Most texts on quantum mechanics include a chapter on scattering theory which derives equations for the scattering amplitude. For example, $$f_k( \mathbf {\hat r}) = -\frac{\sqrt{2\pi}\mu}{\hbar^2}\int exp(-ik\mathbf {\hat r}\cdot \mathbf r’)V(\mathbf r')\psi_k(\mathbf r’)d \mathbf{\tau}' \tag{11.35a}.$$ $$f_k(\theta)=\frac{1}{k}\sum_{l=0}^\infty(2l+1)exp(i\delta_l(k))\sin\delta_l(k)P_l(\cos\theta)\tag{11.59}.$$

However, of interest is the relationship between the phase shift and the potential and radial eigenfunctions. In Merzbacher's Quantum Mechanics (2nd. ed.), the author mentions that a partial wave analysis, which uses the two equations given above, leads to the sought-after relationship, though he provides no derivation $$\sin\delta_l = -k\int_0^\infty j_l(kr')U(r')u_{l,k}(r')r'dr'\tag{11.83}$$ where $$U=\frac {2\mu V}{\hbar^2}.$$

I have tried to derive this equation but have been unsuccessful. I believe its derivation, besides being interesting in its own right, would likely demonstrate the use of concepts important to those learning scattering theory, such as the orthogonality of the complete sets $P_l$ and $j_l$ and their use in demonstrating various relationships.

I would greatly appreciate it if someone would derive the formula above for $\sin\delta_l.$

  • $\begingroup$ Does this answer help? Does Wikipedia help? $\endgroup$
    – Ghoster
    Commented Jan 12, 2023 at 18:52
  • $\begingroup$ But the text does derive it, no? You want a simpler text? Landau & Lifshitz, Ch XVII are quite informative.... $\endgroup$ Commented Jan 12, 2023 at 20:28
  • $\begingroup$ Sakurai & Naplitano, Ch 6, are slightly more explicit; but, frankly, Merzbacher adumbrates any and all glossed-over steps. $\endgroup$ Commented Jan 12, 2023 at 20:38
  • $\begingroup$ I have Landau & Lifshitz and found on p. 517 a derivation that gives a similar result with the Born approximation. Thank you for that reference. However, I'd like to use the approach which, as you say, Merzbacher adumbrates but doesn't explicitly state. I suppose the derivation is trivial, but I just don't see it. I don't have Sakurai & Naplitano. Could you derive the equation for me? $\endgroup$
    – Marcellus
    Commented Jan 12, 2023 at 20:48
  • 1
    $\begingroup$ Thank you. I'll look for that book right away. $\endgroup$
    – Marcellus
    Commented Jan 13, 2023 at 18:41

1 Answer 1


I am answering my own question since two members of the Physics StackExchange (see the comment section) kindly provided me with enough information to derive the equation. Because of the somewhat intricate nature of the derivation, it is easy to make errors. If any are present, they fortuitously cancelled each other out.

Merzbacher indicates that the desired relationship

$\sin\delta_l = -k\int_0^\infty j_l(kr')U(r')u_{l,k}(r')r'dr'\tag{11.83}$

can be obtained from the two equations below by utilizing a partial wave analysis. $ U=\frac {2\mu V}{\hbar^2}$, and $\delta_l$ is understood to be a function of k.

$$f_k( \mathbf {\hat r}) = -\frac{\sqrt{2\pi}\mu}{\hbar^2}\int exp(-ik\mathbf {\hat r}\cdot \mathbf r')V(r')\psi_k(\mathbf r’)d \mathbf{\tau}' \tag{11.35a}$$ $$f_k(\theta)=\frac{1}{k}\sum_{l=0}^\infty(2l+1)exp(i\delta_l)\sin\delta_lP_l(\cos\theta)\tag{11.59}$$

One starts by setting the two equations equal to one another but with only a partial wave taken from Equation 11.59. For consistency, this decision necessitates the use of only a partial wave from the wave equation on the RHS later in this derivation. Alternatively, the summation in 11.59 can be left in place (and in the wave equation) and removed at the end of the derivation, but omitting it now is less cumbersome.

$$\frac{1}{k}exp(i\delta_l)\sin\delta_l P_l(\cos\theta)= -\frac{\sqrt{2\pi}}{2} \int exp(-ik\mathbf {\hat r}\cdot \mathbf r’)U(r’)\psi_k(\mathbf r’) d \mathbf {\tau}' $$

Next, an important identity is substituted for $exp(-ik \mathbf {\hat r}\cdot \mathbf r') $ on the RHS.

$$ = -\frac{\sqrt{2\pi}}{2} \int \sum_{l'=0}^\infty(2l'+1)i^{3l'}j_{l'}(kr')P_{l'}(\mathbf {\hat r}\cdot \mathbf {\hat r’})U(r')\psi_k(\mathbf r’)d \mathbf{\tau}'$$

Another identity is substituted for $P_{l'}(\mathbf {\hat r}\cdot \mathbf {\hat r’})$ (shown in the square brackets below), and a partial wave from the wave equation $\psi(\mathbf r')=\sum _{l=0}^\infty(2l+1)\frac {i^lexp(i\delta_l)}{(2\pi)^{3/2}}P_l(\cos\theta')u_{l,k}(r')/r' $ is substituted for $\psi.$

$$ = -\frac{1}{4\pi} \int \sum_{l'=0}^\infty(2l'+1)i^{3l'}j_{l'}(kr')\mathbf[\sum_{m=-l'}^{l'}\frac{4\pi}{2l'+1}Y_{l'}^{m*}(\theta',\phi')Y_{l'}^m(\theta,\phi) \mathbf]U(r')i^l exp(i\delta_l)P_l(cos \theta')u_{l,k}(r')/r'd \mathbf{\tau}'$$

The integral is now expanded in spherical polar coordinates, $P_l$ and $Y_{l'}^m$ are respectively converted to $Y_{l}^0$ and $P_{l'}^m$, and some rearranging is done.

$$ = -\frac{1}{4\pi} \int_0^\infty \sum_{l'=0}^\infty(2l'+1)i^{3l'+l}j_{l'}(kr')\sum_{m=-l'}^{l'}U(r')exp(i\delta_l)\int_{\phi'=0}^{2\pi} \int_{\theta'=0}^{\pi} \frac{4\pi}{2l'+1}Y_{l'}^{m*}(\theta',\phi')Y_{l}^0(\theta',\phi')[(-1)^m exp(im\phi)\sqrt{\frac{(2l'+1)(l'-m)!}{(2l+1)(l'+m)!}}] P_{l'}^m (cos\theta)sin(\theta')u_{l,k}(r')r'd{\phi'}d{\theta'}dr'$$

After taking orthogonality of terms into account, one obtains

$$ = -\int_0^\infty \sum_{l'=0}^\infty \sqrt {\frac {2l'+1}{2l+1}}i^{3l'+l}\delta_{ll'}j_{l'}(kr')U(r')exp(i\delta_l)P_{l'}(\cos \theta) u_{l,k}(r')r'dr'$$

$$ = -\int_0^\infty j_{l}(kr')U(r')exp(i\delta_l)P_{l}(\cos\theta)u_{l,k}(r')r' dr'$$

The LHS of the equation at the beginning of the derivation and the RHS above are now compared to get $$\sin\delta_l = -k\int_0^\infty j_l(kr')U(r')u_{l,k}(r')r'dr'\tag{11.83}$$


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