half-life'. Half-life or mean lifetime is a

>statement about a sample of identical nuclei, it has no relevance to a

>single nuclei. One has absolutely NO idea when a particular nuclei will

>decay so all one-second time intervals are equally likely. Unless the

>question was devised to expose the gambler's fallacy let me suggest a better

>one.

I think that is the intent of the question. Of course you are correct about

the misapplication of "half-life" to a single nucleus. I read the dependent

clause in the sentence above to be "of a substance known to be unstable

with a half-life of 1 s". The half-life pertains to the substance. That

potential ambiguity is easily resolved by the student who knows too much.

>Given N identical nuclei with mean lifetime = 1 sec what percentage decay in

>the first second. Of those remaining after one second, what percentage

>decay in the next second?

I like your modification, with the exception of the ugly neologism

"percentage".

What's wrong with "how many" or "what fraction"? I still consider the original

question to be utterly kosher.

>The answer is surprising and enlightening.. At least it was to me the first

>time I encountered it.

I agree whole-heartedly. The other shocker is the fact that the mean free

path of a gas molecule is the same number as the mean distance that a molecule

will travel before its next collision, and is also the mean distance it has

traveled since its last collision. The apparent factor of two discrepancy can

keep students actively involved in the intellectual process as well as any

conundrum I know.

A clear understanding of the rudiments of probability is essential to the

understanding of physical theory as it is cast in our culture. I believe the

concepts to be simple, but they can be deceptively hard to teach. They do

yield lots of curious, often counterintuitive, results which can be used to

stimulate enlightening discussion.

Leigh Hunt Palmer

May 21, 1992, 7:33:54 PM5/21/92

to

In article <1992May20...@cubldr.colorado.edu>

Well, my esteemed colleagues, both of you theoreticians: I believe that you

are philosophically challenged! The exponential radioactive decay "law" is

an *empirical* one, like Hooke's Law. The fact that the result is readily

derived on the basis of a poorly founded simple hypothesis has seduced you.

In fact the decay of a sample of radioactive material is not exponential.

It is a stochastic process consisting of discrete events. The exponential

"law" describes the behaviour of a hypothetical ensemble of samples, and if

that is flawed it can probably never be experimentally demonstrated, so

where's the beef?

But I *am* sincerely glad you exposed me to this arcanum, John. On behalf of

all of us who were formerly unaware of this, I thank you.

Leigh

May 22, 1992, 1:34:45 AM5/22/92

to

In article <1992May21....@sfu.ca> pal...@sfu.ca (Leigh Palmer) writes:

>Well, my esteemed colleagues, both of you theoreticians: I believe that you

>are philosophically challenged! The exponential radioactive decay "law" is

>an *empirical* one, like Hooke's Law. The fact that the result is readily

>derived on the basis of a poorly founded simple hypothesis has seduced you.

>In fact the decay of a sample of radioactive material is not exponential.

>It is a stochastic process consisting of discrete events. The exponential

>"law" describes the behaviour of a hypothetical ensemble of samples, and if

>that is flawed it can probably never be experimentally demonstrated, so

>where's the beef?

I'm not quite sure what you mean. My point was simply that the usual

quantum-mechanical "derivations" of the exponential decay law for

excited states (radioactive nuclei, unstable particles, atoms with

excited electrons, etc..) are sloppy, that these calculations are

approximate, and that it's easy to see that they can't be exact.

Moreover it should be possible to experimentally falsify the exponential

law, but I don't think anyone has done so. This is where the beef

is for an experimentalist.

The exponential law, like Hooke's law, could be viewed simply a case

of fitting a curve with the first function that comes to mind. "Going

to zero ever more slowly? Let's try an exponential!" (Did they ever try

a Gaussian?) But ideologies tend to get built up supporting these

simple laws. We all know how physics was wedded to the paradigm of

linearity and how it's struggling to free itself from that now. The

paradigm of stationary processes and exponential decay is a similar sort

of thing.

>But I *am* sincerely glad you exposed me to this arcanum, John. On behalf of

>all of us who were formerly unaware of this, I thank you.

>

>Leigh

Quite welcome!

May 22, 1992, 10:47:51 AM5/22/92

to

pal...@sfu.ca (Leigh Palmer) writes:

[north writes]

>>Given N identical nuclei with mean lifetime = 1 sec what percentage decay in

>>the first second. Of those remaining after one second, what percentage

>>decay in the next second?

>I like your modification, with the exception of the ugly neologism

>"percentage".

>What's wrong with "how many" or "what fraction"? I still consider the original

>question to be utterly kosher.

Nothing. I *hate* it when I do that.

Mark

May 22, 1992, 2:18:37 PM5/22/92

to

In article <1992May22....@galois.mit.edu> jb...@nevanlinna.mit.edu (John C. Baez) writes:

>The exponential law, like Hooke's law, could be viewed simply a case

>of fitting a curve with the first function that comes to mind. "Going

>to zero ever more slowly? Let's try an exponential!" (Did they ever try

>a Gaussian?) But ideologies tend to get built up supporting these

>simple laws. We all know how physics was wedded to the paradigm of

>linearity and how it's struggling to free itself from that now. The

>paradigm of stationary processes and exponential decay is a similar sort

>of thing.

>The exponential law, like Hooke's law, could be viewed simply a case

>of fitting a curve with the first function that comes to mind. "Going

>to zero ever more slowly? Let's try an exponential!" (Did they ever try

>a Gaussian?) But ideologies tend to get built up supporting these

>simple laws. We all know how physics was wedded to the paradigm of

>linearity and how it's struggling to free itself from that now. The

>paradigm of stationary processes and exponential decay is a similar sort

>of thing.

Hmm.

I thought the exponential decay for atoms was at least

theoretical, and the curve-fitting served to verify

the theory, viz, that the rate of decay is directly

proportional to the number of undecayed atoms. The papers

should be about 75 years old, by now.

This feature is fairly obvious, in light of the law of

large numbers. What's also fairly obvious is that for

small numbers the theory will no longer hold, and that

certain configurations of atoms (not all necessarily the

same kind of atoms) can preserve the number of undecayed

atoms, and adding energy or particles can control the

number of undecayed atoms.

--Blair

"Heisenberg's a piker when

it comes to uncertainty;

at least, next to me..."

May 22, 1992, 3:57:06 PM5/22/92

to

>In article <1992May22....@galois.mit.edu> jb...@nevanlinna.mit.edu (John C. Baez) writes:

>>The exponential law, like Hooke's law, could be viewed simply a case

>>of fitting a curve with the first function that comes to mind. "Going

>>to zero ever more slowly? Let's try an exponential!" (Did they ever try

>>a Gaussian?) But ideologies tend to get built up supporting these

>>simple laws. We all know how physics was wedded to the paradigm of

>>linearity and how it's struggling to free itself from that now. The

>>paradigm of stationary processes and exponential decay is a similar sort

>>of thing.

>

>Hmm.

>

>I thought the exponential decay for atoms was at least

>theoretical, and the curve-fitting served to verify

>the theory, viz, that the rate of decay is directly

>proportional to the number of undecayed atoms. The papers

>should be about 75 years old, by now.

>

>>The exponential law, like Hooke's law, could be viewed simply a case

>>of fitting a curve with the first function that comes to mind. "Going

>>to zero ever more slowly? Let's try an exponential!" (Did they ever try

>>a Gaussian?) But ideologies tend to get built up supporting these

>>simple laws. We all know how physics was wedded to the paradigm of

>>linearity and how it's struggling to free itself from that now. The

>>paradigm of stationary processes and exponential decay is a similar sort

>>of thing.

>

>Hmm.

>

>I thought the exponential decay for atoms was at least

>theoretical, and the curve-fitting served to verify

>the theory, viz, that the rate of decay is directly

>proportional to the number of undecayed atoms. The papers

>should be about 75 years old, by now.

>

Exponential decay is in fact a theoretical necessity. It is a generic

quantum mechanical feature of problems in which you have a discrete

state (e.g. an excited atomic or nuclear state) coupled to a continuum

of states (e.g. the atomic or nuclear system in the ground state and

an emitted photon flitting around somewhere). There is nothing ad hoc

about it. The original paper is Weisskopf & Wigner, 1930, Z. Physik,

63, 54. If you can't get a translation from German, (or don't speak

German), see Gasiorowicz, "Quantum Physics" 1974 (Wiley), pp 473-480,

or Cohen-Tannoudji, Diu, & Laloe, "Quantum Mechanics" 1977 (Wiley),

vol 2, pp 1343-1355.

The essence of the result is the effective modification of the energy of

the excited state by a small complex perturbation, E --> E + (dE - i*R/2)

where dE is the small radiative energy correction (Lamb shift) and R

is the decay rate. The time dependent phase factor is thus also modified:

exp(-i*E*t) --> exp[-i*(E+dE)*t]exp[-R*t/2]. This is the source of the decay;

probabilities, which go as the square of the amplitudes, will exhibit a time

dependence exp[-R*t].

Carlo

-----------------------------------------------------------------------------

Carlo Graziani | Warning: some of the above may not be a fully |

Dept. of Physics | correct representation of profound cosmic truth.|

University of Chicago | |

| "It's fun to have fun, but you have to know how!"|

ca...@nu.uchicago.edu | -The Cat in the Hat |

-----------------------------------------------------------------------------

May 22, 1992, 6:34:47 PM5/22/92

to

In article <north.706402084@watop> no...@watop.nosc.mil (Mark North) writes:

>pal...@sfu.ca (Leigh Palmer) writes:

>

>>In article <15MAY92....@uwpg02.uwinnipeg.ca> ch...@uwpg02.uwinnipeg.ca

>>writes:

>>>

>>>Question: A single atom of a substance known to be unstable

>>> with a half-life of 1 s is produced. After an interval of

>>> 1 s the atom has not decayed. Is it now more likely to decay

>>> sometime before the end of the second interval (between

>>> 1 s and 2 s) ? Explain.

>>>

>

>

>>In article <15MAY92....@uwpg02.uwinnipeg.ca> ch...@uwpg02.uwinnipeg.ca

>>writes:

>>>

>>>Question: A single atom of a substance known to be unstable

>>> with a half-life of 1 s is produced. After an interval of

>>> 1 s the atom has not decayed. Is it now more likely to decay

>>> sometime before the end of the second interval (between

>>> 1 s and 2 s) ? Explain.

>>>

>

>The fact that the question is clear to you may be due to the fact that you

>know too much 8^). Let me point out a couple of things. The question is

>misleading in bringing up 'half-life'. Half-life or mean lifetime is a

>statement about a sample of identical nuclei, it has no relevance to a

>single nuclei. One has absolutely NO idea when a particular nuclei will

>decay so all one-second time intervals are equally likely. Unless the

>question was devised to expose the gambler's fallacy let me suggest a better

>one.

>

>Given N identical nuclei with mean lifetime = 1 sec what percentage decay in

>the first second. Of those remaining after one second, what percentage

>decay in the next second?

>

>The answer is surprising and enlightening.. At least it was to me the first

>time I encountered it.

>

>Mark

>know too much 8^). Let me point out a couple of things. The question is

>misleading in bringing up 'half-life'. Half-life or mean lifetime is a

>statement about a sample of identical nuclei, it has no relevance to a

>single nuclei. One has absolutely NO idea when a particular nuclei will

>decay so all one-second time intervals are equally likely. Unless the

>question was devised to expose the gambler's fallacy let me suggest a better

>one.

>

>Given N identical nuclei with mean lifetime = 1 sec what percentage decay in

>the first second. Of those remaining after one second, what percentage

>decay in the next second?

>

>The answer is surprising and enlightening.. At least it was to me the first

>time I encountered it.

>

>Mark

So what is the answer??!!!!!!

May 23, 1992, 4:59:19 AM5/23/92

to

>>>Question: A single atom of a substance known to be unstable

>>> with a half-life of 1 s is produced. After an interval of

>>> 1 s the atom has not decayed. Is it now more likely to decay

>>> sometime before the end of the second interval (between

>>> 1 s and 2 s) ? Explain.

>>> with a half-life of 1 s is produced. After an interval of

>>> 1 s the atom has not decayed. Is it now more likely to decay

>>> sometime before the end of the second interval (between

>>> 1 s and 2 s) ? Explain.

Blair P. Houghton writes:

>Actually, it's a difficult question, since it's dependent

>on whether atomic decay is or is not a stationary stochastic

>process. (A stationary process is one in which the probability

>does not depend on time[....])

>The difficulty is that atomic decay may or may not be a

>stationary process.

John Baez writes:

: Houghton has hit the nail on the head here. The question has

: been hotly disputed, and I've seen lots of articles on it.

: (Unfortunately, I don't know any references off the top of my head.) In

: a sloppy way of speaking, the question is, "does a radioactive nucleus

: know how old it is?"

Does a tossed coin know how many times it previously came up heads/ tails?

(Well, I hope not anyway.)

Does a man who was struck by lightning fear for his life more during the

next thunderstorm?

Of course. Should he?

(Well, I hope not anyway.)

--------------------------------------------------------------------------

| lord snooty @the giant | wolfpack, silver, down to the water |

| poisoned electric head | andrew palfr...@cup.portal.com |

--------------------------------------------------------------------------

May 23, 1992, 6:36:56 AM5/23/92

to

Carlo Graziani writes:

: The time dependent phase factor is thus also modified:

: exp(-i*E*t) --> exp[-i*(E+dE)*t]exp[-R*t/2]. This is the source of the decay;

: probabilities, which go as the square of the amplitudes, will exhibit a time

: dependence exp[-R*t].

: The time dependent phase factor is thus also modified:

: exp(-i*E*t) --> exp[-i*(E+dE)*t]exp[-R*t/2]. This is the source of the decay;

: probabilities, which go as the square of the amplitudes, will exhibit a time

: dependence exp[-R*t].

O Ho. Looks like a non-stationary process to me now. What happened? :-)

May 23, 1992, 10:03:47 AM5/23/92

to

In article <59...@cup.portal.com> lordS...@cup.portal.com (Andrew - Palfreyman) writes:

>John Baez writes:

>: Houghton has hit the nail on the head here. The question has

>: been hotly disputed, and I've seen lots of articles on it.

>: (Unfortunately, I don't know any references off the top of my head.) In

>: a sloppy way of speaking, the question is, "does a radioactive nucleus

>: know how old it is?"

>

>Does a tossed coin know how many times it previously came up heads/ tails?

>(Well, I hope not anyway.)

>John Baez writes:

>: Houghton has hit the nail on the head here. The question has

>: been hotly disputed, and I've seen lots of articles on it.

>: (Unfortunately, I don't know any references off the top of my head.) In

>: a sloppy way of speaking, the question is, "does a radioactive nucleus

>: know how old it is?"

>

>Does a tossed coin know how many times it previously came up heads/ tails?

>(Well, I hope not anyway.)

I hope this response isn't a fancy way of saying that the answer to my

question is obviously NO, because I went on to show the answer was most

likely YES, although the effect is minute.

As for your coin tossing question, here too the answer is likely to be

YES, although again the effect is minute and no doubt utterly swamped by

other effects. If a newly minted coin is thrown and lands heads up, the

"tails" side is probably worn away a little bit from hitting the surface

it landed on. This will affect the probabilities in the future.

In case anyone doubts my sanity, let me repeat that this effect is

obviously utterly negligible. My point is just that stationarity is

almost always an approximation, sometimes an excellent one, sometimes

not so hot. There is a whole subject of failure in which one studies

the rate at which lightbulbs burn out, car parts break, etc., and this

provides lots of interesting examples of nonstationary stochastic

processes.

To digress...

In Knuth's book Concrete Mathematics it says that if you SPIN a brand-new penny

it is noticeably more likely to land heads up than tails up. It needs

to be new. Supposedly this is a way to make money in bars. I urge

anyone to verify this before quitting school.

May 23, 1992, 11:39:17 AM5/23/92

to

In article <1992May22.1...@midway.uchicago.edu> jf...@midway.uchicago.edu writes:

>Exponential decay is in fact a theoretical necessity. It is a generic

>quantum mechanical feature of problems in which you have a discrete

>state (e.g. an excited atomic or nuclear state) coupled to a continuum

>of states (e.g. the atomic or nuclear system in the ground state and

>an emitted photon flitting around somewhere). There is nothing ad hoc

>about it. The original paper is Weisskopf & Wigner, 1930, Z. Physik,

>63, 54. If you can't get a translation from German, (or don't speak

>German), see Gasiorowicz, "Quantum Physics" 1974 (Wiley), pp 473-480,

>or Cohen-Tannoudji, Diu, & Laloe, "Quantum Mechanics" 1977 (Wiley),

>vol 2, pp 1343-1355.

>

>The essence of the result is the effective modification of the energy of

>the excited state by a small complex perturbation, E --> E + (dE - i*R/2)

>where dE is the small radiative energy correction (Lamb shift) and R

>is the decay rate. The time dependent phase factor is thus also modified:

>exp(-i*E*t) --> exp[-i*(E+dE)*t]exp[-R*t/2]. This is the source of the decay;

>probabilities, which go as the square of the amplitudes, will exhibit a time

>dependence exp[-R*t].

This is indeed the conventional wisdom. Let me begin by saying:

1) I agree that the exponential decay law is backed up by theory in this

sort of situation and is far from an ad hoc "curve fitting" sort of

thing.

2) The exponential law is apparently an excellent approximation, and as

far as I know no deviations from it have ever been observed. Here I

am not talking about the (necessary) deviations due to finite sample

size. I am talking about deviations present in the limit as the sample

size approaches infinity.

3) If you ever wanted someone to actually calculate a decay rate for

you, I'm sure Graziani would do a whole lot better job than I would.

What follows has nothing to do with the important job of getting an

answer that's good enough for all practical purposes. It is a matter of

principal (my specialty). There's no real conflict.

Okay. So, Graziani has offered the conventional wisdom, what everyone

knows about radioactive decay, that it is a "theoretical necessity".

It's precisely because this is so well-entrenched that I thought I

should point out that one can easily prove that quantum-mechanical decay

processes cannot be EXACTLY exponential. There are approximations in

all of the arguments Graziani cites.

Let me just repeat the proof that decay processes aren't exactly

exponential. It uses one mild assumption, but if the going gets rough I

imagine someone will raise questions about this assumption. It'd be

nice to get a proof with even weaker assumptions; I vaguely recall that

one could use the fact that the Hamiltonian is bounded below to do so.

This is just the proof that Robert Israel gave a while ago (an improved

version of mine).

Let psi be the wavefunction of a "new-born radioactive nucleus",

together with whatever fields that are involved in the decay. Let P be

the projection onto the space of states in which the nucleus has NOT

decayed. Let H be the Hamiltonian, a self-adjoint operator. The

probability that at time t the system will be observed to have NOT

decayed is

||P exp(-itH) psi||^2

The claim is that this function cannot be of the form exp(-kt) for all

t>0, where k is some positive constant.

Just differentiate this function with respect to t and set t = 0.

First, rewrite the function as

<exp(-itH) psi, P exp(-itH) psi>,

and then differentiate to get

<-iH exp(-itH) psi, P exp(-itH) psi> +

<exp(-itH) psi, -iPH exp(-itH) psi>

and set t = 0 to get

<-iH psi, psi> + <psi, -iH psi> = 0

Here we are using P psi = psi. Since we get zero, the function could

not have been equal to exp(-kt) for k nonzero.

That should satisfy any physicist. A mathematician will worry about why

we can differentiate the function. This is a simple issue if you know

about unbounded self-adjoint operators. (Try Reed and Simon's Methods

of Modern Mathematical Physics vol. I: Functional Analysis, and vol. II:

Fourier Analysis and Self-Adjointness.) For the function to be

differentiable it suffices that psi is in the domain of H. For

physicists, this condition means that ||H psi|| < infinity.

[Let me put in a digression only to be read by the most nitpicky of

nitpickers, e.g. myself. An excited state psi, while presumably an

eigenvector for some "free" Hamiltonian which neglects the interactions

causing the decay, is not an eigenvector for the true Hamiltonian H,

which of course is why it doesn't just sit there. One might worry,

then, that the eigenvector psi of the "free" Hamiltonian is not in the

domain of the true Hamiltonian H. This is a standard issue in

perturbation theory and the answer depends on how singular the

perturbation is. Certainly for perturbations that can be treated by

Kato-Rellich perturbation theory any eigenvector of the free Hamiltonian

is in the domain of the true Hamiltonian H, cf. Thm X.13 vol. II R&S.

But I claim that this issue is a red herring, the real point being

that any state we can *actually prepare* has ||H psi|| < infinity.

Instead of arguing about this, I would hope that any mathematical

physicists would just come up with a theorem with weaker hypotheses.]

As Israel pointed out, this argument shows what's going on: when you are

SURE the nucleus has not decayed yet (i.e., it's "new-born"), the decay

rate must be zero; the decay rate then can "ramp up" very rapidly to the

value obtained by the usual approximate calculations.

Physicists occaisionally mistrust mathematicians on matters such as

these. Arcane considerations about the domains of unbounded

self-adjoint operators probably only serve to enhance this mistrust,

which is ironic, of course, since the mathematicians are simply trying

to avoid sloppiness. In any event, just to show that this isn't

something only mathematicians believe in, let me cite the paper:

Time scale of short time deviations from exponential decay, Grotz and

Klapdor, Phys Rev. C 30 (1984), 2098-2100.

"In this Brief Report we discuss critically whether such quantum

mechanically rigorously demanded deviations from the usual decay

formulas may lead to observable effects and give estimates using the

Heisenberg uncertainty relation.

It is easily seen that the exponential decay law following from a

statistical ansatz is only an approximation in a quantum mechanical

description. [Gives essentially the above argument.] So for very small

times, the decay rate is not constant as characteristic for an

exponential decay law, but varies proportional to t. [....]

Equations (2) and (3) tell us that for sufficiently short times, the

decay rate is whatever [arbitrarily - these guys are German] small.

However, to make any quantitative estimate is extremely difficult.

Peres uses the threshold effect to get a quantitative estimate for the

onset of the exponential decay [...] Applying this estimate to double

beta decay yields approximately 10^{-21} sec, which is much too small to

give any measurable effect. [They then go on to argue with Peres.]"

This is all I want to say about this, unless someone has some nice theorems

about the allowed behavior of the function

||P exp(-itH) psi||^2

when H is bounded below and psi is not necessarily in the domain of H.

(This would probably involve extending t to a complex half-plane.)

May 23, 1992, 12:58:21 PM5/23/92

to

>>>>Question: A single atom of a substance known to be unstable

>>>> with a half-life of 1 s is produced. After an interval of

>>>> 1 s the atom has not decayed. Is it now more likely to decay

>>>> sometime before the end of the second interval (between

>>>> 1 s and 2 s) ? Explain.

>

>>>> with a half-life of 1 s is produced. After an interval of

>>>> 1 s the atom has not decayed. Is it now more likely to decay

>>>> sometime before the end of the second interval (between

>>>> 1 s and 2 s) ? Explain.

>

Much to big of a deal has been made out of this problem. In order to solve it, first find the decay constant which is equal to the probability for each nuclus to decay (in this case there is only one).

1/2A(0) = A(0)exp[-kt], t= 1 sec

kt=ln2=0.693

k =0.693 = decay constant

The probability for the atom to decay in the second interval if it does not decay in the first interval is therefore

(1 - 0.693)(0.693) = 0.213

Therefore it is less likely to decay in the second interval since it did not decay in the first interval. Mind you, these are only probabilities- of course the atom does not "know" how old it is.

n

n

n

n

Therefore it is less likely to decay in the second interval since it did not decay in the first interval. Mind you, these are only probabilities- of course the atom does not "know"

D

A

A

May 23, 1992, 3:10:59 PM5/23/92

to

In article <1992May22.1...@midway.uchicago.edu>, jf...@quads.uchicago.edu (Carlo Graziani) writes:

>>

>

> Exponential decay is in fact a theoretical necessity. It is a generic

> quantum mechanical feature of problems in which you have a discrete

> state (e.g. an excited atomic or nuclear state) coupled to a continuum

> of states (e.g. the atomic or nuclear system in the ground state and

> an emitted photon flitting around somewhere). There is nothing ad hoc

> about it. The original paper is Weisskopf & Wigner, 1930, Z. Physik,

> 63, 54. If you can't get a translation from German, (or don't speak

> German), see Gasiorowicz, "Quantum Physics" 1974 (Wiley), pp 473-480,

> or Cohen-Tannoudji, Diu, & Laloe, "Quantum Mechanics" 1977 (Wiley),

> vol 2, pp 1343-1355.

>>

>

> Exponential decay is in fact a theoretical necessity. It is a generic

> quantum mechanical feature of problems in which you have a discrete

> state (e.g. an excited atomic or nuclear state) coupled to a continuum

> of states (e.g. the atomic or nuclear system in the ground state and

> an emitted photon flitting around somewhere). There is nothing ad hoc

> about it. The original paper is Weisskopf & Wigner, 1930, Z. Physik,

> 63, 54. If you can't get a translation from German, (or don't speak

> German), see Gasiorowicz, "Quantum Physics" 1974 (Wiley), pp 473-480,

> or Cohen-Tannoudji, Diu, & Laloe, "Quantum Mechanics" 1977 (Wiley),

> vol 2, pp 1343-1355.

No, exponential decay is only an extremely accurate approximation.

The Weisskopf-Wigner model is

unphysical because it assumes that the energy spectrum of the continuum

is unbounded. As long as there is a lower bound to the spectrum - an energy

zero - the exponential will turn into a power law on a timescale

3*tau*Ln[E*tau/hbar], where tau is the lifetime and E is the energy of

the state, measured from the physical energy zero.

See:

L. Khalfin, JETP 6, 1053 (1958)

R. G. Winter, Phys. REv. _123_, 1503 (1961).

P. T. Greenland, _Nature_ (News and Views) _335_, 298 (1988).

The last article points out an interesting twist, which Blair Houghton

might enjoy: in order to get the nonexponential decay in the laboratory,

you must set up the experiment so that the decay products can recombine.

So if the crossover takes place on a 3 week timescale, you'll need an

awfully big chamber! (And don't watch the pot...) The result _sensitively_

depends upon the "innocent" assumption that the quantum system is isolated.

However, it is not totally irrelevant to the real world - one can

study decay of resonances very close to threshold, so that E is small.

There have been experiments (see the _Nature_ article) but no luck so far.

Robert

May 23, 1992, 4:30:04 PM5/23/92

to

In article <1992May23.1...@galois.mit.edu> jb...@littlewood.mit.edu (John C. Baez) writes:

>In article <59...@cup.portal.com> lordS...@cup.portal.com (Andrew - Palfreyman) writes:

>>John Baez writes:

>>: Houghton has hit the nail on the head here. The question has

>In article <59...@cup.portal.com> lordS...@cup.portal.com (Andrew - Palfreyman) writes:

>>John Baez writes:

>>: Houghton has hit the nail on the head here. The question has

Reminds me of my days as a stage carpenter. I think I'll

install a garage-door opener tomorrow afternoon...

>>: a sloppy way of speaking, the question is, "does a radioactive nucleus

>>: know how old it is?"

>>

>>Does a tossed coin know how many times it previously came up heads/ tails?

>>(Well, I hope not anyway.)

In the case given, in which ten trials provided ten heads

events, a statistical analysis give no reason even to believe

that the coin possesses a tails side...

Bet heads.

--Blair

"Y' move six-teen tons and whaddaya get?"

May 23, 1992, 5:34:30 PM5/23/92

to

Cheri Anaclerio writes:

: Much to big of a deal has been made out of this problem. In order to solve

: it, first find the decay constant which is equal to the probability for each

: nucleus to decay (in this case there is only one).

:

: 1/2A(0) = A(0)exp[-kt], t= 1 sec

: kt=ln2=0.693

: k =0.693 = decay constant

:

: The probability for the atom to decay in the second interval if it does not

: decay in the first interval is therefore

:

: (1 - 0.693)(0.693) = 0.213

:

: Therefore it is less likely to decay in the second interval since it did not

: decay in the first interval. Mind you, these are only probabilities- of

: course the atom does not "know" how old it is.

: Much to big of a deal has been made out of this problem. In order to solve

: it, first find the decay constant which is equal to the probability for each

:

: 1/2A(0) = A(0)exp[-kt], t= 1 sec

: kt=ln2=0.693

: k =0.693 = decay constant

:

: The probability for the atom to decay in the second interval if it does not

: decay in the first interval is therefore

:

: (1 - 0.693)(0.693) = 0.213

:

: Therefore it is less likely to decay in the second interval since it did not

: decay in the first interval. Mind you, these are only probabilities- of

: course the atom does not "know" how old it is.

There are two errors in here (well, three if you count >75 chars/line :).

1. In the second interval, the probabilty of decay (given no decay in

the first interval) is also <k>. You have to divide your k(1-k) by the

total probability in that state. Since the total probability (given no

first decay) is {decays-in-2nd + does-not-decay-in-2nd} = k(1-k) + (1-k)^2

which is (1-k).

2. Your conclusion does not follow from your maths. Because if

"it is less likely to decay in the second" were true, then you *could*

conclude that the atom "knows" how old it is.

May 24, 1992, 11:35:43 AM5/24/92

to

In article <1992May23...@cubldr.colorado.edu> pars...@cubldr.colorado.edu (Robert Parson) writes:

> No, exponential decay is only an extremely accurate approximation.

> The Weisskopf-Wigner model is

> unphysical because it assumes that the energy spectrum of the continuum

> is unbounded. As long as there is a lower bound to the spectrum - an energy

> zero - the exponential will turn into a power law on a timescale

> 3*tau*Ln[E*tau/hbar], where tau is the lifetime and E is the energy of

> the state, measured from the physical energy zero.

> See:

> L. Khalfin, JETP 6, 1053 (1958)

> R. G. Winter, Phys. REv. _123_, 1503 (1961).

> P. T. Greenland, _Nature_ (News and Views) _335_, 298 (1988).

Thanks immensely. This sounds like the kind of thing I was hoping for,

a proof of the nonexistence of exponential decay using the fact that the

Hamiltonian is bounded below. It's neat how here we have the

*long-time* departures from the exponential law, as opposed to

*short-time* departures as in the work I was discussing. These may have

a chance of being seen?

> (And don't watch the pot...)

Heh. Actually, the article I cited did consider the question of

stopping radioactive decay via the quantum Zeno effect, and concluded it

was impractical. It's amusing that the (German) authors called it the

"zenon" effect, and the reference in Reviews of Physics (or whatever

it's called) corrected that to the "xenon" effect.

> There have been experiments (see the _Nature_ article) but no luck so far.

Too bad!

May 24, 1992, 12:16:57 PM5/24/92

to

jb...@littlewood.mit.edu (John C. Baez) writes:

>In article <1992May23...@cubldr.colorado.edu> pars...@cubldr.colorado.edu (Robert Parson) writes:

>> No, exponential decay is only an extremely accurate approximation.

[...]

>> See:

>> L. Khalfin, JETP 6, 1053 (1958)

>> R. G. Winter, Phys. REv. _123_, 1503 (1961).

>> P. T. Greenland, _Nature_ (News and Views) _335_, 298 (1988).

>Thanks immensely. This sounds like the kind of thing I was hoping for,

>a proof of the nonexistence of exponential decay using the fact that the

>Hamiltonian is bounded below. It's neat how here we have the

>*long-time* departures from the exponential law, as opposed to

>*short-time* departures as in the work I was discussing. These may have

>a chance of being seen?

Rolf Winter taught the third semester of undergrad quantum mechanics

at William and Mary, and he discussed this phenomenon once, mentioning

that he likes to ask all kinds of experimentalists whether they think

they might be able to observe the long-time deviations from exponential

decay. He said that (as of 1990) he hadn't yet gotten an affirmative

answer.

Sidney Coleman claimed in his quantum field theory course at Harvard that

some statements about long-time deviations from exponential decay were

either wrong or tautological, but I couldn't follow his argument, and I

don't know whether he was actually referring to this work.

>> There have been experiments (see the _Nature_ article) but no luck so far.

>Too bad!

--

Matt McIrvin mci...@husc.harvard.edu Grad student, Department of

Clean .sig regularly with a lint-free cloth. Physics, Harvard University

May 24, 1992, 8:54:30 PM5/24/92

to

There seems to be an errror in your answer (well, two if you count "maths"):

1. According to your line of reasoning, if you have a fair coin and want

to know the probability of tossing a heads on the first throw and a

tails on the second throw, that probability would be(1 - 1/2)(1/2)

divided by the total probability in the second state (throw a tails

on the second + don't throw a tails on the second: (1 - 1/2)(1/2) +

(1 - 1/2)(1 - 1/2)) which would give you a final answer of 1/2.

Anyone who has taken an undergraduate course in probability knows

that this answer is incorrect. The probability of throwing a heads

and then a tails is (1/2)(1/2) = 1/4. Likewise, in the problem at

hand, in order to find the probability that the atom does not decay

in the first interval but does decay in the second interval is the

probability that it does not decay in the first interval * the proba-

bility of decaying in the second interval which is still k. The

answer, .213, reflects that the atom has a greater likelihood of

decaying in the first interval than not.

May 25, 1992, 12:21:49 AM5/25/92

to

Cheri Anaclerio:

: There seems to be an errror in your answer (well, two if you count "maths"):

British usage. My apologies!

: There seems to be an errror in your answer (well, two if you count "maths"):

British usage. My apologies!

: 1. According to your line of reasoning, if you have a fair coin and want

: to know the probability of tossing a heads on the first throw and a

: tails on the second throw, that probability would be(1 - 1/2)(1/2)

: divided by the total probability in the second state (throw a tails

: on the second + don't throw a tails on the second: (1 - 1/2)(1/2) +

: (1 - 1/2)(1 - 1/2)) which would give you a final answer of 1/2.

Sorry, but it would not. You are calculating the joint probability of

two independent events and therefore don't need the normalisation.

If you calculate the probability of an event in isolation ("k decays in the

second interval") then the sum of this probability and the other possible

outcome(s) in the second interval have to sum to 1. This is what we want to

do when we need to answer the question "does the atom know how old it is?".

Because of your simple model, you should discover a stationary process - the

answer should be "no". Since you don't, there must be a flaw.

So your two errors are:

1. Finding a time-dependency in decay rate (not there for exponential decay).

2. Interpreting this as a stationary process (not, by definition).

As we've seen, the process is in fact non-stationary, and requires a

sophistication beyond exponential decay to evince this property.

: Likewise, in the problem at

: hand, in order to find the probability that the atom does not decay

: in the first interval but does decay in the second interval is the

: probability that it does not decay in the first interval * the proba-

: bility of decaying in the second interval which is still k. The

: answer, .213, reflects that the atom has a greater likelihood of

: decaying in the first interval than not.

Yes indeed, but only if the atom has a memory! - the point at hand.

@lS

May 25, 1992, 5:27:30 AM5/25/92

to

In article <1992May24.1...@galois.mit.edu>

jb...@littlewood.mit.edu (John C. Baez) writes:

jb...@littlewood.mit.edu (John C. Baez) writes:

>> (And don't watch the pot...)

>

>Heh. Actually, the article I cited did consider the question of

>stopping radioactive decay via the quantum Zeno effect, and concluded it

>was impractical. [...]

>> There have been experiments (see the _Nature_ article) but no luck so far.

>

>Too bad!

It has been assumed implicitly in the above, but I think that it should

be made explicit: the assumed Zeno effect involves some sort of belief

in the (controversial) Collapse Postulate. If the frequency f with which

your detector reports "not decayed ... not decayed ... " is high enough,

the decay should become slower (the decay rate goes to zero near t=0),

since the system is assumed to be projected onto the initial state with

the frequency f. (Also assume time translational invariance of the

Hamiltonian.)

(I tend to believe that such a sophisticated measuring device would

noticeably modify the whole interaction dynamics.)

May 25, 1992, 9:16:12 AM5/25/92

to

In article <1992May23.1...@galois.mit.edu> jb...@littlewood.mit.edu (John C. Baez) writes:

>In Knuth's book Concrete Mathematics it says that if you SPIN a

>brand-new penny it is noticeably more likely to land heads up than tails

>up. It needs to be new. Supposedly this is a way to make money in

>bars. I urge anyone to verify this before quitting school.

>In Knuth's book Concrete Mathematics it says that if you SPIN a

>brand-new penny it is noticeably more likely to land heads up than tails

>up. It needs to be new. Supposedly this is a way to make money in

>bars. I urge anyone to verify this before quitting school.

A couple of people said via email that I got this backward. It's

supposed to be more likely to land tails up, they say. I just spun a

1991 penny nine times and it landed tails up only 3 times, which doesn't

prove much. One of them also said if you simply stand it on edge and

let if fall over, it tends to land HEADS up. I have trouble doing this

without feeling like I can make it fall over either way I want. In any

event, I am not planning to earn a living making bets on this trick in bars.

May 25, 1992, 10:11:08 AM5/25/92

to

>It has been assumed implicitly in the above, but I think that it should

>be made explicit: the assumed Zeno effect involves some sort of belief

>in the (controversial) Collapse Postulate.

I don't think so. I believe in the Zeno effect but not in wavefunction

collapse. I darn well BETTER believe in the Zeno effect, by the way,

because it has been observed.

>(I tend to believe that such a sophisticated measuring device would

>noticeably modify the whole interaction dynamics.)

In the experiment which found the Zeno effect I believe this is the

case. An electron in an atom could be in either a high or low energy

state. Actually, the high energy state really consisted of two very

close energy levels. By bouncing back and forth between these the

electron emitted (and absorbed???) visible light, so one could actually

see it glowing when it was in the high energy state. I am vague about

the details of this, but I guess you have to shine visible light at the

atom to "look" and see which state it's in. By "looking" frequently one

can arrest its decay from the high energy state to the low energy state.

If I have the story right, it does seem that "looking" is noticeably

modifying the interaction dynamics.

I wouldn't mind becoming a lot clearer about this experiment than I am

now. I will not field questions like "so why do YOU think the Zeno

effect is happening if its not wavefunction collapse" until someone

tells me the details of the experiment.

One thing I think is a dangerously murky aspect of the way people talk

about QM is the business of "measurement must disturb the system". To

my mind, the old "every action causes an equal and opposite reaction,"

or "if A affects B, B affects A," even combined with quantized action,

is rather different than the uncertainty principle, which follows from

noncommutativity. (By the way, "if A affects B, B affects A" was on a

famous list of heretical propositions issued in Paris in the 1200's.)

For example, we can imagine a toy system with an electron spin up or

down, ^ or v, and a detector either ^, v, or "blank," o. We can imagine

a scattering operator as follows. States will be written with the state

of the electron followed by the state of the detector. The scattering

operator is:

^ x o -> ^ x ^

v x o -> v x v

^ x ^ -> ^ x o

v x ^ -> v x ^

^ x v -> ^ x v

v x v -> v x o

Note that the tensor product system has a basis of 2x3=6 states and I

have just described a UNITARY operator on this space. The point is

this: 1) if one wants to measure the electron's spin, put the detector

in the o state and wait; then the detector will indicate the electron's

spin, 2) the initial state of the detector does NOT affect the final

state of the electron - the spin of the electron does not change. To be

quite pedantic, we may say that ALL the spin operators, S_x, S_y, and

S_z for the electron commute with the above scattering operator. So

there is no kind of Newtonian "if A affects B, B affects A" stuff going

on here: the electron affects the detector but not vice versa. Of

course, this is a mathematical toy model and I don't know if it could be

implemented. If it couldn't, I'd like to know why.

Let's assume for the moment that it can be implemented. Okay, now

prepare an electron in the state (^ + v)/sqrt(2) and run it through the

detector, and see what you get... :-)

I don't think there's any paradox. All I wanted to note was the logical

distinction between "if A affects B, B affects A" and the weirdnesses

specific to QM.

May 25, 1992, 10:47:47 AM5/25/92

to

In the Netherlands one-guilder coins with queen Juliana on them

falls with "tails" (the value) up much more than 50/50. I just

completed a run of 16 tails, 4 heads. Condition is that

you spin the coin fast in a vertical position on a smooth table.

Remarkably the effect is just opposite with old style 2,50 guilder

coins (they come up heads more often).

New style Dutch coins (with the more flat "Beatrix" design) don't

show the phenomenon.

It is said that with practice the percentage "tails" with old style

guilders can be lowered to 10%, but I did not succeed to come

significantly below 30%.

JWN

May 25, 1992, 11:49:47 AM5/25/92