GR0177 #16

Problem

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Lab Methods $\Rightarrow$ }Sample

The mean of the ten number is $\bar{x}=2$ . Thus, the standard deviation of the sample is $\sqrt{\bar{x}}\approx 1.44$ . (Search for Poisson Distribution on the site for another problem similar to this.)

If the student wants to obtain an uncertainty of 1 percent, then

$\sigma/\bar{x^{'}}=1/100=\sqrt{2 C}/(2C)=1/(\sqrt{2C})$ ,

where one assumes the average scales uniformly and C is the time to count. (Note: a good approximation of the uncertainty is given by the ratio of the standard deviation to the average, since that represents the deviation.)

Thus, one has $\sqrt{2 C}\approx 100$ . Thus the student should count C=5000 s.

A Sneak Peek of The Not-So-Boring Review of Undergrad Physics
(To be published in the to-be-posted library section of http://GREPhysics.NET in Feb 2006.)

The Poisson Distribution is intimately related to the raising and lowering operators of the (quantum mechanical) simple harmonic oscillator (SHO). When you hear the phrase ``simple harmonic oscillator," you should immediately recall the number operator $N=a^{\dagger}a$ , as well as the characteristic relations for the raising $a^{\dagger}$ and lowering $a$ and $a^{\dagger} |n\rangle = \sqrt{n+1} |n+1\rangle$ . And, don\'t forget the commutation relations that you should know by heart by now, $[a,a^{\dagger}]=1$ . (That\'s all part of the collective consciousness of being a physics major.)

Now, here\'s some quasi-quantum magic applied to the Poisson Distribution. I\'m going to show you how to arrive at the result for standard deviation, i.e., $\Delta n \equiv \sqrt{\bar{n}}$ from using the SHO operators.

Let\'s start with something easy to help jog your memory: The mean or average number in the distribution is just the expectation value of the Number operator, $\langle N \rangle = \langle n | N | n \rangle = \langle n | a^{\dagger}a|n\rangle \equiv \bar{n}$

Okay! So, on with the fun stuff: the standard deviation is given by the usual definition, $(\Delta n)^2 = \langle N^2 \rangle - \langle N \rangle^2$ .

The second term is already determined from the above expression for the mean, $\langle n \rangle^2 = \bar{n}^2$ .

The first term can be calculated from $\langle N^2 \rangle = \langle n | a^{\dagger}aa^{\dagger}a | n \rangle$ . Now, the commutation relation gives, $[a,a^{\dagger}]=1=aa^{\dagger}-a^{\dagger}a \Rightarrow aa^{\dagger}=a^{\dagger}a+1$ . Replacing the middle two of the four a's with that result, the expression becomes $\langle N^2 \rangle = \langle n | a^{\dagger} (a^{\dagger}a+1) a |n \rangle = \langle n | a^{\dagger}a^{\dagger}aa + a^{\dagger}a |n \rangle = \langle n | a^{\dagger}a^{\dagger}aa | n \rangle + \langle n | a^{\dagger}a | n \rangle = \langle N \rangle^2 + \langle N \rangle = \bar{n}^2 + \bar{n}$ .

Plugging the above results into the standard deviation, I present to you, this: $(\Delta n)^2 = \bar{n}^2 - ( \bar{n}^2+\bar{n} ) = \bar{n} \Rightarrow (\Delta n) = \sqrt{\bar{n}}$ .

It\'s no coincidence that the above works. The secret lies in the energy eigenfunction that you might not remember...

\subsection{The Poisson Distribution Function is just $P_x(n)=|\langle x | n\rangle|^2$ The Poisson Distribution for a parameter $m$ is given by $P_x(\lambda)\propto e^{-m}\frac{m^\lambd}{\lambda!}$ . But, wait, doesn\'t that look a wee bit too familiar? Indeed, the Poisson Distribution is merely the probability of obtaining $n$ photons at position $x$ : $P(n)=|\langle x | n \rangle|^2$ .\footnote{Note that since $E_n=\hbar\omega\left(n+\frac{1}{2}\right)$ , each time n increases, it is like we've created an extra photon, since the energy of a photon is $E=h\nu=\hbar\omega$ . Thus, $n$ represents the quanta.}

Why? So you ask. Well...

The energy eigenfunctions of the SHO is given by $|n\rangle = \frac{(a^{\dagger})^n}{(n!)^{1/2}} |0\rangle$ , where $\langle x |0\rangle \propto e^{-\bar{n}/2}$ .

This result can be arrived at in the position basis as follows: $\langle x | 0 \rangle = \left(\frac{m\omeg}{\pi\hbar}\right)^{1/4} e^{\frac{-m\omega x^2}{2\hbar}}$ . The $x^2$ in the exponent can be re-expressed by the relation $x=\frac{\hbar}{2m\omeg}(a^\dagger+a)$ . Thus, $\langle x^2 \rangle = \frac{\hbar}{2m\omeg}\langle a^\dagger a + a a^\dagger \rangle = \frac{\hbar}{2m\omeg}(2\bar{n})$ , where the other terms like $\langle n |a^2| n\rangle=0$ and we\'ve used the result for $\langle n \rangle = \langle a a^\dagger \rangle=\bar{n}$ from above with the implicit assumption that its complex conjugate is the same, as the average photon number, $\bar{n}$ , is an observable. Thus, the exponent becomes $\langle x | 0\rangle=e^{-\frac{m\omega x^2}{2\hbar}} = e^{-\bar{n}/2}$ .

The probability in the $x$ -basis is thus, $P(n)=|\langle x | n \rangle|^2=\frac{\langle aa^\dagger \rangle}{n!}^{n} e^{-\bar{n}}=\frac{\bar{n}}{n!}^{n} e^{-\bar{n}}$ , where using the definition $\langle \bar{n} \rangle=\langle aa^\dagger \rangle$ , we\'ve recovered exactly the Poisson Distribution for a parameter $\bar{n}$ .

Making the following associations, $m\rightarrow \bar{n}$ and $\lambda \rightarrow n$ , you carve the first etchings in the Rosetta Stone between probability and photon statistics...

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Comments

wittensdog
2009-10-08 21:45:38

The language of this problem is indeed pretty vague, so, here is one solution based on the way I interpreted it. I know there has already been a lot of talk on this, I hope maybe I can help sort things out a little...

First, if you just average all of those values, you get 2. So now I guess we just postulate that that should be close enough to the true average for us to get an idea of how long we should count for.

Now, in a Poisson distribution (which describes radioactive phenomenon, or most of it), we know that the standard deviation is the square root of the average. So we can take the standard deviation of this distribution to be sqrt(2). I don't know what the SD is if you actually calculate it for those numbers, but anyone who goes trying to calculate standard deviations from data on the GRE is completely insane.

Now, for reasons that can be seen if you take a course in statistics, the error on the mean is generally taken to be the standard deviation of the measurements divided by the square root of the number of measurements (this stems from the central limit theorem). I believe this is what is meant by uncertainty here. They state an uncertainty of one percent. I don't know exactly what it is that we want one percent of, but I'm guessing they mean 1% of the mean value, aka, the error on the mean should be plus or minus 1 percent of the mean. I don't know what else they would be referencing.

Since we are taking the mean as 2, or at least assuming it should be something in that ballpark, one percent of that would be 0.02. So if we know the standard deviation, the uncertainty we want, and the formula for the uncertainty on the mean, then we get,

uncert = SD / sqrt (n) ==>

0.02 = sqrt(2) / sqrt(n) ==>

4e-4 = 2/n ==>

n = 0.5 e +4 ==>

n = 5,000

So we want to make 5,000 measurements, and since each measurement is one second long, this corresponds to 5,000 seconds.

I hope this manages to help someone (and that I'm actually doing it right!).

Prologue
2009-11-05 10:13:11

Thank you!

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tensorwhat
2009-03-19 20:39:42

This is way more simple....

$\sqrt{N}$ /N = 1% = 1E-2

$\Rightarrow$ Solve for N

$\Rightarrow$ N = 1/1E-4 = 10,000 s

$\Rightarrow$ There are 2 counts per second, so 10,000 s/2 = 5,000 s

Done.

AER
2009-04-02 16:14:24

Where 2 is still the average of the ten measurements, and each measurement was 1 sec long.

ajkp2557
2009-11-07 05:03:37

Small typo: 10,000 should be number of counts (unitless), not measured in seconds. Your units will come from the fact that you're dividing by 2 counts / sec.

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eshaghoulian
2007-09-29 19:34:12

I think if you piece together everyone's comments, you'll have a final solution. Here is my thought progression:

The number of counts N is 20 for a time T of 10 seconds, giving a rate R of N/T = 2. Here we invoke the $\sqrt N$ rule without justification, which allows us to say that the uncertainty of an N-count distribution is $\sqrt N$ . We use the formula for fractional uncertainty $\delta R/R = \frac{\delta N/T}{N/T} = \delta N/N$ which motivates rampancy's form of the uncertainty.

So, for X seconds, we have a total number of counts 2X, and we use the equation above to get $\delta R/R=\delta N/N=sqrt {2X}/2X = 0.01 \Rightarrow X = 5000s$

Notice that the fractional uncertainty of the rate is just the fractional uncertainty of the total number of counts. I am not sure about the language here; I want to say that ETS's use of the term "uncertainty" in the question is at best vague, but I am not familiar with this type of experiment (reminiscent of the Q factor, which has as many definitions as it has occurences in physics).

See section "Counting Statistics" in link below for a little more detail
http://www.colorado.edu/physics/phys1140/phys1140_sp05/Experiments/O1Fall04.pdf

ericimo
2007-10-27 14:33:55

Correct, except there IS justification.

Since all we know from the problem is that it involves radiation detection, the vague nature allows us to assume that the distribution will follow the most common distribution in radiation detection.

And for most radiation measurements, the distribution is a Poisson distribution (hence Yosun's inclusion of the Poission discussion) which is where the employed rule for uncertainty comes into play.

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michealmas
2006-12-27 19:09:12

Sorry for the formatting screw-up:

Trying to clear up Yosun's solution - Yosun's formula for uncertainty is wrong. He claims it's:

$\sigma$ /AverageCounts

when actually it's

$\sigma$ /TotalCounts

you can correct Yosun's equation by multiplying the denominator by the total seconds, or C as Yosun calls it. That is what he does, though without explanation.

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michealmas
2006-12-27 19:06:57

Trying to clear up Yosun's solution - Yosun's formula for uncertainty is wrong. He claims it's:

$\sigma/AverageCounts<br /><br /> when actually it's:<br /><br />$ \sigma/TotalCounts

you can correct Yosun's equation by multiplying the denominator by the total seconds, or C as Yosun calls it. That is what he does, though without explanation.

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simpsoxe
2006-11-30 21:43:30

if you claim that $\overline{x}=2$ is the average and $\sigma=\sqrt{\overline{x}}$ is the standard deviation, then how do you go from there to get that $\sigma/\overline{x}=\sqrt{2C}/2C$ is the answer? I'm confused as to how the C's get in there

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rampancy
2006-11-02 00:50:14

That explanation makes no sense, and seems needlessly complicated.

The total number of counts in 10 seconds is 20. The error in that is Sqrt(20).

Counts = 20 +/- sqrt(20).

The average number of counts is 2, so in N seconds, we should see,

2N +/- sqrt(2N) counts.

We want the fractional error to be .01, so,

sqrt(2N)/(2N) = .01

So N = 5000.

nitin
2006-11-13 14:23:09

I agree with rampancy. Yosun, your solution is nonsense, and it seems you don't even know what you're talking about.

mr_eggs
2009-08-16 18:21:53

You don't have to be a jerk, nitin. This is an open community site to help those trying to get into grad school. If you don't have anything intelligent to add, then fuck off.rnrnA little late.. haha..

mr_eggs
2009-08-16 18:31:54

You don't have to be a jerk, nitin. This is an open community site to help those trying to get into grad school. If you don't have anything intelligent to add, then fuck off.

A little late.. haha..

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yosun
2005-11-27 01:50:12

Poisson Distribution, the way it was meant to be.

Blake7
2007-09-19 06:02:24

It's beautiful, Yosun! How can I get a copy of your wonderful book?

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Poisson Distribution, the way it was meant to be.

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