Does Dirac manipulate his Delta function sensibly?

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  • #1
lugita15
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In the Principles of Quantum Mechanics, Dirac derives an identity involving his delta function: xδ(x)=0. From this he concludes that if we have an equation A=B and we want to divide both sides by x, we can take care of the possibility of dividing by zero by writing A/x = B/x + Cδ(x), because our original equation is equivalent to A=B+Cxδ(x).

Now Dirac was writing in the 30's, back before they had a rigorous theory of distributions, so he was freely manipulating the Dirac delta. The identity he derived is certainly correct, but is adding a constant multiple of the delta function when you divide by x really mathematically justified? If it is, why isn't this technique more widely adopted, like when solving Laplace's equation or doing quantum mechanics, places where delta functions usually abound?

Any help would be greatly appreciated.

Thank You in Advance.
 

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  • #2
lugita15
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Just to clarify what Dirac is doing, I have attached the relevant excerpt from his Principles of Quantum Mechanics. What I'm also interested in is his assertion that [itex]\frac{d}{dx} ln(x) = \frac{1}{x} - i\piδ(x)[/itex]. Does that make any sense?
 

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  • #3
CompuChip
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Note that Dirac also realises that all the identities are only valid under an integral sign.
So [itex]x \delta(x)[/itex] is not really zero, but it is true that
[tex]\int x \delta(x) \, dx = 0[/tex]

If you know anything about measure theory, what he is actually saying is that the equalities (7) - (11) hold almost everywhere ("a.e.") - in the strict measure theoretical sense of the word.
 
  • #4
Sina
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I have always wondered when proving that ∫f(x)δ'(x) = ∫f'(x)δ(x) = f'(0) (for f 0 at boundaries) how does it make sense to use integration by parts and argue that f(x)δ(x), the residual term is zero at the boundaries when we can not talk about a function δ(x).
 
  • #5
Mute
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I have always wondered when proving that ∫f(x)δ'(x) = ∫f'(x)δ(x) = f'(0) (for f 0 at boundaries) how does it make sense to use integration by parts and argue that f(x)δ(x), the residual term is zero at the boundaries when we can not talk about a function δ(x).

The identity

[tex]\int dx~ f(x)\delta'(x) = -\int dx~ f'(x) \delta(x) = - f'(0)[/tex]

(note you missed a minus sign) is not derived. Rather, it is a definition made to be consistent with the integration by parts notation.
 
  • #6
Sina
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Well that seems plausible given that it is strange from the start to try taking derivatives of dirac-delta something.

but in great many number of books and one of the exams that I had taken way back in undergraduate and in numerous homeworks infact it is asked for you to show this property and integration by parts is the employed method. so there is still a great confusion about dirac delta distributions :p
 
  • #7
Mute
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Well, that's because integration by parts gives you the "right" answer (by design) if you can be convinced to throw away the boundary terms, and you don't have to understand the theory of generalized functions to get the result!
 
  • #8
Hurkyl
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So [itex]x \delta(x)[/itex] is not really zero, but it is true that
[tex]\int x \delta(x) \, dx = 0[/tex]
But by the very definition of Schwartz distributions, the fact that
[tex]\int x \delta(x) \varphi(x) \, dx = 0[/tex]​
for every test function [itex]\varphi[/itex] means that [itex]x \delta(x) = 0[/itex] is an equality of generalized functions.

(a similar statement is true for measure-theoretic distributions)
 
  • #9
CompuChip
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I didn't know that, never properly learned about distributions. Thanks for that!
 
  • #10
Sina
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Well, that's because integration by parts gives you the "right" answer (by design) if you can be convinced to throw away the boundary terms, and you don't have to understand the theory of generalized functions to get the result!

This made me wonder, is there a formal definition of the derivative of a distribution?
 
  • #11
CompuChip
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There is, see equations (2) and (3) here.
 
  • #12
Sina
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ah well same thing as above then. I was kind of hoping a more geometric definition from which the above property could be derived
 
  • #13
Teethwhitenin
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ah well same thing as above then. I was kind of hoping a more geometric definition from which the above property could be derived

Aggree with sina.. please .. i need more geometric definitions
 
  • #14
Hurkyl
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The derivative is, I believe, a continuous operator on the space of tempered distributions. You could, if you wanted, define the distributional derivative as the continuous extension of the ordinary derivative applied to test functions.

Every tempered distribution is a limit (in the space of tempered distributions) of test functions, so this would define the derivative of any tempered distribution.


But the choice of which property to call "definition" doesn't really matter....
 

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