I'm trying to compute to quantities with Hamiltonian and Dirac delta function but I don't how to do it properly. I'm stuck calculating the following quantity

$$ \frac{d}{dE} \left[ \theta(E-H(x,p;V)) \right]. \tag{1}$$ I know that the derivative of the Heaviside step function $\theta(x)$ is the dirac delta function $\delta(x)$ $$ \frac{d\theta(x)}{dx}= \delta(x)$$

So my guess would be the following:

$$ \frac{d\theta(E-H)}{dE}= \delta(E-H).$$

However if I use the chain rule I get: $$ \frac{d\theta(E-H)}{d(E-H)}\frac{d(E-H)}{dE}= \delta(E-H) \left(1-\frac{dH}{dE}\right).$$

The text I'm studying about areas in phase planes gives the following but leaves out a lot of intermediate steps:

$$ \frac{d}{dE} \left[ \theta(E-H(x,p;V)) \right]=\delta (p^2/2m+\phi-E).$$

Where $H=p^2/2m+\phi(x;V).$

The next step makes me equally confused, it gives the following equation:

$$ \int dp \delta (p^2/2m+\phi-E)=2m/p(x) \ \ \ \ \ \ \ (2)$$

Where the text tells it uses the formula $$\delta(f(p)) = \sum_i \delta(p-p_i)/|f'(p_i)|.$$

Questions How to understand the following equations:

$$ \frac{d}{dE} \left[ \theta(E-H(x,p;V)) \right] =\delta (p^2/2m+\phi-E)) \ \ \ \ \ \ \ \ \ \ (1)$$

$$ \int dp \delta (p^2/2m+\phi-E)=2m/p(x) \ \ \ \ \ \ \ (2)$$


1 Answer 1


The first is simple I think: $$\frac{d}{dE}\theta[E-H(x,p;V)]=\delta[E-H(x,p;V)]=\delta[H(x,p;V)-E]= \delta\Big[\frac{p^2}{2m}+\phi(x;V)-E\Big]$$ where in the third step I used the symmetric property of the $\delta$ function, i.e. $\delta(-x)=\delta(x)$.

As far as the second question is concerned, one can proceed in the following way $$\int dp\delta(p^2/2m+\phi-E)=\int dp \Big[\frac{1}{|p_+(x)/m|}\delta(p_+^2/2m+\phi-E) +\frac{1}{|p_-(x)/m|}\delta(p_-^2/2m+\phi-E)\Big]$$ where $p_{\pm}=\pm\sqrt{2m(E-\phi)}$ are the two solutions to the equation $f(p)=p^2/2m+\phi-E=0$. The $x$ dependence of $p$ indicates that $p$ is a function of $x$ through $\phi$. Performing the integrals and identifying that the magnitude of $p_+(x)$ equals the magnitude of $p_-(x)$, yields $$\int dp\delta(p^2/2m+\phi-E)=\Big[\frac{1}{|p_+(x)/m|} +\frac{1}{|p_-(x)/m|}\Big]=\frac{2m}{p(x)}$$ where $p(x)=|p_+(x)|=|p_-(x)|=\sqrt{2m(E-\phi)}$.

I hope this helps! If anything is unclear, please comment!

  • $\begingroup$ Thank you very much for your quick and lucid answer. I have one question, what is gained by swapping $\delta(E-H)=\delta(H-E)$? Or is this a common thing to do? $\endgroup$ Jul 31, 2022 at 9:12
  • 1
    $\begingroup$ Hi! I spotted a mistake in my answer and I have edited the answer! I actually substituted $p_+(x)$ with $|p_+(x)|$ and $p_-(x)$ with $|p_-(x)|$! As far as your question is concerned, nothing is gained by writting $\delta(E-H)$ instead of $\delta(H-E)$, as they are the same! I do not know why one would prefer one instead of the other, but the reasons must be aesthetic or random probably, because it does not matter which one is preferred... I hope this helps $\endgroup$
    – schris38
    Jul 31, 2022 at 9:43

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