# I Dirac-Delta Function, Different Integration Variable

#### Ibraheem

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?

Related Quantum Physics News on Phys.org

#### Orodruin

Staff Emeritus
Homework Helper
Gold Member
2018 Award
The correct relation is
$$\int \delta’(x) \psi(x) dx = -\psi’(0).$$
This follows directly from the distributional derivative definition
$$f’[\psi] = -f[\psi’].$$

#### vanhees71

Gold Member
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 $\delta'$.

#### Orodruin

Staff Emeritus
Homework Helper
Gold Member
2018 Award
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.

#### vanhees71

Gold Member
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.

#### Orodruin

Staff Emeritus
Homework Helper
Gold Member
2018 Award
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.

#### vanhees71

Gold Member
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.

#### Orodruin

Staff Emeritus
Homework Helper
Gold Member
2018 Award
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 $\psi(x')$.

#### Ibraheem

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)

#### Orodruin

Staff Emeritus
Homework Helper
Gold Member
2018 Award
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?

#### vanhees71

Gold Member
Than it's a typo in the book.

#### Orodruin

Staff Emeritus
Homework Helper
Gold Member
2018 Award
... and you should not be surprised that books as extensive as Shankar has typos in it ...

#### Ibraheem

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:

#### vanhees71

Gold Member
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
$$\hat{p} \psi(x)=-\mathrm{i} \, \mathrm{d}_x \langle x|\psi \rangle=-\mathrm{i} \, \mathrm{d}_x \psi(x).$$

"Dirac-Delta Function, Different Integration Variable"

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving