Integral form of the Schrödinger equation: ground state of hydrogen

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 11.9.

We can check that the ground state of the hydrogen atom satisfies the integral form of the Schrödinger equation:

$\displaystyle \psi\left(\mathbf{r}\right)=\psi_{0}\left(\mathbf{r}\right)-\frac{m}{2\pi\hbar^{2}}\int\frac{e^{ik\left|\mathbf{r}-\mathbf{r}_{0}\right|}}{\left|\mathbf{r}-\mathbf{r}_{0}\right|}V\left(\mathbf{r}_{0}\right)\psi\left(\mathbf{r}_{0}\right)d^{3}\mathbf{r}_{0} \ \ \ \ \ (1)$

For the ground state of hydrogen

 $\displaystyle \psi_{100}\left(\mathbf{r}\right)$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{\pi a^{3}}}e^{-r/a}\ \ \ \ \ (2)$ $\displaystyle a$ $\displaystyle \equiv$ $\displaystyle \frac{4\pi\epsilon_{0}\hbar^{2}}{me^{2}}\ \ \ \ \ (3)$ $\displaystyle V\left(r\right)$ $\displaystyle =$ $\displaystyle -\frac{e^{2}}{4\pi\epsilon_{0}r}=-\frac{\hbar^{2}}{mar}\ \ \ \ \ (4)$ $\displaystyle E_{1}$ $\displaystyle =$ $\displaystyle -\frac{\hbar^{2}}{2ma^{2}}\ \ \ \ \ (5)$ $\displaystyle k$ $\displaystyle =$ $\displaystyle \frac{\sqrt{2mE_{1}}}{\hbar}=\frac{i}{a} \ \ \ \ \ (6)$

In this case, we’re not considering scattering, so the incident plane wave (the free particle) is not present, so ${\psi_{0}=0}$ in 1, and the integral equation becomes

 $\displaystyle \psi\left(\mathbf{r}\right)$ $\displaystyle =$ $\displaystyle -\frac{m}{2\pi\hbar^{2}}\left(-\frac{\hbar^{2}}{ma}\right)\frac{1}{\sqrt{\pi a^{3}}}\int_{0}^{\infty}\int_{0}^{\pi}\int_{0}^{2\pi}\frac{e^{-\left|\mathbf{r}-\mathbf{r}_{0}\right|/a}e^{-r_{0}/a}}{\left|\mathbf{r}-\mathbf{r}_{0}\right|r_{0}}\sin\theta r_{0}^{2}d\phi d\theta dr_{0}\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2\pi^{3/2}a^{5/2}}\int_{0}^{\infty}\int_{0}^{\pi}\int_{0}^{2\pi}\frac{e^{-\left|\mathbf{r}-\mathbf{r}_{0}\right|/a}e^{-r_{0}/a}}{\left|\mathbf{r}-\mathbf{r}_{0}\right|}\sin\theta r_{0}d\phi d\theta dr_{0}\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{\pi a^{5}}}\int_{0}^{\infty}\int_{0}^{\pi}\frac{e^{-\left|\mathbf{r}-\mathbf{r}_{0}\right|/a}e^{-r_{0}/a}}{\left|\mathbf{r}-\mathbf{r}_{0}\right|}\sin\theta r_{0}d\theta dr_{0} \ \ \ \ \ (9)$

where we’ve taken the ${z}$ axis to be parallel to ${\mathbf{r}}$, since for the purposes of the integral, ${\mathbf{r}}$ is constant.

We have

$\displaystyle \left|\mathbf{r}-\mathbf{r}_{0}\right|=\sqrt{r^{2}+r_{0}^{2}-2rr_{0}\cos\theta} \ \ \ \ \ (10)$

so the integral becomes

$\displaystyle \frac{1}{\sqrt{\pi a^{5}}}\int_{0}^{\infty}\int_{0}^{\pi}\frac{e^{-\sqrt{r^{2}+r_{0}^{2}-2rr_{0}\cos\theta}/a}e^{-r_{0}/a}}{\sqrt{r^{2}+r_{0}^{2}-2rr_{0}\cos\theta}}\sin\theta r_{0}d\theta dr_{0}=\frac{1}{\sqrt{\pi a^{3}}}e^{-r/a} \ \ \ \ \ (11)$

where we did the integral using Maple. This is just the original wave function 2 so the integral equation works out.

If you want to do the integral by hand, we do the ${\theta}$ integral first since, despite its appearance, it’s actually quite simple:

$\displaystyle \int_{0}^{\pi}\frac{e^{-\sqrt{r^{2}+r_{0}^{2}-2rr_{0}\cos\theta}/a}e^{-r_{0}/a}}{\sqrt{r^{2}+r_{0}^{2}-2rr_{0}\cos\theta}}\sin\theta r_{0}d\theta=-\frac{ae^{-r_{0}/a}}{r}\left.e^{-\sqrt{r^{2}+r_{0}^{2}-2rr_{0}\cos\theta}/a}\right|_{0}^{\pi} \ \ \ \ \ (12)$

The value of the integral depends on whether ${r or ${r>r_{0}}$:

$\displaystyle -\frac{ae^{-r_{0}/a}}{r}\left.e^{-\sqrt{r^{2}+r_{0}^{2}-2rr_{0}\cos\theta}/a}\right|_{0}^{\pi}=\begin{cases} \frac{a}{r}\left(e^{2r/a}-1\right)e^{-\left(2r_{0}+r\right)/a} & rr_{0} \end{cases} \ \ \ \ \ (13)$

Using these results, we can split the integral over ${r_{0}}$ into two parts (0 to ${r}$ and ${r}$ to ${\infty}$). It is just a simple integral over exponential functions so the answer comes out fairly easily.

Integral form of the Schrödinger equation

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 11.8.

As a prelude to the Born approximation in quantum scattering, we need to look at the integral form of the time-independent Schrödinger equation. The equation in its original differential equation form is

$\displaystyle -\frac{\hbar^{2}}{2m}\nabla^{2}\psi+V\psi=E\psi \ \ \ \ \ (1)$

which can be written as

 $\displaystyle \left(\nabla^{2}+k^{2}\right)\psi$ $\displaystyle =$ $\displaystyle Q\ \ \ \ \ (2)$ $\displaystyle k$ $\displaystyle \equiv$ $\displaystyle \frac{\sqrt{2mE}}{\hbar}\ \ \ \ \ (3)$ $\displaystyle Q$ $\displaystyle \equiv$ $\displaystyle \frac{2m}{\hbar^{2}}V\psi \ \ \ \ \ (4)$

To convert this to an integral equation, we need to define a Green’s function ${G\left(\mathbf{r}\right)}$ which satisfies the differential equation

$\displaystyle \left(\nabla^{2}+k^{2}\right)G\left(\mathbf{r}\right)=\delta^{3}\left(\mathbf{r}\right) \ \ \ \ \ (5)$

Using this function we can write ${\psi}$ as an integral equation

$\displaystyle \psi\left(\mathbf{r}\right)=\int G\left(\mathbf{r}-\mathbf{r}_{0}\right)Q\left(\mathbf{r}_{0}\right)d^{3}\mathbf{r}_{0} \ \ \ \ \ (6)$

We can show this works by plugging in ${G}$ from 5:

 $\displaystyle \left(\nabla^{2}+k^{2}\right)\psi\left(\mathbf{r}\right)$ $\displaystyle =$ $\displaystyle \int\left(\nabla^{2}+k^{2}\right)G\left(\mathbf{r}-\mathbf{r}_{0}\right)Q\left(\mathbf{r}_{0}\right)d^{3}\mathbf{r}_{0}\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int\delta\left(\mathbf{r}-\mathbf{r}_{0}\right)Q\left(\mathbf{r}_{0}\right)d^{3}\mathbf{r}_{0}\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle Q\left(\mathbf{r}\right) \ \ \ \ \ (9)$

which gives us back 2.

This isn’t a solution of the Schrödinger equation, of course, because ${Q}$ contains ${\psi}$, so we’d need to actually know ${\psi}$ in advance in order to work out the integral with the Green’s function. Rather, it’s just a different way of writing the Schrödinger equation which proves useful in scattering theory.

Because 5 doesn’t depend on the potential ${V}$, we can work out the Green’s function which is valid for every potential. The process is rather involved, but Griffiths goes through the details in section 11.4.1, so I won’t reproduce them here, apart from noting that the solution uses what is, to me, one of the most beautiful theorems in mathematics: Cauchy’s theorem on contour integration. Maybe I’ll return to it later.

Anyway, the Green’s function turns out to be

$\displaystyle G\left(\mathbf{r}\right)=-\frac{e^{ikr}}{4\pi r} \ \ \ \ \ (10)$

We can verify this is in fact a solution by plugging it back into 5. We need the Laplacian of ${G}$ which we can get by calculating the divergence of the gradient. Taking the gradient first, we use the product rule for gradients:

$\displaystyle \nabla\left(fg\right)=f\nabla g+g\nabla f \ \ \ \ \ (11)$

We get

 $\displaystyle \nabla G$ $\displaystyle =$ $\displaystyle -\frac{1}{4\pi}\left(\frac{1}{r}\nabla e^{ikr}+e^{ikr}\nabla\frac{1}{r}\right) \ \ \ \ \ (12)$

To calculate the divergence of the gradient, we use the identity for the divergence of the product of a scalar and a vector:

$\displaystyle \nabla\cdot\left(f\mathbf{A}\right)=\mathbf{A}\cdot\nabla f+f\nabla\cdot\mathbf{A} \ \ \ \ \ (13)$

We therefore have

 $\displaystyle \nabla^{2}G$ $\displaystyle =$ $\displaystyle -\frac{1}{4\pi}\left[\left(\nabla\frac{1}{r}\right)\cdot\left(\nabla e^{ikr}\right)+\frac{1}{r}\nabla^{2}e^{ikr}+\left(\nabla e^{ikr}\right)\cdot\left(\nabla\frac{1}{r}\right)+e^{ikr}\nabla^{2}\frac{1}{r}\right] \ \ \ \ \ (14)$

The last term turns out to be a delta function:

$\displaystyle \nabla^{2}\frac{1}{r}=\nabla\cdot\left(\nabla\frac{1}{r}\right)=-\nabla\cdot\left(\frac{\hat{\mathbf{r}}}{r^{2}}\right)=-4\pi\delta^{3}\left(\mathbf{r}\right) \ \ \ \ \ (15)$

To work out the second term, we use the formula for the Laplacian in spherical coordinates, for a function that depends only on ${r}$:

$\displaystyle \nabla f\left(r\right)=\frac{1}{r^{2}}\frac{\partial}{\partial r}\left(r^{2}\frac{\partial f}{\partial r}\right) \ \ \ \ \ (16)$

We get

 $\displaystyle \nabla^{2}e^{ikr}$ $\displaystyle =$ $\displaystyle \frac{1}{r^{2}}\frac{\partial}{\partial r}\left(ikr^{2}e^{ikr}\right)\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2ik}{r}e^{ikr}-k^{2}e^{ikr} \ \ \ \ \ (18)$

Putting this back into 14 we get

 $\displaystyle \nabla^{2}G$ $\displaystyle =$ $\displaystyle -\frac{1}{4\pi}\left[2\left(\nabla\frac{1}{r}\right)\cdot\left(\nabla e^{ikr}\right)+\frac{2ik}{r^{2}}e^{ikr}-\frac{k^{2}}{r}e^{ikr}-4\pi\delta^{3}\left(\mathbf{r}\right)e^{ikr}\right]\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{1}{4\pi}\left[-\frac{2ik}{r^{2}}e^{ikr}+\frac{2ik}{r^{2}}e^{ikr}-\frac{k^{2}}{r}e^{ikr}-4\pi\delta^{3}\left(\mathbf{r}\right)e^{ikr}\right]\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{1}{4\pi}\left[-\frac{k^{2}}{r}e^{ikr}-4\pi\delta^{3}\left(\mathbf{r}\right)\right]\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{k^{2}}{4\pi r}e^{ikr}+\delta^{3}\left(\mathbf{r}\right)\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -k^{2}G\left(\mathbf{r}\right)+\delta^{3}\left(\mathbf{r}\right)\ \ \ \ \ (23)$ $\displaystyle \left(\nabla^{2}+k^{2}\right)G\left(\mathbf{r}\right)$ $\displaystyle =$ $\displaystyle \delta^{3}\left(\mathbf{r}\right) \ \ \ \ \ (24)$

where we dropped the ${e^{ikr}}$ from the last term in the third line since the delta function is zero except when ${\mathbf{r}=0}$.

Using this Green’s function, the integral form of the Schrödinger equation is

$\displaystyle \psi\left(\mathbf{r}\right)=\psi_{0}\left(\mathbf{r}\right)-\frac{m}{2\pi\hbar^{2}}\int\frac{e^{ik\left|\mathbf{r}-\mathbf{r}_{0}\right|}}{\left|\mathbf{r}-\mathbf{r}_{0}\right|}V\left(\mathbf{r}_{0}\right)\psi\left(\mathbf{r}_{0}\right)d^{3}\mathbf{r}_{0} \ \ \ \ \ (25)$

where ${\psi_{0}}$ is a solution of the free particle Schrödinger equation

$\displaystyle \left(\nabla^{2}+k^{2}\right)\psi_{0}\left(\mathbf{r}\right)=0 \ \ \ \ \ (26)$