Dirac-Delta Function, Different Integration Variable

In summary, the conversation discusses the derivative of the Dirac-delta function and its interaction with a given function along the x' axis. The correct relation for the integral of the delta distribution and the given function is $$\int \delta'(x) \psi(x) dx = -\psi'(0).$$ However, there is some disagreement about the second expression in #1 and whether it is a typo or not. Some believe that the constant function 1 can be treated as a test function in the distributional sense, while others argue that it is not a proper test function and the integral would be undefined. Regardless, it is agreed upon that the result of the integral would be 0. Overall, it is suggested to be cautious
  • #1
Ibraheem
51
2
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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  • #2
The correct relation is
$$
\int \delta’(x) \psi(x) dx = -\psi’(0).
$$
This follows directly from the distributional derivative definition
$$
f’[\psi] = -f[\psi’].
$$
 
  • #3
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'##.
 
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  • #4
vanhees71 said:
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'##.
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.
 
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  • #5
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.
 
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  • #6
vanhees71 said:
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.
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.
 
  • #7
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.
 
  • #8
vanhees71 said:
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.
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')##.
 
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  • #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)
 
  • #10
Ibraheem said:
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)
Did you reproduce the statement exactly as Shankar put it?
 
  • #11
Orodruin said:
Did you reproduce the statement exactly as Shankar put it?
Yes.
 
  • #12
Than it's a typo in the book.
 
  • #13
... and you should not be surprised that books as extensive as Shankar has typos in it ...
 
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  • #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'.
 
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  • #15
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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What is the Dirac-Delta Function?

The Dirac-Delta Function, also known as the unit impulse function, is a mathematical concept used in calculus and signal processing. It is defined as a function that is equal to zero everywhere except at one specific point (usually at the origin), where it is infinite. It is often represented by the symbol δ(x) and is commonly used to model a point mass or impulse in physical systems.

How is the Dirac-Delta Function used in integration?

The Dirac-Delta Function is used in integration as a tool to represent functions that are highly concentrated at a single point. It acts as a weighting function, allowing us to integrate functions that are not continuous or differentiable at a specific point. This is known as the Dirac Delta method of integration.

What is the relationship between the Dirac-Delta Function and the Kronecker-Delta Function?

The Dirac-Delta Function and the Kronecker-Delta Function are both mathematical concepts that have a value of zero everywhere except at one specific point. However, the Kronecker-Delta Function is discrete and is only defined for integer values, while the Dirac-Delta Function is continuous and can be defined for any real value. They are also used for different purposes, with the Kronecker-Delta Function often used in summation and the Dirac-Delta Function used in integration.

Can the Dirac-Delta Function be integrated with respect to a different variable?

Yes, the Dirac-Delta Function can be integrated with respect to a different variable, as long as the limits of integration include the point where the Dirac-Delta Function is non-zero. This is known as the Different Integration Variable method and is commonly used in solving differential equations and Fourier transforms.

What are some applications of the Dirac-Delta Function?

The Dirac-Delta Function has many applications in physics, engineering, and mathematics. It is used to model point sources of energy or mass, such as in quantum mechanics and electromagnetism. It is also used in signal processing to represent impulsive signals. Additionally, the Dirac-Delta Function has applications in solving differential equations, Fourier transforms, and in the study of distribution theory.

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