Quantum TheoryThornton and Rex, Ch. 6Matter can behave like waves.1) What is the wave equation?2) How do we interpret the wavefunction y(x,t)?Light WavesPlane wave: y(x,t) = A cos(kx-wt)wave (w,k) ¤ particle (E,p):1) Planck: E = hn = hw2) De Broglie: p = h/l = hkA particle relation:3) Einstein: E = pcA wave relation:4) w = ck (follows from (1),(2), and (3))The wave equation:† 1c2∂2y∂t2=∂2y∂x2Matter WavesEquations (1) and (2) must hold. (wave (w,k) ¤ particle (E,p))Particle relation (non-relativistic, noforces):3’) E = 1/2 m v2 = p2/2mThe wave relation:4’) hw = (hk)2/2m(follows from (1),(2), and (3’))fi Require a wave equation that isconsistent with (4’) for plane waves.The Schrodinger Equation1925, Erwin Schrodinger wrote down theequation:† ih∂y∂t=-h22m∂2y∂x2+ V (x)yThe Schrodinger Equation(Assume V is constant here.)A general solution isy(x,t) = A ei(kx-wt) = A ( cos(kx-wt) + i sin(kx-wt) ) h2k2fi hw = + V 2m p2fi E = + V 2m† ih∂y∂t=-h22m∂2y∂x2+ V (x)yWhat is y(x,t)?Double-slit experiment for light:I(x) ≈ |E(x)|2 + |B(x)|2Probability of photon at x is µ I(x).Max Born suggested:|y(x,t)|2 is the probability of finding amatter particle (electron) at a place xand time t.Probability and NormalizationThe probability of a particle beingbetween x and x + dx is:P(x) dx = |y(x,t)|2 dx = y*(x,t)y(x,t) dxThe probability of being between x1 andx2 is:P = ∫ y*y dxThe wavefunction must be normalized:P = ∫ y*y dx = 1x1x2∞-∞Properties of valid wave functions1. y must be finite everywhere.2. y must be single valued.3. y and dy/dx must be continuous forfinite potentials (so that d2y/dx2remains single valued).4. y Æ 0 as x Æ ± ∞ .These properties (boundary conditions)are required for physically reasonablesolutions.Heisenberg’s Uncertainty PrincipleIndependently, Werner Heisenberg,developed a different approach toquantum theory. It involved abstractquantum states, and it was based ongeneral properties of matrices.Heisenberg showed that certain pairs ofphysical quantities (“conjugate variables”)could not be simultaneously determined toany desired accuracy.We have already seen: Dx Dp ≥ h/2It is impossible to specify simultaneouslyboth the position and momentum of aparticle.Other conjugate variables are (E,t) (L,q): Dt DE ≥ h/2 Dq DL ≥ h/2Expectation ValuesConsider the measurement of a quantity(example, position x). The average valueof many measurements is: ∑ Ni xix = ∑ NiFor continuous variables: ∫ P(x) x dxx = ∫ P(x) dxwhere P(x) is the probability density forobserving the particle at x.ii∞∞-∞-∞Expectation Values (cont’d)In QM we can calculate the “expected”average: ∫ y*y x dx x = = ∫ y*y x dx ∫ y*y dx The expectation value.Expectation value of any function g(x) is: g(x) = ∫ y* g(x) y dx∞∞-∞-∞∞-∞What are the expectation values of p or E?First, represent them in terms of x and t:∂ y ∂ ip = (Aei(kx-wt)) = ik y = y∂ x ∂ x h ∂ yfi p y = -ih ∂ xDefine the momentum operator: ∂p = -ih ∂ xThen: ∂ y p = ∫ y* p y dx = -ih ∫ y* dx ∂ x∞-∞-∞∞Similarly,∂ y -iE = -iw y = y∂ t h ∂ yfi E y = ih ∂ tso the Energy operator is: ∂E = ih ∂ tand ∂ y E = ∫ y* E y dx = ih ∫ y* dx ∂ t∞-∞-∞∞A self-consistency check:The classical system obeys: p2 E = K + V = + V 2mReplace E and p by their respectiveoperators and multiplying by y: p2 E y = ( + V ) y 2mThe Schrodinger Equation!Time-independent Schrodinger EquationIn many (most) cases the potential V willnot depend on time.Then we can write:y(x,t) = y(x) e-iwt ∂ yih = ih (-iw) y = hw y = E y ∂ tThis gives the time-independent S. Eqn:The probability density and distributionsare constant in time:y*(x,t)y(x,t) = y*(x)eiwt y(x)e-iwt = y*(x) y(x)† -h22md2y(x)dx2+ V (x)y(x) = Ey(x)The infinite square well potentialThe particle is constrained to 0 < x < L.Outside the “well” V = ∞,fi y = 0for x<0 or x>L.Vx=0x=LV(x) = ∞for x<0 x>LV(x)=0for 0<x<LInside the well, the t-independent S Eqn:-h2 d2y = E y2m d x2 d2y -2mEfi = y = - k2 y d x2 h2(with k = √2mE/h2 )A general solution is: y(x) = A sin kx + B cos kxContinuity at x=0 and x=L givey(x=0) = 0fi B=0and y(x=L) = 0fi y(x) = A sin kx with kL=np n=1,2,3, . . .Normalization condition givesfi A = √2/Lso the normalized wave functions are:yn(x) = √2/L sin(npx/L) n=1,2,3, . . .with kn = np/L = √2mEn/h2 n2p2h2fi En = 2mL2The possible energies (Energy levels) arequantized with n the quantum number.Finite square well potentialConsider a particle of energy E < V0.Classically, it will be bound inside the wellQuantum Mechanically, there is a finiteprobability of it being outside of the well(in regions I or III).V(x) = V0for x<0 x>LV(x)=0for 0<x<LRegion IIIIIIx=0x=LV00Regions I and III:-h2 d2y = (E-V0) y2m d x2fi d2y/dx2 = a2 y with a2 = 2m(V0-E)/h2 > 0The solutions are exponential decays:yI(x) = A eax Region I (x<0)yIII(x) = B e-ax Region III (x>L)In region II (in the well) the solution isyII(x) = C cos(kx) + D sin(kx)with k2 = 2mE/ h2 as before.Coefficients determined by matchingwavefunctions and derivatives atboundaries.The 3-dimensional infinite square wellS. time-independent eqn in 3-dim:fiThe solution:y = A sin(k1x) sin(k2y) sin(k3z)with k1=n1p/L1, k2=n2p/L2, k3=n3p/L3.Allowed energies: p2 h2 n12 n22 n32E = ( + + ) 2m L12 L22 L32† -h22m∂2y∂x2+∂2y∂y2+∂2y∂z2Ê Ë Á ˆ ¯ ˜ + Vy= Ey† -h22m—2y+ Vy= EyL1L2L3If L1= L2=L3=L (a cubical box)the energies are: p2 h2E = ( n12 + n22 + n32 ) 2m L2The ground state (n1=n2=n3=1) energy is: 3 p2 h2E0 = 2m L2The first excited state can have(n1, n2, n3) = (2,1,1) or (1,2,1) or (1,1,2)There are 3 different wave functionswith the same energy: 6 p2 h2E1 = 2m L2The 3 states are degenerate.Simple Harmonic OscillatorSpring force: F = - k x 1fi V(x) = k x2 2A particle of energy E in this potential:-h2