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My apologies if this question has already been asked. I've tried using the search function but could not find the answer.

I'm looking to solve the following integral:

$$ \int^\infty_0 \frac{1}{q} e^{-aq^2} dq, $$ where $a >0 $ and $q$ denotes the magnitude of the momentum. Now obviously this integral does not converge on the given domain but I am looking for find a way to extract a sensible approximation from this equation. The context is that I am calculating decoherence rates for various quantum mechanical interactions and when calculating the decoherence rate for møller scattering I am left with a bunch of constants and this integral. I would really appreciate any thoughts on this. Thank you in advance!

Edit_1: In response to the question of the wider physical context I will briefly outline the problem below.

I am calculating the rate at which several electromagnetic interactions cause a loss of entanglement in the spatial superposition of a charged particle in free-fall.

The particles are entangled with their environment such that their reduced density matrix becomes (not quite sure how the bra-ket notation works here as the braket latex package is not loading):

$$ \rho_{\mathcal{S}} = \sum_{n,m = 1}^{N} c_{n}c_{m}^{*} |\psi_{n}>< \psi_{m}| <E_{m}|E_{n}>. $$

Since the entanglement is generated using spin-coupling to a magnetic field we can express this using singlet spin states as:

$$ \rho_{\mathcal{S}} = \frac{1}{2} \begin{pmatrix} 1 & <E_{2}|E_{1}> \\ <E_{2}|E_{1}> & 1 \end{pmatrix}. $$

After several steps this gives us the time evolution of the off-diagonal terms as:

$$ \rho_\mathcal{S}(\textbf{x},\textbf{x}^{\prime},t) = \rho_\mathcal{S}(\textbf{x},\textbf{x}^{\prime},0)e^{-\gamma t} $$

This $\gamma$ is called the decoherence rate and is given by $\Gamma \equiv \int dq \varrho(q) v(q) \sigma_{tot}(q)$.

The calculation of this $\Gamma$ is giving me problems. Taking møller scattering (electron-electron scattering) as the relevant interaction we find that:

$$ \sigma \approx \frac{14\pi\alpha^2\hbar^2c^2m^2}{4\textbf{q}^4}, $$

$\varrho$ is given by the Maxwell-Boltzmann distribution:

$$ \varrho = \frac{N}{V} 4\pi q^2 \left( \frac{1}{2\pi m k_b T}\right)^{3/2} exp\left[-\frac{q^2}{2mk_bT}\right], $$

and $v = \frac{\textbf{q}}{m}$. This gives:

$$ \Gamma = \frac{N}{V} 14 m \pi^2 \alpha^2 \hbar^2 c^2 \left( \frac{1}{2\pi m k_b T2}\right)^{3/2} \int_0^\infty dq \frac{1}{q} exp\left[-\frac{q^2}{2mk_bT}\right]. $$

Where the final integral is the one which I am asking the question about. Since q is here is dependent on the thermal velocity (the environmental particles are in a gas) and the temperatures are low, on the order of 1-5K, we have that q is likely limited to not be extremely large. Perhaps this helps in understanding the problem.

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    $\begingroup$ Welcome. What would feel like a sensible approximation for you? Perhaps if you show the wider context of the physical equations, we can find ways to avoid divergent integrals, or at least decide which divergent regularisation methods are appropriate $\endgroup$
    – FShrike
    Jun 8, 2022 at 15:49
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    $\begingroup$ As it is, this integral diverges. $\endgroup$
    – Z Ahmed
    Jun 8, 2022 at 18:24
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    $\begingroup$ $q^{-1}$ is not an integrable singularity without more work (see above). $\endgroup$ Jun 8, 2022 at 19:22
  • $\begingroup$ Hi, thanks for responding. Please see the edited post for more physical context. $\endgroup$
    – Michael
    Jun 10, 2022 at 9:53
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    $\begingroup$ Thank you for posting the physical context, it allowed me to solve the issue you were having. You were setting up the integral incorrectly. @CalvinKhor it's been a while but have a look if you're curious to see the resolution. $\endgroup$ Aug 1, 2023 at 7:33

3 Answers 3

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Given how your Maxwell-Boltzmann distribution has a factor of $(k_BT)^{\frac{3}{2}}$, by Equipartition theorem this integral takes place in $3$ dimensional momentum phase space, and thus your integral set up is incorrect. Putting together the functions you have, you should instead get

$$\Gamma = k \int_{\Bbb{R}^3}d^3\mathbf{q}\:\frac{e^{-aq^2}}{q}$$

To properly compute the integral, use spherical coordinates to obtain

$$\Gamma = 4\pi k \int_0^\infty dq\:qe^{-aq^2} = \frac{2\pi k}{a}$$

since the spherical Jacobian means $d^3\mathbf{q} = q^2 dq\:d\Omega$, not $d^3\mathbf{q} = dq$ as you have been using.

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    $\begingroup$ Sadly it seems like OP may never see this answer since they haven't been on MSE since they posed the question. But hopefully this can help other people who make a similar mistake on setting up physical integrals. $\endgroup$ Aug 1, 2023 at 7:39
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As @Avnish Singh answered $$\int\frac{1}{q} e^{-aq^2}\, dq=\frac 12 \int \frac{e^{-t}}{ t}\,dt=\frac 12 \text{Ei}(-t)$$

Expanded as a series around $t=0$ gives for $$\text{Ei}(-t)=\log (t)+\gamma -t+\frac{t^2}{4}-\frac{t^3}{18}+O\left(t^4\right)$$ which is a very good approximation.

So $$\int_\epsilon^\infty\frac{1}{q} e^{-aq^2}\, dq=\frac{\log (a)+\gamma}{2}+\log(\epsilon)-\frac a 2 \epsilon^2+O\left(\epsilon^4\right)$$

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  • $\begingroup$ Hi! Thank you for providing the approximation with regards to the exponential integral. If you look at the edited OP you can see some more physical context for my problem. Unfortunately $\epsilon$ should really be 0 in my case. But that would obviously give an infinity. Perhaps I can indeed take some small non-zero epsilon. If you don't mind me asking, what is $\gamma$ in the expression? $\endgroup$
    – Michael
    Jun 10, 2022 at 9:56
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    $\begingroup$ @Michael Euler Mascheroni constant, one of the super constants. One of the many ways of defining it is as: $$\lim_{n\to\infty}\left[\sum_{k=1}^n\frac{1}{k}-\ln n\right]$$ $\endgroup$
    – FShrike
    Jun 10, 2022 at 9:59
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Take $aq^2 = t$

Now, $dt$ = $2aq \ dq$

$\implies \frac{dq}{q} = \frac{dt}{2aq^2}$

$\implies \frac{dq}{q} = \frac{dt}{2t}$

Thus the integration is simplified to

$\int \frac{e^{-t}}{2t} \ dt$ which is an exponential integral.

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  • $\begingroup$ Hi! Thank you for your answer. I indeed found the exponential integral as the answer too using Mathematica. My problem here is that I am unsure how to work with this integral as I don't think this has been covered in any of my courses. If you don't mind please see the edited post for some more physical context. $\endgroup$
    – Michael
    Jun 10, 2022 at 9:54

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