Particle Moving in a Spherically Symmetric Potential
Many quantum systems such as the hydrogen atom possess spherical symmetry. In such systems, the potential energy depends only on the radial distance from the origin.
\[ V = V(r) \]Since the potential is independent of the angular coordinates \(\theta\) and \(\phi\), the Schrödinger equation is conveniently solved in spherical coordinates.
Time-Independent Schrödinger Equation
The three-dimensional time-independent Schrödinger equation is
$$ \boxed{ \nabla^2\psi+\frac{2m}{\hbar^2}(E-V)\psi=0 } $$where
- \(\psi\) = wave function
- \(m\) = mass of the particle
- \(E\) = total energy
- \(V\) = potential energy
- \(\hbar\) = reduced Planck constant
Schrödinger Equation in Spherical Coordinates
The Laplacian operator in spherical coordinates is
$$ \boxed{ \begin{aligned} \nabla^2 &= \frac{1}{r^2}\frac{\partial}{\partial r} \left(r^2\frac{\partial}{\partial r}\right) + \frac{1}{r^2\sin\theta} \frac{\partial}{\partial\theta} \left(\sin\theta\frac{\partial}{\partial\theta}\right) \\[6pt] &\quad+ \frac{1}{r^2\sin^2\theta} \frac{\partial^2}{\partial\phi^2} \end{aligned} } $$Substituting into the Schrödinger equation gives
$$ \begin{aligned} &\frac{1}{r^2}\frac{\partial}{\partial r} \left(r^2\frac{\partial\psi}{\partial r}\right) + \frac{1}{r^2\sin\theta} \frac{\partial}{\partial\theta} \left(\sin\theta\frac{\partial\psi}{\partial\theta}\right) \\[6pt] &+ \frac{1}{r^2\sin^2\theta} \frac{\partial^2\psi}{\partial\phi^2} + \frac{2m}{\hbar^2}(E-V)\psi = 0 \end{aligned} $$Separation of Variables
Assume that the wave function can be written as the product of three functions:
\[ \psi(r,\theta,\phi) = R(r)\Theta(\theta)\Phi(\phi) \]where
- \(R(r)\) = radial function
- \(\Theta(\theta)\) = polar function
- \(\Phi(\phi)\) = azimuthal function
Assume a Product Solution
To separate the variables, assume that the wave function can be expressed as the product of three independent functions:
\[ \psi(r,\theta,\phi) = R(r)\Theta(\theta)\Phi(\phi) \]where
- R(r) depends only on the radial coordinate r.
- Θ(θ) depends only on the polar angle θ.
- Φ(φ) depends only on the azimuthal angle φ.
Find the Partial Derivatives
Derivative with Respect to r
\[ \frac{\partial\psi}{\partial r} = \Theta\Phi\frac{dR}{dr} \] Therefore, \[ \frac{1}{r^2} \frac{\partial}{\partial r} \left( r^2\frac{\partial\psi}{\partial r} \right) = \frac{\Theta\Phi}{r^2} \frac{d}{dr} \left( r^2\frac{dR}{dr} \right) \]Derivative with Respect to θ
\[ \frac{\partial\psi}{\partial\theta} = R\Phi\frac{d\Theta}{d\theta} \] Hence, \[ \frac{1}{r^2\sin\theta} \frac{\partial}{\partial\theta} \left( \sin\theta \frac{\partial\psi}{\partial\theta} \right) = \frac{R\Phi}{r^2\sin\theta} \frac{d}{d\theta} \left( \sin\theta \frac{d\Theta}{d\theta} \right) \]Derivative with Respect to φ
\[ \frac{\partial^2\psi}{\partial\phi^2} = R\Theta \frac{d^2\Phi}{d\phi^2} \] Hence, \[ \frac{1}{r^2\sin^2\theta} \frac{\partial^2\psi}{\partial\phi^2} = \frac{R\Theta}{r^2\sin^2\theta} \frac{d^2\Phi}{d\phi^2} \]Substitute into Schrödinger Equation
Substituting all the derivatives into the Schrödinger equation gives
\[ \frac{\Theta\Phi}{r^2} \frac{d}{dr} \left( r^2\frac{dR}{dr} \right) + \frac{R\Phi}{r^2\sin\theta} \frac{d}{d\theta} \left( \sin\theta \frac{d\Theta}{d\theta} \right) \] \[ + \frac{R\Theta}{r^2\sin^2\theta} \frac{d^2\Phi}{d\phi^2} + \frac{2m}{\hbar^2}(E-V) R\Theta\Phi = 0 \]Divide by RΘΦ
Divide the entire equation by RΘΦ:
\[ \frac{1}{Rr^2} \frac{d}{dr} \left( r^2\frac{dR}{dr} \right) + \frac{1}{r^2\Theta\sin\theta} \frac{d}{d\theta} \left( \sin\theta \frac{d\Theta}{d\theta} \right) \] \[ + \frac{1}{r^2\sin^2\theta\Phi} \frac{d^2\Phi}{d\phi^2} + \frac{2m}{\hbar^2}(E-V) = 0 \]Now each term contains only one unknown function.
Multiply by r²sin²θ
Multiply the entire equation by
\[ r^2\sin^2\theta \]to obtain
\[ \frac{\sin^2\theta}{R} \frac{d}{dr} \left( r^2\frac{dR}{dr} \right) + \frac{\sin\theta}{\Theta} \frac{d}{d\theta} \left( \sin\theta \frac{d\Theta}{d\theta} \right) \] \[ + \frac{2m}{\hbar^2} (E-V)r^2\sin^2\theta = -\frac{1}{\Phi} \frac{d^2\Phi}{d\phi^2} \]Separate the φ Variable
The left-hand side depends only on r and θ, while the right-hand side depends only on φ.
Since the variables are independent, both sides must be equal to the same constant. Let that constant be m².
\[ -\frac{1}{\Phi} \frac{d^2\Phi}{d\phi^2} = m^2 \] or \[ \frac{d^2\Phi}{d\phi^2} = -m^2\Phi \]This is the azimuthal equation.
Substitute Back
Removing the φ-dependent part gives
\[ \frac{1}{R} \frac{d}{dr} \left( r^2\frac{dR}{dr} \right) + \frac{2m}{\hbar^2} (E-V)r^2 \] \[ = -\frac{1}{\Theta\sin\theta} \frac{d}{d\theta} \left( \sin\theta \frac{d\Theta}{d\theta} \right) + \frac{m^2}{\sin^2\theta} \]Separate r and θ Variables
The left-hand side depends only on r, whereas the right-hand side depends only on θ.
Therefore, both sides must equal another constant, denoted by λ.
\[ \frac{1}{R} \frac{d}{dr} \left( r^2\frac{dR}{dr} \right) + \frac{2m}{\hbar^2} (E-V)r^2 = \lambda \] and \[ -\frac{1}{\Theta\sin\theta} \frac{d}{d\theta} \left( \sin\theta \frac{d\Theta}{d\theta} \right) + \frac{m^2}{\sin^2\theta} = \lambda \]Thus, the original Schrödinger equation has been separated into independent radial and angular equations.
Substituting into the Schrödinger equation and dividing by \(R\Theta\Phi\), we obtain
$$ \begin{aligned} \frac{\sin^2\theta}{R} \frac{d}{dr} \left(r^2\frac{dR}{dr}\right) + \frac{\sin\theta}{\Theta} \frac{d}{d\theta} \left(\sin\theta\frac{d\Theta}{d\theta}\right) \\[6pt] + \frac{2m}{\hbar^2}[E-V(r)]r^2\sin^2\theta = -\frac{1}{\Phi} \frac{d^2\Phi}{d\phi^2} \end{aligned} $$Solution of the \(\phi\)-Equation
The solution is
\[ \Phi(\phi) = Ae^{im\phi} \]where \(A\) is a constant.
For a physically acceptable wave function,
\[ \Phi(\phi)=\Phi(\phi+2\pi) \]Substituting the solution,
\[ Ae^{im\phi} = Ae^{im(\phi+2\pi)} \] \[ 1=e^{i2\pi m} \]This condition is satisfied only when
\[ m=0,\pm1,\pm2,\pm3,\ldots \]The integer \(m\) is called the magnetic quantum number.
Normalization of \(\Phi(\phi)\)
The normalization condition is
\[ \int_0^{2\pi}\Phi^*\Phi\,d\phi=1 \]Substituting
\[ \Phi=Ae^{im\phi} \]gives
\[ |A|^2\int_0^{2\pi}d\phi=1 \] \[ |A|^2(2\pi)=1 \]Hence,
\[ A=\frac{1}{\sqrt{2\pi}} \]The normalized azimuthal wave function becomes
\[ \Phi(\phi) = \frac{1}{\sqrt{2\pi}} e^{im\phi} \]Separation of Radial and Polar Parts
After separating the \(\phi\)-part, we obtain
\[ \frac{1}{R} \frac{d}{dr} \left(r^2\frac{dR}{dr}\right) + \frac{2m}{\hbar^2}(E-V)r^2 = -\frac{1}{\Theta\sin\theta} \frac{d}{d\theta} \left(\sin\theta\frac{d\Theta}{d\theta}\right) + \frac{m^2}{\sin^2\theta} \]The left side depends only on \(r\), while the right side depends only on \(\theta\). Therefore, both sides must equal a constant \(\lambda\).
Polar Equation
\[ \frac{1}{\sin\theta} \frac{d}{d\theta} \left( \sin\theta \frac{d\Theta}{d\theta} \right) + \left( \lambda-\frac{m^2}{\sin^2\theta} \right)\Theta = 0 \]This equation leads to the associated Legendre functions.
Radial Equation
\[ \frac{1}{r^2} \frac{d}{dr} \left( r^2\frac{dR}{dr} \right) + \frac{2m}{\hbar^2} [E-V(r)]R -\frac{\lambda}{r^2}R = 0 \]Final Result
The three-dimensional Schrödinger equation is separated into three one-dimensional equations.
Azimuthal Equation
\[ \frac{d^2\Phi}{d\phi^2} = -m^2\Phi \]Polar Equation
\[ \frac{1}{\sin\theta} \frac{d}{d\theta} \left( \sin\theta\frac{d\Theta}{d\theta} \right) + \left( \lambda-\frac{m^2}{\sin^2\theta} \right)\Theta = 0 \]Radial Equation
\[ \frac{1}{r^2} \frac{d}{dr} \left( r^2\frac{dR}{dr} \right) + \frac{2m}{\hbar^2} [E-V(r)]R -\frac{\lambda}{r^2}R = 0 \]Thus, the complicated three-dimensional Schrödinger equation is reduced to three simpler equations, which form the basis for the quantum mechanical treatment of the hydrogen atom and the introduction of quantum numbers.
Hydrogen Atom
The hydrogen atom is the simplest atom in nature and plays a fundamental role in quantum mechanics. It consists of a positively charged nucleus (proton) and a negatively charged electron. Since the hydrogen atom contains only one electron, the Schrödinger equation can be solved exactly.
1. Coulomb Potential Energy
The electron and nucleus attract each other through the Coulomb force. The potential energy of interaction is:
$$ V(r)=-\frac{kZe^2}{r} $$where
- \(k\) = Coulomb constant
- \(Z\) = Atomic number
- \(e\) = Electronic charge
- \(r\) = Distance between electron and nucleus
The negative sign indicates an attractive interaction.
2. Schrödinger Equation for Hydrogen Atom
The time-independent Schrödinger equation is
$$ \left[ -\frac{\hbar^2}{2\mu}\nabla^2 -\frac{kZe^2}{r} \right]\psi(\mathbf r) = E\psi(\mathbf r) $$where
- \(\hbar = \frac{h}{2\pi}\)
- \(\mu\) = Reduced mass
- \(\nabla^2\) = Laplacian operator
- \(E\) = Energy of the electron
- \(\psi\) = Wave function
This equation contains three spatial variables and is difficult to solve directly.
3. Why Use Spherical Coordinates?
Notice that the potential energy depends only on the distance \(r\):
$$ V(r)=-\frac{kZe^2}{r} $$Since the potential is spherically symmetric, spherical polar coordinates are the natural choice.
The coordinates are:
$$ (r,\theta,\phi) $$- \(r\) = Radial distance
- \(\theta\) = Polar angle
- \(\phi\) = Azimuthal angle
4. Separation of Variables
To simplify the equation, assume that the wave function can be written as a product of three functions:
$$ \psi(r,\theta,\phi) = R(r)\Theta(\theta)\Phi(\phi) $$This method is called separation of variables.
Each function depends on only one coordinate:
- \(R(r)\) → Radial part
- \(\Theta(\theta)\) → Polar-angle part
- \(\Phi(\phi)\) → Azimuthal-angle part
Substituting this expression into the Schrödinger equation separates the problem into angular and radial equations.
5. Radial Equation
After separating variables, the radial equation becomes
$$ \frac{1}{r^2} \frac{d}{dr} \left( r^2\frac{dR}{dr} \right) + \frac{2\mu}{\hbar^2} \left[ E - \frac{l(l+1)\hbar^2}{2\mu r^2} + \frac{kZe^2}{r} \right]R = 0 $$Meaning of Each Term
First term
$$ \frac{1}{r^2} \frac{d}{dr} \left( r^2\frac{dR}{dr} \right) $$Represents the radial variation of the wave function.
Second term
$$ E $$Represents the total energy of the electron.
Third term
$$ \frac{l(l+1)\hbar^2}{2\mu r^2} $$Represents the angular momentum contribution. This behaves like a centrifugal barrier.
Fourth term
$$ \frac{kZe^2}{r} $$Represents the Coulomb attraction between electron and nucleus.
6. Angular Solution
The angular part of the solution is given by the spherical harmonics:
$$ Y_{lm}(\theta,\phi) $$These determine the shape of atomic orbitals.
| Orbital | l Value |
|---|---|
| s | 0 |
| p | 1 |
| d | 2 |
| f | 3 |
7. Introducing a New Variable
The radial equation is still complicated. To simplify it, introduce a dimensionless variable \(\rho\):
$$ \rho = \sqrt{ \frac{-8\mu E}{\hbar^2} } \,r $$This converts the radial distance into a dimensionless quantity.
8. Introducing the Constant λ
Define
$$ \lambda = \frac{kZe^2}{\hbar} \sqrt{ \frac{\mu}{-2E} } $$This combines several physical constants into a single parameter.
9. Why Must E Be Negative?
For a bound electron:
$$ E<0 p=""> Negative energy means the electron is trapped inside the atom. Energy must be supplied to remove it completely.Because \(E\) is negative,
$$ \sqrt{\frac{-8\mu E}{\hbar^2}} $$is real, making both \(\rho\) and \(\lambda\) real quantities.
10. Simplified Radial Equation
Substituting \(\rho\) and \(\lambda\) into the radial equation gives
$$ \frac{d^2R}{d\rho^2} + \frac{2}{\rho} \frac{dR}{d\rho} + \left[ -\frac14 - \frac{l(l+1)}{\rho^2} + \frac{\lambda}{\rho} \right]R = 0 $$This equation is much easier to solve.
11. Physical Meaning of the Terms
Energy Term
$$ -\frac14 $$Originates from the total energy of the electron.
Angular Momentum Term
$$ -\frac{l(l+1)}{\rho^2} $$Represents the centrifugal effect due to angular motion.
Coulomb Attraction Term
$$ \frac{\lambda}{\rho} $$Represents the attractive force between nucleus and electron.
12. Final Result
Solving the radial equation and applying boundary conditions yields the allowed energy levels:
$$ E_n = -\frac{\mu k^2 Z^2 e^4} {2\hbar^2 n^2} $$For hydrogen (\(Z=1\)):
$$ E_n = -\frac{13.6}{n^2}\; \text{eV} $$Examples:
$$ E_1=-13.6\;\text{eV} $$ $$ E_2=-3.4\;\text{eV} $$ $$ E_3=-1.51\;\text{eV} $$ $$ E_4=-0.85\;\text{eV} $$Solution of the Hydrogen Atom Radial Equation
After separating the Schrödinger equation in spherical coordinates, the radial part can be written in dimensionless form as
$$ \rho^2 \frac{d^2R}{d\rho^2} + 2\rho \frac{dR}{d\rho} + \left[ \lambda \rho -\frac{\rho^2}{4} -l(l+1) \right]R=0 \tag{1} $$where
$$ \rho=\alpha r, $$ $$ \lambda=\frac{Ze^2}{4\pi\varepsilon_0} \sqrt{\frac{\mu}{-2E\hbar^2}}, $$and \(l\) is the orbital angular momentum quantum number.
Examine the Behaviour for Large \(\rho\)
To understand the nature of the solution, consider the limit
$$ \rho \rightarrow \infty. $$For very large \(\rho\), the dominant term inside the bracket is
$$ -\frac{\rho^2}{4}. $$Therefore Equation (1) becomes approximately
$$ \rho^2 \frac{d^2R}{d\rho^2} -\frac{\rho^2}{4}R=0. $$Dividing by \(\rho^2\),
$$ \frac{d^2R}{d\rho^2} -\frac14 R=0. \tag{2} $$Solve the Asymptotic Equation
Assume a trial solution
$$ R=e^{m\rho}. $$ Then $$ \frac{d^2R}{d\rho^2}=m^2 e^{m\rho}. $$ Substituting into Equation (2), $$ m^2e^{m\rho} -\frac14 e^{m\rho}=0. $$ Cancelling \(e^{m\rho}\), $$ m^2-\frac14=0. $$ Hence $$ m=\pm\frac12. $$ Therefore, $$ R=e^{-\rho/2} \qquad \text{or}\qquad R=e^{+\rho/2}. $$Since
$$ e^{+\rho/2}\rightarrow\infty \quad \text{as}\quad \rho\rightarrow\infty, $$it is physically unacceptable because the wave function would diverge. Therefore the acceptable behavior is
$$ R(\rho)\propto e^{-\rho/2}. $$Introduce a New Function \(F(\rho)\)
To extract the exponential decay explicitly, write
$$ R(\rho)=e^{-\rho/2}F(\rho). \tag{3} $$The unknown function \(F(\rho)\) will now contain the remaining part of the solution.
Compute the Derivatives
Using the product rule, $$ \frac{dR}{d\rho} = e^{-\rho/2} \left( \frac{dF}{d\rho} -\frac12F \right). $$ Differentiating again, $$ \frac{d^2R}{d\rho^2} = e^{-\rho/2} \left( \frac{d^2F}{d\rho^2} -\frac{dF}{d\rho} +\frac14F \right). $$Substitute into the Radial Equation
Substituting these derivatives into Equation (1), $$ \rho^2 \left( F''-F'+\frac14F \right) + 2\rho \left( F'-\frac12F \right) + \left( \lambda\rho -\frac{\rho^2}{4} -l(l+1) \right) F = 0. $$ Expanding, $$ \rho^2F'' -\rho^2F' +\frac{\rho^2}{4}F + 2\rho F' -\rho F + \lambda\rho F -\frac{\rho^2}{4}F -l(l+1)F = 0. $$ Notice that $$ +\frac{\rho^2}{4}F -\frac{\rho^2}{4}F=0. $$ Therefore, $$ \rho^2F'' + \rho(2-\rho)F' + \left[ \lambda\rho -\rho -l(l+1) \right]F = 0. \tag{4} $$ This is the differential equation satisfied by \(F(\rho)\).Behaviour Near \(\rho=0\)
Set $$ \rho=0 $$ in Equation (4). The terms containing \(\rho\) vanish, leaving $$ -l(l+1)F(0)=0. $$ Therefore $$ l(l+1)F(0)=0. $$ For $$ l\neq0, $$ we obtain $$ F(0)=0. \tag{5} $$Thus \(F(\rho)\) cannot contain a constant term.
Assume a Power Series Solution
Let $$ F(\rho) = \sum_{k=0}^{\infty} a_k \rho^{c+k}. \tag{6} $$ where \(c\) is to be determined. Differentiating, $$ F' = \sum_{k=0}^{\infty} a_k(c+k)\rho^{c+k-1}, $$ and $$ F'' = \sum_{k=0}^{\infty} a_k(c+k)(c+k-1)\rho^{c+k-2}. $$Substitute the Series into Equation (4)
Substituting \(F\), \(F'\), and \(F''\) into Equation (4) and collecting powers of \(\rho\), $$ \begin{aligned} & \sum_k a_k (\lambda-1-c-k) \rho^{c+k+1} \\[6pt] &+ \sum_k a_k \left[ c^2+2ck+k^2+c+k-l^2-l \right] \rho^{c+k} = 0 \end{aligned} \tag{7} $$ Since this equation must hold for every value of \(\rho\), the coefficient of each power must vanish separately.Determine the Value of \(c\)
The coefficient of the lowest power \(\rho^c\) is $$ a_0 \left( c^2+c-l^2-l \right)=0. $$ Since $$ a_0\neq0, $$ we obtain $$ c^2+c-l^2-l=0. $$ Factorizing, $$ (c-l)(c+l+1)=0. $$ Hence $$ c=l \qquad \text{or}\qquad c=-(l+1). \tag{8} $$Choose the Physical Solution
If $$ c=-(l+1), $$ the first term becomes $$ a_0\rho^{-(l+1)}. $$ As $$ \rho\rightarrow0, $$ this tends to infinity. Therefore this solution is not physically acceptable. Hence, $$ c=l. \tag{9} $$Obtain the Recurrence Relation
Setting the coefficient of $$ \rho^{l+k+1} $$ equal to zero gives $$ a_{k+1} = \frac{l+k+1-\lambda} {(k+1)(k+2l+2)} a_k. \tag{10} $$ This is the recurrence relation for the series coefficients.Final Result
The radial solution is $$ R(\rho) = e^{-\rho/2} \sum_{k=0}^{\infty} a_k \rho^{l+k}. $$ The series terminates only when $$ \lambda=n, $$ where $$ n=1,2,3,\dots $$ leading to the quantized energy levels of the hydrogen atom: $$\boxed{ E_n = -\frac{\mu Z^2 e^4} {2(4\pi\varepsilon_0)^2\hbar^2 n^2}} $$ This is the origin of the hydrogen atom energy spectrum obtained from the radial Schrödinger equation.Summary
- Start with the Coulomb potential \(V(r)=-kZe^2/r\).
- Substitute it into Schrödinger's equation.
- Convert to spherical coordinates.
- Separate variables using \(\psi=R\Theta\Phi\).
- Obtain the radial equation.
- Solve the angular equation using spherical harmonics \(Y_{lm}\).
- Introduce \(\rho\) and \(\lambda\) to simplify the radial equation.
- Solve the resulting differential equation.
- Apply boundary conditions.
- Obtain quantized energy levels of the hydrogen atom.