Two Particles in Harmonic Potential

It is a 1D, two-body problem with an interacting Hamiltonian

H(x_1, x_2) = -\half{\partial^2\over\partial x_1^2}
-\half{\partial^2\over\partial x_2^2}
    +{1\over|x_1 - x_2|} + \half\omega^2 x_1^2 + \half\omega^2 x_2^2

and it can be solved analytically. The Schrödinger equation is

\left(-\half{\partial^2\over\partial x_1^2} -\half{\partial^2\over\partial x_2^2}
    +{1\over|x_1 - x_2|} + \half\omega^2 x_1^2 + \half\omega^2 x_2^2
\right)\Psi(x_1, x_2) = E \Psi(x_1, x_2)

we use the substitution:

u = {1\over\sqrt2}(x_1 - x_2)

v = {1\over\sqrt2}(x_1 + x_2)


{\partial^2\over\partial x_1^2} + {\partial^2\over\partial x_2^2}=
{\partial^2\over\partial u^2} + {\partial^2\over\partial v^2}

|x_1 - x_2| = \sqrt2|u|

x_1^2 + x_2^2 = u^2 + v^2


\left(-\half{\partial^2\over\partial u^2} -\half{\partial^2\over\partial v^2}
    +{1\over\sqrt2 |u|} + \half\omega^2 u^2 + \half\omega^2 v^2
\right)\Psi(u, v) = E \Psi(u, v)

Note also the symmetry of the Hamiltonian H(x_1, x_2) = H(x_2, x_1) which after substitution is equivalent to H(u, v) = H(-u, v). Now we can separate the equation:

\Psi(u, v) = f(u)g(v)

\left(-\half{\d^2\over\d u^2} +{1\over\sqrt2 |u|} + \half\omega^2 u^2
\right)f_k(u) = \epsilon_k f_k(u)

\left(-\half{\d^2\over\d v^2} + \half\omega^2 v^2
\right)g_l(v) = \epsilon_l g_l(v)

E_{kl} = \epsilon_k + \epsilon_l

the solution of the second equation is:

g_l(v) = {1\over\sqrt{2^l l!}} \left(\omega\over\pi\right)^{1\over4}
    e^{-{\omega v^2\over2}} H_l(\sqrt\omega v)

\epsilon_l = \omega(l+\half)\quad\mbox{for $l=0, 1, 2, \dots$}

where H_n(x) are the Hermite polynomials:

H_n(x) = (-1)^n e^{x^2} {\d^n\over\d x^n} e^{-x^2}

The solution to the first equation can be approximated around the minimum of the potential, which occurs at point u=u_0 (since the potential is symmetric with respect to u, we only treat the branch u>0):

V(u) = {1\over\sqrt2 |u|} + \half\omega^2 u^2 =
    \left(2^{-{1\over 3}} + 2^{-{4\over3}}\right) \omega^{2\over3}
    +{3\over 2}\omega^2(u-u_0)^2

u_0 = 2^{-{1\over6}}\omega^{-{2\over3}}

So the first few states can be approximated by the harmonic oscillator solution with frequency \sqrt3\omega:

f_k(u) = {1\over\sqrt{2^k k!}} \left(\sqrt3\omega\over\pi\right)^{1\over4}
    e^{-{\sqrt3\omega (u-u_0)^2\over2}} H_k(3^{1\over4}\sqrt\omega (u-u_0))

\epsilon_k =
    \left(2^{-{1\over 3}} + 2^{-{4\over3}}\right) \omega^{2\over3}
    \sqrt3\omega(k+\half)\quad\mbox{for $k=0, 1, 2, \dots$}

The final solution is then:

\Psi_{kl}(u, v) = f_k(u) g_l(v) =

    {1\over\sqrt{2^k k!}} \left(\sqrt3\omega\over\pi\right)^{1\over4}
    e^{-{\sqrt3\omega (u-u_0)^2\over2}} H_k(3^{1\over4}\sqrt\omega (u-u_0))
    {1\over\sqrt{2^l l!}} \left(\omega\over\pi\right)^{1\over4}
    e^{-{\omega v^2\over2}} H_l(\sqrt\omega v)

E_{kl} = \epsilon_k + \epsilon_l =
    \left(2^{-{1\over 3}} + 2^{-{4\over3}}\right) \omega^{2\over3}

Quantum Harmonic Oscillator

The quantum harmonic oscillator for one particle in 1D is:

i\hbar{\partial\over\partial t}\psi(x, t)= -{\hbar^2\over2m}{\partial^2\over\partial x^2}\psi(x,t)+V(x)\psi(x,t)


This is a partial differential equation for the time evolution of the wave function \psi(x, t), but one method to solve it is the eigenvalues expansion:

\psi(x,t) = \sum_E c_E\psi_E(x)e^{-{i\over\hbar}Et}

where the sum goes over the whole spectrum (for continuous spectrum the sum turns into an integral), the c_E coefficients are determined from the initial condition and \psi_E(x) satisfies the one dimensional one particle time independent Schrödinger equation:

-{\hbar^2\over2m}{\d^2\over\d x^2}\psi_E(x)+V(x)\psi_E(x)=E\psi_E(x)

and this is just an ODE and thus can be solved with Hermes1D. There can be many types of boundary conditions for this equation, depending on the physical problem, but in our case we simply have \lim_{x\to\pm\infty}\psi_E(x)=0 and the normalization condition \int_{-\infty}^\infty|\psi_E(x)|^2\d x=1.

We can set m=\hbar=1 and from now on we’ll just write \psi(x) instead of \psi_E(x):

-{1\over2}{\d^2\over\d x^2}\psi(x)+V(x)\psi(x)=E\psi(x)

and we will solve it on the interval (a, b) with the boundary condition \psi(a)=\psi(b)=0. The weak formulation is

\int_a^b{1\over2}{\d\psi(x)\over\d x}{\d v(x)\over\d x}+V(x)\psi(x)v(x)\,\d x -\left[{\d\psi(x)\over\d x}v(x)\right]^a_b =E\int_a^b\psi(x)v(x)\,\d x

but due to the boundary condition v(a)=v(b)=0 so \left[\psi'(x)v(x)\right]^a_b=0 and we get

\int_a^b{1\over2}{\d\psi(x)\over\d x}{\d v(x)\over\d x}+V(x)\psi(x)v(x)\,\d x =E\int_a^b\psi(x)v(x)\,\d x

And the finite element formulation is then \psi(x)=\sum_j y_j\phi_j(x) and v=\phi_i(x):

\left(\int_a^b{1\over2}\phi_i'(x)\phi_j'(x)+V(x)\phi_i(x)\phi_j(x)\,\d x\right) y_j =E\int_a^b\phi_i(x)\phi_j(x)\,\d x\ y_j

which is a generalized eigenvalue problem:



A_{ij}=\int_a^b{1\over2}\phi_i'(x)\phi_j'(x)+V(x)\phi_i(x)\phi_j(x)\,\d x

B_{ij}=\int_a^b\phi_i(x)\phi_j(x)\,\d x

Radial Schrödinger Equation

Another important example is the three dimensional one particle time independent Schrödinger equation for a spherically symmetric potential:

-{1\over2}\nabla^2\psi({\bf x})+V(r)\psi({\bf x})=E\psi({\bf x})

The way to solve it is to separate the equation into radial and angular parts by writing the Laplace operator in spherical coordinates as:

\nabla^2f =  {\partial^2 f\over\partial\rho^2} +{2\over \rho}{\partial^2 f\over\partial\rho^2} -{L^2\over \rho^2}

L^2= -{\partial^2 f\over\partial\theta^2} -{1\over\sin^2\theta}{\partial^2 f\over\partial\phi^2} -{1\over\tan\theta}{\partial f\over\partial\theta}

Substituting \psi({\bf x})=R(\rho)Y(\theta,\phi) into the Schrödinger equation yields:



Using the fact that L^2Y=l(l+1)Y we can cancel Y and we get the radial Schrödinger equation:


The solution is then:

\psi({\bf x})=\sum_{nlm}c_{nlm}R_{nl}(r)Y_{lm}\left({\bf x}\over r\right)

where R_{nl}(r) satisfies the radial Schrödinger equation (from now on we just write R(r)):

-{1\over2}R''(r)-{1\over r}R'(r)+\left(V+{l(l+1)\over2r^2}\right)R(r)=ER(r)

Again there are many types of boundary conditions, but the most common case is \lim_{r\to\infty}R(r)=0 and R(0)=1 or R(0)=0. One solves this equation on the interval (0, a) for large enough a.

The procedure is similar to the previous example, only we need to remember that we always have to use covariant integration (in the previous example the covariant integration was the same as the coordinate integration), in this case r^2\sin\theta \d
r\d\theta\d\phi, so the weak formulation is:

\int \left(-{1\over2}R''(r)-{1\over r}R'(r)+\left(V+{l(l+1)\over2r^2}\right)R(r)\right)v(r)r^2\sin\theta \d r\d\theta\d\phi=

=\int ER(r) v(r)r^2\sin\theta \d r\d\theta\d\phi

Integrating over the angles gives 4\pi which we cancel out at both sides and we get:

\int_0^a \left(-{1\over2}R''(r)-{1\over r}R'(r)+\left(V+{l(l+1)\over2r^2}\right)R(r)\right)v(r)r^2 \d r=

=E\int_0^a R(r) v(r)r^2 \d r

We apply per partes to the first two terms on the left hand side:

\int_0^a \left(-{1\over2}R''(r)-{1\over r}R'(r)\right)v(r)r^2 \d r =\int_0^a -{1\over2r^2}\left(r^2 R'(r)\right)'v(r)r^2 \d r=

=\int_0^a -{1\over2}\left(r^2 R'(r)\right)'v(r) \d r =\int_0^a {1\over2}r^2 R'(r)v'(r) \d r-{1\over2} [r^2R'(r)v(r)]_0^a=

=\int_0^a {1\over2} R'(r)v'(r) r^2\d r -{1\over2} a^2R'(a)v(a)

We used the fact that \lim_{r\to0} r^2 R'(r) = 0. If we also prescribe the boundary condition R'(a)=0, then the boundary term vanishes completely. The weak formulation is then:

\int_0^a {1\over2}R'(r)v'(r)r^2+ \left(V+{l(l+1)\over2r^2}\right)R(r)v(r)r^2\,\d r = E\int_0^aR(r)v(r)r^2\,\d r


\int_0^a {1\over2}R'(r)v'(r)r^2+ V(r)R(r)v(r)r^2+{l(l+1)\over2} R(r)v(r)\,\d r = E\int_0^aR(r)v(r)r^2\,\d r

Another approach

Another (equivalent) approach is to write a weak formulation for the 3D problem in cartesian coordinates:

\int_\Omega{1\over2}\nabla\psi({\bf x})\nabla v({\bf x})+V(r)\psi({\bf x})v({\bf x})\,\d^3 x =E\int_\Omega\psi({\bf x})v({\bf x})\,\d^3 x

and only then transform to spherical coordinates:

\int_0^{2\pi}\d\varphi\int_0^\pi\d\theta\int_0^a\d r \left({1\over2}\nabla\psi({\bf x})\nabla v({\bf x})+V(r)\psi({\bf x})v({\bf x})\right)r^2\sin\theta=

= E\int_0^{2\pi}\d\varphi\int_0^\pi\d\theta\int_0^a\d r\, \psi({\bf x})v({\bf x})r^2\sin\theta

The 3d eigenvectors \psi({\bf x}) however are not spherically symmetric. Nevertheless we can still proceed by choosing our basis as

v_{ilm}({\bf x})=\phi_{il}(r)Y_{lm}(\theta, \varphi)

and seek our solution as

\psi({\bf x})=\sum_{jlm}y_{jlm}\phi_{jl}(r)Y_{lm}(\theta, \varphi)

Using the properties of spherical harmonics and the gradient:

\int Y_{lm} Y_{l'm'} \sin\theta\,\d\theta\,\d\varphi= \delta_{ll'}\delta_{mm'}

\int r^2\nabla Y_{lm} \nabla Y_{l'm'} \sin\theta\,\d\theta\,\d\varphi= l(l+1)\delta_{ll'}\delta_{mm'}

\nabla f = {\partial f\over \partial r}\boldsymbol{\hat r} + {1\over r} {\partial f\over\partial\theta}\boldsymbol{\hat\theta}+{1\over r\sin\theta} {\partial f\over\partial\phi}\boldsymbol{\hat\phi}

the weak formulation becomes:

\left(\int_0^a {1\over2}r^2\phi_{il}'(r)\phi_{jl}'(r)+ {1\over2}X+ {l(l+1)\over2}\phi_{il}(r)\phi_{jl}(r)+ r^2V(r)\phi_{il}(r)\phi_{jl}(r)\,\d r\right)y_{jlm}=

= E\int_0^ar^2 \phi_{il}(r)\phi_{jl}(r)\,\d r\ y_{jlm}

where both l and m indices are given by the indices of the particular base function v_{ilm}. The X term is (schematically):

X=\int r^2\sin\theta(r)Y_{lm}(\theta,\varphi) (\phi_{il}\nabla\phi_{jl}+\nabla\phi_{il}\phi_{jl}) \nabla Y_{lm}

There is an interesting identity:

\int r{\bf \hat r} Y_{lm} \nabla Y_{l'm'} \sin\theta\,\d\theta\,\d\varphi= 0

But it cannot be applied, because we have one more r in the expression. Nevertheless the term is probably zero, as can be seen when we compare the weak formulation to the one we got directly from the radial equation.

How Not To Derive The Weak Formulation

If we forgot that we have to integrate covariantly, this section is devoted to what happens if we integrate using the coordinate integration. We would get:

\int_0^a {1\over2}R'(x)v'(x)-{1\over r}R'(x)v(x)+ \left(V+{l(l+1)\over2r^2}\right)R(x)v(x)\,\d x = E\int_0^aR(x)v(x)\,\d x

Notice the matrix on the left hand side is not symmetric. There is another way of writing the weak formulation by applying per-partes to the R'(r)v(r) term:

-\int_0^a{1\over r}R'(x)v(x)\d x=

=\int_0^a{1\over r}R(x)v'(x)\d x -\int_0^a{1\over r^2}R(x)v(x)\d x -\left[{1\over r}R'(x)v'(x)\right]_0^a +\left[{1\over r^2}R'(x)v(x)\right]_0^a

We can use v(a)=0 and R'(a)=0 to simplify a bit:

-\int_0^a{1\over r}R'(x)v(x)\d x=

=\int_0^a{1\over r}R(x)v'(x)\d x -\int_0^a{1\over r^2}R(x)v(x)\d x +\lim_{r\to0}\left({R'(x)v'(x)\over r}-{R'(x)v(x)\over r^2}\right)

Since R(x)\sim r^l near r=0, we can see that for l\ge3 the limits on the right hand side are zero, but for l=0, 1, 2 they are not zero and need to be taken into account. Let’s assume l\ge3 for now, then our weak formulation looks like:

\int_0^a {1\over2}R'(x)v'(x)+{1\over r}R(x)v'(x)+ \left(V+{l(l+1)\over2r^2}-{1\over r^2}\right)R(x)v(x)\,\d x = E\int_0^aR(x)v(x)\,\d x


\int_0^a {1\over2}R'(x)v'(x)+{1\over r}R(x)v'(x)+ \left(V+{(l-2)(l+1)\over2r^2}\right)R(x)v(x)\,\d x = E\int_0^aR(x)v(x)\,\d x

The left hand side is also not symmetric, however we can now take an average of our both weak formulations to get a symmetric weak formulation:

\int_0^a {1\over2}R'(x)v'(x)+{R(x)v'(x)-R'(x)v(x)\over 2r}+ \left(V+{l(l+1)-1\over2r^2}\right)R(x)v(x)\,\d x =

= E\int_0^aR(x)v(x)\,\d x

Keep in mind, that this symmetric version is only correct for l\ge3. For l<3 we need to use our first nonsymmetric version.

As you can see, this is something very different to what we got in the previous section. First there were lots of technical difficulties and second the final result is wrong, since it doesn’t correspond to the 3D Schrödinger equation.

Scattering in radial potential

If V=0, the radial equation is:

-{1\over2}R''(r)-{1\over r}R'(r)+{l(l+1)\over2r^2} R(r)=ER(r)

The general solution is a linear combination of the spherical Bessel functions j_l(kr) and n_l(kr), whose asymptotic expansion for r\to\infty is:

j_l(kr)\to{1\over kr}\sin\left(kr-{l\pi\over 2}\right)

n_l(kr)\to{1\over kr}\cos\left(kr-{l\pi\over 2}\right)

so we get for large r:

R_l(kr) = A_l {1\over kr}\sin\left(kr-{l\pi\over 2}\right)
    + B_l{1\over kr}\cos\left(kr-{l\pi\over 2}\right) =

= \sqrt{A_l^2 + B_l^2} {1\over kr}\sin\left(kr-{l\pi\over 2}+\delta_l\right)
= C_l {1\over kr}\sin\left(kr-{l\pi\over 2}+\delta_l\right)


\delta_l = \atan2(B_l, A_l)

C_l = \sqrt{A_l^2 + B_l^2}

We can then compare this to \phi \approx e^{ikz} + f(\theta, \phi)
{e^{ikr}\over r}, by expanding e^{ikz} = e^{ikr\cos\theta}=
\sum (2l+1) i^l j_l(kr)P_l(\cos\theta):

C_l = {e^{i\delta_l}\over k}

f(\theta, \phi) = {1\over 2ik} \sum (2l+1)(e^{2i\delta_l}-1)P_l(\cos\theta)

Since \sigma(\theta) = |f(\theta)|^2 and integrating over \omega we get the total cross section:

\sigma = {4\pi\over k}\sum (2l+1)\sin^2\delta_l

In order to find the phase shifts \delta_l, we solve the radial equation for the full potential

-{1\over2}R''(r)-{1\over r}R'(r)+\left(V+{l(l+1)\over2r^2}\right)R(r)=ER(r)

and then fit it to the above asymptotic solution for V=0. We require that the value and the slope must be continuous. In particular, we take the logarithmic derivative ((\log u)'={u'\over u}) at the point r=a:

\gamma_l \equiv \left.{\d\over\d r} \log u\right|_{r=a} =
\left.{\d\over\d r} \log R_l(kr)\right|_{r=a}

expressing R_l(kr) using \delta_l and solving for it we get:

\tan\delta_l = {k j_l'(ka) - \gamma_l j_l(ka)\over k n_l'(ka)-\gamma_l

Now we can use these \delta_l in the formula for the total cross section.

The problem can now be formulated in two ways. Either to solve the radial equation for a potential with finite reach and then “measure” those phase shifts in the solution. Or by prescribing those phase shifts and we now need to calculate the solutions (e.g. the energies) from the radial equation.