Quantum Mechanics in 3D Notes

Quantum Mechanics in Three Dimensions

Hamiltonian and Energy Eigenstates

• The Hamiltonian operator is crucial for understanding the time development of a quantum system.
• Energy eigenstates allow us to express any state as a linear combination, enabling us to figure out how the wave function evolves in time.
• When measurements are made, the wave function collapses into an eigenfunction of the corresponding operator. These eigenstates can be expressed in terms of energy eigenstates.
• The ability to express any function, including eigenstates of operators like momentum or angular momentum, in terms of energy eigenfunctions is vital for determining time evolution.

Hamiltonian in Three Dimensions

• The Hamiltonian in three dimensions is given by:
  $H = -\frac{{\hbar^2}}{{2m}} \left( \frac{{\partial^2}}{{\partial x^2}} + \frac{{\partial^2}}{{\partial y^2}} + \frac{{\partial^2}}{{\partial z^2}} \right) + V(x, y, z)$
  Where V can be in terms of Cartesian coordinates $(x, y, z)$ or spherical coordinates $(r, \theta, \phi)$.

• The eigenvalue problem for the Hamiltonian in 3D is: $H \Psi<em>E = E \Psi</em>E$, where $\Psi_E$ is the energy eigenfunction.

• The solutions are no longer functions of just $x$, but of $(x, y, z)$.

Free Particle Case (V=0)

• For a free particle, the potential $V = 0$.
• We use separation of variables to solve the differential equation.

Separation of Variables

• Assume the energy eigenfunction can be written as a product of three functions, each dependent on one coordinate:
  $\Psi(x, y, z) = X(x) Y(y) Z(z)$
• Substitute this solution into the Schrodinger equation:
  $-\frac{{\hbar^2}}{{2m}} \left( \frac{{\partial^2}}{{\partial x^2}} + \frac{{\partial^2}}{{\partial y^2}} + \frac{{\partial^2}}{{\partial z^2}} \right) X(x)Y(y)Z(z) = E X(x)Y(y)Z(z)$
• Divide all terms by $X(x)Y(y)Z(z)$.
• The equation becomes:
  $\frac{{1}}{{X(x)}} \frac{{\partial^2 X(x)}}{{\partial x^2}} + \frac{{1}}{{Y(y)}} \frac{{\partial^2 Y(y)}}{{\partial y^2}} + \frac{{1}}{{Z(z)}} \frac{{\partial^2 Z(z)}}{{\partial z^2}} = -\frac{{2mE}}{{\hbar^2}}$
• Each term is a function of only one coordinate (x, y, or z), and the sum is equal to a constant. Therefore, each term must be equal to a constant separately.
  • $f(x) = \frac{1}{X(x)} \frac{\partial^2 X(x)}{\partial x^2} = C_1$
  • $g(y) = \frac{1}{Y(y)} \frac{\partial^2 Y(y)}{\partial y^2} = C_2$
  • $h(z) = \frac{1}{Z(z)} \frac{\partial^2 Z(z)}{\partial z^2} = C_3$
  • $-\frac{{2mE}}{{\hbar^2}} = C<em>1 + C</em>2 + C_3$

Solving for X(x)

• Consider the x term: $\frac{{\partial^2 X(x)}}{{\partial x^2}} = C_1 X(x)$
• The solution is of the form: $X(x) = e^{i \sqrt{C_1} x}$
• $C<em>1$ must be negative for the solution to work: C1 < 0
• The solution can be $e^{+i \sqrt{C<em>1} x}$ or $e^{-i \sqrt{C</em>1} x}$.

General Solution

• The general solution is a combination of complex exponentials:
  $X(x) = b<em>1 e^{i \sqrt{C</em>1} x} + b<em>2 e^{-i \sqrt{C</em>1} x}$
• Similarly for y and z:
  $Y(y) = b<em>3 e^{i \sqrt{C</em>2} y} + b<em>4 e^{-i \sqrt{C</em>2} y}$
  $Z(z) = b<em>5 e^{i \sqrt{C</em>3} z} + b<em>6 e^{-i \sqrt{C</em>3} z}$
• The overall solution is the product of these:
  $\Psi(x, y, z) = X(x) Y(y) Z(z)$

Energy Components

• We can write the total energy $E$ as a sum of energies associated with each dimension:
  $E = E<em>1 + E</em>2 + E_3$
• Where each energy component is:
  • $C<em>1 = -\frac{{2mE</em>1}}{{\hbar^2}}$

Simplified Solution

• To simplify, consider only solutions moving in one direction (positive k).
• Set $b<em>2 = b</em>4 = b_6 = 0$.
• Then $X(x) = b<em>1 e^{i k</em>x x}$, where $k<em>x = \sqrt{\frac{{2mE</em>1}}{{\hbar^2}}}$.
• Similarly, define $k<em>y = \sqrt{\frac{{2mE</em>2}}{{\hbar^2}}}$ and $k<em>z = \sqrt{\frac{{2mE</em>3}}{{\hbar^2}}}$.

Momentum Vector

• In k-space, consider a sphere with radius k. We focus on a free particle with momentum in a specific direction.
• The k-vector is: $\vec{k} = k<em>x \hat{x} + k</em>y \hat{y} + k_z \hat{z}$
• There are eight possible solutions with the same energy, but we ignore the negative solutions for simplicity.

Full Solution in Terms of Energies

• The full solution is:
  $\Psi(x, y, z) = Ce^{i \sqrt{\frac{{2mE<em>1}}{{\hbar^2}}} x} e^{i \sqrt{\frac{{2mE</em>2}}{{\hbar^2}}} y} e^{i \sqrt{\frac{{2mE_3}}{{\hbar^2}}} z}$
• The total energy $E = E<em>1 + E</em>2 + E_3$.

Energy in Terms of Wave Numbers

• We can express each energy component in terms of wave numbers:
  $E<em>1 = \frac{{\hbar^2 k</em>x^2}}{{2m}}$, $E<em>2 = \frac{{\hbar^2 k</em>y^2}}{{2m}}$, $E<em>3 = \frac{{\hbar^2 k</em>z^2}}{{2m}}$
• The total energy is: $E = \frac{{\hbar^2}}{{2m}} (k<em>x^2 + k</em>y^2 + k_z^2)$
• This can be written as: $E = \frac{{\hbar^2 k^2}}{{2m}}$, where $k = |\vec{k}| = \sqrt{k<em>x^2 + k</em>y^2 + k_z^2}$.

Compact Notation

• The solution can be written compactly as:
  $\Psi(x, y, z) = C e^{i \vec{k} \cdot \vec{r}}$
• Where $\vec{k} \cdot \vec{r} = k<em>x x + k</em>y y + k_z z$ and $\vec{r}$ is the position vector with components $(x, y, z)$.

Plane Waves

• The solution represents a plane wave which means wave is a series of sheets of the same phase.
• An electromagnetic wave can be visualized along the x-axis, with the electric field component $E_y(x, t)$.
• A plane wave consists of sheets where all po ints have the same phase and the electric field is constant.
• The solution fills all of space.

Plane Wave in k Direction

• The solution represents a plane wave propagating in the direction of the k-vector.
• Example: If $\vec{k} = 3 \hat{x} - 2 \hat{y}$, the wave propagates in the xy-plane where x is positive and y is negative.
• The vector plays the same role as $k_x$ in one dimension.
• We want to think of sheets perpendicular to this direction.

Wavelength

• The wavelength $\lambda$ is the spacing between planes with the same phase:
  $\lambda = \frac{{2\pi}}{{k}} = \frac{{2\pi}}{{\sqrt{k<em>x^2 + k</em>y^2 + k_z^2}}}$

Field Quantity

• The solution is a field quantity with values at every coordinate in space and time:
  $\Psi(x, y, z, t) = C e^{i k<em>x x} e^{i k</em>y y} e^{i k_z z} e^{-\frac{{iE}}{{\hbar}} t}$
• Given $x$, $y$, and $z$, you can calculate the value of the function.
• This generalizes the one-dimensional case where the particle was confined to the x-axis.

Infinite Degeneracy

• The plane wave solution has infinite degeneracy.
• For a given energy level, there are many combinations of $k<em>x$, $k</em>y$, and $k_z$ on a sphere that yield the same energy.
  • $E = \frac{{\hbar^2k^2}}{{2m}}$
• Increasing energy corresponds to increasing the momentum of the particle, but there are infinitely many states with that energy.
• All points have the same energy.

Mixed Momentum Eigenstates

• We can have mixed states with a linear combination of different momentum directions.

Eigenstates of Momentum Operator

• The solutions are also eigenstates of the momentum operator in any direction.

• The momentum operator is: $\hat{p} = -i\hbar \frac{{\partial}}{{\partial x}}$

• Applying the momentum operator to the wave function yields:

  $-i\hbar \frac{{\partial}}{{\partial x}} C e^{i \vec{k} \cdot \vec{r}} = \text{constant} \cdot C e^{i \vec{k} \cdot \vec{r}}$

• Since the Hamiltonian commutes with the momentum operator, they share eigenfunctions.

• We verify that it works: $\hat{p}<em>x \Psi = p</em>x \Psi$, So, $\hat{p}<em>x = \hbar k</em>x$

Commutation Relations

• The commutation relation between $p<em>x$ and $p</em>y$ is zero: $[p<em>x, p</em>y] = 0$.
• Four quantities can be specified simultaneously: $p<em>x$, $p</em>y$, $p_z$, and the total energy $E$.

Probability Density

• The probability of finding the particle in a small volume $dV = dx dy dz$ is:
  $P(x, y, z) dV = |\Psi(x, y, z)|^2 dV$
• For a free particle, $|\Psi(x, y, z)|^2$ is a constant, so the particle is equally likely to be found anywhere in space.
• This is consistent with the Heisenberg uncertainty principle.
• If $\Delta p_x = 0$, then $\Delta x = \infty$.

Particle in a Box

• Consider a particle in a box with dimensions $L<em>x$, $L</em>y$, and $L_z$.
• The potential is infinite outside the box.
• Using the same technique as the free particle problem, but with boundary conditions, we find the solution:
  $\Psi(x, y, z, t) = C \sin(\frac{{n<em>x \pi x}}{{L</em>x}}) \sin(\frac{{n<em>y \pi y}}{{L</em>y}}) \sin(\frac{{n<em>z \pi z}}{{L</em>z}}) e^{-\frac{{i E(n<em>x, n</em>y, n_z)}}{{\hbar}} t}$
• The sine function is also an eigenfunction of the Hamiltonian when the potential is zero.

Energy Levels

• The energy levels are discrete:
  $E(n<em>x, n</em>y, n<em>z) = \frac{{\hbar^2 \pi^2}}{{2m}} (\frac{{n</em>x^2}}{{L<em>x^2}} + \frac{{n</em>y^2}}{{L<em>y^2}} + \frac{{n</em>z^2}}{{L_z^2}})$
• The energy depends on three quantum numbers ($n<em>x, n</em>y, n_z$).
• Unlike the free particle, the energy is now discrete.
• If any of the $n$ values are zero, the function is zero, which is not a valid solution.

Lowest Energy State

• The lowest energy state is E(1, 1, 1).
• The next energy state depends on the values of $L<em>x, L</em>y, L_z$.
• The energy ordering depends on the dimensions of the box.

Particle in a Cubic Box (Lx = Ly = Lz = L)

• If the box is a cube, the energies simplify to:
  $E(n<em>x, n</em>y, n<em>z) = \frac{{\hbar^2 \pi^2}}{{2mL^2}} (n</em>x^2 + n<em>y^2 + n</em>z^2)$
• This is similar to the three-dimensional n-space in the black body problem.

Degeneracy in a Cubic Box

• The lowest energy state is E(1, 1, 1).
• The next state has a threefold degeneracy: E(2, 1, 1), E(1, 2, 1), E(1, 1, 2).
• Degeneracy means multiple eigenfunctions have the same energy.
• Further energy levels: E(2, 2, 1), E(2, 1, 2), E(1, 2, 2) and E(1, 1, 3), E(1, 3, 1), E(3, 1, 1)
• Degeneracies: onefold, threefold, and sixfold.

Mixed States

• For both problems, we can have mixed states as linear combinations of pure energy eigenstates.
• In the free particle problem, we can have mixed states with different energy levels or different k-vectors.

Hamiltonian in Spherical Coordinates

• The Hamiltonian can be transformed from Cartesian coordinates to spherical coordinates.
• Coordinates:
  • $x = r \sin(\theta) \cos(\phi)$
  • $y = r \sin(\theta) \sin(\phi)$
  • $z = r \cos(\theta)$

Hamiltonian Form

• The transformation yields a complicated expression:
  $H = \frac{{1}}{{r}} \frac{{\partial^2}}{{\partial r^2}} r + \text{terms involving } \theta \text{ and } \phi$

Radial and Tangential Kinetic Energy

• The Hamiltonian can be thought of as having radial and tangential kinetic energy components.
• The kinetic energy can be broken up into $p<em>r^2/2m$ (radial) and $p</em>t^2/2m$ (tangential).
• The full momentum vector: $p^2 = p<em>r^2 + p</em>t^2$
• So we can say that the Kinectic Energy $KE = \frac{p^2}{2m}$
• Only points on the sphere have the same energy.
• The first term represents the radial kinetic energy, and the other terms the tangential kinetic energy.

Angular Momentum

• The tangential component is related to the angular momentum:
  • $L = r \times p$ and so $\vec{L} = r p \sin(\theta)$
• The kinetic energy can be written as:
  $KE = \frac{{p_r^2}}{{2m}} + \frac{{L^2}}{{2mr^2}}$
• The Hamiltonian contains a radial part and a term related to the angular momentum operator.

Shrodinger Equation in Spherical Coordinates

• The Hamiltonian in the Shrodinger Equation is made up of two operators:

  1. Radial component of the momentum
  2. Angular momentum squared operator

• Coming up Next the Separation of variables can be used again with the following form:

  $\Psi(x, y, z) = R(r) \Theta(\theta) \Phi(\phi)$