We continue this perfunctory review of quantum mechanics by solving the wave function of the electron in a hydrogen atom. It has not only historical significance, but the fine tructure of the hydrogen atom also provides a great instance of perturbation theory. In the presence of the Coulumb potential, with \(q\) the elementary charge, and \(\mu\) the mass of proton (because \(m\) is used up for magnetic quantum numbers), the hydrogen atom has this Hamiltonian:
Hamiltonian:
\(H\)\(\;\equiv\;\)\(\dfrac{-\, \hslash \vphantom{ \hslash}^{2}}{2 \mu}\)\(\nabla \vphantom{ \nabla}^{2}\)\(\,-\,\)\(\dfrac{q \vphantom{ q}^{2}}{4 \pi \varepsilon \vphantom{ \varepsilon}_{0}}\)\(\dfrac{1}{r}\)
By the formula of Laplacian in spherical coodinates:
\(\;=\;\)\(\,-\,\)\(\dfrac{\hslash \vphantom{ \hslash}^{2}}{2 \mu}\)\(\dfrac{1}{r \vphantom{ r}^{2}}\)\(\dfrac{\partial}{\partial r}\)\(r \vphantom{ r}^{2}\)\(\dfrac{\partial}{\partial r}\)\(\,-\,\)\(\dfrac{\hslash \vphantom{ \hslash}^{2}}{2 \mu}\)\(\dfrac{1}{r \vphantom{ r}^{2}}\)\(\dfrac{1}{\mathrm{Sin}\, \vartheta}\)\(\dfrac{\partial}{\partial \vartheta}\)\(\left(\vphantom{\left(\,\mathrm{Sin}\, \vartheta\right) \dfrac{\partial}{\partial \vartheta}}\right.\)\(\left(\vphantom{\,\mathrm{Sin}\, \vartheta}\right.\)\(\,\mathrm{Sin}\,\)\(\vartheta\)\(\left.\vphantom{\,\mathrm{Sin}\, \vartheta}\right)\)\(\dfrac{\partial}{\partial \vartheta}\)\(\left.\vphantom{\left(\,\mathrm{Sin}\, \vartheta\right) \dfrac{\partial}{\partial \vartheta}}\right)\)\(\,-\,\)\(\dfrac{\hslash \vphantom{ \hslash}^{2}}{2 \mu}\)\(\dfrac{1}{r \vphantom{ r}^{2} \mathrm{Sin} \left[\vartheta\right] \vphantom{\mathrm{Sin} \left[\vartheta\right]}^{2}}\)\(\dfrac{\partial \vphantom{ \partial}^{2}}{\partial \varphi \vphantom{ \varphi}^{2}}\)\(\,-\,\)\(\dfrac{q \vphantom{ q}^{2}}{4 \pi \varepsilon \vphantom{ \varepsilon}_{0}}\)\(\dfrac{1}{r}\)\(;\,\)
By symmetry, the wave function must have the form \(\psi\)\(\;=\;\)\(\rho\)\(\left[\vphantom{r}\right.\)\(r\)\(\left.\vphantom{r}\right]\)\(\chi\)\(\left[\vphantom{\varphi ,\, \vartheta}\right.\)\(\varphi\)\(,\,\)\(\vartheta\)\(\left.\vphantom{\varphi ,\, \vartheta}\right]\), where \(\varphi\) is azimuth and \(\vartheta\) zenith. Hence the time-independent Schrödinger equation reads:
\(H\)\(\rho\)\(\left[\vphantom{r}\right.\)\(r\)\(\left.\vphantom{r}\right]\)\(\chi\)\(\left[\vphantom{\varphi ,\, \vartheta}\right.\)\(\varphi\)\(,\,\)\(\vartheta\)\(\left.\vphantom{\varphi ,\, \vartheta}\right]\)\(\;=\;\)\(E\)\(\rho\)\(\left[\vphantom{r}\right.\)\(r\)\(\left.\vphantom{r}\right]\)\(\chi\)\(\left[\vphantom{\varphi ,\, \vartheta}\right.\)\(\varphi\)\(,\,\)\(\vartheta\)\(\left.\vphantom{\varphi ,\, \vartheta}\right]\)\(;\,\)
We want to relate the expression of \(H\) with the total angular momentum \(L\). Here \(L\) is given as:
Angular_momentum:
\(\hslash \vphantom{ \hslash}^{-\, 2}\)\(L \vphantom{ L}^{2}\)\(\psi\)\(\;=\;\)\(\,-\,\)\(\left(\vphantom{\left(\mathbf{r} \,\times\, \nabla\right) \,\cdot\, \left(\mathbf{r} \,\times\, \nabla\right)}\right.\)\(\left(\vphantom{\mathbf{r} \,\times\, \nabla}\right.\)\(\mathbf{r}\)\(\,\times\,\)\(\nabla\)\(\left.\vphantom{\mathbf{r} \,\times\, \nabla}\right)\)\(\,\cdot\,\)\(\left(\vphantom{\mathbf{r} \,\times\, \nabla}\right.\)\(\mathbf{r}\)\(\,\times\,\)\(\nabla\)\(\left.\vphantom{\mathbf{r} \,\times\, \nabla}\right)\)\(\left.\vphantom{\left(\mathbf{r} \,\times\, \nabla\right) \,\cdot\, \left(\mathbf{r} \,\times\, \nabla\right)}\right)\)\(\psi\)
By noting \(\nabla\) either acts on the second \(\mathbf{r}\) or on \(\psi\) :
\(\;=\;\)\(\,-\,\)\(\left|\mathbf{r} \,\times\, \nabla\right| \vphantom{ \left|\mathbf{r} \,\times\, \nabla\right|}^{2}\)\(\psi\)\(\,+\,\)\(\mathbf{r}\)\(\,\cdot\,\)\(\nabla\)\(\psi\)
\(\;=\;\)\(\,-\,\)\(r \vphantom{ r}^{2}\)\(\nabla \vphantom{ \nabla}^{2}\)\(\psi\)\(\,+\,\)\(\left(\mathbf{r} \,\cdot\, \nabla\right) \vphantom{ \left(\mathbf{r} \,\cdot\, \nabla\right)}^{2}\)\(\psi\)\(\,+\,\)\(\mathbf{r}\)\(\,\cdot\,\)\(\nabla\)\(\psi\)
Again by the formula of Laplacian in spherical coodinates:
\(\;=\;\)\(\,-\,\)\(\dfrac{\partial}{\partial r}\)\(r \vphantom{ r}^{2}\)\(\dfrac{\partial}{\partial r}\)\(\psi\)\(\,-\,\)\(\dfrac{1}{\mathrm{Sin}\, \vartheta}\)\(\dfrac{\partial}{\partial \vartheta}\)\(\left(\vphantom{\left(\,\mathrm{Sin}\, \vartheta\right) \dfrac{\partial}{\partial \vartheta} \psi}\right.\)\(\left(\vphantom{\,\mathrm{Sin}\, \vartheta}\right.\)\(\,\mathrm{Sin}\,\)\(\vartheta\)\(\left.\vphantom{\,\mathrm{Sin}\, \vartheta}\right)\)\(\dfrac{\partial}{\partial \vartheta}\)\(\psi\)\(\left.\vphantom{\left(\,\mathrm{Sin}\, \vartheta\right) \dfrac{\partial}{\partial \vartheta} \psi}\right)\)\(\,-\,\)\(\dfrac{1}{\mathrm{Sin} \left[\vartheta\right] \vphantom{\mathrm{Sin} \left[\vartheta\right]}^{2}}\)\(\dfrac{\partial \vphantom{ \partial}^{2}}{\partial \varphi \vphantom{ \varphi}^{2}}\)\(\psi\)\(\,+\,\)\(\left(r \dfrac{\partial}{\partial r}\right) \vphantom{ \left(r \dfrac{\partial}{\partial r}\right)}^{2}\)\(\psi\)\(\,+\,\)\(r\)\(\dfrac{\partial}{\partial r}\)\(\psi\)
\(\;=\;\)\(\,-\,\)\(\dfrac{1}{\mathrm{Sin}\, \vartheta}\)\(\dfrac{\partial}{\partial \vartheta}\)\(\left(\vphantom{\left(\,\mathrm{Sin}\, \vartheta\right) \dfrac{\partial}{\partial \vartheta} \psi}\right.\)\(\left(\vphantom{\,\mathrm{Sin}\, \vartheta}\right.\)\(\,\mathrm{Sin}\,\)\(\vartheta\)\(\left.\vphantom{\,\mathrm{Sin}\, \vartheta}\right)\)\(\dfrac{\partial}{\partial \vartheta}\)\(\psi\)\(\left.\vphantom{\left(\,\mathrm{Sin}\, \vartheta\right) \dfrac{\partial}{\partial \vartheta} \psi}\right)\)\(\,-\,\)\(\dfrac{1}{\mathrm{Sin} \left[\vartheta\right] \vphantom{\mathrm{Sin} \left[\vartheta\right]}^{2}}\)\(\dfrac{\partial \vphantom{ \partial}^{2}}{\partial \varphi \vphantom{ \varphi}^{2}}\)\(\psi\)\(;\,\)
Now we observe a key fact:
\(\,\mathrm{Comm}\,\)\(\left[\vphantom{L \vphantom{ L}^{2} ,\, H}\right.\)\(L \vphantom{ L}^{2}\)\(,\,\)\(H\)\(\left.\vphantom{L \vphantom{ L}^{2} ,\, H}\right]\)\(\;=\;\)\(0\)\(;\,\)
To see this, note that \(L \vphantom{ L}^{2}\) are completely angular, which both commute with the two radial terms of \(H\) , and the two angular terms of \(H\) are precisely \(L \vphantom{ L}^{2}\), since both came from the Laplacian. Moreover, from the relation between azimuth angle and angular momentum:
\(L \vphantom{ L}_{z}\)\(\psi\)\(\;=\;\)\(\,-\,\)\(\mathring{\imath}\)\(\hslash\)\(\dfrac{\partial}{\partial \varphi}\)\(\psi\)\(;\,\)
And it is also seen that:
\(\,\mathrm{Comm}\,\)\(\left[\vphantom{L \vphantom{ L}^{2} ,\, L \vphantom{ L}_{z}}\right.\)\(L \vphantom{ L}^{2}\)\(,\,\)\(L \vphantom{ L}_{z}\)\(\left.\vphantom{L \vphantom{ L}^{2} ,\, L \vphantom{ L}_{z}}\right]\)\(\;=\;\)\(0\)\(;\,\)
\(\,\mathrm{Comm}\,\)\(\left[\vphantom{H ,\, L \vphantom{ L}_{z}}\right.\)\(H\)\(,\,\)\(L \vphantom{ L}_{z}\)\(\left.\vphantom{H ,\, L \vphantom{ L}_{z}}\right]\)\(\;=\;\)\(0\)\(;\,\)
Thus, we may assume that \(H\)\(,\,\)\(L \vphantom{ L}^{2}\)\(,\,\)\(L \vphantom{ L}_{z}\) share eigenstates. Indeed we have concluded exactly the same from our discussion on commutator relations!
In fact, solving the radial part of Hamiltonian equation deals with the one radial degree of freedom, and solving the angular momentum equation deals with the two angular degree of freedom. Let us just do that.
The \(\varphi\) dependence is the easiest; it follows that, for eigenvalue \(m\) :
\(\chi\)\(\left[\vphantom{\vartheta ,\, \varphi}\right.\)\(\vartheta\)\(,\,\)\(\varphi\)\(\left.\vphantom{\vartheta ,\, \varphi}\right]\)\(\;=\;\)\(e \vphantom{ e}^{\mathring{\imath} m \varphi \,/\, \hslash}\)\(\chi\)\(\left[\vphantom{\vartheta ,\, 0}\right.\)\(\vartheta\)\(,\,\)\(0\)\(\left.\vphantom{\vartheta ,\, 0}\right]\)\(;\,\)
The \(\vartheta\) dependence is characterized by eigenvalue equation \(L \vphantom{ L}^{2}\)\(\chi\)\(\;=\;\)\(\lambda\)\(\chi\); this can be cast for eigenvalue \(\lambda\) as follows:
\(\,-\,\)\(1\)\(\;\leq\;\)\(u\)\(\left[\vphantom{\vartheta}\right.\)\(\vartheta\)\(\left.\vphantom{\vartheta}\right]\)\(\;\equiv\;\)\(\,\mathrm{Cos}\,\)\(\vartheta\)\(\;\leq\;\)\(1\)\(;\,\)
Azimuthal:
\(\left(\vphantom{1 \,-\, u \vphantom{ u}^{2}}\right.\)\(1\)\(\,-\,\)\(u \vphantom{ u}^{2}\)\(\left.\vphantom{1 \,-\, u \vphantom{ u}^{2}}\right)\)\(\dfrac{\mathrm{d}}{\mathrm{d} u}\)\(\left(\vphantom{1 \,-\, u \vphantom{ u}^{2}}\right.\)\(1\)\(\,-\,\)\(u \vphantom{ u}^{2}\)\(\left.\vphantom{1 \,-\, u \vphantom{ u}^{2}}\right)\)\(\dfrac{\mathrm{d}}{\mathrm{d} u}\)\(\chi \vphantom{ \chi}_{l ,\, m}\)\(\,+\,\)\(\left(\vphantom{\,-\, m \vphantom{ m}^{2} \,+\, \lambda \left(1 \,-\, u \vphantom{ u}^{2}\right)}\right.\)\(\,-\,\)\(m \vphantom{ m}^{2}\)\(\,+\,\)\(\lambda\)\(\left(\vphantom{1 \,-\, u \vphantom{ u}^{2}}\right.\)\(1\)\(\,-\,\)\(u \vphantom{ u}^{2}\)\(\left.\vphantom{1 \,-\, u \vphantom{ u}^{2}}\right)\)\(\left.\vphantom{\,-\, m \vphantom{ m}^{2} \,+\, \lambda \left(1 \,-\, u \vphantom{ u}^{2}\right)}\right)\)\(\chi \vphantom{ \chi}_{l ,\, m}\)\(\;=\;\)\(0\)\(;\,\)
For our purpose, it is most important to determine \(\lambda\). If the leading degree of the series expansion of \(\chi \vphantom{ \chi}_{l ,\, 0}\) is infinity, then its derivative blows up. If we accept that \(\chi \vphantom{ \chi}_{l ,\, 0}\) is a polynomial with finitely many terms, then by observing the leading coefficient, we know:
Eigenvalue_angular:
\(\lambda\)\(\;=\;\)\(l\)\(\left(\vphantom{l \,+\, 1}\right.\)\(l\)\(\,+\,\)\(1\)\(\left.\vphantom{l \,+\, 1}\right)\)\(,\,\)\(l\)\(\;=\;\)\(0\)\(,\,\)\(1\)\(,\,\)\(2\)\(,\,\)\(\,\dotsc\,\)\(;\,\)
Particularly, when \(m\)\(\;=\;\)\(0\) it reads:
Legendre:
\(\left(\vphantom{1 \,-\, u \vphantom{ u}^{2}}\right.\)\(1\)\(\,-\,\)\(u \vphantom{ u}^{2}\)\(\left.\vphantom{1 \,-\, u \vphantom{ u}^{2}}\right)\)\(\dfrac{\mathrm{d} \vphantom{ \mathrm{d}}^{2} \chi \vphantom{ \chi}_{l ,\, 0}}{\mathrm{d} u \vphantom{ u}^{2}}\)\(\,-\,\)\(2\)\(u\)\(\dfrac{\mathrm{d} \chi \vphantom{ \chi}_{l ,\, 0}}{\mathrm{d} u}\)\(\,+\,\)\(\lambda\)\(\chi \vphantom{ \chi}_{l ,\, 0}\)\(\;=\;\)\(0\)\(;\,\)
This is known as Legendre equation. The closed form of \(\chi \vphantom{ \chi}_{l ,\, 0}\) is given by the Rodrigues formula:
Rodrigues_Legendre:
\(\chi \vphantom{ \chi}_{l ,\, 0}\)\(\left[\vphantom{\vartheta}\right.\)\(\vartheta\)\(\left.\vphantom{\vartheta}\right]\)\(\;=\;\)\(\dfrac{1}{2 \vphantom{ 2}^{n} n !}\)\(\left(\dfrac{\mathrm{d}}{\mathrm{d} u}\right) \vphantom{ \left(\dfrac{\mathrm{d}}{\mathrm{d} u}\right)}^{l}\)\(\left(u \vphantom{ u}^{2} \,-\, 1\right) \vphantom{ \left(u \vphantom{ u}^{2} \,-\, 1\right)}^{n}\)\(\left[\vphantom{u \:\leftarrow\: \,\mathrm{Cos}\, \vartheta}\right.\)\(u\)\(\:\leftarrow\:\)\(\,\mathrm{Cos}\,\)\(\vartheta\)\(\left.\vphantom{u \:\leftarrow\: \,\mathrm{Cos}\, \vartheta}\right]\)\(;\,\)
It can be shown to satisfy Legendre, and its normalization may be checked. Either by the axiom on quantum observables or by an integration by parts, we see the left side of Legendre is hermitian. So we have orthogonality of \(\chi \vphantom{ \chi}_{l ,\, 0}\) on the inner product space \(\left\{\vphantom{\,-\, 1 \;\leq\; u \;\leq\; 1}\right.\)\(\,-\,\)\(1\)\(\;\leq\;\)\(u\)\(\;\leq\;\)\(1\)\(\left.\vphantom{\,-\, 1 \;\leq\; u \;\leq\; 1}\right\}\). (Details on this part don’t aid physical understanding much, which I only sketch here.)
In summary, the angular equation is solved to be:
\(\chi \vphantom{ \chi}_{l ,\, m}\)\(\left[\vphantom{\vartheta ,\, \varphi}\right.\)\(\vartheta\)\(,\,\)\(\varphi\)\(\left.\vphantom{\vartheta ,\, \varphi}\right]\)\(\;=\;\)\(e \vphantom{ e}^{\mathring{\imath} m \varphi \,/\, \hslash}\)\(\chi \vphantom{ \chi}_{l ,\, 0}\)\(\left[\vphantom{\vartheta}\right.\)\(\vartheta\)\(\left.\vphantom{\vartheta}\right]\)\(;\,\)
It remains to find the radial dependence. In view of Hamiltonian, AngMom, and EigAng, we have
Radial:
\(r\)\(\dfrac{\mathrm{d} \vphantom{ \mathrm{d}}^{2}}{\mathrm{d} r \vphantom{ r}^{2}}\)\(\rho\)\(\,+\,\)\(2\)\(\dfrac{\mathrm{d}}{\mathrm{d} r}\)\(\rho\)\(\,+\,\)\(\left(\vphantom{\dfrac{2 \mu E}{\hslash \vphantom{ \hslash}^{2}} r \,+\, \dfrac{q \vphantom{ q}^{2} \mu}{2 \pi \hslash \vphantom{ \hslash}^{2} \varepsilon \vphantom{ \varepsilon}_{0}} \,-\, \dfrac{l \left(l \,+\, 1\right)}{r}}\right.\)\(\dfrac{2 \mu E}{\hslash \vphantom{ \hslash}^{2}}\)\(r\)\(\,+\,\)\(\dfrac{q \vphantom{ q}^{2} \mu}{2 \pi \hslash \vphantom{ \hslash}^{2} \varepsilon \vphantom{ \varepsilon}_{0}}\)\(\,-\,\)\(\dfrac{l \left(l \,+\, 1\right)}{r}\)\(\left.\vphantom{\dfrac{2 \mu E}{\hslash \vphantom{ \hslash}^{2}} r \,+\, \dfrac{q \vphantom{ q}^{2} \mu}{2 \pi \hslash \vphantom{ \hslash}^{2} \varepsilon \vphantom{ \varepsilon}_{0}} \,-\, \dfrac{l \left(l \,+\, 1\right)}{r}}\right)\)\(\rho\)\(\;=\;\)\(0\)\(;\,\)
Let \(v\)\(\;=\;\)\(\surd\!\)\(\left(\vphantom{8 \mu E}\right.\)\(8\)\(\mu\)\(E\)\(\left.\vphantom{8 \mu E}\right)\)\(r\)\(\,/\,\)\(\hslash\). When \(r\)\(\:\rightarrow\:\)\(\infty\) it may be guessed to be \(\rho\)\(\;\propto\;\)\(e \vphantom{ e}^{-\, v \,/\, 2}\); when \(r\)\(\:\rightarrow\:\)\(0\) it may be guessed to be \(\rho\)\(\;\propto\;\)\(v \vphantom{ v}^{l}\). Introduce:
\(w\)\(\left[\vphantom{v}\right.\)\(v\)\(\left.\vphantom{v}\right]\)\(\;\equiv\;\)\(v \vphantom{ v}^{-\, l}\)\(e \vphantom{ e}^{v \,/\, 2}\)\(\rho\)\(\left[\vphantom{v}\right.\)\(v\)\(\left.\vphantom{v}\right]\)\(;\,\)
After a bunch of manipulation:
\(n\)\(\;\equiv\;\)\(\dfrac{q \vphantom{ q}^{2} \surd\! \left(\mu\right)}{\surd\! \left(3 2\right) \pi \hslash \varepsilon \vphantom{ \varepsilon}_{0} \surd\! \left(E\right)}\)\(;\,\)
Laguerre:
\(v\)\(\dfrac{\mathrm{d} \vphantom{ \mathrm{d}}^{2}}{\mathrm{d} v \vphantom{ v}^{2}}\)\(w\)\(\,+\,\)\(\left(\vphantom{2 l \,-\, v \,+\, 2}\right.\)\(2\)\(l\)\(\,-\,\)\(v\)\(\,+\,\)\(2\)\(\left.\vphantom{2 l \,-\, v \,+\, 2}\right)\)\(\dfrac{\mathrm{d}}{\mathrm{d} v}\)\(w\)\(\,+\,\)\(\left(\vphantom{n \,-\, l \,-\, 1}\right.\)\(n\)\(\,-\,\)\(l\)\(\,-\,\)\(1\)\(\left.\vphantom{n \,-\, l \,-\, 1}\right)\)\(w\)\(\;=\;\)\(0\)\(;\,\)
This is known as generalized Laguerre equation, which has orthogonal solutions called Laguerre polynomials and are given by another Rodrigues formula. After given back the asymptotic part, the answer is:
Rodrigues_Laguerre:
\(\rho \vphantom{ \rho}_{n ,\, l}\)\(\left[\vphantom{r}\right.\)\(r\)\(\left.\vphantom{r}\right]\)\(\;\propto\;\)\(v \vphantom{ v}^{-\, l \,-\, 1}\)\(e \vphantom{ e}^{v \,/\, 2}\)\(\left(\dfrac{\mathrm{d}}{\mathrm{d} v}\right) \vphantom{ \left(\dfrac{\mathrm{d}}{\mathrm{d} v}\right)}^{n \,-\, l \,-\, 1}\)\(\left(\vphantom{v \vphantom{ v}^{n \,+\, l} e \vphantom{ e}^{-\, v}}\right.\)\(v \vphantom{ v}^{n \,+\, l}\)\(e \vphantom{ e}^{-\, v}\)\(\left.\vphantom{v \vphantom{ v}^{n \,+\, l} e \vphantom{ e}^{-\, v}}\right)\)\(\left[\vphantom{v \:\leftarrow\: \dfrac{\surd\! \left(8 \mu E\right)}{\hslash} r}\right.\)\(v\)\(\:\leftarrow\:\)\(\dfrac{\surd\! \left(8 \mu E\right)}{\hslash}\)\(r\)\(\left.\vphantom{v \:\leftarrow\: \dfrac{\surd\! \left(8 \mu E\right)}{\hslash} r}\right]\)\(;\,\)
For brevity we don’t normalize the wave function. (Again we omit the details, which isn’t crucial for phyical understanding.)
From consideration of continuity, we must have \(m\)\(\;=\;\)\(0\)\(,\,\)\(1\)\(,\,\)\(2\)\(,\,\)\(\,\dotsc\,\); while Legendre requires:
\(\left|\vphantom{m}\right.\)\(m\)\(\left.\vphantom{m}\right|\)\(\;\leq\;\)\(l\)\(;\,\)
Furthermore, Laguerre requires:
\(n\)\(\,-\,\)\(l\)\(\,-\,\)\(1\)\(\;\geq\;\)\(0\)\(;\,\)
This restricts the energy levels to be:
\(E\)\(\;=\;\)\(\dfrac{q \vphantom{ q}^{4} \mu}{3 2 \pi \vphantom{ \pi}^{2} \hslash \vphantom{ \hslash}^{2} \varepsilon \vphantom{ \varepsilon}^{2}_{0}}\)\(\dfrac{1}{n \vphantom{ n}^{2}}\)\(;\,\)
We summarize the conclusion on quantum numbers as thus:
\(n\)\(\;=\;\)\(1\)\(,\,\)\(2\)\(,\,\)\(\,\dotsc\,\)\(;\,\)
\(l\)\(\;=\;\)\(0\)\(,\,\)\(1\)\(,\,\)\(2\)\(,\,\)\(\,\dotsc\,\)\(,\,\)\(n\)\(\,-\,\)\(1\)\(;\,\)
\(m\)\(\;=\;\)\(\,-\,\)\(l\)\(,\,\)\(\,\dotsc\,\)\(\,-\,\)\(1\)\(,\,\)\(0\)\(,\,\)\(1\)\(,\,\)\(\,\dotsc\,\)\(,\,\)\(l\)\(;\,\)
❧ September 4, 2021