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Homework Help: Worded Problem, Probability Density, quantum mechanics

  1. May 18, 2013 #1
    1. The problem statement, all variables and given/known data

    The needle on a broken car speedometer is free to swing and bounces off perfectly off the pins at either end, so that if you gave it a flick it's equally likely to come to rest at 0 and ## \pi ##

    What is the probability density, ## p(x) ##?

    2. Relevant equations

    [tex]\int^{\pi}_{0} p(x) dx = \int^{\pi}_{0} |\Psi (x,t)|^2 dx = 1[/tex]

    3. The attempt at a solution

    This may sound really dumb but I was thinking that since that it's equally likely to be from anywhere between 0 and pi, then it'll just be a straight horizontal line, p(x) = 1.
    But since it has to be normalised from 0 to pi, then p(x) will just be 1/pi.

    But that didn't seem right because it had to drop off to 0 at x=0 and x=pi so I was thinking of a sine curve, and maybe p(x) = sin(x)

    How does that work though? It says it's equally likely to land anywhere from 0 to pi, but with a sine curve it's obvious that it'll have a way higher chance of landing in the middle...

    Attached Files:

  2. jcsd
  3. May 18, 2013 #2

    Simon Bridge

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    Why do you need ##p(x=0)=0## and ##p(x=\pi)=0## ?

    Note: if p(x) is a probability density function, then $$P(a<x<b)=\int_a^bp(x)dx$$
  4. May 18, 2013 #3
    Oh sorry I was thinking of a gaussian curve and how it drops off to zero at infinity.

    So I was right about ## p(x) = \dfrac{1}{\pi} ##??

    Then if I evaluate the integral ## \displaystyle P(a < x < b) = \int^{\pi}_{0} \dfrac{1}{\pi} dx = 1 ##

    If I evaluate it from ## \displaystyle \theta ## to ## d \theta ## I'll always get the same number, no matter what theta is: ## \displaystyle \int^{\theta + d \theta}_{\theta} \dfrac{1}{\pi} dx = Constant ##

    It's the only function that I can think of that has the same probability no matter what theta is.
  5. May 18, 2013 #4

    Simon Bridge

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    OK you are having a bit of trouble having confidence in your deductions...

    The top-hat distribution from x=0 to x=L would be: ##p(x)=\frac{1}{L}( h(x)-h(x-L) )## where ##h(x)## is the Heaviside step. Notice that the requirement ##\int_{-infty}^\infty p(x)dx## means that##P(0<x<L)=1## and so does not need the points x=0 and x=L to be included?

    I can write that out:$$p(x)=\left \{ \begin{array}{ll}
    \frac{1}{L} & : \; 0<x<L\\
    0 & : \; x\leq 0\\
    0 & : \; x\geq L
    \end{array} \right . $$
    ... but does it matter to the statistics if we do include 0 and L in the first inequalities?
    (note: if you look up the Heaviside function, you'll see different people use different values at the limits.)

    The question says that the probability of finding the needle in any arbitrarily small region inside 0≤x≤π is a constant - so far so good - and, intuitively, you'd figure that the probability of finding the needle exactly at x=0 or x=π would be zero. Does the top-hat distribution give you that?

    Incidentally: since you mentioned QM, you could check to see if ##\psi(x)=\frac{1}{\sqrt{L}}( h(x)-h(x-L) )## is a solution to the Scrödinger equation ;)
    Last edited: May 18, 2013
  6. May 19, 2013 #5


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    Is the problem statement exactly as it was given to you? I ask because as you typed it, it doesn't say that the needle is likely to be anywhere between 0 and ##\pi##; it said it's equally likely to be 0 or ##\pi##.
  7. May 19, 2013 #6
    I'm pretty sure I typed it out exactly as the book had it.
    I took a photo of the original problem (In the book Quantum Mechanics Second Edition by David J Griffiths)

    It's near the very start of the book so I'm presuming it doesn't require that much knowledge about quantum mechanics.

    Sorry I don't know what a top-hat distribution is, nor what a Heaviside step. (I quickly googled it a top-hat distribution and I see that it would be the shape I need for equal probability.)

    I get that:

    $$p(x)=\left \{ \begin{array}{ll}
    \frac{1}{L} & : \; 0<x<L\\
    0 & : \; x\leq 0\\
    0 & : \; x\geq L
    \end{array} \right . $$

    where I presume L is ## \pi ## in this case.

    I don't know if the top hat distribution would give me 0 for x=0 and x=pi, we haven't needed to study it for this course.
    From the diagrams it seems like it'll give me 1/pi for when x=0 and x=pi, because it looks like the infinite square well. But does it really matter what x=0 and x=pi are?
    I guess the speedometer needle angle can never be zero because there's a pin there that stops it, and same with the angle pi.
    But I don't quite get the top hat equation is. I google it and can't seem to find a general form of it.
  8. May 19, 2013 #7


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    Oh, I see you left out a couple of words when you typed up the statement that completely change the meaning.
  9. May 19, 2013 #8

    Simon Bridge

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    ... you didn;t type it out the way your attached picture tells it - but never mind, I followed the attachment and not what you wrote.

    But you do know how to compute a probability from a probability density function don't you?

    Anyway - you have seen it before - probably just not under that name.
    Unless... where abouts in your education are you at right now?

    How did you compute that?

    You mean p(x)? - um - no it's pretty much the opposite of a square well ... very little like an infinite one. Did you try plotting it?

    That was my question to you ;)
    Note: you are the one who brought it up.

    <puzzled> I gave you a general form of it - it's the piece-wise function with the L in it. Really generally, a uniform pdf : a<x<b is $$p(x) = \frac{1}{b-a}[h(x-a)-h(x-b)]$$ where h(x) is the Heaviside step function.

    I googled for "top hat probability density function" and came up with lots of examples, i.e.
    http://www.jhu.edu/virtlab/course-info/ei/notes/uncertainty_notes.pdf (p4)
    http://www.pha.jhu.edu/~neufeld/numerical/lecturenotes5.pdf (slide 104)

    It's one of the first distributions you get taught studying probability - the discrete version is the distribution obeyed by dice. It's one of the simplest around: try "uniform distribution":

    That help?

    But you don't need to know the name ... just the maths.
    It is looking almost like you don't understand what a probability density function does.
    Last edited: May 19, 2013
  10. May 19, 2013 #9
    Oh sorry I thought the hint part wasn't that important and that the all I needed in the question was the main part.

    I'm guessing it's just the area under the curve of the probability density function, ##\displaystyle P(a \le x \le b) = \int_{a}^{b} p(x)dx ##

    I don't think I've seen it before, We've only studied the basics of quantum mechanics.
    I'm studyng at Otago University, New Zealand.
    Second year physics (PHSI231), doing a bit of classical, quantum and thermal physics.

    I didn't really compute it, I just looked at the pictures of what a top-hat distribution looked like.

    Oh yea, like the opposite of an infinite square well.
    How would you plot it, something like this?
    Code (Text):

         |     |  
         |     |  
         |     |        
    _____|     |_____
    I'm guessing since the integral I used to calculate it ## \displaystyle \int_{0}^{\pi}p(x) dx = 1 ## then it would have to include 0 and pi, because they are the bounds of the integral and so p(x) = 1/pi and also p(0) = 1/pi

    Ah that Wikipedia page helped me heaps, thanks. It didn't show up when I googled stuff but I guess I just didn't know what to search.

    So then the probability density would just be the top hat distribution (I don't really understand what a Heaviside step function is, and the wikipedia article on it is quite confusing for me):
    $$ p(x) = \frac{1}{\pi}[h(x)-h(x-\pi)] $$

    $$p(x)=\left \{ \begin{array}{ll}
    \frac{1}{\pi} & : \; 0 \le x \le \pi \\
    0 & : \; x < 0\\
    0 & : \; x > \pi
    \end{array} \right . $$

    I presume this is what I want? It satisfies all the conditions,

    [tex]\displaystyle \int_{0}^{\pi} p(x) dx = 1[/tex]

    [tex]\displaystyle P(0 \le \theta < d \theta \le \pi) = \int_{\theta}^{\theta + d \theta} p(x) dx = \text{A Constant}[/tex]

    Haha sorry, I'm not really 'with it' when it comes to understanding physics concepts. I'm more of a Math Major. I do get that the area under the probability density function from a to b tells us that's the probability of finding the particle within that region. And that the wave function squared gives us a probability density function.
  11. May 19, 2013 #10

    Simon Bridge

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    That makes sense.

    At level 2 you are expected top bring all your learning to bear on the course and not just the stuff specifically mentioned in the lectures. You are expected to have a core knowledge base to build on. What has happened here is that a prior familiarity with probability and statistics (part of mathematics), that you do not possess, has been expected of you.

    In New Zealand - you get introduced to probability and statistics as part of the core math course at secondary school ... you will have met probabilities with dice by year 9.
    You will have learned that the probability of a particular outcome is (by definition) the number of ways to get that outcome divided by the number of ways to get anything else.

    When you specialize - you take a calculus or a statistics track in maths, but the core curriculum for both tracks includes basic probability and statistics - so you should have seen at least the normal distribution, and, by extension, continuous probability distributions, before you left secondary school. The continuous uniform distribution may not have been taught in the calculus track - that would depend on the school.

    Otago does not require a statistics paper at stage 1 as a prerequisite for stage 2 QM as the paper does not require an understanding of stats above year 12 - but it is advisable if you did not take the stats track at secondary level.

    So it is possible that you have done no probability or statistics since graduating secondary school.
    If this is the case - then you should add a statistics paper to your course.

    You say you have "got" the basic math for probability density functions but you had trouble with basic questions involving that knowledge... you "guess"ed about how to get a probability from a density function for eg, and were confused about what was part of physics and what part of math, you appeared to have trouble picturing the shapes of functions, and you don't seem to recall secondary math

    I suspect you are used to the "applying formulas" part of math - when what you need is the "understanding" part.
    Last edited: May 19, 2013
  12. May 19, 2013 #11

    Simon Bridge

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    Seeing you are at level 2 physics, then you probably have not met the Heaviside function in so many words yet - another function you will see a lot of but have yet to meet will be the Dirac delta function ##\delta(x)## - though you have probably seen the Kronecker delta ##\delta_{jk}##.

    The Heaviside function is simply a way of writing a piece-wise function on one line ...
    h(x)=0: x<0 and h(x)=1: x>0 ... what happens at x=0 depends on who you ask.
    I usually put h(0)=0, some people put h(0)=1/2 or 1. For the purposes above it doesn't matter.

    You seem to have got it in the end though.
    It shouldn't be this hard for you - this is math, not physics, and you say you are "math oriented".
  13. May 20, 2013 #12
    I remember being introduced to basic probability with coins and dice but not as much statistics. We only did like the mean, median and mode sort of stuff in High School.
    I specialised in calculus, not statistics and never saw a normal distribution in High School. Right now I am taking a first year statistics paper, and we're only getting into the basics of a normal distribution.

    You seem to know quite a lot about NZ and Otago university, which is weird. Not many people know much about New Zealand.

    I am still quite confused about understanding physics concepts so that's probably why it seems like I'm lost.
    Starting from the top, the whole concept of a particle not being in a particular location is new and surprising to me. I always thought Laplace's[/PLAIN] [Broken] Demon was sorta true and that if you measure the precise location and momentum of every particle you can predict the past present and future. Then they introduce this whole thing called Heisenberg's uncertainty principle to me and then everything's just changed.
    And then there's the whole thing of a potential and a stray particle. Do they mean a particle as in a proton (Hydrogen ion)?

    I'm not confused about shapes of functions (at least not continuous ones). In calculus we've only ever dealt with continuous functions and so the top hat distribution is still a bit weird for me.

    I guess the NZ High School education system isn't that great at preparing you for university. I got straight into second year physics because of my 'strong' mathematics background (and I took a basic first year university maths paper and got A+).

    Nope, never heard of the Heaviside function, but that (how you explained it) makes it so much simpler. I don't know why the Wikipedia article on it is so much more confusing to understand.
    Last edited by a moderator: May 6, 2017
  14. May 20, 2013 #13

    Simon Bridge

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    This is basically what I'd come to suspect.

    More than you'd suspect - Kiwis have been getting around and seem to have a pretty good international reputation.

    I am very familiar with NZ education system, having works in it for decades across all the levels.

    From what you'g got now, you should be able to cope with the maths of continuous probability distributions - bear in mind that p(x) can be any function at all so long as the total area under the graph can be normalized.

    It's something everyone has to go through ... you get used to it.

    The definition of a particle is deliberately left vague at this stage - just think of it as building upon what you already know about classical particles: a purely abstract idea which has application in the real world.
    A good RL example of a particle is an electron.
    The idea applies to any object whose internal structure and size does not affect it's dynamics.

    The top-hat distribution is continuous in the sense that you saw in calc class.

    Oh OK - that would explain a lot then. People who get the accelerated courses tend to struggle.

    The stage 1 papers mostly concentrate on going over everything you were supposed to get by the end of secondary school ... makes sure everyone is on the same page. This is really great for solidifying what you already know and fills in a lot of gaps where you may have forgotten something or the teachers ran out of time or something.

    You need a good crash course then - you should try to get hold of NZQA primer books - the sorts of things that prepare you for external exams. The NZQA website has past exam papers and exemplars that can make a useful study guide.

    Wikipedia math articles tend to be marginal in usefulness because they have been written by mathematicians who care more about being mathematically rigorous than being easy for a novice to understand.

    Don't let the names of things throw you - focus on what the thing actually is.
    Last edited by a moderator: May 6, 2017
  15. May 23, 2013 #14
    I'm trying to get A+ in this paper and it seems like it's near impossible! I have 2 weeks before my exams and I'm studying this at least 3 hours a day for this and reading the books and going through the questions but I'm still lacking a lot of base knowledge.
    I think the best thing is to go through more questions so I understand how to do them.

    Okay, so I have answered the first part of the question, which is:

    $$p(\theta)=\left \{ \begin{array}{ll}
    \frac{1}{\pi} & \text{for} \; 0 \le \theta \le \pi \\
    0 & \text{for} \; \theta < 0\\
    0 & \text{for} \; \theta > \pi
    \end{array} \right . $$

    The second part asks "Compute ## < \theta > ##, ## < \theta^2 > ##, and ## < \sigma > ##, for this distribution"
    Exactly as is, no hints or anything.

    So I know that ## | \Psi (x,t) |^2 = p(x) ##

    Therefore: ## \Psi (\theta , t)=\left \{ \begin{array}{ll}
    \frac{1}{\sqrt{\pi}} & \text{for} \; 0 \le \theta \le \pi \\
    0 & \text{for} \; \theta < 0\\
    0 & \text{for} \; \theta > \pi
    \end{array} \right . ##

    The expectation value is just:

    [tex]< \theta > = \displaystyle \int^{\pi}_{0} \Psi (\theta , t)^* (\theta) \Psi (\theta , t) d \theta[/tex]

    [tex]< \theta > = \displaystyle \int^{\pi}_{0} \frac{1}{\sqrt{\pi}} ( \theta) \frac{1}{\sqrt{\pi}} d \theta[/tex]

    [tex]< \theta > = \displaystyle \frac{1}{\pi} \int^{\pi}_{0} \theta d \theta[/tex]

    [tex]< \theta > = \displaystyle \frac{1}{\pi} \left[ \dfrac{\theta ^2}{2} \right]^{\pi}_{0}[/tex]

    [tex]< \theta > = \dfrac{\pi}{2}[/tex]

    This seems correct, because that's the center of probability distribution.

    And then, for the expectation value of ## \theta ^2 ##

    [tex]< \theta ^2 > = \displaystyle \int^{\pi}_{0} \frac{1}{\sqrt{\pi}} ( \theta ^2) \frac{1}{\sqrt{\pi}} d \theta[/tex]

    [tex]< \theta ^2 > = \dfrac{\pi ^2}{3}[/tex]

    And I know that ## \sigma = \sqrt{< \theta ^2> - < \theta >^2} ## so I can just easil work ou sigma from that.

    Is that right?

    The last question then asks for me to compute ##< \sin ( \theta )>##, ## < \cos ( \theta )> ## and ##< \cos ^2 ( \theta ) ##.
    It should be fairly simple right, sort of like above, and for the ## \cos^2 ## I just use the trig identity to simplify it so I can integrate it?
  16. May 25, 2013 #15

    Simon Bridge

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    Doing good so far.
    Just a niggle - in the very first line of math, the limits of the integral should be all space ... you got specific in the next line.

    You are on the right track - notice that the sine and cosine parts are y and x components of the needle's arc?
    For full marks you may want to comment on if <sinA> bears any relation to the <A> etc.
    There is nothing wrong with just looking up such simple integrals.
  17. May 25, 2013 #16
    What do you mean, so I should integrate from negative infinity to infinity? But then how would I write the integral, since the wave function is a piecewise function? Something like this?

    [tex]\displaystyle \int_{- \infty}^{0} 0^2 \theta d \theta + \int_{0}^{\pi} \dfrac{\theta}{\sqrt{\pi}^2}d \theta + \int_{\pi}^{\infty} 0^2 \theta d \theta [/tex]

    I think the sine and cosine things relate to the next question in the book:

    So for this one, the probability density would be like a cosine curve?

    Code (Text):

            | | |
           |  |  |
          |   |   |
        -2r   |    2r
    (Sorry for the horrible cosine curve drawn by the code tags, but you know what a cosine curve looks like)

    So to normalise it, I'd need:

    [tex]\displaystyle \int^{2r}_{-2r} Acos(\dfrac{\pi x}{2r}) dx = 1[/tex]

    Although I don't know what to do with the hint...
  18. May 25, 2013 #17

    Simon Bridge

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    The first line is fine the way you wrote it - but the limits should, technically, be all space. The second line is fine to follow.

    After all, you started from the definition of the expectation value ... then you observe that some sub-areas are zero.

    But, like I said, it's just a niggle.

    That hint is a little bit like how you convert between polar and rectangular coordinates.
    You know ##p(\theta)## of finding the needle between ##\theta## and ##\theta+d\theta## ... can you relate that range to x and x+dx?

    The arclength in that range, for eg, ##rd\theta## for instance.
  19. May 25, 2013 #18
    Oh right I get it, because I wrote ## \Psi (x,t) ## in the first integral and the wave function is always from negative infinity to infinity. And then I changed to the specific piece of the wave function which had a different boundary, limited from 0 to pi.

    So the x position is related to theta by:

    ## x = \cos \left(\dfrac{\pi \theta}{2r} \right) ## for ## 0 \le \theta \le \pi ##

    ## dx = - \dfrac{\pi}{2r} \sin \left(\dfrac{\pi \theta}{2r} \right) d \theta ##

    What do I need to do with these now?


    ## \displaystyle \int_{x}^{x + dx} A \cos \left(\dfrac{\pi \theta}{2r} \right) dx ##

  20. May 25, 2013 #19

    Simon Bridge

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    In fact ##\Psi(x,t)## was overkill - you are only worried about the space part.

    Note: everything inside a trig function should have the same dimensions as angle.

    What you are trying to do is relate ##\psi(x)## with what you know about ##\psi(\theta)##.
    I'd start by sketching it out.
  21. May 26, 2013 #20
    Oh right I think the 2r part is supposed to be outside of the cosine function.

    So something like this: ## x = r \cos( \theta) ##
    And so ## \theta = \arccos \left( \dfrac{x}{r} \right) ##

    So when the angle is zero, the needle's x position is r away, and when the angle is 90 degrees then the needle's x position is at 0.

    And then ## dx = -r \sin (\theta) d \theta ##

    Now do I replace ## \theta ## with x for the probability density?

    p(\theta)=\left \{ \begin{array}{ll}
    \frac{1}{r} & : \; 0< \theta <r \\
    0 & : \; \theta \leq 0\\
    0 & : \; \theta \geq r
    \end{array} \right . ##

    Am I headed in the right direction?
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