# Posts on Physics

## Hydrogen energy levels: an algebraic derivation

Posted by Diego Assencio on 2016.10.15 under Physics (Quantum mechanics)

In this post, we will derive the energy levels of the Hydrogen atom using only operator algebra, i.e., without dealing with wave functions. Our work will mainly consist in defining special operators which will assist us in determining the eigenvalues of the Hamiltonian $H$ directly. Our model of the Hydrogen atom is an electron of mass $m$ under the influence of a Coulomb potential; the associated Hamiltonian should be known by any physics student who has taken a quantum mechanics course: $$\displaystyle H = \frac{{\bf p}^2}{2m} - \frac{e^2}{r} \label{post_d379f7e34709563115ebd4c41241ed5e_hamiltonian}$$ The first term on the right-hand side above represents the kinetic energy of the electron and the second term represents its electrostatic potential energy in the presence of a proton fixed at the origin. On the equation above, ${\bf p}$ and ${\bf x}$ are the momentum and position operators respectively, and $r = \|{\bf x}\|$.

Most of our derivation will rely on the properties of the self-adjoint operator ${\bf A}$ defined below (called the Laplace-Runge-Lenz vector operator): $$\displaystyle{\bf A} = \frac{1}{2}({\bf L}\times{\bf p} - {\bf p}\times{\bf L}) + me^2\frac{\bf x}{r}$$ where ${\bf L} = {\bf x}\times{\bf p}$ is the angular momentum operator. The self-adjoint property of ${\bf A}$ comes from the fact that ${\bf x}$ and $r$ are self-adjoint and also because $({\bf L}\times{\bf p})^{\dagger} = - {\bf p}\times{\bf L}$. In what follows, we will extensively use the Einstein summation notation, i.e., a product $a_ib_i$ represents the sum $\sum_{i=1}^3 a_ib_i$ for two vector quantities ${\bf a} = (a_1, a_2, a_3)$ and ${\bf b} = (b_1, b_2, b_3)$ respectively.

### Basic commutators

First, let us list some well-known commutators and compute some others which we will need later: $$\begin{eqnarray} [x_i, x_j] &=& 0 \\[5pt] [p_i, p_j] &=& 0 \\[5pt] [x_i, p_j] &=& i\hbar\delta_{ij} \\[5pt] [x_i, L_j] &=& [x_i, \epsilon_{jkl}x_k p_l] = \epsilon_{jkl} x_k [x_i, p_l] = \epsilon_{jkl} x_k i\hbar\delta_{il} = i\hbar \epsilon_{jki} x_k = i\hbar \epsilon_{ijk} x_k \label{post_d379f7e34709563115ebd4c41241ed5e_comm_xi_Lj} \\[5pt] [p_i, L_j] &=& [p_i, \epsilon_{jkl}x_k p_l] = \epsilon_{jkl} [p_i, x_k] p_l = \epsilon_{jkl} (-i\hbar)\delta_{ik} p_l = -i\hbar\epsilon_{jil} p_l = i\hbar\epsilon_{ijk} p_k \label{post_d379f7e34709563115ebd4c41241ed5e_comm_pi_Lj} \\[5pt] \end{eqnarray}$$ where above we used multiple properties of the Levi-Civita symbol $\epsilon_{ijk}$. Using the following identity: $$[A, BC] = ABC - BCA + BAC - BAC = [A,B]C + B[A,C] \label{post_d379f7e34709563115ebd4c41241ed5e_comm_A_BC}$$ we have that: $$\begin{eqnarray} [L_i, L_j] &=& [\epsilon_{imn}x_m p_n, \epsilon_{jkl}x_kp_l] \nonumber\\[5pt] &=& \epsilon_{imn}\epsilon_{jkl} [x_m p_n, x_k p_l] \nonumber\\[5pt] &=& \epsilon_{imn}\epsilon_{jkl} \left( [x_m p_n, x_k ] p_l + x_k [x_m p_n, p_l] \right) \nonumber\\[5pt] &=& \epsilon_{imn}\epsilon_{jkl} \left(x_m [p_n, x_k] p_l + x_k [x_m, p_l] p_n\right) \nonumber\\[5pt] &=& \epsilon_{imn}\epsilon_{jkl} \left( x_m (-i\hbar)\delta_{kn} p_l + x_k i\hbar\delta_{ml} p_n\right) \nonumber\\[5pt] &=& i\hbar \left(-\epsilon_{imk}\epsilon_{jkl} x_m p_l + \epsilon_{iln}\epsilon_{jkl} x_k p_n\right) \nonumber\\[5pt] &=& i\hbar \left(\epsilon_{imk}\epsilon_{jlk} x_m p_l - \epsilon_{inl}\epsilon_{jkl} x_k p_n\right) \nonumber\\[5pt] &=& i\hbar \big((\delta_{ij}\delta_{ml} - \delta_{il}\delta_{mj}) x_m p_l - (\delta_{ij}\delta_{nk} - \delta_{ik}\delta_{nj}) x_k p_n\big) \nonumber\\[5pt] &=& i\hbar \left(\delta_{ij}x_m p_m - \delta_{il}\delta_{mj} x_m p_l - \delta_{ij}x_n p_n + \delta_{ik}\delta_{nj} x_k p_n\right) \nonumber\\[5pt] &=& i\hbar \left( - \delta_{il}\delta_{mj} x_m p_l + \delta_{ik}\delta_{nj} x_k p_n\right) \nonumber\\[5pt] &=& i\hbar \left( - \delta_{il}\delta_{jm} x_m p_l + \delta_{im}\delta_{jl} x_m p_l\right) \nonumber\\[5pt] &=& i\hbar \left( \delta_{im}\delta_{jl} - \delta_{il}\delta_{jm} \right)x_m p_l \nonumber\\[5pt] &=& i\hbar \epsilon_{ijk}\epsilon_{mlk} x_m p_l \nonumber\\[5pt] &=& i\hbar \epsilon_{ijk}\epsilon_{kml} x_m p_l \nonumber\\[5pt] &=& i\hbar \epsilon_{ijk} L_k \label{post_d379f7e34709563115ebd4c41241ed5e_comm_Li_Lj} \end{eqnarray}$$ Additionally, since $r = \sqrt{x_ix_i}$: $$\displaystyle [p_i, f(r)] = -i\hbar \frac{\partial f(r)}{\partial x_i} = -i\hbar \frac{d f(r)}{d r}\frac{\partial r}{\partial x_i} = -i\hbar \frac{d f(r)}{d r}\frac{x_i}{r}$$ we have that: $$\displaystyle \left[p_i, \frac{1}{r}\right] = -i\hbar \frac{d}{d r}\left(\frac{1}{r}\right)\frac{x_i}{r} = -i\hbar \left(-\frac{1}{r^2}\right)\frac{x_i}{r} = i\hbar \frac{x_i}{r^3} \label{post_d379f7e34709563115ebd4c41241ed5e_eq_p1r}$$ Using equation \eqref{post_d379f7e34709563115ebd4c41241ed5e_eq_p1r} and the fact that $[x_i, f(r)] = 0$, we can then show that: $$\left[L_i, \frac{1}{r}\right] = \epsilon_{ijk}\left[x_j p_k, \frac{1}{r}\right] = \epsilon_{ijk}x_j \left[p_k, \frac{1}{r}\right] = \epsilon_{ijk} x_j i\hbar \frac{x_k}{r^3} = \frac{i\hbar}{r^3} \epsilon_{ijk} x_j x_k = 0 \label{post_d379f7e34709563115ebd4c41241ed5e_comm_Li_1r}$$ since $\epsilon_{ijk}$ is an anti-symmetric tensor. Equation \eqref{post_d379f7e34709563115ebd4c41241ed5e_comm_Li_1r} is directly related to the fact that $r$ is invariant under rotations.

### Computation of $[L_i, A_j]$

Using the commutators we computed above, we can compute $[L_i, A_j]$ directly: $$\begin{eqnarray} \displaystyle [L_i, A_j] &=& \left[L_i, \frac{1}{2}\epsilon_{jkl}(L_k p_l - p_k L_l) + me^2 \frac{x_j}{r}\right] \nonumber\\[5pt] &=& \frac{1}{2}\epsilon_{jkl}\left([L_i, L_k p_l] - [L_i,p_k L_l]\right) + me^2 \left[L_i,\frac{x_j}{r}\right] \label{post_d379f7e34709563115ebd4c41241ed5e_comm_Li_Aj_tmp} \end{eqnarray}$$ We will compute each of the terms above separately. From equations \eqref{post_d379f7e34709563115ebd4c41241ed5e_comm_pi_Lj}, \eqref{post_d379f7e34709563115ebd4c41241ed5e_comm_A_BC} and \eqref{post_d379f7e34709563115ebd4c41241ed5e_comm_Li_Lj}, we have that: $$\begin{eqnarray} \epsilon_{jkl}[L_i, L_k p_l] &=& \epsilon_{jkl}([L_i, L_k] p_l + L_k [L_i, p_l]) \nonumber\\[5pt] &=& \epsilon_{jkl}i\hbar\epsilon_{ikm} L_m p_l + \epsilon_{jkl}L_k(-i\hbar)\epsilon_{lim} p_m \nonumber\\[5pt] &=& i\hbar\epsilon_{jlk}\epsilon_{imk} L_m p_l - i\hbar \epsilon_{jkl}\epsilon_{iml} L_k p_m \nonumber\\[5pt] &=& i\hbar\big((\delta_{ji}\delta_{lm} - \delta_{jm}\delta_{li}) L_m p_l - (\delta_{ji}\delta_{km} - \delta_{jm}\delta_{ki}) L_k p_m \big)\nonumber\\[5pt] &=& i\hbar\left(\delta_{ji} L_m p_m - \delta_{jm}\delta_{li} L_m p_l - \delta_{ji}L_m p_m + \delta_{jm}\delta_{ki} L_k p_m \right)\nonumber\\[5pt] &=& i\hbar\left(- \delta_{jm}\delta_{li} L_m p_l + \delta_{jm}\delta_{ki} L_k p_m \right)\nonumber\\[5pt] &=& i\hbar\left(- \delta_{jm}\delta_{li} L_m p_l + \delta_{jl}\delta_{mi} L_m p_l \right)\nonumber\\[5pt] &=& i\hbar\left(\delta_{im}\delta_{jl} - \delta_{il}\delta_{jm} \right) L_m p_l\nonumber\\[5pt] &=& i\hbar \epsilon_{ijk}\epsilon_{mlk} L_m p_l \nonumber\\[5pt] &=& i\hbar \epsilon_{ijk}\epsilon_{kml} L_m p_l \nonumber\\[5pt] &=& i\hbar \epsilon_{ijk} ({\bf L}\times{\bf p})_k \label{post_d379f7e34709563115ebd4c41241ed5e_comm_Li_Lkpl} \end{eqnarray}$$ $$\begin{eqnarray} \epsilon_{jkl}[L_i, p_k L_l] &=& \epsilon_{jkl}([L_i, p_k] L_l + p_k [L_i, L_l]) \nonumber\\[5pt] &=& \epsilon_{jkl}(-i\hbar)\epsilon_{kim} p_m L_l + \epsilon_{jkl} p_k i\hbar\epsilon_{ilm} L_m \nonumber\\[5pt] &=& i\hbar\epsilon_{jlk}\epsilon_{imk} p_m L_l - i\hbar \epsilon_{jkl}\epsilon_{iml} p_k L_m \nonumber\\[5pt] &=& i\hbar\big((\delta_{ji}\delta_{lm} - \delta_{jm}\delta_{li}) p_m L_l - (\delta_{ji}\delta_{km} - \delta_{jm}\delta_{ki}) p_k L_m \big)\nonumber\\[5pt] &=& i\hbar\left(\delta_{ji} p_m L_m - \delta_{jm}\delta_{li} p_m L_l - \delta_{ji} p_m L_m + \delta_{jm}\delta_{ki} p_k L_m \right)\nonumber\\[5pt] &=& i\hbar\left(- \delta_{jm}\delta_{li} p_m L_l + \delta_{jm}\delta_{ki} p_k L_m \right)\nonumber\\[5pt] &=& i\hbar\left(- \delta_{jm}\delta_{li} p_m L_l + \delta_{jl}\delta_{mi} p_m L_l \right)\nonumber\\[5pt] &=& i\hbar\left(\delta_{im}\delta_{jl} - \delta_{il}\delta_{jm} \right) p_m L_l\nonumber\\[5pt] &=& i\hbar \epsilon_{ijk}\epsilon_{mlk} p_m L_l \nonumber\\[5pt] &=& i\hbar \epsilon_{ijk}\epsilon_{kml} p_m L_l \nonumber\\[5pt] &=& i\hbar \epsilon_{ijk} ({\bf p}\times{\bf L})_k \label{post_d379f7e34709563115ebd4c41241ed5e_comm_Li_pkLl} \end{eqnarray}$$ Finally, using equations \eqref{post_d379f7e34709563115ebd4c41241ed5e_comm_xi_Lj} and \eqref{post_d379f7e34709563115ebd4c41241ed5e_comm_Li_1r}, we can show that: $$\displaystyle \left[L_i,\frac{x_j}{r}\right] = \frac{1}{r} [L_i, x_j] = \frac{1}{r} (-i\hbar) \epsilon_{jik} x_k = i\hbar \epsilon_{ijk} \frac{x_k}{r} = i\hbar \epsilon_{ijk}\left(\frac{{\bf x}}{r}\right)_k \label{post_d379f7e34709563115ebd4c41241ed5e_comm_Li_xjr}$$ Using equations \eqref{post_d379f7e34709563115ebd4c41241ed5e_comm_Li_Lkpl}, \eqref{post_d379f7e34709563115ebd4c41241ed5e_comm_Li_pkLl} and \eqref{post_d379f7e34709563115ebd4c41241ed5e_comm_Li_xjr} on equation \eqref{post_d379f7e34709563115ebd4c41241ed5e_comm_Li_Aj_tmp} then yields: $$\begin{eqnarray} \displaystyle [L_i, A_j] &=& \frac{1}{2}\epsilon_{jkl}\left([L_i, L_k p_l] - [L_i,p_k L_l]\right) + me^2 \left[L_i,\frac{x_j}{r}\right] \nonumber\\[5pt] &=& \frac{1}{2}\big(i\hbar \epsilon_{ijk} ({\bf L}\times{\bf p})_k - i\hbar \epsilon_{ijk} ({\bf p}\times{\bf L})_k\big) + me^2 i\hbar \epsilon_{ijk}\left(\frac{{\bf x}}{r}\right)_k \nonumber\\[5pt] &=& i\hbar \epsilon_{ijk} \left( \frac{1}{2}\big( ({\bf L}\times{\bf p})_k - ({\bf p}\times{\bf L})_k\big) + me^2 \left(\frac{{\bf x}}{r}\right)_k \right) \label{post_d379f7e34709563115ebd4c41241ed5e_comm_Li_Aj_tmp2} \end{eqnarray}$$ Therefore: $$\boxed{ [L_i, A_j] = i\hbar \epsilon_{ijk} A_k } \label{post_d379f7e34709563115ebd4c41241ed5e_comm_Li_Aj}$$

### Computation of $[A_i, A_j]$

The computation of $[A_i, A_j]$ is very long and tedious. Since it does not introduce any new techniques beyond what has been presented above, we will merely state the result of this commutator, but the reader is strongly encouraged to try to prove the equation below as its derivation is an excellent way to deeply understand everything we have done so far (a lot of helpful material for that can be found here): $$\boxed{ [A_i, A_j] = -i\hbar 2m \epsilon_{ijk} L_k H }$$ where $H$ is the Hamiltonian given in equation \eqref{post_d379f7e34709563115ebd4c41241ed5e_hamiltonian}.

### The ${\bf K}^+$ and ${\bf K}^-$ operators

Suppose we have an eigenstate of $H$ representing a bound state with energy eigenvalue $E = -\kappa^2 / 2m$ for some $\kappa \gt 0$ (the assumption $E \lt 0$ is correct for a bound state since $E \geq 0$ means the electron has enough kinetic energy to escape the proton's electric potential). Let then ${\bf K}^{\pm}$ be defined as below: $$\displaystyle {\bf K}^{\pm} = \frac{1}{2}{\bf L} \pm \frac{1}{2\kappa}{\bf A}$$ We are interested in computing $[K^+_i, K^-_j]$ and $[K^\pm_i, K^\pm_j]$: $$\begin{eqnarray} \displaystyle [K^+_i, K^-_j] &=& \frac{1}{4}[L_i, L_j] - \frac{1}{4\kappa}[L_i,A_j] + \frac{1}{4\kappa}[A_i,L_j] - \frac{1}{4\kappa^2}[A_i,A_j] \nonumber\\[5pt] &=& \frac{i\hbar}{4}\left(\epsilon_{ijk}L_k - \frac{1}{\kappa}\epsilon_{ijk}A_k - \frac{1}{\kappa} \epsilon_{jik}A_k - \frac{1}{\kappa^2} (-2m) \epsilon_{ijk}L_k H\right) \nonumber\\[5pt] &=& \frac{i\hbar}{4}\left(\epsilon_{ijk}L_k - \frac{1}{\kappa}\epsilon_{ijk}A_k + \frac{1}{\kappa} \epsilon_{ijk}A_k - \frac{1}{\kappa^2} (-2m) \epsilon_{ijk}L_k H\right) \nonumber\\[5pt] &=& \frac{i\hbar}{4} \left(\epsilon_{ijk}L_k - \frac{1}{\kappa^2} (-2m) \epsilon_{ijk}L_k H\right) \end{eqnarray}$$ Applying this result to the energy eigenstate mentioned above, we have $H \rightarrow E = -\kappa^2/2m$, so we get: $$\begin{eqnarray} \displaystyle [K^+_i, K^-_j] &=& \frac{i\hbar}{4}\left(\epsilon_{ijk}L_k - \frac{1}{\kappa^2} (-2m) \epsilon_{ijk}L_k \frac{(-\kappa^2)}{2m}\right) \nonumber\\[5pt] &=& \frac{i\hbar}{4}\left(\epsilon_{ijk}L_k - \epsilon_{ijk}L_k\right) \end{eqnarray}$$ Therefore: $$\boxed{ \displaystyle [K^{+}_i, K^{-}_j] = 0 }$$ Also: $$\begin{eqnarray} \displaystyle [K^\pm_i, K^\pm_j] &=& \frac{1}{4} [L_i, L_j] \pm \frac{1}{4\kappa}[L_i, A_j] \pm \frac{1}{4\kappa}[A_i, L_j] + \frac{1}{4\kappa^2}[A_i,A_j] \nonumber\\[5pt] &=& \frac{i\hbar}{4}\left(\epsilon_{ijk}L_k \pm \frac{1}{\kappa}\epsilon_{ijk}A_k \mp \frac{1}{\kappa} \epsilon_{jik}A_k + \frac{1}{\kappa^2} (-2m) \epsilon_{ijk}L_k H\right) \nonumber\\[5pt] &=& \frac{i\hbar}{4}\left(\epsilon_{ijk}L_k \pm \frac{1}{\kappa}\epsilon_{ijk}A_k \pm \frac{1}{\kappa} \epsilon_{ijk}A_k + \frac{1}{\kappa^2} (-2m) \epsilon_{ijk}L_k H\right) \nonumber\\[5pt] &=& \frac{i\hbar}{4} \epsilon_{ijk} \left( L_k \pm \frac{2}{\kappa} A_k + \frac{1}{\kappa^2}(-2m)L_k \frac{(-\kappa^2)}{2m}\right) \nonumber\\[5pt] &=& \frac{i\hbar}{4} \epsilon_{ijk} \left( 2L_k \pm \frac{2}{\kappa}A_k\right) \nonumber\\[5pt] &=& i\hbar \epsilon_{ijk} \left( \frac{1}{2} L_k \pm \frac{1}{2\kappa}A_k\right) \end{eqnarray}$$ Therefore: $$\boxed{ [K^\pm_i, K^\pm_j] = i\hbar \epsilon_{ijk} K^{\pm}_k } \label{post_d379f7e34709563115ebd4c41241ed5e_comm_Ki_Kj}$$

### Computation of ${\bf A}^2$

$$\begin{eqnarray} {\bf A}^2 &=& {\bf A}\cdot{\bf A} \nonumber\\[5pt] &=& \frac{1}{4}\big(({\bf L}\times{\bf p} )\cdot( {\bf L}\times{\bf p}) - ({\bf L}\times{\bf p})\cdot( {\bf p}\times{\bf L})\big) \nonumber\\[5pt] &+& \frac{1}{4}\big(-({\bf p}\times{\bf L})\cdot({\bf L}\times{\bf p}) + ({\bf p}\times{\bf L})\cdot( {\bf p}\times{\bf L})\big) \nonumber\\[5pt] &+& \frac{me^2}{2}({\bf L}\times{\bf p} - {\bf p}\times{\bf L})\cdot \frac{\bf x}{r} + \frac{me^2}{2}\frac{\bf x}{r} \cdot ({\bf L}\times{\bf p} - {\bf p}\times{\bf L}) \nonumber\\[5pt] &+& m^2e^4\frac{\bf x}{r}\cdot \frac{\bf x}{r} \label{post_d379f7e34709563115ebd4c41241ed5e_AA_tmp} \end{eqnarray}$$ Each of the terms on the equation above will be computed separately: $$\begin{eqnarray} ({\bf L}\times{\bf p} )\cdot( {\bf L}\times{\bf p}) &=& ({\bf L}\times{\bf p} )_i( {\bf L}\times{\bf p})_i \nonumber\\[5pt] &=&(\epsilon_{ijk}L_j p_k)(\epsilon_{ilm}L_l p_m) \nonumber\\[5pt] &=&\epsilon_{jki}\epsilon_{lmi}L_j p_k L_l p_m \nonumber\\[5pt] &=&(\delta_{jl}\delta_{km} - \delta_{jm}\delta_{kl})L_j p_k L_l p_m \nonumber\\[5pt] &=&L_j p_k L_j p_k - L_j p_k L_k p_j \end{eqnarray}$$ Since $\epsilon_{ijk}$ is anti-symmetric, we have that: $$p_k L_k = {\bf p}\cdot{\bf L} = \epsilon_{ijk} p_i x_j p_k = 0$$ The equation above then becomes: $$\begin{eqnarray} ({\bf L}\times{\bf p} )\cdot( {\bf L}\times{\bf p}) &=& L_j p_k L_j p_k \nonumber\\[5pt] &=&L_j (L_j p_k + [p_k, L_j]) p_k \nonumber\\[5pt] &=&L_j L_j p_k p_k + i\hbar\epsilon_{kjm}p_m p_k \nonumber\\[5pt] &=&{\bf L}^2{\bf p}^2 \end{eqnarray}$$ Using a nearly identical sequence of derivation steps, we can also show that: $$\begin{eqnarray} ({\bf L}\times{\bf p})\cdot( {\bf p}\times{\bf L}) &=& -{\bf L}^2{\bf p}^2 \\[5pt] ({\bf p}\times{\bf L})\cdot( {\bf L}\times{\bf p}) &=& -({\bf L}^2 + 4\hbar^2){\bf p}^2 \\[5pt] ({\bf p}\times{\bf L})\cdot( {\bf p}\times{\bf L}) &=& {\bf L}^2{\bf p}^2 \end{eqnarray}$$ Let us now proceed to the other terms: $$\displaystyle({\bf L}\times{\bf p}) \cdot \frac{\bf x}{r} = \epsilon_{ijk} L_j p_k x_i\frac{1}{r} = - L_j \epsilon_{jik} p_k x_i\frac{1}{r} = - L_j L_j^\dagger\frac{1}{r} = -\frac{\bf L^2}{r}$$ $$\displaystyle \frac{\bf x}{r} \cdot ({\bf p}\times{\bf L}) = \left( ({\bf p}\times{\bf L})^\dagger \cdot \left(\frac{\bf x}{r}\right)^\dagger \right)^\dagger = \left( -({\bf L}\times{\bf p}) \cdot \frac{\bf x}{r} \right)^\dagger = \left( \frac{\bf L^2}{r} \right)^\dagger = \frac{\bf L^2}{r}$$ $$\begin{eqnarray} \displaystyle \frac{\bf x}{r} \cdot ({\bf L}\times{\bf p}) &=& \frac{1}{r} \epsilon_{ijk} x_i L_j p_k \nonumber\\[5pt] &=& \frac{1}{r} \epsilon_{ijk} (L_j x_i + [x_i, L_j]) p_k \nonumber\\[5pt] &=& \frac{1}{r} (\epsilon_{ijk} L_j x_i p_k + \epsilon_{ijk} i\hbar \epsilon_{ijm} x_m p_k) \nonumber\\[5pt] &=& \frac{1}{r} (-\epsilon_{jik} L_j x_i p_k + i\hbar \epsilon_{ikj} \epsilon_{imj} x_m p_k) \nonumber\\[5pt] &=& \frac{1}{r} (- L_j L_j + i\hbar (\delta_{ii}\delta_{km} - \delta_{im}\delta_{ki}) x_m p_k) \nonumber\\[5pt] &=& \frac{1}{r} (- {\bf L}^2 + i\hbar (3x_k p_k - x_i p_i)) \nonumber\\[5pt] &=& \frac{1}{r} (- {\bf L}^2 + 2 i\hbar {\bf x}\cdot{\bf p}) \end{eqnarray}$$ $$\begin{eqnarray} \displaystyle ({\bf p}\times{\bf L}) \cdot \frac{\bf x}{r} &=& \left( \left( \frac{\bf x}{r} \right)^\dagger \cdot ({\bf p}\times{\bf L})^\dagger \right)^\dagger \nonumber\\[5pt] &=& \left( \frac{\bf x}{r} \cdot (-{\bf L}\times{\bf p}) \right)^\dagger \nonumber\\[5pt] &=& -\left( \frac{1}{r} (- {\bf L}^2 + 2 i\hbar {\bf x}\cdot{\bf p}) \right)^\dagger \nonumber\\[5pt] &=& -(-{\bf L}^2 - 2 i\hbar {\bf p}\cdot{\bf x}) \frac{1}{r} \nonumber\\[5pt] &=& \left({\bf L}^2 + 2 i\hbar p_i x_i\right) \frac{1}{r} \nonumber\\[5pt] &=& \left({\bf L}^2 + 2 i\hbar (x_i p_i + [p_i, x_i]) \right) \frac{1}{r} \nonumber\\[5pt] &=& \left({\bf L}^2 + 2 i\hbar {\bf x}\cdot{\bf p} + 2 i\hbar (-i\hbar)\delta_{ii}\right) \frac{1}{r} \nonumber\\[5pt] &=& ({\bf L}^2 + 2 i\hbar {\bf x}\cdot{\bf p} + 6\hbar^2) \frac{1}{r} \nonumber\\[5pt] &=& \frac{1}{r} ({\bf L}^2 + 2 i\hbar {\bf x}\cdot{\bf p} + 6\hbar^2) + \left[ 2 i\hbar {\bf x}\cdot{\bf p}, \frac{1}{r}\right] \nonumber\\[5pt] &=& \frac{1}{r} ({\bf L}^2 + 2 i\hbar {\bf x}\cdot{\bf p} + 6\hbar^2) + 2i\hbar x_i \left[ p_i, \frac{1}{r}\right] \nonumber\\[5pt] &=& \frac{1}{r} ({\bf L}^2 + 2 i\hbar {\bf x}\cdot{\bf p} + 6\hbar^2) + 2i\hbar x_i i\hbar \frac{x_i}{r^3} \nonumber\\[5pt] &=& \frac{1}{r} ({\bf L}^2 + 2 i\hbar {\bf x}\cdot{\bf p} + 6\hbar^2) - 2\hbar^2 \frac{1}{r} \nonumber\\[5pt] &=& \frac{1}{r} ({\bf L}^2 + 2 i\hbar {\bf x}\cdot{\bf p} + 4\hbar^2) \end{eqnarray}$$ Putting all the results above on equation \eqref{post_d379f7e34709563115ebd4c41241ed5e_AA_tmp} gives us: $$\begin{eqnarray} {\bf A}^2 &=& \frac{1}{4}({\bf L}^2{\bf p}^2 + {\bf L}^2{\bf p}^2 + {\bf L}^2{\bf p}^2 + 4\hbar^2 {\bf p}^2 + {\bf L}^2{\bf p}^2) \nonumber\\[5pt] &+& \frac{me^2}{2r}\left(-{\bf L}^2 -{\bf L}^2 - 2i\hbar{\bf x}\cdot{\bf p} - 4\hbar^2 -{\bf L}^2 + 2i\hbar{\bf x}\cdot{\bf p} - {\bf L}^2\right) \nonumber\\[5pt] &+& m^2e^4 \nonumber\\[5pt] &=& {\bf L}^2{\bf p}^2 + \hbar^2{\bf p}^2 - \frac{2me^2}{r}{\bf L}^2 - \frac{2me^2\hbar^2}{r} + m^2e^4 \nonumber\\[5pt] &=& ({\bf L}^2 + \hbar^2)\left({\bf p}^2 - \frac{2me^2}{r}\right) + m^2e^4 \nonumber\\[5pt] &=& ({\bf L}^2 + \hbar^2)2mH + m^2e^4 \end{eqnarray}$$ Applying this result to our energy eigenstate, we have $H \rightarrow E = -\kappa^2/2m$, so we get: $$\boxed{ {\bf A}^2 = -\kappa^2({\bf L}^2 + \hbar^2) + m^2e^4 } \label{post_d379f7e34709563115ebd4c41241ed5e_AA}$$

### Determination of $E$ in terms of ${\bf K}^\pm$

First, since ${\bf L}$ and ${\bf A}$ are self-adjoint operators: $${\bf L}\cdot{\bf A} + {\bf A}\cdot{\bf L} = {\bf L}\cdot{\bf A} + ({\bf L}^\dagger\cdot{\bf A}^\dagger)^\dagger = {\bf L}\cdot{\bf A} + ({\bf L}\cdot{\bf A})^\dagger$$ Given that: $$\displaystyle {\bf L}\cdot{\bf A} = {\bf L}\cdot\frac{1}{2}({\bf L}\times{\bf x} - {\bf x}\times{\bf L}) + me^2{\bf L}\cdot\frac{\bf x}{r} = me^2({\bf x}\times{\bf p})\cdot\frac{\bf x}{r} = 0$$ then: $${\bf L}\cdot{\bf A} + {\bf A}\cdot{\bf L} = 0 \label{post_d379f7e34709563115ebd4c41241ed5e_LA_AL}$$ From equations \eqref{post_d379f7e34709563115ebd4c41241ed5e_AA} and \eqref{post_d379f7e34709563115ebd4c41241ed5e_LA_AL}, we have that: $$\begin{eqnarray} \displaystyle ({\bf K}^\pm)^2 &=& {\bf K}^\pm\cdot{\bf K}^\pm \nonumber\\[5pt] &=& \frac{1}{4}{\bf L}\cdot{\bf L} \pm \frac{1}{4\kappa}({\bf L}\cdot{\bf A} + {\bf A}\cdot{\bf L}) + \frac{1}{4\kappa^2}{\bf A}\cdot{\bf A} \nonumber\\[5pt] &=& \frac{1}{4}{\bf L}^2 + \frac{1}{4\kappa^2}( -\kappa^2({\bf L}^2 + \hbar^2) + m^2e^4 ) \nonumber\\[5pt] &=& \frac{1}{4}{\bf L}^2 - \frac{1}{4}{\bf L}^2 - \frac{\hbar^2}{4} + \frac{m^2e^4}{4\kappa^2} \end{eqnarray}$$ Therefore: $$({\bf K}^+)^2 = ({\bf K}^-)^2 = \frac{1}{4}\left(-\hbar^2 + \frac{m^2e^4}{\kappa^2}\right) \label{post_d379f7e34709563115ebd4c41241ed5e_Kp2}$$ With this result, we then have that: $$\boxed{ \displaystyle E = -\frac{\kappa^2}{2m} = -\frac{m^2e^4}{2m}\left(\hbar^2 + 4({\bf K}^\pm)^2\right)^{-1} } \label{%INDEX_E_tmp}$$

### A final expression for $E$

As shown on equation \eqref{post_d379f7e34709563115ebd4c41241ed5e_comm_Ki_Kj}, the operators $K^+_i$ and $K^-_i$ satisfy the commutation relations for angular momentum operators, i.e.: $$\begin{eqnarray} [K^+_i, K^+_j] &=& i\hbar \epsilon_{ijk} K^+_k \\[5pt] [K^-_i, K^-_j] &=& i\hbar \epsilon_{ijk} K^-_k \end{eqnarray}$$ It then follows directly that: $$\begin{eqnarray} [({\bf K}^+)^2, K^+_i] &=& 0 \\[5pt] [({\bf K}^-)^2, K^-_i] &=& 0 \end{eqnarray}$$ Using ladder operators, it is straightforward to prove that the eigenvalues of $({\bf K}^+)^2$ and $({\bf K}^-)^2$ have the form $\hbar^2 j(j+1)$ for values of $j$ which are either integers of half-integers, i.e., $j = (n-1)/2$ for integer values $n \geq 1$. Since we are assuming we have an energy eigenstate with energy $E = -\kappa^2/2m$, and since equation \eqref{post_d379f7e34709563115ebd4c41241ed5e_Kp2} implies that such a state is also an eigenstate of both $({\bf K}^+)^2$ and $({\bf K}^-)^2$, then from equation \eqref{%INDEX_E_tmp} we have that: $$\begin{eqnarray} \displaystyle E &=& -\frac{m^2e^4}{2m}\left(\hbar^2 + 4({\bf K}^\pm)^2\right)^{-1} \nonumber\\[5pt] &=& -\frac{m^2e^4}{2m}\left(\hbar^2 + 4\hbar^2j(j+1)\right)^{-1} \nonumber\\[5pt] &=& -\frac{m^2e^4}{2m}\left(\hbar^2 + 4\hbar^2\frac{n-1}{2}\frac{n+1}{2}\right)^{-1} \nonumber\\[5pt] &=& -\frac{m^2e^4}{2m}\left(\hbar^2 + \hbar^2(n^2 - 1)\right)^{-1} \end{eqnarray}$$ Finally, we have our desired result, which we express as $E_n$ instead of $E$ to make the dependence on $n$ explicit: $$\displaystyle E_n = -\frac{me^4}{2\hbar^2n^2}$$ Since we used $4\pi\epsilon_0 = 1$ on equation \eqref{post_d379f7e34709563115ebd4c41241ed5e_hamiltonian}, we can replace $e^2$ with $e^2 / 4\pi\epsilon_0$ to get the energy levels in SI units: $$\boxed{ \displaystyle E_n = -\frac{me^4}{2\hbar^2(4\pi\epsilon_0)^2n^2} \quad \textrm{for} \quad n = 1, 2, 3, \ldots }$$

## Earth's rotation and the surface of the ocean

Posted by Diego Assencio on 2016.09.16 under Physics (Mechanics)

In this post, we will show that the Earth's rotation alters the surface of its ocean, with "ocean" here meaning the global ocean of the Earth, i.e., the continuous body of water encircling it. Our model will consist of a spherical Earth with radius $R$ rotating at an angular velocity $\omega$ around a fixed axis.

The easiest way to solve this problem is by using a non-inertial frame of reference which rotates with the Earth and with origin at its center. The vertical axis $z'$ of this rotating frame is chosen to be the axis of rotation of the Earth (see figure 1). In this frame, the ocean is at rest since it moves with the Earth.

 Fig. 1: Earth and the surface of the ocean. The reference frame $S$ sees the Earth rotating with angular velocity $\omega$ around the $z$ axis, while in the frame $S'$, the Earth is at rest since $S'$ is also rotating with respect to $S$ with angular velocity $\omega$, i.e., $S'$ is "glued" to the Earth. The origins of both $S$ and $S'$ coincide with the center of the Earth. The effect of the Earth's rotation on the surface of its ocean is greatly exaggerated in the figure.

Let $S$ be a reference frame with origin at the center of the Earth and such that the Earth itself is seen as rotating with angular velocity $\omega$, and let $S'$be the frame which rotates with the Earth as shown in figure 1. As proven in a previous post, the following equation relates the force ${\bf F}_S$ experienced by a fluid element at the ocean with mass $m$ in $S$ and the effective force ${\bf F}^{\textrm{eff}}_{S'}$ experienced by this same fluid element in $S'$ (terms which trivially vanish were omitted below): $${\bf F}^{\textrm{eff}}_{S'} = m{\bf a}_{S'} = {\bf F}_S - m{\pmb\omega} \times ({\pmb\omega}\times{\bf x}') = {\bf F}_S + m\omega^2 \rho\hat{\pmb\rho} \label{post_f9a58faf18bdbd019f06a0f09c123d60_eq_forces1}$$ where ${\bf a}_{S'}$ and ${\bf x}'$ are the acceleration and position of the fluid element as measured in $S'$ respectively, ${\pmb\omega} = \omega\hat{\bf z}$ is the angular velocity of the Earth as seen in $S$, $\rho$ is the distance between the fluid element and the $z$ (or $z'$) axis and $\hat{\pmb\rho}$ is the unit vector which points radially outwards from the $z$ (or $z'$) axis (see figure 2).

 Fig. 2: Forces acting on a fluid element of mass $m$ on the surface of the Earth as seen in $S'$. The figure shows a cross section of the Earth which passes through the $z$ (or $z'$) axis.

The term ${\bf F}_S$ is the sum of the only two forces experienced in the non-rotating frame $S$: the gravitational force $m{\bf g}$ and the force ${\bf F}_{\textrm{fluid}}$ exerted on the fluid element by the surrounding fluid (up to this point, we have not explicitly assumed that the fluid element is on the surface, so the equation below applies to any fluid element on the ocean): $${\bf F}_S = m{\bf g} + {\bf F}_{\textrm{fluid}} \label{post_f9a58faf18bdbd019f06a0f09c123d60_eq_forces2}$$ Since in $S'$ the fluid element is at rest, ${\bf a}_{S'} = {\bf 0}$. This fact, together with equations \eqref{post_f9a58faf18bdbd019f06a0f09c123d60_eq_forces1} and \eqref{post_f9a58faf18bdbd019f06a0f09c123d60_eq_forces2}, yields the following: $${\bf 0} = {\bf F}_{\textrm{fluid}} + m\left({\bf g} - {\pmb\omega} \times ({\pmb\omega}\times{\bf x}')\right) = {\bf F}_{\textrm{fluid}} + m\left({\bf g} + \omega^2 \rho\hat{\pmb\rho}\right) \label{post_f9a58faf18bdbd019f06a0f09c123d60_eq_acc}$$ Figure 2 shows these three forces. In equation \eqref{post_f9a58faf18bdbd019f06a0f09c123d60_eq_acc}, the centrifugal acceleration term $\omega^2 \rho\hat{\pmb\rho}$ can be interpreted as an additional component which, together with the gravitational acceleration ${\bf g}$, yields an effective gravitational acceleration ${\bf g}^{\textrm{eff}}$ which is no longer homogeneous in space: $${\bf g}^{\textrm{eff}} = {\bf g} - {\pmb\omega} \times ({\pmb\omega}\times{\bf x}') = {\bf g} + \omega^2 \rho\hat{\pmb\rho} = {\bf g} + \omega^2 R \sin\theta\hat{\pmb\rho} \label{post_f9a58faf18bdbd019f06a0f09c123d60_effec_gravity}$$ where above we used the fact that $\rho \approx R\sin\theta$ since the surface of the ocean is close enough to the surface of the spherical Earth, and $\theta$ is the angle between the $z'$ axis and a segment connecting the center of the Earth to the fluid element.

The force ${\bf F}_{\textrm{fluid}}$ is a very interesting one since it is always orthogonal to the surface of the ocean for fluid elements on the ocean surface. To understand this point, imagine the case in which there is no rotation, i.e., ${\pmb\omega} = {\bf 0}$. In this case, the frames $S$ and $S'$ coincide and the ocean is at rest on both. From equation \eqref{post_f9a58faf18bdbd019f06a0f09c123d60_eq_acc}, we see that ${\bf F}_{\textrm{fluid}} = - m{\bf g}$, which means ${\bf F}_{\textrm{fluid}}$ is the force on the fluid element exerted by the surrounding fluid against the gravitational force: the fluid element is not moving, so the surrounding fluid must be providing the force ${\bf F}_{\textrm{fluid}} = -m{\bf g}$ to keep it in place. Since the fluid surface is locally horizontal in this case, i.e., orthogonal to ${\bf g}$, then ${\bf F}_{\textrm{fluid}}$ is orthogonal to the fluid surface. When $\omega \neq 0$, ${\bf F}_{\textrm{fluid}}$ plays exactly the same role, but now the effective gravitational force given in equation \eqref{post_f9a58faf18bdbd019f06a0f09c123d60_effec_gravity} is no longer homogeneous in space but depends explicitly on the position ${\bf x}'$: ${\bf F}_{\textrm{fluid}}$ will still be orthogonal to the surface of the ocean for every fluid element on the ocean surface; a tangential force component would set the fluid element in motion since it would not be able to resist this force (but a perpendicular force is resisted by a pressure gradient created inside the water).

Since ${\bf F}_{\textrm{fluid}}$ is always perpendicular to every fluid element on the ocean surface, and since, from equations \eqref{post_f9a58faf18bdbd019f06a0f09c123d60_eq_acc} and \eqref{post_f9a58faf18bdbd019f06a0f09c123d60_effec_gravity}: $${\bf F}_{\textrm{fluid}} = -m\left({\bf g} + \omega^2 R\sin\theta\hat{\pmb\rho}\right) = -m{\bf g}^{\textrm{eff}} \label{post_f9a58faf18bdbd019f06a0f09c123d60_eq_acc2}$$ then the ocean surface is always perpendicular to the effective gravitational acceleration ${\bf g}^{\textrm{eff}}$. As shown in figure 3, even though ${\bf g}$ always points to the center of the Earth, ${\bf g}^{\textrm{eff}}$ does not, meaning the surface of the ocean will not be spherical. At the poles, we have $\theta = 0$ and $\theta = \pi$, so ${\bf g}^{\textrm{eff}} = {\bf g}$, meaning the ocean surface is perpendicular to ${\bf g}$ at those points. Along the equator, $\theta = \pi/2$ and therefore ${\bf g}^{\textrm{eff}} = {\bf g} + \omega^2 R \hat{\pmb\rho}$; since $\hat{\pmb\rho}$ is parallel to ${\bf g}$, the resulting ${\bf g}^{\textrm{eff}}$ still points to the center of the Earth and therefore the ocean is perpendicular to ${\bf g}$ there as well, but the fact that $\omega^2 R \hat{\pmb\rho}$ points away from ${\bf g}$ means ${\bf g}^{\textrm{eff}}$ is smaller in magnitude at the equator than at the poles (where it attains its highest magnitude). But at any other point on the surface of the Earth, $\bf g$ and $\omega^2 R\sin\theta\hat{\pmb\rho}$ are not parallel to each other and therefore the surface of the ocean is in general not perpendicular to ${\bf g}$.

 Fig. 3: Components of ${\bf g}^{\textrm{eff}}$ for fluid elements on diverse points along the surface of the ocean. Except at the poles and at the equator, the surface of the ocean is in general not orthogonal to ${\bf g}$.

A final comment is necessary here: when studying the surface of the ocean, we only considered the rotation of the Earth. In reality, the centrifugal acceleration $\omega^2 \rho \hat{\pmb\rho}$ causes the Earth itself to be shaped more like an oblate spheroid with an equatorial bulge of $42.77\textrm{km}$, and the ocean is affected as a result. The gravitational forces of the Sun and the Moon also significantly change the shapes of both the Earth and the ocean due to the tidal forces which they generate.

For the curious, $\omega^2 R \approx 0.34\textrm{m/s}^2$, so the rotation of the Earth makes objects on the equator be $\omega^2 R / g \approx 0.35\%$ lighter than at the poles (here we used $g = 9.8\textrm{m/s}^2$). In practice, however, the fact that the Earth is an oblate spheroid means the difference is even higher since objects at the equator are farther from the center of the Earth than objects at the poles.

## Elastic forces on a rubber band

Posted by Diego Assencio on 2015.10.14 under Physics (Mechanics)

Consider a uniform elastic band of rest length $L$ and mass $M$ as shown in figure 1a. The rest length $L$ is such that if the elastic band is shaped as a circle of perimeter $L$ (i.e., a circle with radius $R = L/2\pi$), the elastic force over its entire extension is identically zero. In this rest configuration, we parameterize each point of the band using its arclength $s$: for that, we choose a fixed point to be the one with $s = 0$ and all other points of the band are assigned $s$ values which grow in a counterclockwise manner, so $s$ values are in the range $[0,L)$ as shown in figure 1b.

 (a)
 (b)
 Fig. 1: An elastic band. Figure (a) shows the elastic band in an arbitrary configuration described by a function ${\bf X}(s,t)$ and the unit vector $\pmb\tau(s,t)$ which is tangent to the band at the point ${\bf X}(s,t)$. Figure (b) shows how the arclength value $s$ is defined for each point of the interface in the rest configuration.

As the elastic band is allowed to move, we can track the motion of each of its points using the arclength value $s$ assigned to each point in the rest (circular) configuration, i.e., we can define the position of each point of the band through a function ${\bf X}(s,t)$ which defines the location of the point with assigned arclength value $s$ at time $t$ (see figure 1a).

Our goal is to compute the force per unit length ${\bf f}(s,t)$ at the point ${\bf X}(s,t)$ for an arbitrary configuration of the elastic band. For that, let us first model the band as a set of $n$ small masses (each one with mass $M/n$) connected by identical springs of elastic coefficient $k$ and rest length $\Delta{s} = L/n$ as shown in figure 2. In what follows, we will refer to this model as the "discrete elastic band".

 Fig. 2: A discrete elastic band consisting of masses connected by springs.

The force at the mass located at ${\bf x}_i$ is the sum of the elastic forces produced by the two springs connected to it. Let $\Delta{\bf x}_i = {\bf x}_{i+1} - {\bf x}_i$ and let $\pmb\eta_{i}$ be a unit vector parallel to $\Delta{\bf x}_i$, i.e.: $$\pmb\eta_{i} = \displaystyle\frac{{\bf x}_{i+1} - {\bf x}_i}{\|{\bf x}_{i+1} - {\bf x}_i\|} = \displaystyle\frac{\Delta{\bf x}_i}{\|\Delta{\bf x}_i\|}$$ With these definitions, the force at the mass located at ${\bf x}_i$ is given by: $$\Delta{\bf F}_i = k \left(\|\Delta{\bf x}_i\| - \Delta{s}\right)\pmb\eta_{i} + k\left(\|\Delta{\bf x}_{i-1}\| - \Delta{s}\right)(-\pmb\eta_{i-1}) \label{post_be83d9c57ef2d0c138c880ce222471eb_force_discrete_band}$$ The reason for the minus sign before $\pmb\eta_{i-1}$ on the second term comes from the fact that if the spring which connects ${\bf x}_i$ and ${\bf x}_{i-1}$ is stretched beyond its rest length $\Delta{s}$, the force on the mass at ${\bf x}_i$ points towards the mass at ${\bf x}_{i-1}$, i.e., it is parallel to $(-\pmb\eta_{i-1})$. Since we are interested in computing the force per unit length ${\bf f}_i$ on ${\bf x}_i$, we can divide equation \eqref{post_be83d9c57ef2d0c138c880ce222471eb_force_discrete_band} by the band length $\Delta{L}_i$ associated with the mass at ${\bf x}_i$: $$\Delta{L}_i = \displaystyle\frac{\|\Delta{\bf x}_i\| + \|\Delta{\bf x}_{i-1}\|}{2} \label{post_be83d9c57ef2d0c138c880ce222471eb_eq_delta_L}$$ We then get: $$\begin{eqnarray} \displaystyle {\bf f}_i &=& \frac{\Delta{\bf F}_i}{\Delta{L}_i} \nonumber\\[5pt] &=& \frac{\Delta{s}}{\Delta{L}_i} \frac{\Delta{\bf F}_i}{\Delta{s}}\nonumber\\[5pt] &=& \frac{\Delta{s}}{\Delta{L}_i} k \left[\left(\frac{\|\Delta{\bf x}_i\|}{\Delta{s}} - 1\right)\pmb\eta_{i} - \left(\frac{\|\Delta{\bf x}_{i-1}\|}{\Delta{s}} - 1\right)\pmb\eta_{i-1}\right] \nonumber\\[5pt] &=& (k\Delta{s}) \frac{\Delta{s}}{\Delta{L}_i}\frac{1}{\Delta{s}}\left[\left(\frac{\|\Delta{\bf x}_i\|}{\Delta{s}} - 1\right)\pmb\eta_{i} - \left(\frac{\|\Delta{\bf x}_{i-1}\|}{\Delta{s}} - 1\right)\pmb\eta_{i-1}\right] \label{post_be83d9c57ef2d0c138c880ce222471eb_force_density_discrete_band} \end{eqnarray}$$ As we increase the number of masses $n$ in the the discrete band while keeping the total mass equal to $M$, the discrete force density ${\bf f}_i$ will converge to the force density ${\bf f}(s,t)$ we are trying to compute. To obtain ${\bf f}(s,t)$, we must analyze each term on the last line of equation \eqref{post_be83d9c57ef2d0c138c880ce222471eb_force_density_discrete_band} and see how it behaves as $n \rightarrow \infty$. Since the elastic band is parameterized by a function ${\bf X}(s,t)$ as described earlier, and since $\Delta{s} \rightarrow ds$ as $n \rightarrow \infty$, we have that: $$\displaystyle \frac{\|\Delta{\bf x}_i\|}{\Delta{s}} = \left\| \frac{\Delta{\bf x}_i}{\Delta{s}} \right\| \longrightarrow \left\|\frac{\partial{\bf X}}{\partial s}\right\| \label{post_be83d9c57ef2d0c138c880ce222471eb_eq_dx_ds1}$$ Similarly: $$\displaystyle \frac{\|\Delta{\bf x}_{i-1}\|}{\Delta{s}} = \left\| \frac{\Delta{\bf x}_{i-1}}{\Delta{s}} \right\| \longrightarrow \left\|\frac{\partial{\bf X}}{\partial s}\right\| \label{post_be83d9c57ef2d0c138c880ce222471eb_eq_dx_ds2}$$ With the definition of $\Delta{L}_i$ given in equation \eqref{post_be83d9c57ef2d0c138c880ce222471eb_eq_delta_L} as well as equations \eqref{post_be83d9c57ef2d0c138c880ce222471eb_eq_dx_ds1} and \eqref{post_be83d9c57ef2d0c138c880ce222471eb_eq_dx_ds2}, we have that: $$\displaystyle \frac{\Delta{L}_i}{\Delta{s}} = \displaystyle \frac{1}{2}\left(\frac{\|\Delta{\bf x}_i\|}{\Delta{s}} + \frac{\|\Delta{\bf x}_{i-1}\|}{\Delta{s}}\right) \longrightarrow \left\|\frac{\partial{\bf X}}{\partial s}\right\| \label{post_be83d9c57ef2d0c138c880ce222471eb_dL_ds}$$ Additionally: $$\pmb\eta_i = \displaystyle\frac{\Delta{\bf x}_i}{\|\Delta{\bf x}_i\|} = \left(\displaystyle\frac{\Delta{\bf x}_i}{\Delta{s}}\right) \Big/ \left(\frac{\|\Delta{\bf x}_i\|}{\Delta{s}}\right)$$ and therefore (the same result is true for $\pmb\eta_{i-1}$): $$\pmb\eta_i \longrightarrow \frac{\partial{\bf X}}{\partial s} \Big/ \left\|\frac{\partial{\bf X}}{\partial s}\right\| = \pmb\tau(s,t) \label{post_be83d9c57ef2d0c138c880ce222471eb_tau_converges}$$ where $\pmb\tau(s,t)$ is the unit vector which is tangent to the elastic band at the point ${\bf X}(s,t)$ as shown in figure 1a. From equations \eqref{post_be83d9c57ef2d0c138c880ce222471eb_eq_dx_ds1}, \eqref{post_be83d9c57ef2d0c138c880ce222471eb_eq_dx_ds2} and \eqref{post_be83d9c57ef2d0c138c880ce222471eb_tau_converges}, we see that the term: $$\frac{1}{\Delta{s}}\left[\left(\frac{\|\Delta{\bf x}_i\|}{\Delta{s}} - 1\right)\pmb\eta_{i} - \left(\frac{\|\Delta{\bf x}_{i-1}\|}{\Delta{s}} - 1\right)\pmb\eta_{i-1}\right]$$ converges to: $$\frac{\partial}{\partial{s}}\left[ \left(\left\|\frac{\partial{\bf X}}{\partial{s}}\right\| - 1\right) \frac{\partial{\bf X}}{\partial s} \Big/ \left\|\frac{\partial{\bf X}}{\partial s}\right\|\right] = \frac{\partial}{\partial{s}}\left[ \left(\left\|\frac{\partial{\bf X}}{\partial{s}}\right\| - 1\right) \pmb\tau\right] \label{post_be83d9c57ef2d0c138c880ce222471eb_partial_term}$$ The last component we have to analyze is the elastic constant $k$. What does it converge to when $n \rightarrow \infty$? The answer to this question lies on equation \eqref{post_be83d9c57ef2d0c138c880ce222471eb_force_density_discrete_band} itself. Since all other terms converge to fixed quantities, $k\Delta{s}$ must converge to a fixed quantity as well. Given that $k\Delta{s}$ is physically associated with the elasticity of the band, we must have $k\Delta{s} \rightarrow \kappa$ for a fixed constant $\kappa$, meaning $k$ must grow proportionally to $1/\Delta{s}$ as $\Delta{s} \rightarrow 0$ in order for the assumption ${\bf f}_i \rightarrow {\bf f}(s,t)$ to hold. The constant $\kappa$ represents the elastic coefficient of the continuous elastic band. Using this fact as well as equations \eqref{post_be83d9c57ef2d0c138c880ce222471eb_dL_ds} and \eqref{post_be83d9c57ef2d0c138c880ce222471eb_partial_term}, we have that the force density ${\bf f}(s,t)$ on the elastic band at the point ${\bf X}(s,t)$ is given by: $$\boxed{ \displaystyle {\bf f}(s,t) = \frac{\partial}{\partial{s}}(T\pmb\tau) \Big/ \left\|\frac{\partial{\bf X}}{\partial{s}}\right\| }$$ where $T(s,t)$ is the tension on the elastic band at the point ${\bf X}(s,t)$: $$\boxed{ \displaystyle T(s,t) = \kappa\left(\left\|\frac{\partial{\bf X}}{\partial{s}}\right\| - 1\right) }$$ Finally, notice that our derivation did not require the elastic band to be really closed, i.e., the results above apply to elastic fibers in general. Elastic fibers are commonly used in numerical simulations of fluid-structure interactions. For instance, see the immersed boundary method developed by Charles S. Peskin to simulate blood flow inside the human heart.