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Uncertainty principle for time-energy & momentum-position

  1. Mar 6, 2015 #1
    1. The problem statement, all variables and given/known data
    Very short laser pulses can be produced, on the order of a few femtoseconds. (look it up.) If a laser has a center wavelength of 800nm, and lasts for 20 femtoseconds, (2*10-14s) what spread of wavelengths must it have to be compatible with the time-energy uncertainty principle?

    Minimize the total energy (kinetic + potential) of a particle with mass m that is oscillating on a spring of spring constant k, subject to the uncertainty principle. (The distance from equilibrium can't be smaller than the position uncertainty, and the momentum can't be smaller than the momentum uncertainty.)

    2. Relevant equations
    deltaE/E = -deltalambda/lambda


    3. The attempt at a solution
    plugging in for deltaT and h
    obviously this answer is too small.
    my teacher told me to use the deltaE/E = -deltalambda/lambda equation to plug in for E but then I get a number way smaller :
    deltaE=-deltahclambda/hcdeltalamda If using E for 8*10^-7 E=2.4825*10^-19J or 3.972*10^-38
    Where am I going wrong?

    2)I dont even know where to start but I know to get the minimum I just take the derivative. All the equations I wrote down are just ones that I think MIGHT pertain, but not sure which ones to use where and when ...

    any help appreciated. Thanks!
  2. jcsd
  3. Mar 6, 2015 #2


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    1) If ##d{1\over\lambda}## is not the same as ## d{\lambda}##.

    "plugging in for deltaT and h": ##\Delta f = {1\over 4\pi \Delta t}## gives me something else !

    2) You have ##\Delta p\Delta x## So expressing the minimum sum of kinetic and potential energies can be done in one variable, e.g. ##\Delta x##. ##{d\over d(\delta x)} = 0## gives you the minimum. The corresponding E is no surprise.
    Last edited: Mar 6, 2015
  4. Mar 6, 2015 #3
    Δf ≠ 1/4πΔt it is supposed to be Δf = h/4πΔt I thought??
  5. Mar 6, 2015 #4


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    If E = hf then f = E/h and if
    ΔE Δt = h/4π​
    Δf Δt = 1/4π​
    Δf= 1/(4πΔt)​

    Not that complicated.

    And I completely forgot, but nevertheless:

    Hello mt, welcome to PF ! :smile:
  6. Mar 6, 2015 #5
    Thank you, this is all a pretty new experience and I have been out of physics for years, jumping right into modern ..
    I realized my math error that shortly after I posted, Too many hours doing work in front of the computer, Sorry haha, But is this really a reasonable answer for Δλ...
    Δƒ=Δ(c/λ) >>> 1/Δλ=1/4πΔtc
    plug in and Δλ=2513nm ... now it seems larger?
    If I'm using the ΔE/E=-Δλ/λ in this problem
    I get an answer of .81m ... which is even bigger ....

    when you say d/d(∂x) you mean d^2/dx^2 ??
    I worked it this is what I'm getting:
    beginning equations:
    E=hc/λ, Δp=h/λ > Δp=E/c
    E=½mv^2+½kx^2 , v=P/m >>>>> E=p^2/2m +½kx^2
    d/dx ( Δx = h/4π(E/c)) ⇒ 0=d/dx (h/(4π (p^2/2m +½kx^2)))
    0= chkm^2x / π(kmx^2+p^2)^2
    d/dx above expression
    0=( chkm^2(p^2-3kmx^2))/π(kmx^2+p^2)^3

    does that seem like a reasonable equation ? it has m, k, and I used the uncertainty principle
    BUT, am I really just supposed to be solving for E ? from the Etot=KE+PE ?
    not sure what the instructor is asking for, an equation for E or an equation=0? (If you understand a sample answer would be nice so that maybe I could retry it on my own again )

    thank you so much for taking the time to read this!
    i checked my derivations with wolframalpha and they were correct, but if I'm missing something let me know! thanks
    Last edited: Mar 6, 2015
  7. Mar 7, 2015 #6


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    In that case, welcome back to the wonderful world of physics too :smile: !

    Good. But now we are back to ##
    d{1\over\lambda} \ne d\lambda##

    Δƒ=Δ(c/λ) >>> 1/Δλ=1/4πΔtc is not correct !

    ( apart from that, I really can't reverse engineer 2500 nm from what you write. I get 9 x 10-5 m. Are you careful enough when calculating things ? )​

    Δƒ=Δ(c/λ) >>> 1/Δλ=1/4πΔtc is not correct ! Δƒ=Δ(c/λ) >>> Δ(1/λ) = 1/(4π Δt c)

    Advice: don't plug in numbers, but symbols only until you have a final single expression. Then check the dimensions. Then check the order of magnitude without a calculator. Only then finish off with a numerical calculation. Many advantages: you weed out a lot of errors and scoring is maximized (especially when human physics teachers do the scoring).

    2) "when you say d/d(∂x) you mean d^2/dx^2 ??"

    No. Sorry for my casual writing. Instead of ##{d\over d(\delta x)} = 0\ ## with a lower case \delta I should have typed ##
    {d\over d(\Delta x)} = 0\ ## (with an uppercase \Delta).

    You are supposed to get an expression in Δx . The idea is to find a minimum for this expression by treating it as a function of Δx : differentiating to this Δx and looking for a value of Δx that makes the derivative zero. (substitute y = Δx if this looks weird to you).

    And now I see a slight derailment: forget part 1 when working in part 2. There is no ##\lambda## of relevance here. Work from Heisenberg ##\Delta p\Delta x \ge {\hbar\over 2}##, and realize that p2 has to be at least Δp2.
    Last edited: Mar 7, 2015
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