Daily Archives: Sun, 14 June 2015

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<r_{0}} 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} & r<r_{0}\\ \frac{a}{r}\left(e^{2r_{0}/a}-1\right)e^{-\left(2r_{0}+r\right)/a} & r>r_{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)