Vacuum case states

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

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

the equation becomes a Bessel equation for ${\displaystyle J(\rho )\ {\stackrel {\mathrm {def} }{=}}\ {\sqrt {\rho }}R(r)}$:

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

where regular solutions for positive energies are given by so-called Bessel functions of the first kind ${\displaystyle J_{\ell +1/2}(\rho )}$ so that the solutions written for ${\displaystyle R(r)}$ are the so-called spherical Bessel function

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

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 ${\displaystyle [0,\infty )}$:

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

These solutions represent states of definite angular momentum, rather than of definite (linear) momentum, which are provided by plane waves ${\displaystyle \exp(i\mathbf {k} \cdot \mathbf {r} )}$.

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.^[2]

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:

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

Note that for bound states, ${\displaystyle V_{0}<E<0}$. Bound states bring the novelty as compared to the vacuum case now that ${\displaystyle E<0}$. This, along with the third constraint, selects the Hankel function of the first kind as the only converging solution at infinity (the singularity at the origin of these functions does not matter since we are now outside the sphere):

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

The second constraint on continuity of ${\displaystyle \psi }$ at ${\displaystyle r=r_{0}}$ along with normalization allows the determination of constants ${\displaystyle A}$ and ${\displaystyle B}$. Continuity of the derivative (or logarithmic derivative for convenience) requires quantization of energy.

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:

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

so that the problem is reduced to the computations of these zeros ${\displaystyle u_{\ell ,k}}$, typically by using a table or calculator, as these zeros are not solvable for the general case.

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

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

An N-dimensional isotropic harmonic oscillator has the energies

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

i.e., ${\displaystyle n}$ is a non-negative integral number; ${\displaystyle \omega }$ is the (same) fundamental frequency of the ${\displaystyle N}$ modes of the oscillator. In this case ${\displaystyle N=3}$, so that the radial Schrödinger equation becomes,

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

Introducing

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

and recalling that ${\displaystyle u(r)=rR(r)}$, we will show that the radial Schrödinger equation has the normalized solution,

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

where the function ${\displaystyle L_{k}^{(\alpha )}(\gamma r^{2})}$ is a generalized Laguerre polynomial in ${\displaystyle yr^{2}}$ of order ${\displaystyle k}$.

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

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

This is the same result as the quantum harmonic oscillator, with ${\displaystyle \gamma =2\nu }$.

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

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

and then the equation becomes

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

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

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

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

This substitution transforms the differential equation to

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

where we divided through with ${\displaystyle y^{\ell +1}e^{-y^{2}/2}}$, which can be done so long as y is not zero.

Transformation to Laguerre polynomials

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

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

and

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

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

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

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

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 wavefunction 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

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

where we used the definition of the double factorial. Hence, the normalization constant is also given by

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

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:

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

where

• 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,

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

Two classes of solutions of this equation exist: (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

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

which are generalized Laguerre polynomials of order ${\displaystyle k}$. The energy becomes

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

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

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

with normalization which absorbs extra terms from ${\textstyle {\frac {1}{r}}}$

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

via^[3]

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

The corresponding energy is

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