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I Dirac-Delta Function, Different Integration Variable

  1. Nov 29, 2017 #1
    Hello,
    I know that the derivative of Dirac-delta function (##\delta'(x-x') = \frac{d}{dx} (\delta(x-x')))## does the following:

    ##\int_{-\infty}^{\infty}\psi(x')*\delta'(x-x') dx' = \frac{d\psi(x)}{dx}##
    it is easy to visualize how the delta function and the function ##\psi(x')## interact along the x' axis to give the derivative with respect to x ( i.e scaling ##\psi(x\pm \varepsilon ) ## at each bump ##\pm \varepsilon ##..etc)
    However, I am stuck at a situation where

    ##\int_{-\infty}^{\infty}\psi(x')*\delta'(x-x') dx ##
    I am not sure what to do here. But what I think it should equal to is ##\frac{-d\psi(x')}{dx'}## just by applying the same way of thinking as when the integration variable is the same as that of of the function ##\psi##. But here the function I think acts as constant in this integral which makes me doubt my answer. Is this answer correct?
     
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  3. Nov 29, 2017 #2

    Orodruin

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    The correct relation is
    $$
    \int \delta’(x) \psi(x) dx = -\psi’(0).
    $$
    This follows directly from the distributional derivative definition
    $$
    f’[\psi] = -f[\psi’].
    $$
     
  4. Nov 29, 2017 #3

    vanhees71

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    Well, the original formula was also right, if one interprets the prime in a specific way (it's of course often misleading to use a prime for a derivative when many independent variables are involved). I read the equation as follows:
    $$\int_{\mathbb{R}} f(x') \delta'(x-x') = \int_{\mathbb{R}} f(x') \partial_x \delta(x-x') \\
    = \int_{\mathbb{R}} f(x') [-\partial_{x'} \delta(x-x')]=+\int_{\mathbb{R}} \mathrm{d} x' f'(x') \delta(x-x')=+f'(x).$$
    The 2nd expression in #1 doesn't make sense, because you cannot integrate the ##\delta## distribution or its derivatives without multiplying it with a test function. In the 2nd term, the ##\psi(x')## is just a constant that you can move out of the integral, and then you are left with an undefined integral over ##\delta'##.
     
  5. Nov 29, 2017 #4

    Orodruin

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    You are right. The point is that ##\delta'## is an anti-symmetric distribution, which means that it makes a difference whether one writes ##\delta(x-x')## or ##\delta(x'-x)##. So, this solves the sign issue. I believe the 2nd expression in #1 to be a typo. Regardless, the constant function 1 is sufficiently nice to be able to act on it with the delta distribution and the delta distribution derivative. If one wants to be more stringent, one could look at it as the convolution of two distributions.
     
  6. Nov 29, 2017 #5

    vanhees71

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    You have to be careful with the domain of the distribution. Usually the test-function space is either the Schwartz space of quickly falling ##C^{\infty}(\mathbb{R})## functions (i.e., going to 0 faster than any power for ##|x| \rightarrow \infty##). Then the constant function is not a test function. Usually the integral over a distribution without being multiplied by a proper test function is an undefined expression.
     
  7. Nov 29, 2017 #6

    Orodruin

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    But this was my point. I do not see where the disagreement lies. The exact same definition of the delta distribution works on a set of test-functions that does not necessarily going to zero as ##|x| \to \infty##, in other words it is a distribution on the test functions in ##C^\infty(\mathbb R)## without the additional requirements of being in the Schwartz class. Naturally, the set of distributions on this space is smaller, but the delta distribution does belong to it as well. Furthermore, I did qualify that the given integral could also be interpreted as a convolution of distributions. In that case it should work perfectly well with test functions in the Schwartz class as well.
     
  8. Nov 29, 2017 #7

    vanhees71

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    Ok, if this is the case, the 2nd expression in #1 is simply 0, because
    $$\int_{\mathbb{R}} \mathrm{d} x \delta'(x-x')=\int_{\mathbb{R}} \mathrm{d} x \delta(x-x') \mathrm{d}_x 1 =0.$$
    That makes also intuitive sense, because you can define the ##\delta## distribution as a weak limit of symmetric analytical test functions and its derivative by the derivative of these test functions.
     
  9. Nov 29, 2017 #8

    Orodruin

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    Sure, I agree with this. This would be the result regardless of which of the possible interpretations you make. Then again, I continue to suspect a typo in the 2nd expression and that the integrand should really contain ##\psi(x)## and not ##\psi(x')##.
     
  10. Nov 29, 2017 #9
    So the second equation is wrong. That's odd since this is what Shankar says , at least according to my understanding, in page 64(Principals of Quantum Mechanics)
     
  11. Nov 29, 2017 #10

    Orodruin

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    Did you reproduce the statement exactly as Shankar put it?
     
  12. Nov 29, 2017 #11
    Yes.
     
  13. Nov 29, 2017 #12

    vanhees71

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    Than it's a typo in the book.
     
  14. Nov 29, 2017 #13

    Orodruin

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    ... and you should not be surprised that books as extensive as Shankar has typos in it ...
     
  15. Nov 29, 2017 #14
    I actually think I misquoted him. However in either case, this is what made me look into it. I was trying to insert the identity operator in ##XP|\psi>##, where X and P are the position and momentum operators respectively. I first inserted it between X and ##|\psi>## and I got the following:
    ##\int_{-\infty}^{\infty}PX|x><x|\psi>dx##
    ##=\int_{-\infty}^{\infty}x\psi(x)P|x>dx##
    Then I again inserted the identity operator between ##\psi(x)## and P and got the following:
    ##=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}x\psi(x)|x'><x'|P|x>dx'dx##
    so from here I substituted for ##<x'|P|x>## and this is what I got:
    ##=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}x\psi(x)|x'>(-i\hbar\delta'(x'-x))dx'dx##

    here is where I got stuck. I am not sure if inserting the identity like this is okay. If the integral of the delta function goes to zero then the whole thing goes to zero? which doesn't make sense.I am missing something here. The only way to make sense of it is to integrate first with respect to dx and not dx' in which case I used the following fact that the derivative of the delta function is odd:

    ##\delta'(x-x')= -\frac{d}{dx}(\delta(x'-x))## so the integral becomes:

    ##=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}x\psi(x)|x'>(i\hbar\delta'(x'-x))dxdx'##

    which evaluates to :

    ##-i\hbar \int_{-\infty}^{\infty}\frac{d}{dx}(x'\psi(x')) dx##

    which I hope is correct. However I am not sure how will this be if I integrate first with respect to x'.
     
    Last edited: Nov 29, 2017
  16. Nov 30, 2017 #15

    vanhees71

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    Well, as I said before, instead of writing a prime just clearly specify with respect to which variable you are taking the derivative. The matrix element of the momentum operator in position representation is
    $$\langle x'|\hat{p}|x \rangle=-\mathrm{i} \partial_{x'} \delta(x-x').$$
    To prove this, you need
    $$u_p(x)=\langle x|p \rangle=\frac{1}{\sqrt{2 \pi}} \exp(\mathrm{i} p x).$$
    To prove this you only need the commutator relation
    $$[\hat{x},\hat{p}]=\mathrm{i}.$$
    To that end define
    $$\hat{x}(\xi)=\exp(\mathrm{i} \xi \hat{p}) \hat{x} \exp(-\mathrm{i} \xi \hat{p}).$$
    Then you find
    $$\mathrm{d}_{\xi} \hat{x}(\xi)=\mathrm{i} \exp(\mathrm{i} \xi \hat{p}) [\hat{p},\hat{x}] \exp(-\mathrm{i} \xi \hat{p})=\mathbb{1}.$$
    Integrating this wrt. to ##\xi## again, you get
    $$\hat{x}(\xi)=\hat{x} + \xi \mathbb{1}.$$
    This implies that
    $$|x \rangle=\exp(-\mathrm{i} x \hat{p}) |x=0 \rangle$$
    and thus
    $$\langle p|x \rangle=\exp(-\mathrm{i} x p) \langle p|x=0 \rangle=N_{p} \exp(-\mathrm{i} x p).$$
    To get
    $$\langle p'|p \rangle=\delta(p-p'),$$
    you have to specify ##N_{p}##. That's easy by inserting a unit operator
    $$\langle p'|p \rangle=\int_{\mathbb{R}} \mathrm{d} x \langle p'|x \rangle \langle x|p \rangle=N_{p'} N_{p}^* \int_{\mathbb{R}} \mathrm{d} x \exp[\mathrm{i} x(p-p')]=|N_{p}|^2 2 \pi \delta(p-p'),$$
    which implies that (up to an unimportant phase)
    $$N_p=\frac{1}{\sqrt{2 \pi}}.$$
    Since
    $$|x \rangle=\exp(-\mathrm{i} x \hat{p}) |x=0 \rangle,$$
    you find
    $$\mathrm{d}_x |x \rangle=-\mathrm{i} \hat{p} \exp(-\mathrm{i} x \hat{p}) |x=0 \rangle = -\mathrm{i} \hat{p} |x \rangle,$$
    and thus by taking the Hermitean adjoint
    $$\mathrm{d}_x \langle x| =+\mathrm{i} \langle x|\hat{p},$$
    and thus finally
    $$\hat{p} \psi(x)=-\mathrm{i} \, \mathrm{d}_x \langle x|\psi \rangle=-\mathrm{i} \, \mathrm{d}_x \psi(x).$$
     
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