In quantum mechanics, a particle in a spherically symmetric potential is a system with a potential that depends only on the distance between the particle and a center. A particle in a spherically symmetric potential can be used as an approximation, for example, of the electron in a hydrogen atom or of the formation of chemical bonds.^[1]

In the general time-independent case, the dynamics of a particle in a spherically symmetric potential are governed by a Hamiltonian of the following form:

$${\displaystyle {\hat {H))={\frac ((\hat {p))^{2)){2m_{0))}+V({r})}$$

${\displaystyle m_{0))$

${\displaystyle {\hat {p))}$

${\displaystyle V(r)}$

To describe a particle in a spherically symmetric system, it is convenient to use spherical coordinates ${\displaystyle r}$, ${\displaystyle \theta }$ and ${\displaystyle \phi }$. The time-independent Schrödinger equation for the system is then a separable, partial differential equation. This means solutions to the angular dimensions of the equation can be found independently of the radial dimension. This leaves an ordinary differential equation in terms only of the radius, ${\displaystyle r}$, which determines the eigenstates for the particular potential, ${\displaystyle V(r)}$.

Structure of the eigenfunctions

If solved by separation of variables, the eigenstates of the system will have the form:

$${\displaystyle \psi (r,\theta ,\phi )=R(r)\Theta (\theta )\Phi (\phi )}$$

${\displaystyle \theta }$

${\displaystyle \phi }$

${\displaystyle \psi }$

$${\displaystyle \psi (r,\theta ,\phi )=R(r)Y_{\ell m}(\theta ,\phi ).}$$

The differential equation which characterises the function ${\displaystyle R(r)}$ is called the radial equation.

Derivation of the radial equation

The kinetic energy operator in spherical polar coordinates is:

$${\displaystyle {\frac ((\hat {p))^{2)){2m_{0))}=-{\frac {\hbar ^{2)){2m_{0))}\nabla ^{2}=-{\frac {\hbar ^{2)){2m_{0}\,r^{2))}\left[{\frac {\partial }{\partial r))\left(r^{2}{\frac {\partial }{\partial r))\right)-{\hat {L))^{2}\right].}$$

$${\displaystyle {\hat {L))^{2}Y_{\ell m}(\theta ,\phi )\equiv \left\{-{\frac {1}{\sin ^{2}\theta ))\left[\sin \theta {\frac {\partial }{\partial \theta ))\left(\sin \theta {\frac {\partial }{\partial \theta ))\right)+{\frac {\partial ^{2)){\partial \phi ^{2))}\right]\right\}Y_{\ell m}(\theta ,\phi )=\ell (\ell +1)Y_{\ell m}(\theta ,\phi ).}$$

$${\displaystyle {\frac {1}{r^{2))}{\frac {d}{dr))\left(r^{2}{\frac {dR}{dr))\right)-{\frac {\ell (\ell +1)}{r^{2))}R+{\frac {2m_{0)){\hbar ^{2))}\left[E-V(r)\right]R=0.}$$

${\displaystyle R(r)=u(r)/r}$

${\displaystyle u(r)}$

$${\displaystyle {d^{2}u \over dr^{2))+{\frac {2m_{0)){\hbar ^{2))}\left[E-V_{\mathrm {eff} }(r)\right]u=0}$$

$${\displaystyle V_{\mathrm {eff} }(r)=V(r)+{\hbar ^{2}\ell (\ell +1) \over 2m_{0}r^{2)),}$$

${\displaystyle r\in [0,\infty )}$

If ${\textstyle \lim _{r\to 0}r^{2}V(r)=0}$, then near the origin, ${\displaystyle R\sim r^{\ell ))$.

Spherically symmetric Hamiltonians

Since the Hamiltonian is spherically symmetric, it is said to be invariant under rotation, ie:

${\displaystyle {U(R)}^{\dagger }{\hat {H))U(R)={\hat {H))}$

Since angular momentum operators are generators of rotation, applying the Baker-Campbell-Hausdorff Lemma we get:

${\displaystyle {U(R)}^{\dagger }{\hat {H))U(R)=e^{\frac {i{\hat {n))\cdot {\vec {L))\theta }{\hbar )){\hat {H))e^{\frac {-i{\hat {n))\cdot {\vec {L))\theta }{\hbar ))={\hat {H))+i{\frac {\theta }{\hbar ))\left[{\hat {n))\cdot {\vec {L)),{\hat {H))\right]+{\frac {1}{2!))\left(i{\frac {\theta }{\hbar ))\right)^{2}\left[{\hat {n))\cdot {\vec {L)),\left[{\hat {n))\cdot {\vec {L)),{\hat {H))\right]\right]+\cdots ={\hat {H))}$

Since this equation holds for all values of ${\displaystyle \theta }$, we get that ${\displaystyle [{\hat {n))\cdot {\vec {L)),{\hat {H))]=0}$, or that every angular momentum component commutes with the Hamiltonian.

Since ${\displaystyle L_{z))$ and ${\displaystyle L^{2))$ are such mutually commuting operators that also commute with the Hamiltonian, the wavefunctions can be expressed as ${\displaystyle |\alpha ;\ell ,m\rangle }$ or ${\displaystyle \psi _{\alpha ;\ell ,m}(r,\theta ,\phi )}$ where ${\displaystyle \alpha }$ is used to label different wavefunctions.

Since ${\textstyle L_{\pm }=L_{x}\pm iL_{y))$ also commutes with the Hamiltonian, the energy eigenvalues in such cases are always independent of ${\displaystyle m}$.

$${\displaystyle {\hat {H))|\alpha ;\ell ,m\rangle =E_{\alpha ;\ell }|\alpha ;\ell ,m\rangle }$$

Combined with the fact that ${\textstyle L_{\pm ))$ differential operators only act on the functions of ${\displaystyle \theta }$ and ${\displaystyle \phi }$, it shows that if the solutions are assumed to be separable as ${\textstyle \psi (r,\theta ,\phi )=R(r)Y(\theta ,\phi )}$, the radial wavefunction ${\displaystyle R(r)}$ can always be chosen independent of ${\displaystyle m}$ values. Thus the wavefunction is expressed as:^[2]

$${\displaystyle \psi _{\alpha ;\ell ,m}(r,\theta ,\phi )=R(r)_{\alpha ;\ell }Y_{\ell m}(\theta ,\phi ).}$$

Solutions for potentials of interest

There are five cases of special importance:

1. ${\displaystyle V(r)=0}$, or solving the vacuum in the basis of spherical harmonics, which serves as the basis for other cases.
2. ${\displaystyle V(r)=V_{0))$ (finite) for ${\displaystyle r<r_{0))$ and zero elsewhere.
3. ${\displaystyle V(r)=0}$ for ${\displaystyle r<r_{0))$ and infinite elsewhere, the spherical equivalent of the square well, useful to describe bound states in a nucleus or quantum dot.
4. ${\displaystyle V(r)\propto r^{2))$for the three-dimensional isotropic harmonic oscillator.
5. ${\displaystyle V(r)\propto {\frac {1}{r))}$ to describe bound states of hydrogen-like atoms.

The solutions are outlined in these cases, which should be compared to their counterparts in cartesian coordinates, cf. particle in a box.

Vacuum case states

Let us now consider ${\displaystyle V(r)=0}$. Introducing the dimensionless variables

$${\displaystyle \rho \ {\stackrel {\mathrm {def} }{=))\ kr,\qquad k\ {\stackrel {\mathrm {def} }{=))\ {\sqrt {2m_{0}E \over \hbar ^{2))},}$$

${\displaystyle J(\rho )\ {\stackrel {\mathrm {def} }{=))\ {\sqrt {\rho ))R(r)}$

$${\displaystyle \rho ^{2}{\frac {d^{2}J}{d\rho ^{2))}+\rho {\frac {dJ}{d\rho ))+\left[\rho ^{2}-\left(\ell +{\frac {1}{2))\right)^{2}\right]J=0}$$

${\displaystyle J_{\ell +1/2}(\rho )}$

${\displaystyle R(r)}$

${\textstyle R(r)=j_{l}(kr)\ {\stackrel {\mathrm {def} }{=))\ {\sqrt {\pi /(2kr)))J_{\ell +1/2}(kr)}$.

The solutions of the Schrödinger equation in polar coordinates in vacuum are thus labelled by three quantum numbers: discrete indices ℓ and m, and k varying continuously in ${\displaystyle [0,\infty )}$:

$${\displaystyle \psi (\mathbf {r} )=j_{\ell }(kr)Y_{\ell m}(\theta ,\phi )}$$

${\displaystyle \exp(i\mathbf {k} \cdot \mathbf {r} )}$

Sphere with finite "square" potential

Consider the potential ${\displaystyle V(r)=V_{0))$ for ${\displaystyle r<r_{0))$ and ${\displaystyle V(r)=0}$ elsewhere - that is, inside a sphere of radius ${\displaystyle r_{0))$ the potential is equal to ${\displaystyle V_{0))$ and it is zero outside the sphere. A potential with such a finite discontinuity is called a square potential.^[3]

We first consider bound states, i.e. states which display the particle mostly inside the box (confined states). Those have an energy ${\displaystyle E}$ less than the potential outside the sphere, i.e., they have negative energy. Also worth noticing is that unlike Coulomb potential, featuring an infinite number of discrete bound states, the spherical square well has only a finite (if any) number because of its finite range.

The resolution essentially follows that of the vacuum case above with normalization of the total wavefunction added, solving two Schrödinger equations — inside and outside the sphere — of the previous kind, i.e., with constant potential. The following constraints must hold for a normalizable, physical wavefunction:

1. The wavefunction must be regular at the origin.
2. The wavefunction and its derivative must be continuous at the potential discontinuity.
3. The wavefunction must converge at infinity.

The first constraint comes from the fact that Neumann and Hankel functions are singular at the origin. The physical requirement that ${\displaystyle \psi }$ must be defined everywhere selected Bessel function of the first kind over the other possibilities in the vacuum case. For the same reason, the solution will be of this kind inside the sphere:

$${\displaystyle R(r)=Aj_{\ell }\left({\sqrt {\frac {2m_{0}(E-V_{0})}{\hbar ^{2))))r\right),\qquad r<r_{0}.}$$

${\displaystyle V_{0}<E<0}$

${\displaystyle E<0}$

$${\displaystyle R(r)=Bh_{\ell }^{(1)}\left(i{\sqrt {\frac {-2m_{0}E}{\hbar ^{2))))r\right),\qquad r>r_{0))$$

${\displaystyle \psi }$

${\displaystyle r=r_{0))$

${\displaystyle A}$

${\displaystyle B}$

Sphere with infinite "square" potential

In case where the potential well is infinitely deep, so that we can take ${\displaystyle V_{0}=0}$ inside the sphere and ${\displaystyle \infty }$ outside, the problem becomes that of matching the wavefunction inside the sphere (the spherical Bessel functions) with identically zero wavefunction outside the sphere. Allowed energies are those for which the radial wavefunction vanishes at the boundary. Thus, we use the zeros of the spherical Bessel functions to find the energy spectrum and wavefunctions. Calling ${\displaystyle u_{\ell ,k))$ the k^th zero of ${\displaystyle j_{\ell ))$, we have:

$${\displaystyle E_{kl}={\frac {u_{\ell ,k}^{2}\hbar ^{2)){2m_{0}r_{0}^{2))))$$

${\displaystyle u_{\ell ,k))$

In the special case ${\displaystyle \ell =0}$ (spherical symmetric orbitals), the spherical Bessel function is ${\textstyle j_{0}(x)={\frac {\sin x}{x))}$, which zeros can be easily given as ${\displaystyle u_{0,k}=k\pi }$. Their energy eigenvalues are thus:

$${\displaystyle E_{k0}={\frac {(k\pi )^{2}\hbar ^{2)){2m_{0}r_{0}^{2))}={\frac {k^{2}h^{2)){8m_{0}r_{0}^{2))))$$

3D isotropic harmonic oscillator

The potential of a 3D isotropic harmonic oscillator is

$${\displaystyle V(r)={\frac {1}{2))m_{0}\omega ^{2}r^{2}.}$$

$${\displaystyle E_{n}=\hbar \omega \left(n+{\frac {N}{2))\right)\quad {\text{with))\quad n=0,1,\ldots ,\infty ,}$$

${\displaystyle n}$

${\displaystyle \omega }$

${\displaystyle N}$

${\displaystyle N=3}$

$${\displaystyle \left[-{\frac {\hbar ^{2)){2m_{0))}{\frac {d^{2)){dr^{2))}+{\hbar ^{2}\ell (\ell +1) \over 2m_{0}r^{2))+{\frac {1}{2))m_{0}\omega ^{2}r^{2}-\hbar \omega \left(n+{\tfrac {3}{2))\right)\right]u(r)=0.}$$

Introducing

$${\displaystyle \gamma \equiv {\frac {m_{0}\omega }{\hbar ))}$$

${\displaystyle u(r)=rR(r)}$

$${\displaystyle R_{n\ell }(r)=N_{n\ell }\,r^{\ell }\,e^{-{\frac {1}{2))\gamma r^{2))\;L_((\frac {1}{2))(n-\ell )}^{\scriptscriptstyle \left(\ell +{\frac {1}{2))\right)}(\gamma r^{2}),}$$

${\displaystyle L_{k}^{(\alpha )}(\gamma r^{2})}$

${\displaystyle yr^{2))$

${\displaystyle k}$

The normalization constant ${\displaystyle N_{nl))$ is,

$${\displaystyle N_{n\ell }=\left[{\frac {2^{n+\ell +2}\,\gamma ^{\ell +{\frac {3}{2)))){\pi ^{\frac {1}{2))))\right]^{\frac {1}{2))\left[{\frac {\left[{\frac {1}{2))(n-\ell )\right]!\;\left[{\frac {1}{2))(n+\ell )\right]!}{(n+\ell +1)!))\right]^{\frac {1}{2)).}$$

The eigenfunction ${\displaystyle R_{nl}(r)}$ is associated with energy ${\displaystyle E_{n))$, where

$${\displaystyle \ell =n,n-2,\ldots ,\ell _{\min }\quad {\text{with))\quad \ell _{\min }={\begin{cases}1&{\text{if))~n~{\text{odd))\\0&{\text{if))~n~{\text{even))\end{cases))}$$

${\displaystyle \gamma =2\nu }$

Derivation

First we transform the radial equation by a few successive substitutions to the generalized Laguerre differential equation, which has known solutions: the generalized Laguerre functions. Then we normalize the generalized Laguerre functions to unity. This normalization is with the usual volume element r² dr.

First we scale the radial coordinate

$${\displaystyle y={\sqrt {\gamma ))r\quad {\text{with))\quad \gamma \equiv {\frac {m_{0}\omega }{\hbar )),}$$

$${\displaystyle \left[{\frac {d^{2)){dy^{2))}-{\frac {\ell (\ell +1)}{y^{2))}-y^{2}+2n+3\right]v(y)=0}$$

${\textstyle v(y)=u\left(y/{\sqrt {\gamma ))\right)}$

Consideration of the limiting behavior of v(y) at the origin and at infinity suggests the following substitution for v(y),

$${\displaystyle v(y)=y^{\ell +1}e^{-y^{2}/2}f(y).}$$

$${\displaystyle \left[{\frac {d^{2)){dy^{2))}+2\left({\frac {\ell +1}{y))-y\right){\frac {d}{dy))+2n-2\ell \right]f(y)=0,}$$

${\displaystyle y^{\ell +1}e^{-y^{2}/2))$

Transformation to Laguerre polynomials

If the substitution ${\displaystyle x=y^{2))$ is used, ${\displaystyle y={\sqrt {x))}$, and the differential operators become

$${\displaystyle {\frac {d}{dy))={\frac {dx}{dy)){\frac {d}{dx))=2y{\frac {d}{dx))=2{\sqrt {x)){\frac {d}{dx)),}$$

$${\displaystyle {\frac {d^{2)){dy^{2))}={\frac {d}{dy))\left(2y{\frac {d}{dx))\right)=4x{\frac {d^{2)){dx^{2))}+2{\frac {d}{dx)).}$$

The expression between the square brackets multiplying ${\displaystyle f(y)}$ becomes the differential equation characterizing the generalized Laguerre equation (see also Kummer's equation):

$${\displaystyle x{\frac {d^{2}g}{dx^{2))}+\left(\left(\ell +{\frac {1}{2))\right)+1-x\right){\frac {dg}{dx))+{\frac {1}{2))(n-\ell )g(x)=0}$$

${\displaystyle g(x)\equiv f({\sqrt {x)))}$

Provided ${\displaystyle k\equiv (n-\ell )/2}$ is a non-negative integral number, the solutions of this equations are generalized (associated) Laguerre polynomials

$${\displaystyle g(x)=L_{k}^{\scriptscriptstyle \left(\ell +{\frac {1}{2))\right)}(x).}$$

From the conditions on ${\displaystyle k}$ follows: (i) ${\displaystyle n\geq \ell }$ and (ii) ${\displaystyle n}$ and ${\displaystyle l}$ are either both odd or both even. This leads to the condition on ${\displaystyle l}$ given above.

Recovery of the normalized radial wavefunction

Remembering that ${\displaystyle u(r)=rR(r)}$, we get the normalized radial solution:

$${\displaystyle R_{n\ell }(r)=N_{n\ell }\,r^{\ell }\,e^{-{\frac {1}{2))\gamma r^{2))\;L_((\frac {1}{2))(n-\ell )}^{\scriptscriptstyle \left(\ell +{\frac {1}{2))\right)}(\gamma r^{2}).}$$

The normalization condition for the radial wave function is:

$${\displaystyle \int _{0}^{\infty }r^{2}|R(r)|^{2}\,dr=1.}$$

Substituting ${\displaystyle q=\gamma r^{2))$, gives ${\displaystyle dq=2\gamma r\,dr}$ and the equation becomes:

$${\displaystyle {\frac {N_{n\ell }^{2)){2\gamma ^{\ell +{\frac {3}{2))))}\int _{0}^{\infty }q^{\ell +{1 \over 2))e^{-q}\left[L_((\frac {1}{2))(n-\ell )}^{\scriptscriptstyle \left(\ell +{\frac {1}{2))\right)}(q)\right]^{2}\,dq=1.}$$

By making use of the orthogonality properties of the generalized Laguerre polynomials, this equation simplifies to:

$${\displaystyle {\frac {N_{n\ell }^{2)){2\gamma ^{\ell +{\frac {3}{2))))}\cdot {\frac {\Gamma {\left[{\frac {1}{2))(n+\ell +1)+1\right])){\left[{\frac {1}{2))(n-\ell )\right]!))=1.}$$

Hence, the normalization constant can be expressed as:

$${\displaystyle N_{n\ell }={\sqrt {\frac {2\,\gamma ^{\ell +{\frac {3}{2))}\,\left({\frac {n-\ell }{2))\right)!}{\Gamma {\left({\frac {n+\ell }{2))+{\frac {3}{2))\right)))))}$$

Other forms of the normalization constant can be derived by using properties of the gamma function, while noting that ${\displaystyle n}$ and ${\displaystyle l}$ are both of the same parity. This means that ${\displaystyle n+l}$ is always even, so that the gamma function becomes:

$${\displaystyle \Gamma \left[{\frac {1}{2))+\left({\frac {n+\ell }{2))+1\right)\right]={\frac ((\sqrt {\pi ))(n+\ell +1)!!}{2^((\frac {n+\ell }{2))+1))}={\frac ((\sqrt {\pi ))(n+\ell +1)!}{2^{n+\ell +1}\left[{\frac {1}{2))(n+\ell )\right]!)),}$$

$${\displaystyle N_{n\ell }=\left[{\frac {2^{n+\ell +2}\,\gamma ^{\ell +{3 \over 2))\left[{1 \over 2}(n-\ell )\right]!\left[{1 \over 2}(n+\ell )\right]!}{\;\pi ^{1 \over 2}(n+\ell +1)!))\right]^{1 \over 2}={\sqrt {2))\left({\frac {\gamma }{\pi ))\right)^{1 \over 4}\,({2\gamma })^{\ell \over 2}\,{\sqrt {\frac {2\gamma (n-\ell )!!}{(n+\ell +1)!!))}.}$$

Hydrogen-like atoms

A hydrogenic (hydrogen-like) atom is a two-particle system consisting of a nucleus and an electron. The two particles interact through the potential given by Coulomb's law:

$${\displaystyle V(r)=-{\frac {1}{4\pi \varepsilon _{0))}{\frac {Ze^{2)){r))}$$

• ε₀ is the permittivity of the vacuum,
• Z is the atomic number (eZ is the charge of the nucleus),
• e is the elementary charge (charge of the electron),
• r is the distance between the electron and the nucleus.

In order to simplify the Schrödinger equation, we introduce the following constants that define the atomic unit of energy and length:

$${\displaystyle E_{\textrm {h))=\mu \left({\frac {e^{2)){4\pi \varepsilon _{0}\hbar ))\right)^{2}\quad {\text{and))\quad a_{0}=((4\pi \varepsilon _{0}\hbar ^{2)) \over {\mu e^{2))}.}$$

where ${\displaystyle \mu \approx m_{e))$ is the reduced mass in the ${\displaystyle m_{e}\ll m_{nucleus))$ limit. Substitute ${\displaystyle y=Zr/a_{0))$ and ${\displaystyle W=E/(Z^{2}E_{\textrm {h)))}$ into the radial Schrödinger equation given above. This gives an equation in which all natural constants are hidden,

$${\displaystyle \left[-{\frac {1}{2)){\frac {d^{2)){dy^{2))}+{\frac {1}{2)){\frac {\ell (\ell +1)}{y^{2))}-{\frac {1}{y))\right]u_{\ell }=Wu_{\ell }.}$$

(i) ${\displaystyle W}$ is negative, the corresponding eigenfunctions are square-integrable and the values of ${\displaystyle W}$ are quantized (discrete spectrum).

(ii) ${\displaystyle W}$ is non-negative, every real non-negative value of ${\displaystyle W}$ is physically allowed (continuous spectrum), the corresponding eigenfunctions are non-square integrable. Considering only class (i) solutions restricts the solutions to wavefunctions which are bound states, in contrast to the class (ii) solutions that are known as scattering states.

For class (i) solutions with negative W the quantity ${\displaystyle \alpha \equiv 2{\sqrt {-2W))}$ is real and positive. The scaling of ${\displaystyle y}$, i.e., substitution of ${\displaystyle x\equiv \alpha y}$ gives the Schrödinger equation:

$${\displaystyle \left[{\frac {d^{2)){dx^{2))}-{\frac {\ell (\ell +1)}{x^{2))}+{\frac {2}{\alpha x))-{\frac {1}{4))\right]u_{\ell }=0,\quad {\text{with ))x\geq 0.}$$

For ${\displaystyle x\rightarrow \infty }$ the inverse powers of x are negligible and the normalizable (and therefore, physical) solution for large ${\displaystyle x}$ is ${\displaystyle \exp[-x/2]}$. Similarly, for ${\displaystyle x\rightarrow 0}$ the inverse square power dominates and the physical solution for small ${\displaystyle x}$ is x^ℓ+1. Hence, to obtain a full range solution we substitute

$${\displaystyle u_{l}(x)=x^{\ell +1}e^{-x/2}f_{\ell }(x).}$$

The equation for ${\displaystyle f_{l}(x)}$ becomes,

$${\displaystyle \left[x{\frac {d^{2)){dx^{2))}+(2\ell +2-x){\frac {d}{dx))+(n-\ell -1)\right]f_{\ell }(x)=0\quad {\text{with))\quad n=(-2W)^{-1/2}={\frac {2}{\alpha )).}$$

Provided ${\displaystyle n-\ell -1}$ is a non-negative integer, this equation has polynomial solutions written as

$${\displaystyle L_{k}^{(2\ell +1)}(x),\qquad k=0,1,\ldots ,}$$

${\displaystyle k}$

$${\displaystyle W=-{\frac {1}{2n^{2))}\quad {\text{with))\quad n\equiv k+\ell +1.}$$

The principal quantum number ${\displaystyle n}$ satisfies ${\displaystyle n\geq \ell +1}$. Since ${\displaystyle \alpha =2/n}$, the total radial wavefunction is

$${\displaystyle R_{n\ell }(r)={\frac {u(r)}{r))=N_{n\ell }\left({\frac {2Zr}{na_{0))}\right)^{\ell }\;e^{-{\tfrac {Zr}{na_{0))))\;L_{n-\ell -1}^{(2\ell +1)}\left({\frac {2Zr}{na_{0))}\right),}$$

${\textstyle {\frac {1}{r))}$

$${\displaystyle N_{n\ell }=\left[\left({\frac {2Z}{na_{0))}\right)^{3}\cdot {\frac {(n-\ell -1)!}{2n[(n+\ell )!]^{3))}\right]^{1 \over 2},}$$

$${\displaystyle \int _{0}^{\infty }x^{2\ell +2}e^{-x}\left[L_{n-\ell -1}^{(2\ell +1)}(x)\right]^{2}dx={\frac {2n(n+\ell )!}{(n-\ell -1)!)).}$$

The corresponding energy is

$${\displaystyle E=-{\frac {Z^{2)){2n^{2))}E_{\textrm {h)),\qquad n=1,2,\ldots .}$$