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Quantum Mechanics for Several Dimensions • Solutions for Set #10 Are Posted Quantum Mechanics for Several Dimensions • Solutions for set #10 are posted. • Still plan a Tuesday 1-3pm session. • Some Material Covered today is not in the book- particle in ring – 2D Square Well in Chapter 8 – Coulomb Potential • Homework Set #11 is available per our vote last Friday – it is due Dec. 2. • Simple Harmonic Oscillator • Particle in a Ring -- idea of energy degeneracy • Two Dimensional Square Well – again energy degeneracy • Multiparticles in 3D http://www.colorado.edu/physics/phys2170/ Physics 2170 – Fall 2013 1 The wave functions of the first three states are Where ω = (k/m)–½ is the classical angular frequency, and n is the quantum number http://www.colorado.edu/physics/phys2170/ Physics 2170 – Fall 2013 2 http://www.colorado.edu/physics/phys2170/ Physics 2170 – Fall 2013 3 Clicker question 1 Set frequency to AD What happens to the node spacing away from x=0 for higher energy values in the harmonic oscillator potential? Think about the deBroglie wavelength. A. The node spacing doesn’t change. B. The node spacing decreases. C. The node spacing increases. http://www.colorado.edu/physics/phys2170/ Physics 2170 – Fall 2013 4 Clicker question 1 Set frequency to AD What happens to the node spacing away from x=0 for higher energy values in the harmonic oscillator potential? Think about the deBroglie wavelength. A. The node spacing doesn’t change. B. The node spacing decreases. C. The node spacing increases. http://www.colorado.edu/physics/phys2170/ Physics 2170 – Fall 2013 5 Clicker question 2 Set frequency to AD What happens to the probability density away from the equilibrium position at x=0 in the harmonic oscillator potential? A. The probability density is uniformly spaced in x. B. The probability density decreases away from x=0. C. The probability density increases away from x=0. http://www.colorado.edu/physics/phys2170/ Physics 2170 – Fall 2013 6 Clicker question 2 Set frequency to AD What happens to the probability density away from the equilibrium position at x=0 in the harmonic oscillator potential? A. The probability density is uniformly spaced in x. B. The probability density decreases away from x=0. C. The probability density increases away from x=0. Classically, a particle spends Most of its time where it is moving the slowest. http://www.colorado.edu/physics/phys2170/ Physics 2170 – Fall 2013 7 Particle in a Ring • Consider a particle constrained to a ring of radius r in the x-y plane – it is like the particle in the box. The wavefunction only depends on the angular coordinate. • x = r cos(θ) ; y = r sin(θ) - 2 d 2ψ − 2 = Eψ 2I dθ What are the wave functions and what are the boundary conditions? http://www.colorado.edu/physics/phys2170/€ Physics 2170 – Fall 2013 8 Particle in a Ring d 2 ( ) 2IE 2mr 2 E ψ θ Note I=mr2 2 = − 2 ψ(θ) = − 2 ψ(θ) dθ 2 2 2mr E ±iκθ Let κ = 2 Solutions are: ψ(θ) = Ce € In order for this wavefunction to be physically acceptable, it must be single-valued, so ψ ( θ ) = ψ ( θ + 2 π ) . € € Can only be true if e − i κ 2 π = 1 + e + i κ 2 π = 1 κ is an integer € € € http://www.colorado.edu/physics/phys2170/ Physics 2170 – Fall 2013 9 Ring cont. • Also require the wavefunction is normalized. 2 2 2mr E κ = 2 € Note e iκθ and e − iκθ are mutually orthogonal wavefunctions € http://www.colorado.edu/physics/phys2170/ Physics 2170 – Fall 2013 10 The Solution of Particle Ring 1 ψ(θ) = e±iκθ 2π κ 22 € Eκ = 2 κ = 0, ±1, ± 2, ± 3, ... 2mr Energy Levels are degenerate! Two different quantum numbers give the same energy. € http://www.colorado.edu/physics/phys2170/ Physics 2170 – Fall 2013 11 Comments on Ring Example Note there are two degenerate quantum states for every value of κ > 0, e±ικθ. Good example to study in the quantization of angular momentum for an electron orbiting the nucleus. Any wavefunction for particle on a ring can be written as a superposition of energy eigenfunctions. This model can be used to find approximate energy levels of some ring molecules in organic chemistry. The above explains why the ring behaves like circular waveguide, with the valence electrons orbiting in both directions. http://www.colorado.edu/physics/phys2170/ Physics 2170 – Fall 2013 12 Angular Momentum is quantized 1 dψ ψ(θ) = e±iκθ Lˆ ψ(θ) = = (±iκ)ψ = ±κψ(θ) 2π z i dθ i L = κ κ = 0, ±1, ± 2, ...... € Angular momentum is restricted to integer multiples of hbar! http://www.colorado.edu/physics/phys2170/€ Physics 2170 – Fall 2013 13 Two Dimensional Square Well Let’s consider a rectangular "infinite square well". The potential is 0 inside a rectangle with diagonal points of the origin and (Lx,Ly) and infinite outside the rectangle. http://www.colorado.edu/physics/phys2170/ Physics 2170 – Fall 2013 14 2D Square Well Let lengths be Lx and Ly Use separation of variables to solve the problem, namely ψ(x,y) = X(x)Y(y) ∂ 2ψ d 2 X ∂ 2ψ d 2Y X''(x) Y''(y) 2ME Y(y) X(x) 2 = 2 2 = 2 + = − 2 ∂x dx ∂y dy X(x) Y(y) € http://www.colorado.edu/physics/phys2170/ Physics 2170 – Fall 2013 15 € € € 2D cont. X''(x) Y''(y) 2ME X''(x) 2 Y''(y) 2ME + = − 2 = −kx = − − 2 X(x) Y(y) X(x) Y(y) Function of x equal to function of y + constant -- this can only be true if function of x is a constant (separation constant) € € 2 X’’(x)/X(x) = constant = -k x Now just like 1 D case – match boundary conditions 2 2 X(x) = sin(nxπx /Lx ) Y(y) = sin(nyπy /Ly ) Lx Ly kx = nxπ /Lx ky = nyπ /Ly €http://www.colorado.edu/physics/phys2170/ € Physics 2170 – Fall 2013 16 € € 2D cont. Wave function is ⎛ ⎞ ⎛ ⎞ 4 nxπx nyπy ψ(x,y) = sin⎜ ⎟ sin⎜ ⎟ LxLy ⎝ Lx ⎠ ⎝ Ly ⎠ ∂ 2ψ ∂ 2ψ 2M 2 + 2 = − 2 Eψ € ∂ x ∂y Find Energy eigenvalues € 2 2 2 2 nxπ nyπ 2ME € 2 + 2 = 2 Lx Ly 2 2 2 2 2 n π nyπ E = E = ( x + ) nx ny 2M L2 L2 € x y http://www.colorado.edu/physics/phys2170/€ Physics 2170 – Fall 2013 17 2 D Degeneracy ⎛ ⎞ ⎛ ⎞ 4 nxπx nyπy ψ(x,y) = sin⎜ ⎟ sin⎜ ⎟ LxLy ⎝ Lx ⎠ ⎝ Ly ⎠ If Lx and Ly are equal, then we find two different wave functions can give the same energy, namely. € 2 2 π 2 2 E = E n n = (nx + ny ) E = E x y 2ML2 1,2 2,1 2 ⎛ π ⎞ ⎛ 2π ⎞ Different probability densities ψ(x,y) = sin⎜ x⎟ sin⎜ y⎟ € L ⎝ L ⎠ ⎝€ L ⎠ which means they are 2 ⎛ 2πx ⎞ ⎛ πy ⎞ experimentally distinguishable, ψ(x,y) = sin⎜ ⎟ sin⎜ ⎟ L ⎝ L ⎠ ⎝ L ⎠ but they have the same energy http://www.colorado.edu/physics/phys2170/ Physics 2170 – Fall 2013 18 € 2D Wave Functions Here is a plot of the probability density for a wavefunction with nx=2, ny=4, for Lx=4, Ly=3: http://www.colorado.edu/physics/phys2170/ Physics 2170 – Fall 2013 19 Multiparticle Schrodinger Eq. • Let’s consider N particles whose positions are defined by a set of rectangular coordinates x, y, z. The classical expression for the total energy of the system is N 1 p2 + p2 + p2 + V (x ,y ,z ,......,x ,y ,z ,t) = E ∑ ( x j y j z j ) 1 1 1 N N N j=1 2m j • Now replace momenta, E with differential operators € N 2 ⎛ ∂ 2 ∂ 2 ∂ 2 ⎞ ∂ − ⎜ + + ⎟ + V(x ,y ,z ,......,x ,y ,z ,t) = i ∑ 2m ⎜ ∂x 2 ∂y 2 ∂z2 ⎟ 1 1 1 N N N ∂t j=1 j ⎝ j j j ⎠ • Multiply both sides by the wavefunction € http://www.colorado.edu/physics/phys2170/ Physics 2170 – Fall 2013 20 Laplacian 2 2 2 • We use the symbol 2 ∂ ∂ ∂ ∇ j ≡ 2 + 2 + 2 ∂x j ∂y j ∂z j which is called the Laplacian in spherical coordinates of the jth particle + write the Schrodinger equation as N € 2 ∂Ψ − ∇ 2 Ψ + VΨ = i ∑ 2m j ∂t j=1 j So long as V does not depend on t, then we can write Ψ(x1,....,zN ,t) =ψ(x1,....,zN )φ(t) where€ −iEt / φ (t) = e € http://www.colorado.edu/physics/phys2170/ Physics 2170 – Fall 2013 21 € Schrodinger Eq. for Coulomb • Get time independent Schrodinger Equation N 2 − ∇ 2ψ + Vψ = Eψ ∑ 2m j j=1 j • For One-Electron Atom; let Ze be the charge on the Nucleus€ of mass m1, and –e is charge on electron of mass m2. The Coulomb potential energy is Ze2 V(x ,....,z ) = 1 2 2 2 2 (x1 − x2 ) + (y1 − y2 ) + (z1 − z2 ) where x1,y1,z1 are coordinates of nucleus and x2,y2, z2 are coordinates of the electron. € http://www.colorado.edu/physics/phys2170/ Physics 2170 – Fall 2013 22 One Electron Atom 2 ⎛ 2 2 2 ⎞ 2 ⎛ 2 2 2 ⎞ ∂ ψT ∂ ψT ∂ ψT ∂ ψT ∂ ψT ∂ ψT − ⎜ + + ⎟ − ⎜ + + ⎟ + V (x1,......,z2)ψT = ETψT 2m ∂x 2 ∂y 2 ∂z2 2m ∂x 2 ∂y 2 ∂z2 1 ⎝ 1 1 1 ⎠ 2 ⎝ 2 2 2 ⎠ Reason for the subscript T will become apparent in a minute. € We want to change variables corresponding to the translational motion of the center of mass (x,y,z) and the motion of the two particles relative to each other (r,θ,ϕ). m1x1 + m2 x2 x = rsin(θ)cos(φ) = x2 − x1 m1 + m2 m1y1 + m2 y2 y = rsin(θ)sin(φ) = y2 − y1 m + m 1 2 € m1z1 + m2z2 € z = rcos(θ) = z2 − z1 m1 + m2 http://www.colorado.edu/physics/phys2170/ Physics€ 2170 – Fall 2013 23 € € € .
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