Quantum Theory. Thornton and Rex, Ch. 6

Quantum Theory Thornton and Rex, Ch. 6

Matter can behave like waves. 1) What is the wave equation? 2) How do we interpret the wave function y(x,t)?

Light Waves Plane wave: y(x,t) = A cos(kx-wt) wave (w,k) particle (E,p): 1) Planck: E = hn = hw 2) De Broglie: p = h/l = hk A particle relation: 3) Einstein: E = pc A wave relation: 4) w = ck (follows from (1),(2), and (3)) The wave equation: 1 2 y c 2 t = 2 y 2 x 2

Matter Waves Equations (1) and (2) must hold. (wave (w,k) particle (E,p)) Particle relation (non-relativistic, no forces): 3 ) E = 1/2 m v 2 = p 2 /2m The wave relation: 4 ) hw = (hk) 2 /2m (follows from (1),(2), and (3 )) fi Require a wave equation that is consistent with (4 ) for plane waves.

The Schrodinger Equation 1925, Erwin Schrodinger wrote down the equation: ih y t = -h 2 2m 2 y x 2 + V (x)y

Q uick uiz In Schrödinger s equation, the variable ψ represents the following property of a wave:- (A) (B) (C) (D) (E) the frequency the wavelength the period the amplitude the phase

The Schrodinger Equation ih y t = -h 2 2m 2 y x 2 + V (x)y (Assume V is constant here.) A general solution is y(x,t) = A e i(kx-wt) = A ( cos(kx-wt) + i sin(kx-wt) ) fi hw = h 2 k 2 2m + V fi E = p 2 2m + V

Complex Numbers i = (-1) e iθ = cosθ + i sinθ All complex numbers have a polar form: z = x+iy = re iθ x = rcosθ, y = rsinθ z * = x-iy, zz * = x 2 +y 2 = r 2 Plane wave: ψ = Ae i(kx-ωt) Classical Physics: use only real or imaginary part In Quantum Mechanics, the waves can be complex

Q uick uiz If z = 4 + 3i what is the value of zz*? (A) 5 (B) 7 (C) 9 (D) 16 (E) 25

What is y(x,t)? Double-slit experiment for light: I(x) E(x) 2 + B(x) 2 Probability of photon at x is µ I(x). Max Born suggested: y(x,t) 2 is the probability of finding a matter particle (electron) at a place x and time t.

Probability and Normalization The probability of a particle being between x and x + dx is: P(x) dx = y(x,t) 2 dx = y*(x,t)y(x,t) dx The probability of being between x 1 and x 2 is: P = y*y dx x 1 x 2 The wavefunction must be normalized: P = y*y dx = 1 -

Properties of valid wave functions 1. y must be finite everywhere. 2. y must be single valued. 3. y and dy/dx must be continuous for finite potentials (so that d 2 y/dx 2 remains single valued). 4. y Æ 0 as x Æ ±. These properties (boundary conditions) are required for physically reasonable solutions.

Heisenberg s Uncertainty Principle Independently, Werner Heisenberg, developed a different approach to quantum theory. It involved abstract quantum states, and it was based on general properties of matrices. Heisenberg showed that certain pairs of physical quantities ( conjugate variables ) could not be simultaneously determined to any desired accuracy.

We have already seen: Dx Dp h/2 It is impossible to specify simultaneously both the position and momentum of a particle. Other conjugate variables are (E,t) (L,q): Dt DE h/2 Dq DL h/2

Expectation Values Consider the measurement of a quantity (example, position x). The average value of many measurements is: x = N i x i i N i For continuous variables: x = i P(x) x dx - P(x) dx - where P(x) is the probability density for observing the particle at x.

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 = (Ae i(kx-wt) ) = ik y = y x x h y fi p y = -ih x Define the momentum operator: p = -ih x Then: y p = y* p y dx = -ih y* dx - - x

Similarly, y t -ie = -iw y = y h y fi E y = ih t so the Energy operator is: E = ih t and y E = y* E y dx = ih y* dx - - t

A Self-Consistency Check Non-relativistically we have E = K + V = p2 2m + V 2 ˆp! Ê" = ( 2m + V )"!i! "# "t = ($i! " "x )2 2m # + V#!i! "# "t = $!2 2m " 2 "x 2 # + V# The Schrödinger Equation!!

Time-independent Schrodinger Equation In many (most) cases the potential V will not depend on time. Then we can write: y(x,t) = y(x) e -iwt ih y t = ih (-iw) y = hw y = E y This gives the time-independent S. Eqn: -h 2 2m d 2 y(x) dx 2 + V(x)y(x) = Ey(x) The probability density and distributions are constant in time: y*(x,t)y(x,t) = y*(x)e iwt y(x)e -iwt = y*(x) y(x)

The infinite square well potential V V(x) = for x<0 x>l V(x)=0 for 0<x<L x=0 x=l The particle is constrained to 0 < x < L. Outside the well V =, fi y = 0 for x<0 or x>l.

Inside the well, the t-independent S Eqn: -h 2 d 2 y = E y 2m d x 2 d 2 y -2mE fi = y = - k 2 y d x 2 h 2 (with k = 2mE/h 2 ) A general solution is: y(x) = A sin kx + B cos kx Continuity at x=0 and x=l give y(x=0) = 0 fi B=0 and y(x=l) = 0 fi y(x) = A sin kx with kl=np n=1,2,3,...

Normalization condition gives fi A = 2/L so the normalized wave functions are: y n (x) = 2/L sin(npx/l) n=1,2,3,... with k n = np/l = 2mE n /h 2 n 2 p 2 h 2 fi E n = 2mL 2 The possible energies (Energy levels) are quantized with n the quantum number.

Finite square well potential V 0 0 Region I II III x=0 x=l V(x) = V 0 for x<0 x>l V(x)=0 for 0<x<L Consider a particle of energy E < V 0. Classically, it will be bound inside the well Quantum Mechanically, there is a finite probability of it being outside of the well (in regions I or III).

Regions I and III: -h 2 d 2 y = (E-V0 ) y 2m d x 2 fi d 2 y/dx 2 = a 2 y with a 2 = 2m(V 0 -E)/h 2 > 0 The solutions are exponential decays: y I (x) = A e ax y III (x) = B e -ax Region I (x<0) Region III (x>l) In region II (in the well) the solution is y II (x) = C cos(kx) + D sin(kx) with k 2 = 2mE/ h 2 as before. Coefficients determined by matching wavefunctions and derivatives at boundaries.

The 3-dimensional infinite square well S. time-independent eqn in 3-dim: -h 2 2m Ê 2 y x + 2 y 2 y + 2 y ˆ Á Ë 2 z 2 + Vy = Ey fi -h 2 2m 2 y + Vy = Ey The solution: y = A sin(k 1 x) sin(k 2 y) sin(k 3 z) L 2 L 3 L 1 with k 1 =n 1 p/l 1, k 2 =n 2 p/l 2, k 3 =n 3 p/l 3. Allowed energies: p 2 h 2 n 1 2 n 2 2 n 3 2 E = ( + + ) 2m L 1 2 L 2 2 L 3 2

If L 1 = L 2 =L 3 =L (a cubical box) the energies are: p 2 h 2 E = ( n 1 2 + n 2 2 + n 3 2 ) 2m L 2 The ground state (n 1 =n 2 =n 3 =1) energy is: 3 p 2 h 2 E 0 = 2m L 2 The first excited state can have (n 1, n 2, n 3 ) = (2,1,1) or (1,2,1) or (1,1,2) There are 3 different wave functions with the same energy: 6 p 2 h 2 E 1 = 2m L 2 The 3 states are degenerate.

Simple Harmonic Oscillator Spring force: F = - k x 1 fi V(x) = k x 2 2 A particle of energy E in this potential: V(x) E -a a -h 2 d 2 y 1 + k x 2 y = E y 2m d x 2 2

d 2 y d x 2 = ( a 2 x 2 - b ) y with 2m E m k b = and a 2 = h h 2 The solutions are: y n (x) = H n (x) e -ax 2 /2 Hermite Polynomial (oscillates at small x) Gaussian (exponential decay at large x)

Polynomials wiggle more as n increases

Approach Classical as n increases Eventually wiggles too narrow to resolve

The energy levels are E n = ( n + 1/2 ) h w where w 2 = k/m is the classical angular frequency. The minimum energy (n=0) is E 0 = h w /2 This is known as the zero-point energy.

Q uick uiz If the lowest energy state in a quantum harmonic oscillator is E 0 = 2 ev, what is the value of the next energy state, E 1? (A) (B) (C) (D) (E) 0.5 ev 1 ev 3 ev 4 ev 6 ev

Q uick uiz If the lowest energy state in a quantum harmonic oscillator is E 0 = 2 ev, what is the value of the next energy state, E 1? (A) (B) (C) (D) (E) 0.5 ev 1 ev 3 ev 4 ev 6 ev E n = (n+1/2)hω ==> E 0 = hω/2 = 2 ev ==> E 1 = 3/2 hω = 6 ev

The lowest energy state saturates the Heisenberg uncertainty bound: Dx Dp = h/2 We can use this to calculate E 0. In the SHM, PE = K = E/2. fi k x 2 /2 = p 2 /(2m) = E/2 fi k Dx 2 = Dp 2 /m = E fi Dx = Dp / m k From uncertainty principle: Dx = h/(2dp) So E = k Dx Dx = k (Dp / m k )(h/(2dp)) = (h/2) k / m = h w /2

This calculation is done in Example 6.11 on page 221-222. A general definition for the uncertainties Δx and Δp. They are defined to be the root mean square deviations:-

Barriers and Tunneling A particle of total energy E approaches a change in potential V 0. Assume E>V 0 : E V 0 Region I II III Classically, the particle slows down over the barrier, but it always makes it past into region III. Quantum Mechanically, there are finite probabilities of the particle being reflected as well as transmitted.

Incident Intermediate Reflected Transmitted Region I II III Solutions: Incident Reflected I y I = A e ikx + B e -ikx II y II = C e ik x + D e -ik x Intermediate III y III = F e ikx Transmitted with k = 2mE/h 2 k = 2m(E-V 0 )/h 2

Q uick uiz If the transmission probability for a particular barrier is 1/4 (= 25%), what is the probability of reflection? (A) 1/8 (B) 1/4 (C) 1/2 (D) 3/4 (E) 1.0

Q uick uiz If the transmission probability for a particular barrier is 1/4 (= 25%), what is the probability of reflection? (A) 1/8 (B) 1/4 (C) 1/2 (D) 3/4 (E) 1.0 R = 1 - T

Applications of tunneling: 1) Scanning Tunneling Microscope Consider two metals separated by vacuum: metal L metal Potential work function f Apply a voltage difference between the metals: E The current ~ Tunneling Probability ~ e -al Very sensitive to distance L With thin metal probe can map the surface contours of the other metal at the atomic level.

2) Nuclear a-decay The potential seen by the a-particle looks like: V(r) electrostatic repulsion ~ 1/r E r strong force attraction Because the decay is limited by the probability of tunneling, small changes in the potential or energy of the a particle lead to large changes in the nuclear lifetime. 232 Th t 2x10 8 yr 212 Po t 4x10-7 s fi 24 orders of magnitude!