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{{Quantum Mechanics A}}
{{Quantum Mechanics A}}
[[Image:H_atom.jpg|thumb|300px|left]]
[[Image:H_atom.jpg|thumb|300px|left|Illustration of the hydrogen atom.]]


The [[Schrödinger equation]] for the particle moving in central potential can be represented in a [[Spherical Coordinates|spherical coordinate]] system as follows:
We now discuss the solution of the [[Schrödinger Equation|Schrödinger equation]] for the hydrogen atom.  We are especially interested in this system because it is of direct physical interest and because it is possible to solve it exactly.
:<math>\left( -\frac{\hbar^2}{2\mu}\frac{1}{r}\frac{\partial^2}{\partial r^2}r+\frac{L^2}{2\mu r^2}+V(r)\right) \psi(r,\theta,\phi)=E\psi(r,\theta,\phi)</math>
where <math>L\!</math> is the angular momentum operator and <math>\mu\!</math> is the reduced mass.


In this case, being invariant under the rotation, the Hamiltonian, <math>H\!</math>, commutes with both <math>L^2\!</math> and <math>L_z\!</math>. Furthermore, <math>L^2\!</math> and <math>L_z\!</math> commute with each other.  Therefore, the energy eigenstates can be chosen to be simultaneously the eigenstates of <math>H\!</math>, <math>L^2\!</math> and <math>L_z\!</math>. Such states can be expressed as the following:
The effective one-dimensional problem for this system is
:<math>\psi(r,\theta,\phi)=R(r)Y_{lm}(\theta,\phi)=\frac{u_l(r)}{r}Y_{lm}(\theta,\phi)</math>
where <math>Y_l^m(\theta,\phi)</math> is the spherical harmonic, which is the simultaneous eigenstates of <math>L^2\!</math> and <math>L_z\!</math>, and the <math>\frac{u_l(r)}{r}</math> substitution is made for simplification.


Substituting <math>\psi(r,\theta,\phi)=\frac{u_l(r)}{r}Y_{lm}(\theta,\phi) </math> into the Schrödinger equation, and taking into account the fact that:
<math>\left[ -\frac{\hbar^2}{2\mu}\frac{d^2}{dr^2}+\frac{\hbar^2}{2\mu}\frac{l(l+1)}{r^2}-\frac{Ze^2}{r}\right] u_{nl}=Eu_{nl},</math>
:<math>L^2\psi(r,\theta,\phi)=L^2[R(r)Y_{lm}(\theta,\phi)]=R(r)L^2Y_{lm}(\theta,\phi)=R(r)\hbar^2l(l+1)Y_{lm}(\theta,\phi)
=\hbar^2l(l+1)\psi(r,\theta,\phi)</math>
we have the equation for <math>u_l(r)\!</math>:
:<math>\left[ -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial r^2}+\frac{\hbar^2}{2m}\frac{l(l+1)}{r^2}+V(r)\right] u_l(r)=Eu_l(r)</math>


In the hydrogen atom or single electron ion, the potential is the Coulomb potential between the electron and the nucleus:
where <math>Z=1\!</math> represents the hydrogen atom, <math>Z=2\!</math> represents the helium ion, <math>He^+,\!</math> and so on.  Here, <math>\mu\!</math> is the effective mass of the atom.
:<math>V(r)=-\frac{Ze^2}{r}</math>
where <math>Z=1\!</math> for the hydrogen, <math>Z=2\!</math> for helium ion <math>He^+\!</math>, etc.


Plugging this potential into the Schrödinger equation we obtain
We will focus on the bound states here.  To solve the above equation, let us begin by writing down its solutions in the limits of small and large <math>r\!.</math>  
:<math>\left[ -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial r^2}+\frac{\hbar^2}{2m}\frac{l(l+1)}{r^2}-\frac{Ze^2}{r}\right] u_l(r)=Eu_l(r)</math>


Since we are only concentrating on the bound states, we can write down the solutions to this equation in the asymptotic limits.
In the limit of small <math>r;\!</math> i.e., <math>r \to 0,</math> the equation becomes


In the <math>r \to 0 </math> limit the equation becomes
<math>\left[ -\frac{\hbar^2}{2\mu}\frac{d^2}{dr^2}+\frac{\hbar^2}{2\mu}\frac{l(l+1)}{r^2}\right] u_{nl}=0.</math>
:<math>\left[ -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial r^2}+\frac{\hbar^2}{2m}\frac{l(l+1)}{r^2}\right] u_l(r)=0</math>


and the solution in this limit is <math>u_l(r)\rightarrow r^{l+1}</math>.
The only solution to this equation that is not divergent as <math>r\to 0</math> is <math>u_{nl}(r)\sim r^{l+1}.\!</math>


In the <math>r \to \infty </math> limit the equation becomes
In the opposite limit, <math>r \to \infty,</math> we obtain
:<math>\frac{\partial^2 u_l(r)}{\partial r^2}+\frac{2mE}{\hbar^2}u_l(r)=0</math>


and the solution in this limit is <math>u_l(r)\rightarrow e^{-\frac{r}{a}}</math>.
<math>\frac{d^2 u_{nl}}{dr^2}+\frac{2\mu E}{\hbar^2}u_{nl}=0.</math>


where  
The only solution to this equation that does not diverge as <math>r\to\infty</math> is <math>u_{nl}(r)\sim e^{-r/a},\!</math> where <math>a=\sqrt{\frac{-\hbar^2}{2\mu E}}.</math>
:<math>a=\sqrt{\frac{-\hbar^2}{2m E}}</math>


If we allow <math>\kappa=a^{-1}\!</math>, then the large limit of <math>r\!</math> can be expressed as
Let us now define <math>\kappa=a^{-1}.\!</math> The above asymptotic limit for large <math>r\!</math> can now be written as
:<math>u_l(r)\sim e^{-\kappa r}\!</math>


Using the limits of <math>u(r)\!</math>, the wavefunction can be expressed as the following
<math>u_{nl}(r)\sim e^{-\kappa r}.\!</math>
:<math>u_l(r)=(\kappa r)^{l+1}e^{-\kappa r}W(\kappa r)\!</math>


To simplify the equation, make a substitution <math>\rho=\kappa r\!</math>. The equation now turns into
We now assume that the full expression for <math>u_{nl}(r)\!</math> has the form, <math>u_{nl}(r)=(\kappa r)^{l+1}e^{-\kappa r}W(\kappa r).\!</math>
:<math>u_l(\rho)=\rho^{l+1}e^{-\rho}W(\rho)\!</math>


Plugging this equation into the Schrödinger equation and simplifying, it turns into
To simplify the equation, let us introduce the dimensionless radial coordinate, <math>\rho=\kappa r,\!</math> so that <math>u_{nl}(r)\!</math> may now be written as <math>u_{nl}(\rho)=\rho^{l+1}e^{-\rho}W(\rho).\!</math>
:<math>\frac{d^2W(\rho)}{d\rho^2}+2\left(\frac{l+1}{\rho}-1\right)\frac{dW(\rho)}{d\rho}+\left(\frac{\rho_0}{\rho}-\frac{2(l+1)}{\rho}\right)W(\rho)=0</math>        where <math>\rho _{0}=Ze^{2}\sqrt {\frac{2\mu }{\hbar ^{2}E}} </math>


Here <math>W(\rho)\!</math> can be expressed as an expansion of polynomials;
Substituting this into the effective [[Schrödinger Equation|Schrödinger equation]] and simplifying, it becomes
:<math>W(\rho)=a_0+a_1 \rho+a_2\rho^2+...=\sum_{k=0}^\infty a_k \rho^k</math>


The Schrödinger equation is then expressed as
<math>\frac{d^2W}{d\rho^2}+2\left(\frac{l+1}{\rho}-1\right)\frac{dW}{d\rho}+\left(\frac{\rho_0}{\rho}-\frac{2(l+1)}{\rho}\right)W=0,</math>
:<math>\sum_{k=0}^\infty (a_{k}k(k-1)\rho^{k-2}+2(l+1)k\rho^{k-2}a_k-2\rho^{k-1}a_k k)+\sum_{k=0}^\infty(\rho_0 a_k\rho^{k-1}-2(l+1)a_k\rho^{k-1})=0</math>


And simplified into
where <math>\rho_{0}=Ze^{2}\sqrt{\frac{2\mu }{\hbar^{2}E}}.</math>
:<math>\sum_{k=0}^\infty (a_{k+1}(k+1)k\rho^{k-1}+2(l+1)(k+1)\rho^{k-1}a_{k+1}-2\rho^{k-1}a_k k+(\rho_0-2(l+1))a_k\rho^{k-1})=0</math>


Bring all <math>\rho\!</math>'s to the same power
Let us now try a series solution for <math>W(\rho):\!</math>
:<math>k(k+1)a_{k+1}+2(l+1)(k+1)a_{k+1}-2ka_k+(\rho_0-2(l+1))a_k=0\!</math>
which can be expressed in the simplest fractional form as
:<math>\frac{a_{k+1}}{a_k}=\frac{2(k+l+1)-\rho_0}{(k+1)(k+2l+2)}</math>


where <math>\rho_0=2(N+l+1)\!</math> and <math>N=0,1,2...\!</math> and <math>l=0,1,2,...\!</math>
<math>W(\rho)=a_0+a_1 \rho+a_2\rho^2+...=\sum_{k=0}^\infty a_k \rho^k</math>


In the limit of large k
Our equation now becomes
:<math>\lim_{k\to\infty}\frac{a_{k+1}}{a_k}\rightarrow\frac{2}{k}</math>
:<math>a_{k+1}=\frac{2}{k}a_k\sim\frac{2^k}{k!}</math>
this will make
:<math>\psi(r) \sim e^{kr} \rightarrow\infty</math>
so we need to break, and make
:<math>a_{k+1}=0\!</math>
from this, we get the energy spectrum.
The fractional form can be expressed as a confluent hypergeometric function
:<math>_1F_1(a,c,z)=1+\frac{a}{c}\frac{z}{1!}+\frac{a(a+1)}{c(c+1)}\frac{z^2}{2!}+...=\sum_{k=0}^\infty a_k z^k</math>
It should be noted that a confluent hypergeometric function is a solution of Kummer's equation, which is:
:<math>z\frac{d^2w}{dz^2} + (b-z)\frac{dw}{dz} - aw = 0.\,\!</math>
Derriving the recursion relations we have:
:<math>\frac{a_k}{a_{k-1}}=\frac{\frac{a(a+1)(a+2)...(a+k-1)}{c(c+1)(c+2)...(c+k-1)k!}}{\frac{a(a+1)(a+2)...(a+k-2)}{c(c+1)(c+2)...(c+k-2)(k-1)!}}=\frac{a+k-1}{(c+k-1)k}</math>
:<math>\frac{a_{k+1}}{a_k}=\frac{a+k}{(c+k)(k+1)}</math>
by comparison, we find that
:<math>c=2(l+1)\!</math>
:<math>a=-N\!</math>
:<math>z=2\rho\!</math>


So the solution of <math>W(\rho)\!</math> is
<math>\sum_{k=0}^\infty (a_{k}k(k-1)\rho^{k-2}+2(l+1)k\rho^{k-2}a_k-2\rho^{k-1}a_k k)+\sum_{k=0}^\infty(\rho_0 a_k\rho^{k-1}-2(l+1)a_k\rho^{k-1})=0,</math>
:<math>W(\rho)=constant _1F_1(-N,2(l+1),2\rho)\!</math>
 
where :<math>\rho=\kappa r=\sqrt{\frac{-2\mu E}{\hbar^2}}r</math>
or, upon simplification,
Full wavefunction solution with normalization is
 
:<math>\psi_{n,l,m}(r,\theta,\phi)=\frac{e^{-\kappa r}(2\kappa r)^l}{(2l+1)!}\sqrt{\frac{(2\kappa)^3(n+l)!}{2n(n-l+1)}}\  _1F_1(-n+l+1,2(l+1),2\kappa r)Y_{lm}(\theta,\phi)</math>
<math>\sum_{k=0}^\infty (a_{k+1}(k+1)k\rho^{k-1}+2(l+1)(k+1)\rho^{k-1}a_{k+1}-2\rho^{k-1}a_k k+(\rho_0-2(l+1))a_k\rho^{k-1})=0.</math>
 
Setting the coefficients of all powers of <math>\rho\!</math> to zero, we obtain
 
<math>k(k+1)a_{k+1}+2(l+1)(k+1)a_{k+1}-2ka_k+(\rho_0-2(l+1))a_k=0,\!</math>
 
or
 
<math>\frac{a_{k+1}}{a_k}=\frac{2(k+l+1)-\rho_0}{(k+1)(k+2l+2)}.</math>
 
In the limit of large <math>k,\!</math> the recursion relation becomes
 
<math>\frac{a_{k+1}}{a_k}=\frac{2}{k},</math>
 
or
 
<math>a_{k+1}=\frac{2}{k}a_k=\frac{2^k}{k!}.</math>
 
These are the coefficients of the series for <math>e^{2\rho}.\!</math>  We therefore see that, unless we terminate the series at a finite order (i.e., we make <math>W(\rho)\!</math> a polynomial), the wave function will diverge as <math>r\to\infty.</math>  If we wish to terminate the series at <math>N^{\text{th}}\!</math> order, we set <math>a_{N+1}=0.\!</math>  This yields the condition,
 
<math>\rho_0=2(N+l+1).\!</math>
 
If we now solve this for the energy, we obtain the energy eigenvalues of the system,
 
<math>E_{Nl}=-\frac{Z^2\mu e^4}{2\hbar^2(N+l+1)^2}.</math>
 
The recursion relation now becomes
 
<math>\frac{a_{k+1}}{a_k}=\frac{2(k-N)}{(k+1)(k+2l+2)}.</math>
 
The function, <math>W(\rho),\!</math> can be expressed in terms of the [http://mathworld.wolfram.com/ConfluentHypergeometricFunctionoftheFirstKind.html confluent hypergeometric function],
 
<math>_1F_1(a,c,z)=1+\frac{a}{c}\frac{z}{1!}+\frac{a(a+1)}{c(c+1)}\frac{z^2}{2!}+...=\sum_{k=0}^\infty a_k z^k,</math>
 
which is a solution of [http://mathworld.wolfram.com/ConfluentHypergeometricDifferentialEquation.html Kummer's equation],
 
<math>z\frac{d^2w}{dz^2} + (b-z)\frac{dw}{dz} - aw = 0.\,\!</math>
 
If we assume a power series solution as before, then the recursion relations are
 
<math>\frac{a_{k+1}}{a_k}=\frac{a+k}{(c+k)(k+1)}.</math>
 
Comparing this form to the recursion relations for our solution for <math>W(\rho),\!</math> we see that
 
<math>c=2(l+1),\!</math>
 
<math>a=-N,\!</math>
 
and
 
<math>z=2\rho.\!</math>
 
Therefore, we may write <math>W(\rho)\!</math> as
 
<math>W(\rho)=C _1F_1(-N,2(l+1),2\rho),\!</math>
 
where <math>\rho=\kappa r=\sqrt{\frac{-2\mu E}{\hbar^2}}r.</math>
 
The full normalized wave function is given by
 
<math>\psi_{n,l,m}(r,\theta,\phi)=\frac{e^{-\kappa r}(2\kappa r)^l}{(2l+1)!}\sqrt{\frac{(2\kappa)^3(n+l)!}{2n(n-l+1)}}\  _1F_1(-n+l+1,2(l+1),2\kappa r)Y_{l}^m(\theta,\phi),</math>
 
where we have introduced the more commonly-used principal quantum number <math>n=N+l+1.\!</math>


[[Image:Energy levels of H atom.jpeg|thumb|450px]]
[[Image:Energy levels of H atom.jpeg|thumb|450px]]


The first couple of normalized wavefunctions for the hydrogen atom are as follows
The first few normalized wave functions for the hydrogen atom are as follows
 
<math>\psi_{100}= \dfrac{e^{-r/a}}{\sqrt{\pi a^3}}</math>


<math>\psi_{100}= \dfrac{e^{-r/a_o}}{\sqrt{\pi a_o^3}}</math>
<math>\psi_{200}= \dfrac{e^{-r/2a}}{\sqrt{32\pi a^3}} \left( 2-\dfrac{r}{a} \right)</math>


<math>\psi_{200}= \dfrac{e^{-r/2a_o}}{\sqrt{32\pi a_o^3}} \left( 2-\dfrac{r}{a_o} \right)</math>
<math>\psi_{210}= \dfrac{e^{-r/2a}}{\sqrt{32\pi a^3}} \left( \dfrac{r}{a} \right) \cos(\theta)</math>


<math>\psi_{210}= \dfrac{e^{-r/2a_o}}{\sqrt{32\pi a_o^3}} \left( \dfrac{r}{a_o} \right) cos(\theta)</math>
<math>\psi_{2 \pm 10}= \dfrac{e^{-r/2a}}{ \sqrt{64\pi a^3}} \left( \dfrac{r}{a} \right) \sin(\theta) e^{\pm i \phi}</math>


<math>\psi_{2 \pm 10}= \dfrac{e^{-r/2a_o}}{ \sqrt{64\pi a_o^3}} \left( \dfrac{r}{a_o} \right) sin(\theta) e^{\pm i \phi}</math>
The energy may be written as


The energy is then found to be
<math>E=-\frac{Z^2\mu e^4}{2\hbar^2 n^2}=-\frac{Z^2}{n^2}\,\text{Ry},</math>
:<math>E=-\frac{Z^2 e^4 \mu}{2\hbar^2 n^2}=-\frac{Z^2}{n^2}Ry</math>


where <math>Ry=13.6 eV\!</math> for the hydrogen atom and <math>n=1,2,3,...\!</math> and the degeneracy for each level is <math>n^2\!</math>.
where the Rydberg <math>\text{Ry}=13.6\,\text{eV}\!</math> for the hydrogen atom and <math>n=1,2,3,...\!</math> The degeneracy of each energy level is <math>n^2.\!</math>


To the side is a chart depicting the energy levels for hydrogen atom for <math>n=1, 2, 3\!</math> in units of <math>Ry</math>. (The parenthesis indicates the degeneracy due to possibile values of the magnetic quantum number <math>m</math> from <math>-l</math> to <math>+l</math>).
To the side is a chart that depicts the energy levels for the hydrogen atom graphically for <math>n=1,2,3\!</math> in units of <math>\text{Ry}\!</math>. The parenthesis indicates the degeneracy due to possibile values of the magnetic quantum number <math>m</math> from <math>-l</math> to <math>+l.</math>





Revision as of 23:23, 1 September 2013

Quantum Mechanics A
SchrodEq.png
Schrödinger Equation
The most fundamental equation of quantum mechanics; given a Hamiltonian Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathcal{H}} , it describes how a state Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\Psi\rangle} evolves in time.
Basic Concepts and Theory of Motion
UV Catastrophe (Black-Body Radiation)
Photoelectric Effect
Stability of Matter
Double Slit Experiment
Stern-Gerlach Experiment
The Principle of Complementarity
The Correspondence Principle
The Philosophy of Quantum Theory
Brief Derivation of Schrödinger Equation
Relation Between the Wave Function and Probability Density
Stationary States
Heisenberg Uncertainty Principle
Some Consequences of the Uncertainty Principle
Linear Vector Spaces and Operators
Commutation Relations and Simultaneous Eigenvalues
The Schrödinger Equation in Dirac Notation
Transformations of Operators and Symmetry
Time Evolution of Expectation Values and Ehrenfest's Theorem
One-Dimensional Bound States
Oscillation Theorem
The Dirac Delta Function Potential
Scattering States, Transmission and Reflection
Motion in a Periodic Potential
Summary of One-Dimensional Systems
Harmonic Oscillator Spectrum and Eigenstates
Analytical Method for Solving the Simple Harmonic Oscillator
Coherent States
Charged Particles in an Electromagnetic Field
WKB Approximation
The Heisenberg Picture: Equations of Motion for Operators
The Interaction Picture
The Virial Theorem
Commutation Relations
Angular Momentum as a Generator of Rotations in 3D
Spherical Coordinates
Eigenvalue Quantization
Orbital Angular Momentum Eigenfunctions
General Formalism
Free Particle in Spherical Coordinates
Spherical Well
Isotropic Harmonic Oscillator
Hydrogen Atom
WKB in Spherical Coordinates
Feynman Path Integrals
The Free-Particle Propagator
Propagator for the Harmonic Oscillator
Differential Cross Section and the Green's Function Formulation of Scattering
Central Potential Scattering and Phase Shifts
Coulomb Potential Scattering
Illustration of the hydrogen atom.

We now discuss the solution of the Schrödinger equation for the hydrogen atom. We are especially interested in this system because it is of direct physical interest and because it is possible to solve it exactly.

The effective one-dimensional problem for this system is

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left[ -\frac{\hbar^2}{2\mu}\frac{d^2}{dr^2}+\frac{\hbar^2}{2\mu}\frac{l(l+1)}{r^2}-\frac{Ze^2}{r}\right] u_{nl}=Eu_{nl},}

where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Z=1\!} represents the hydrogen atom, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Z=2\!} represents the helium ion, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle He^+,\!} and so on. Here, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mu\!} is the effective mass of the atom.

We will focus on the bound states here. To solve the above equation, let us begin by writing down its solutions in the limits of small and large Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r\!.}

In the limit of small Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r;\!} i.e., Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r \to 0,} the equation becomes

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left[ -\frac{\hbar^2}{2\mu}\frac{d^2}{dr^2}+\frac{\hbar^2}{2\mu}\frac{l(l+1)}{r^2}\right] u_{nl}=0.}

The only solution to this equation that is not divergent as Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r\to 0} is Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u_{nl}(r)\sim r^{l+1}.\!}

In the opposite limit, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r \to \infty,} we obtain

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{d^2 u_{nl}}{dr^2}+\frac{2\mu E}{\hbar^2}u_{nl}=0.}

The only solution to this equation that does not diverge as Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r\to\infty} is Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u_{nl}(r)\sim e^{-r/a},\!} where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a=\sqrt{\frac{-\hbar^2}{2\mu E}}.}

Let us now define Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \kappa=a^{-1}.\!} The above asymptotic limit for large Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r\!} can now be written as

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u_{nl}(r)\sim e^{-\kappa r}.\!}

We now assume that the full expression for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u_{nl}(r)\!} has the form, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u_{nl}(r)=(\kappa r)^{l+1}e^{-\kappa r}W(\kappa r).\!}

To simplify the equation, let us introduce the dimensionless radial coordinate, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \rho=\kappa r,\!} so that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u_{nl}(r)\!} may now be written as Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u_{nl}(\rho)=\rho^{l+1}e^{-\rho}W(\rho).\!}

Substituting this into the effective Schrödinger equation and simplifying, it becomes

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{d^2W}{d\rho^2}+2\left(\frac{l+1}{\rho}-1\right)\frac{dW}{d\rho}+\left(\frac{\rho_0}{\rho}-\frac{2(l+1)}{\rho}\right)W=0,}

where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \rho_{0}=Ze^{2}\sqrt{\frac{2\mu }{\hbar^{2}E}}.}

Let us now try a series solution for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle W(\rho):\!}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle W(\rho)=a_0+a_1 \rho+a_2\rho^2+...=\sum_{k=0}^\infty a_k \rho^k}

Our equation now becomes

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{k=0}^\infty (a_{k}k(k-1)\rho^{k-2}+2(l+1)k\rho^{k-2}a_k-2\rho^{k-1}a_k k)+\sum_{k=0}^\infty(\rho_0 a_k\rho^{k-1}-2(l+1)a_k\rho^{k-1})=0,}

or, upon simplification,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{k=0}^\infty (a_{k+1}(k+1)k\rho^{k-1}+2(l+1)(k+1)\rho^{k-1}a_{k+1}-2\rho^{k-1}a_k k+(\rho_0-2(l+1))a_k\rho^{k-1})=0.}

Setting the coefficients of all powers of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \rho\!} to zero, we obtain

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k(k+1)a_{k+1}+2(l+1)(k+1)a_{k+1}-2ka_k+(\rho_0-2(l+1))a_k=0,\!}

or

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{a_{k+1}}{a_k}=\frac{2(k+l+1)-\rho_0}{(k+1)(k+2l+2)}.}

In the limit of large Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k,\!} the recursion relation becomes

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{a_{k+1}}{a_k}=\frac{2}{k},}

or

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a_{k+1}=\frac{2}{k}a_k=\frac{2^k}{k!}.}

These are the coefficients of the series for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle e^{2\rho}.\!} We therefore see that, unless we terminate the series at a finite order (i.e., we make Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle W(\rho)\!} a polynomial), the wave function will diverge as Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r\to\infty.} If we wish to terminate the series at Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle N^{\text{th}}\!} order, we set Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a_{N+1}=0.\!} This yields the condition,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \rho_0=2(N+l+1).\!}

If we now solve this for the energy, we obtain the energy eigenvalues of the system,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E_{Nl}=-\frac{Z^2\mu e^4}{2\hbar^2(N+l+1)^2}.}

The recursion relation now becomes

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{a_{k+1}}{a_k}=\frac{2(k-N)}{(k+1)(k+2l+2)}.}

The function, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle W(\rho),\!} can be expressed in terms of the confluent hypergeometric function,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle _1F_1(a,c,z)=1+\frac{a}{c}\frac{z}{1!}+\frac{a(a+1)}{c(c+1)}\frac{z^2}{2!}+...=\sum_{k=0}^\infty a_k z^k,}

which is a solution of Kummer's equation,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle z\frac{d^2w}{dz^2} + (b-z)\frac{dw}{dz} - aw = 0.\,\!}

If we assume a power series solution as before, then the recursion relations are

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{a_{k+1}}{a_k}=\frac{a+k}{(c+k)(k+1)}.}

Comparing this form to the recursion relations for our solution for we see that

and

Therefore, we may write as

where

The full normalized wave function is given by

where we have introduced the more commonly-used principal quantum number

Energy levels of H atom.jpeg

The first few normalized wave functions for the hydrogen atom are as follows

The energy may be written as

where the Rydberg for the hydrogen atom and The degeneracy of each energy level is

To the side is a chart that depicts the energy levels for the hydrogen atom graphically for in units of . The parenthesis indicates the degeneracy due to possibile values of the magnetic quantum number from to


Problems

Problem 1
Problem 2