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Calculating uncertainty of energy if all we know is dimension of box

  1. Jul 28, 2013 #1
    1. The problem statement, all variables and given/known data
    2. Relevant equations
    \begin{align}
    \Delta p \Delta x &\geq \frac{\hbar}{2}\\
    \langle p \rangle &= \sqrt{\langle p^2 \rangle - \langle p \rangle^2}\\
    \langle x \rangle &= \sqrt{\langle x^2 \rangle - \langle x \rangle^2}
    \end{align}
    I allso know that a wavefunction of a particle in an infinite square well is: ##\psi=\sqrt{\frac{2}{d}}\sin\left(\frac{N\pi}{d}x\right)##, but i don't know ##N##. From this it can be derived that ##\langle p\rangle = 0## and ##\langle p^2\rangle = \frac{\hbar^2\pi^2}{d^2}N^2=E2m##. Too bad i don't know ##N## again...

    3. The attempt at a solution
    I first assumed that uncertainty in position ##\Delta x=d## and then calculated a momentum ##\Delta p=\frac{\hbar}{2\Delta x}=9.845MeV/c##?

    If I try to calculate uncertainty in full energy relativisticaly I start with a Lorentz invariance and apply standard uncertainty equation which i found in this PDF.
    \begin{align}
    E^2&=p^2c^2+{E_0}^2\\
    E&=\sqrt{p^2c^2+{E_0}^2}\\
    &\left\downarrow\substack{\text{Because energy $E$ is a function}\\\text{of only one variable $p$, we use}\\\text{standard formula for calculating}\\\text{uncertainty}.}\right. \quad \boxed{\Delta q = \frac{dq}{dp}\Delta p}\\
    \Delta E &= \frac{d}{dp}\left(\sqrt{p^2c^2+{E_0}^2}\right)\cdot\Delta p\\
    \Delta E &= \frac{1}{2}\frac{1}{\sqrt{p^2c^2+{E_0}^2}}2c^2p\cdot \Delta p\\
    \Delta E &= \frac{c^2p}{\sqrt{p^2c^2+{E_0}^2}}\cdot \Delta p\\
    \end{align}
    I did all by the book but I can't seem to calculate ##\Delta E## as I have to get rid of ##p## (which is unknowne) somehow. I get even more lost when trying to calculate the uncertainty ##\Delta E_k##:
    \begin{align}
    E^2 &= p^2c^2 + {E_0}^2\\
    E &= \sqrt{p^2c^2 + {E_0}^2}\\
    E_k + E_0 &= \sqrt{p^2c^2 + {E_0}^2}\\
    E_k &= \sqrt{p^2c^2 + {E_0}^2} - E_0\\
    &\downarrow\\
    \Delta E_k &= \frac{d}{dp}\left(\sqrt{p^2c^2 + {E_0}^2} - E_0\right) \cdot \Delta p\\
    \Delta E_k &= \frac{d}{dp}\left(\sqrt{p^2c^2 + {E_0}^2} - E_0\right) \cdot \Delta p\\
    \Delta E_k &= \frac{1}{2}\frac{1}{\sqrt{p^2c^2+{E_0}^2}}2c^2p\cdot \Delta p\\
    \Delta E_k &= \frac{c^2p}{\sqrt{p^2c^2+{E_0}^2}}\cdot \Delta p\\
    \end{align}
    From neither of the equations I could calculate ##\Delta E## or ##\Delta E_k## but I have noticed that they are the same. Which makes sense.

    What could I do to be able to calculate ##\Delta E## or ##\Delta E_k##?
     
  2. jcsd
  3. Jul 28, 2013 #2

    mfb

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    2016 Award

    Staff: Mentor

    I think you are supposed to calculate the minimal uncertainty. You can use the ground state.

    This formula requires that the variation of the parameter is small - that is not true here.
    <p[sup2[/sup]> gives you the expected kinetic energy. You can calculate ##\Delta (p^2)## with the help of this, or just use that value as expected variance.
     
  4. Jul 29, 2013 #3
    Ok i will use ##N=1##.

    Because the absolute uncertainty for position ##\Delta x## is as big as the best estimate ##x\equiv a##?

    You mean the clasicall aproximation when ##W_k=p^2/2m##? Can i use standard formula here or still no? What I mean is can i use:
    \begin{align}
    \scriptsize
    \frac{\Delta W_k}{W_k} &= \frac{d}{d p^2}\left(\frac{p^2}{2m}\right) \cdot \Delta (p^2)\\
    \frac{\Delta W_k}{W_k} &= \frac{1}{2m} \cdot \Delta (p^2)
    \end{align}

    This is bugging me because if i use uncertainty principle to calculate ##\Delta p=9.845Mev/c##. It is bigger than the rest energy of an electron but not proton. But still it is the same order of magnitude and isnt much smaller ##\ll##. Can i use clasicall approximation?

    I know that variance is ##\sigma^2 = \langle p^2 \rangle - \langle p \rangle^2## but can i insert ##\Delta p## instead of ##p## and write the variance like this: ##\langle \Delta p^2 \rangle = \langle (\Delta p)^2 \rangle - \langle \Delta p \rangle^2##? This seems weird...
     
    Last edited: Jul 29, 2013
  5. Jul 29, 2013 #4
    I think i have found where the solution.

    FOR ELECTRON
    If we take ##d=\Delta x## as the absolute uncertainty and calculate ##\Delta p=9.845MeV/c## we notice that ##\Delta pc < E_{0e}## but ##\Delta p## is supposed to be small compared to ##p## which means ##p## must be huge! So i can with a confidence say that ##pc\gg E_{0e}## ant this means i can deal with this problem ultrarelativistically!

    So i can write the Lorentz invariance:
    \begin{align}
    E^2 &= p^2c^2 + {E_{0e}}^2\\
    (E_k+e_{0e})^2 &= p^2c^2 + {E_{0e}}^2\longleftarrow \substack{\text{Here i use the ultrarelativistic approach and neglect the $E_{0e}$}}\\
    E_k &= pc\\
    \end{align}
    I got a fairly simple relation between momentum and energy and i can use standard formula for propagation of uncertainty on it like this:
    \begin{align}
    \Delta E_k &= \frac{d E_k}{dp}\Delta p\\
    \Delta E_k &= c \Delta p\\
    \Delta E_k &= 9.845MeV\\
    \end{align}
    The result matches with my book's result! So i understand this problem now. Or so i thought untill i try to redo it for the proton

    FOR PROTON
    Well if i compare ##\Delta p c## with a rest energy of a proton ##E_{0p}=933.41MeV## i find that ##\Delta p c > E_{0p}## but the ##p## is much larger than than ##\Delta p## and therefore i think i can say that ##pc \approx E_{0p}## which means i have to deal with this problem relativistically which again means i cannot solve it:

    \begin{align}
    E^2 &= p^2c^2 + {E_0}^2\\
    E &= \sqrt{p^2c^2 + {E_0}^2}\\
    E_k + E_0 &= \sqrt{p^2c^2 + {E_0}^2}\\
    E_k &= \sqrt{p^2c^2 + {E_0}^2} - E_0\\
    &\downarrow\\
    \Delta E_k &= \frac{d}{dp}\left(\sqrt{p^2c^2 + {E_0}^2} - E_0\right) \cdot \Delta p\\
    \Delta E_k &= \frac{d}{dp}\left(\sqrt{p^2c^2 + {E_0}^2} - E_0\right) \cdot \Delta p\\
    \Delta E_k &= \frac{1}{2}\frac{1}{\sqrt{p^2c^2+{E_0}^2}}2c^2p\cdot \Delta p\\
    \Delta E_k &= \frac{c^2p}{\sqrt{p^2c^2+{E_0}^2}}\cdot \Delta p\\
    \end{align}

    Please help me to understand how to solve this for a proton as well. The result for a proton from the book is ##\Delta E_k = 53keV##.

    Will i have to use the variances and expectation values that we've discussed before? I am not sure how to connect expectation values with an absolute uncertainty. Are those the same? So can i write ##\Delta p = \langle\Delta p \rangle##?
     
    Last edited: Jul 29, 2013
  6. Jul 29, 2013 #5

    mfb

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    Staff: Mentor

    For the proton, ##\Delta p c \ll E_0## (two orders of magnitude smaller)
     
  7. Jul 29, 2013 #6
    Yes this is for the uncertainty ##\Delta p##, but isn't a momentum ##p## (no delta) much larger than a ##\Delta p##. This is why i assumed ##pc \approx E_{0p}##...

    If I try do it like you sugested i have to work in a clasicall aproximation where Lorentz invariant simplifies into:

    \begin{align}
    E^2 &= p^2c^2 + {E_{0p}}^2\\
    (E_k + E_{0p})^2 &= p^2c^2 + {E_{0p}}^2\\
    E_k + E_{0p} &= \sqrt{p^2c^2 + {E_{0p}}^2}\\
    E_k &= \sqrt{p^2c^2 + {E_{0p}}^2} - E_{0p}\\
    E_k &= {E_{0p}}\sqrt{1 + \frac{p^2c^2}{{E_{0p}}^2}} - E_{0p}\longleftarrow\substack{\text{I use binomial developement and}\\\text{neglect all but first two parts - i can}\\\text{do that because i use a clasicall}\\\text{aproximation $\boxed{\tiny pc \ll E_{0p}}$.}} \\
    E_k &= {E_{0p}} \left(1 + \frac{1}{2}\frac{p^2c^2}{{E_{0p}}^2} \underbrace{+ \dots}_{\text{neglected}}\right) - E_{0p}\\
    E_k &= {E_{0p}} \left(1 + \frac{1}{2}\frac{p^2c^2}{{E_{0p}}^2} \right) - E_{0p}\\
    E_k &= {E_{0p}} + \frac{1}{2}\frac{p^2c^2}{{E_{0p}}} - E_{0p}\\
    E_k &= \frac{1}{2}\frac{p^2c^2}{{E_{0p}}}\\
    E_k &= \frac{p^2}{2m}\\
    \end{align}

    So now that i know the equation (which is a basic kinetic energy equation) i can spot that ##E_k## is in fact a function of ##p^2## can you tell me if i am allowed to calculate ##\Delta E_k## like this:
    \begin{align}
    \Delta E &= \underbrace{\frac{d E_k}{dp^2}}_{\llap{\text{I used $p^2$ !!!}}} \Delta p^2= \frac{d}{dp}\left(\frac{p^2}{2m}\right) \Delta p^2 = \frac{1}{2m}\Delta p^2
    \end{align}

    And i have one more unclarity about how to calculate ##\Delta p^2##. Well I know that ##\Delta p \geq 9.845Mev/c = 5.275\cdot10^{-18}kgm/s##. But do i only need to square the later to get ##\Delta p^2##? I mean is this ok ##\Delta p^2 = (5.275\cdot10^{-18}kgm/s)^2##.
     
    Last edited: Jul 29, 2013
  8. Jul 29, 2013 #7

    mfb

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    Staff: Mentor

    For the proton in the ground state, both Δp and p are much smaller than its mass.

    ##\Delta p^2 = (\Delta p)^2## is probably fine - I would expect a small (close to 1) numerical prefactor between them, but in those problems that is usually not relevant.
     
  9. Jul 29, 2013 #8

    TSny

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    Gold Member

    If you assume the particle is in the ground state, shouldn't you be able to state the uncertainty in the energy without any calculation (or, at most, very little calculation)?
     
  10. Jul 29, 2013 #9
    NOW i am totaly confused about this problem. First i have to clear some basic things out. If i understand corectly there are 2 ways to approach this problem.

    1. We can directly state that ##\Delta x = d=10fm## and then directly calculate the uncertainty in momentum using the Heisenberg's uncertainty relation like this: ##\Delta p \geq \frac{\hbar}{2\Delta x}##. A lot of people solve this problem like this including my professor.
    2. Or we can look at this particle from a QM perspective and deal with it like it is a particle in an infinite square well for whom i know the wavefunction ##\psi## and can use it to calculate expectation values ##\langle p \rangle## and ##\langle p^2\rangle##. Later on i can use theese to calculate uncertainty (standard deviation) ##\Delta p=\sqrt{\langle p^2 \rangle - \langle p\rangle^2} = \frac{\hbar\pi}{d}N##.

    I tried to use the first way which is easier and the result for electron makes sense while for proton it is confusing. I like the second QM way better, but the results for ##\Delta p## are totaly different! I already hate equation ##\Delta p \Delta x \geq \frac{\hbar}{2}## which our professor derived using his fingers :S How can it return a totaly different result?

    But in this case we have a momentum squared which i can write like this ##p^2 = p \cdot p## and if i want to calculate its absolute uncertainty i have to first obey the rule that at multiplication we have to add relative uncertainties:
    \begin{align}
    \Delta p^2 = \sqrt{\left(\frac{\Delta p}{p}\right)^2 + \left(\frac{\Delta p}{p}\right)^2 } = \sqrt{2}\frac{\Delta p}{p}
    \end{align}
    Here i don't know ##p##... I have read about this here.
     
    Last edited: Jul 29, 2013
  11. Jul 29, 2013 #10

    TSny

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    This method is a very crude way to obtain an order of magnitude approximation for the uncertainty in p. Taking Δx = d is just a quick, rough estimate of the order of magnitude of Δx.

    In this method, you are adopting a definition of uncertainty as the square root of the variance. For the ground state of a particle in a box, you then find Δx ≈ .18d which is quite different than the choice made in method 1; but, it is still roughly of the same order of magnitude. You should not be surprised to find that Δp using the variance definition leads to a different result than the first method.

    When thinking about the uncertainty in the energy for the case where the particle is in the ground state, keep in mind that the ground state is an eigenstate of the energy operator (Hamiltonian).

    [EDIT: Note that your original question did not specify the state of the particle. If the particle were in a superposition of different energy eigenstates, then the uncertainty in energy would depend on the particular superposition. It seems to me that the question should have been clearer here.]
     
    Last edited: Jul 29, 2013
  12. Jul 29, 2013 #11
    Thank you. I presume that second method is better. Too bad it can't be used in this case as i don't know ##p## - i only know ##\Delta p## and therefore this can't be calculated (or is there still any way to do it this way?):
    \begin{align}
    \Delta p^2 = \sqrt{\left(\frac{\Delta p}{p}\right)^2 + \left(\frac{\Delta p}{p}\right)^2 } = \sqrt{2}\frac{\Delta p}{p}
    \end{align}

    Please tell me what do you think about this ##\longrightarrow## Our professor solved this case (in case of a proton) in a completely different manner. He first assumed ##\Delta x = d##, then calculated ##\Delta p \geq \frac{\hbar}{2\Delta x}##. I could accept this but what followed shocked me.

    He then plugged in the ##\Delta p## in the Lorentz invariant and calculated the energy ##E## like this (pretending that ##\Delta p \approx p##):
    \begin{align}
    E^2 &= p^2c^2 + {E_{0p}}^2\\
    (E_k + E_{0p})^2 &= \Delta p^2c^2 + {E_{0p}}^2\\
    E_k &= \sqrt{\Delta p^2c^2 + {E_{0p}}^2} - E_{0p}\\
    E_k &= 51.1keV
    \end{align}

    Why did he do it like he did? Was his intention to just get the order of the kinetic energy by assuming that an uncertainty ##\Delta p## is as big as the ##p##? I wonder allso if we could calculate uncertainty in kinetic energy the same:

    \begin{align}
    E^2 &= p^2c^2 + {E_{0p}}^2\\
    (\Delta E_k + E_{0p})^2 &= \Delta p^2c^2 + {E_{0p}}^2\\
    \Delta E_k &= \sqrt{\Delta p^2c^2 + {E_{0p}}^2} - E_{0p}\\
    \Delta E_k &= 51.1keV
    \end{align}

    I bet that this later attempt is totaly wrong! But please confirm as some of the confused students allso solved this same case like this and i am now totaly confused and need to resolve this in my head.
     
  13. Jul 29, 2013 #12

    TSny

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    This equation can't be correct. The dimension of the left side doesn't match the dimension of the right side.

    Yes, I think that's correct. He is using a crude method of estimating the energy (not the uncertainty in energy) and he's assuming that the magnitude of the momentum must have a value that is at least of the order of the uncertainty in momentum. "Hand waving" at its best.

    I don't think this is valid. In general, you cannot assume that the uncertainty in p2 can be approximated by the square of the uncertainty in p. For example, it's possible for there to be a nonzero uncertainty in p but zero uncertainty in p2. So, you cannot estimate the uncertainty in energy (which depends on p2) by using the uncertainty in p.

    Here's a classical example. Suppose a marble of mass m is bouncing elastically back and forth in 1D inside a box and the marble has a definite speed v. You are not allowed to look in the box. What would you specify as the uncertainty in your knowledge of the momentum? Crudely, it would be 2mv because sometimes the marble is moving to the right with positive momentum p = +mv and sometimes it is moving to the left with negative momentum p = -mv. So, Δp ≈ 2mv. But what would be your uncertainty in the square of the momentum? Well, p2 of the marble always has (except at the instants of collision) a definite value p2 = m2v2. So, there is no uncertainty in p2 (or the energy) even though there is uncertainty in p.
     
    Last edited: Jul 29, 2013
  14. Jul 29, 2013 #13
    I am sorry i misstyped. It should be:
    \begin{align}
    \frac{\Delta p^2}{p^2} = \sqrt{\left(\frac{\Delta p}{p}\right)^2 + \left(\frac{\Delta p}{p}\right)^2 } = \sqrt{2}\frac{\Delta p}{p} \longrightarrow \Delta p^2 = \sqrt{2}\frac{\Delta p}{p} p^2 = \sqrt{2}\Delta p \cdot \underbrace{p}_{\rlap{\text{unfortunately unknown}}}
    \end{align}
    I got this equation from this manual for calculating uncertainties. I think that equation is correct but i can't use it in this case. Now i understand why the professor does what he does to calculate an estimate of energy (which is not an estimate of energy uncertainty). Thank you.
     
  15. Jul 29, 2013 #14

    mfb

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    Check again how to handle powers of the same variable.

    Anyway, this formula would require small relative deviations, something you do not have.
     
  16. Jul 29, 2013 #15
    Thank you for this note!

    I remembered i had to ask 1 more thing. If i assume that use ##\Delta p## is of the same order than ##p## and use it in an equation instead of ##p## to calculate estimate for energy ##E##, do i have to change the sign ##"="## to ##"\geq"## like this:

    \begin{align}
    E^2 &\geq p^2c^2 + {E_{0p}}^2\\
    (E_k + E_{0p})^2 &\geq \Delta p^2c^2 + {E_{0p}}^2\\
    E_k &\geq \sqrt{\Delta p^2c^2 + {E_{0p}}^2} - E_{0p}\\
    E_k &\geq 51.1keV
    \end{align}
     
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