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Gaussian function to derive Heisenberg's uncertainty principle

  1. Feb 24, 2013 #1
    At our QM intro our professor said that we derive uncertainty principle using the integral of plane waves ##\psi = \psi_0(k) e^{i(kx - \omega t)}## over wave numbers ##k##. We do it at ##t=0## hence ##\psi = \psi_0(k) e^{ikx}##

    [itex]
    \psi = \int\limits_{-\infty}^{+\infty} \psi_0\!(k) \cdot e^{ikx} \, \textrm{d} k
    [/itex]

    where ##\psi_0(k)## is a ##k##-dependent normalisation factor (please correct me if I am wrong). This dependency was said to be a Gaussian function

    [itex]
    \psi_0(k)= \psi_0 e^{i(k-k_0)^2/4\sigma k^2}
    [/itex]

    where ##\psi_0## is an ordinary normalisation factor (please correct me if I am wrong).



    QUESTION 1: Why do we choose ##\psi_0(k)## as a gauss function? Why is this function so appropriate in this case?

    QUESTION 2: I don't know how did our professor get a gauss function with an imagnary number ##i## in it. His gauss is nothing like the one on Wikipedia which is

    [itex]
    f(x) = a e^{-(x-b)^2/2c^2}
    [/itex]

    QUESTION 3: We used the first integral i wrote down to calculate the Heisenberg's uncertainty principle like shown below, but it seems to me that most of the steps are missing and this is the reason i don't understand this. Could anyone explain to me step by step how to do this.

    [itex]
    \begin{split}
    \psi &= \int\limits_{-\infty}^{+\infty} \psi_0\!(k) \cdot e^{ikx} \, \textrm{d} k\\
    \psi &= \int\limits_{-\infty}^{+\infty} \psi_0 e^{i(k-k_0)^2/4\sigma k^2} \cdot e^{ikx} \, \textrm{d} k\\
    \psi &= \psi_0 2 \sqrt{\pi} e^{ik_0x} e^{-x/2 \sigma k^2}
    \end{split}
    [/itex]

    I think this is connected to a Gaussian integral, but it doesn't look quite like it to me. Well in the end our professor just says that out of the above it follows that

    [itex]
    \boxed{\delta x \delta k = \frac{1}{2}}
    [/itex]

    I don't understand this neither. It was way too fast or me.
     
  2. jcsd
  3. Feb 24, 2013 #2

    mfb

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    Staff: Mentor

    1 and 2: This choice gives something which looks like a moving particle. Without a careful choice of ##\psi(k)##, you get some wave spread out over the whole space.

    It is one. You can combine those two exponentials to a single one, make a linear substitution and get the gaussian integral. Are you sure that the last exponent in the last line has just an x there?

    For δxδk >> 1/2, the last exponential function should vanish, so you get a significant wave only if the product δxδk is small. A better analysis would give that factor of 1/2.
     
  4. Feb 24, 2013 #3
    uncertainty product is minimum for gaussian wave packet.the inequality becomes the equality.
     
  5. Feb 24, 2013 #4
    So what does this mean our professor was wrong? Could please someone show a correct derivation f possible using equations.

    Do you mean this one?

    [itex]
    \begin{split}
    \psi &= \psi_0 2 \sqrt{\pi} e^{ik_0x} e^{-x/2 \sigma k^2}
    \end{split}
    [/itex]

    Thats how our professor did this. Was he wrong? I am totaly confused.
     
    Last edited: Feb 24, 2013
  6. Feb 24, 2013 #5

    mfb

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    Staff: Mentor

    I'm not sure, but the equation looks not as I would expect it.
     
  7. Feb 24, 2013 #6
    I think my professor was wrong so please tell me how would you expect the equation to be.
     
  8. Feb 24, 2013 #7

    mfb

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    Staff: Mentor

    It is better if you look for a quantum mechanics book with a proper derivation of the uncertainty for a gaussian wave packet.
     
  9. Feb 24, 2013 #8
    Well i am quite weak in mathematics. Is there any nice book suited for me?
     
  10. Feb 24, 2013 #9

    tom.stoer

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    Science Advisor

    The HUP is a property of the Fourier Transformation and applies to every pair f(x), g(p) where g is the FT of f.

    Four Gausssian wave pakets the product Δx Δp is minimized.

    But of course one cannot derive the HUP in general based on a special choice of functions.
     
  11. Feb 24, 2013 #10
    Could you provide such a case and derivaton for HUP?
     
  12. Feb 24, 2013 #11

    tom.stoer

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    Science Advisor

    Try this derivation which does not use Fourier transform at all but only "algebraic" formulas applied to Hilbert space vectors
     
  13. Feb 25, 2013 #12
    In the case of a special wave function as the given gaussian one,one can do get a relation by exploiting the definition of Δx and Δp as
    Δx=√(<x2>-<x>2) and similarly for Δp.After that one can employ
    <x2>=∫ψ*x2ψ dx(in case of noramalized ψ)
    one can evaluate <x2>,<x>,<p>,<p2> by this using first quantized form for momentum operator i.e. while evaluating integral of p2,put -d2/dx2.then one can evaluate Δx and Δp and product of it will give h-/2
    p=h-k,gives ΔxΔp=1/2.
     
  14. Feb 25, 2013 #13

    stevendaryl

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    Staff Emeritus
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    Here's an answer: Define

    [itex](\delta x)^2[/itex] = [itex]\langle(x - \langle x \rangle)^2\rangle[/itex]
    [itex](\delta p)^2[/itex] = [itex]\langle(p - \langle p \rangle)^2\rangle[/itex]

    where [itex]\langle A \rangle = \int \psi^{*}(x) A \psi(x) dx[/itex]

    Then we can prove (using the calculus of variations) that the [itex]\psi(x)[/itex] that minimizes [itex]\delta x \delta p[/itex] is a Gaussian.
     
  15. Feb 25, 2013 #14

    jtbell

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    Staff: Mentor

    I think the second equation should be

    $$\psi_0(k) = \psi_0 e^{-(k-k_0)^2/4\sigma^2}$$

    with a "-" sign instead of an "i" in the exponent, and no "k" in the denominator. That "k" could be a subscript:

    $$\psi_0(k) = \psi_0 e^{-(k-k_0)^2/4\sigma_k^2}$$

    which would be appropriate because ##\sigma## or ##\sigma_k## is the standard deviation of k.
     
  16. Feb 25, 2013 #15
    I think that our professor was lecturing by using this document where they used a suare root of a gauss so this is where he got ##\psi_0(k)##:

    [itex]
    \psi_0(k)= \sqrt{gauss} = \sqrt{\psi_0 e^{-(k-k_0)^2/2\sigma_k^2}} = \psi_0 e^{-(k-k_0)^2/4\sigma_k^2}
    [/itex]

    My question is Why do we take a square root of a gauss?
     
    Last edited: Feb 25, 2013
  17. Feb 25, 2013 #16
    Our professor did a mistake. You are right! Please take a look at my previous post.
     
  18. Feb 25, 2013 #17

    jtbell

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    Staff: Mentor

    As you probably know already, ##\psi(x)## is the probability amplitude for x, which we "complex-square" to get the probability distribution for x: ##P(x) = |\psi(x)|^2##.

    Similarly, your ##\psi_0(k)## is the probability amplitude for k, which we "complex-square" to get the probability distribution for k: ##P_k(k) = |\psi_0(k)|^2##.

    I suspect that your professor wants to make ##P_k(k)## a Gaussian with standard deviation ##\sigma_k##. Some books do it the other way, i.e. they make ##\psi_0(k)## a Gaussian with standard deviation ##\sigma_k##.
     
    Last edited: Feb 25, 2013
  19. Feb 25, 2013 #18
    In the link i provided earlier for example they do both (##\psi_0(k)## as well as ##\psi_0(x)##) a standard deviation - Gauss. Why is that so?
     
    Last edited: Feb 25, 2013
  20. Feb 26, 2013 #19
    it is much easier than this.The schwarz inequality becomes an equality when the two function f and g used are in relation f=λg,where λ is a constant.Also f and g are most appropriately chosen as for deriving uncertainty principle
    f=-ih-∂ψ/∂x,g=ixψ,the condition f=λg does give a gaussian function.
     
  21. Feb 26, 2013 #20
    But i am weak in math and Fourier is more suiable for me. I hae no clue about Schwarz inequaity... Could anyone provide a clean derivation of uncertainty principle using gauss wave packet together with Fourier?
     
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