# Derivative of delta function

I am reading and following along the appendices of "The Physical Principles Of The Quantum Theory", and trying to learn how he derives Schrödinger's Equation from his Matrix Mechanics, but I have run into a bit of trouble. It seems like for his derivation to work, it must be necessary for the integral of a function times the Dirac Delta Function's derivative be: $$\int^{\infty}_{-\infty}f(\xi)\delta'(a-\xi)d\xi=f'(a). \tag{36}$$ But the actual identity is $$\int^{\infty}_{-\infty}f(\xi)\delta'(\xi-a)d\xi=-f'(a).$$ Does anybody care to explain why it is like this in Heisenberg's book, or provide a derivation along the same vein, but with the correct identity for the delta function?

It’s not a typo. The distribution $\delta’$ is odd meaning $\delta’(y-x)=-\delta’(x-y)$.
• Wait, so both $f(a)$ and $-f(a)$ are valid answers to $$\int^{\infty}_{-\infty}f(\xi)\delta'(a-\xi)d\xi?$$ Because through integration my parts I obtain $-f(a)$. Jul 27 '18 at 15:46
• No. You can shift the variable to $x=\xi-a$ if it makes it easier to see. After using $\delta'(a-\xi)=\delta'(\xi-a)$Your first equation then becomes $$-\int_{-\infty}^{\infty} dx \ f(x+a) \delta'(x)= f'(a)$$ Jul 27 '18 at 15:57
• Where $\delta'$ has the property $$\int_{-\infty}^{\infty}dx\ F(x) \delta'(x) = - F'(0)$$ Here you just have $F(x)=-f(x+a)$ Jul 27 '18 at 15:58
• I meant $\delta'(a-\xi)=-\delta'(\xi-a)$ in the above. For some reason I can't edit what I wrote. Jul 27 '18 at 16:07
• I just realized I made a mistake. I meant to write $$\int^{\infty}_{-\infty}f(\xi)\delta(a-\xi)d\xi$$ for the second equation. Is it too lste to edit ot or should I just delete the question and start anew and correct? Jul 28 '18 at 3:52
The Dirac $\;\delta\;$ function is even. This is more clear looking it as limit of proper functions.