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In microcanonical ensembles we have (for one particle in 1 dimension)

$$\int \delta(H(p,q)-E)dpdq=\Omega(E)$$

I am not convinced and believe that this integral diverges. Take for example a harmonic oscillator in one dimension, with $H(p,q)=p^2/2+q^2/2=E$ ($m=k=1$ for simplicity). One can write the above integral as

$$\int f(p,q)dpdq$$

The integrand is $f(p,q)=\infty$ for all points $(p,q)$ with $H(p,q)=E$. Thus, one can think of $f(p,q)$ as a distribution with infinitely many (uncountably infinite) Dirac functions sprinkled around every point $(p,q)$ on the circle $H(p,q)=p^2/2+q^2/2=E$ (see figure). Each Dirac function will contribute 1 to the integral, but there are infinitely many of them on the circle so the integral will diverge. This argument is generalizable to higher dimensions.

I need help in clearing up my confusion as to why 1) this integral does not diverge and 2) it is proportional to $\Omega(E)$, i.e., the circumference of this circle.

enter image description here

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    $\begingroup$ Note that the Dirac delta function isn't equal to infinity when its argument is zero. That's a common kind of intuition people build about the delta function, but it isn't the definition. Rather, it's simply defined in such a way that integrating it picks out the value at which its argument is zero. See this answer physics.stackexchange.com/q/371700 for a lot of detail on integrating dirac deltas with non-trivial functions as arguments. $\endgroup$ – Metropolis May 10 at 13:44
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    $\begingroup$ You should also have a look at the wikipedia page, in particular the section "Properties in n dimensions". $\endgroup$ – Yvan Velenik May 10 at 13:48
  • $\begingroup$ @Metropolis That is not my problem. My main contention is that there are infinitely many Dirac delta functions (one for each point $(p,q)$) on the circle, where each Dirac function which will contribute one to the total integral: $1+1+1+...=\infty$ $\endgroup$ – Omar Nagib May 10 at 14:00
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    $\begingroup$ @OmarNagib No, the $\delta$ "function" does not decompose like that. Read the references suggested. $\endgroup$ – Yvan Velenik May 10 at 14:01
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    $\begingroup$ that's just not how the Dirac $\delta$ works. $\endgroup$ – fqq May 10 at 14:03
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The integrand is $f(p,q)=\infty$ for all points $(p,q)$ with $H(p,q)=E$.

No, it's not.

Thus, one can think of $f(p,q)$ as a distribution with infinitely many (uncountably infinite) Dirac functions sprinkled around every point $(p,q)$ on the circle

No, you can't.

The hand-waving tale that the Dirac delta $\delta(x)$ equals "zero at $x\neq 0$ and infinity at $x=0$" is just that: a hand-waving tale. It is useful for building intuition in one dimension, but that's it.

What the Dirac delta actually represents is a distribution (also sometimes known as a "generalized function"), in the formal sense of distribution theory. A distribution is a function $\varphi: \mathscr F\to \mathbb R$ which acts on the space of well-behaved functions $\mathscr F$ and assigns a real number to each function. (For example, for any function $g$ you can create a distribution that takes $f$ to $\int_{-\infty}^\infty f(x)g(x)dx$.) The Dirac delta distribution is the function \begin{align} \delta_0 : \mathscr F &\to \mathbb R \\ f & \mapsto \delta_0(f) = f(0) \end{align} which takes an arbitrary well-behaved function $f$ and returns the value of $f$ at the origin, $f(0)$.

In your integral, $$\int \delta(H(p,q)-E)dpdq,$$ the Dirac delta is being taken over a one dimensional space, and evaluated at the variable $\epsilon = H(p,q)-E$. If your analysis does not account for that, then it is wrong.

This is easier to handle for the specific example of the harmonic oscillator, where you're trying to calculate $$\mathrm{int}(E) = \iint \delta(\tfrac12(p^2+q^2)-E)dpdq.$$ This now has a Dirac delta evaluated over a one-dimensional energy variable which itself depends on a function of two separate integration variables $-$ which should look like quite a nightmare! To handle this, the thing to do is to make a suitable change of variables (in this case, to radial coordinates) that will separate the two. So, if we define $h=\tfrac12(p^2+q^2)$ and $\phi = \arctan(p/q)$ (so that $q=\sqrt{2h}\cos(\phi)$ and $p=\sqrt{2h}\sin(\phi)$, and $dp\,dq = dh\,d\phi$), which gives us $$\mathrm{int}(E) = \int_0^{2\pi}\int_0^\infty \delta(h-E)dh\,d\phi= \int_0^{2\pi}1 d\phi \times \int_0^\infty \delta(h-E)dh.$$ This separation shows what's really happening: the Dirac delta is indeed singular, and it is being integrated over one dimension to give a finite result. The other integration variable then gives the circumference of the ring of equal-energy states, which is what we want to calculate with $\Omega(E)$.

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  • $\begingroup$ Thank you for your detailed answer. Quick follow up. It seems to me that your last integral $\mathrm{int}(E)=2 \pi$ (for all positive $E$). I thought that it should rather be equal to the circumference of a circle with a radius $\sqrt{E}$, i.e., $\mathrm{int}(E)= \Omega(E)= 2 \pi \sqrt{E}$? $\endgroup$ – Omar Nagib May 10 at 15:50
  • $\begingroup$ @OmarNagib That's appealing, but it is wrong starting with the dimensional analysis. The Dirac delta has dimensions of $[\delta(x)]=[1/x]$, which in this case gives $[\delta(h)]=[1/h]=[1/p^2]=[1/q^2]=[1/pq]$, so $\mathrm{int}(E)$ must be dimensionless. $\endgroup$ – Emilio Pisanty May 10 at 15:58
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    $\begingroup$ From the statistical-mechanics side, $\mathrm{int}(E)$ is counting the number of microstates between energies $E$ and $E+dE$ (heuristically speaking), which forms a circular strip. As $E$ increases, the radius of this circular strip increases, but its width decreases proportionally. For more, see this answer. $\endgroup$ – Emilio Pisanty May 10 at 15:59
  • $\begingroup$ Thank you very much. You have been very helpful. $\endgroup$ – Omar Nagib May 10 at 20:15

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