PH253 Lecture 14: Schrödinger’s equation mechanics with matter waves P. LeClair Department of Physics & Astronomy The University of Alabama Spring 2020 LeClair, Patrick (UA) PH253 Lecture 14 February 12, 2020 1 / 29
electron waves are a thing Single atoms of Co on a Cu single crystal surface. Due to the differing number of electrons per atom, the Co atoms create a standing wave disturbance on the Cu surface. Courtesy O. Kurnosikov (unpublished, ca. 2001) LeClair, Patrick (UA) PH253 Lecture 14 February 12, 2020 2 / 29
Outline Overview of Schrödinger’s equation 1 Free particles 2 Potential Step 3 LeClair, Patrick (UA) PH253 Lecture 14 February 12, 2020 3 / 29
Outline Overview of Schrödinger’s equation 1 Free particles 2 Potential Step 3 LeClair, Patrick (UA) PH253 Lecture 14 February 12, 2020 4 / 29
Time-dependent version: h 2 ∂ 2 ψ − ¯ h ∂ψ ∂ x 2 + V ψ = i ¯ 2 m ∂ t ψ ( x , t ) = wavefunction for object, the “amplitude” 1 This equation gives the time evolution of system, given ψ ( x , t = 0 ) 2 1st order in time, evolution of amplitude deterministic 3 Write out for discrete time steps ( ∂ψ / ∂ t → ∆ ψ / ∆ t ) 4 ∂ 2 ψ ( x , t ) ∆ ψ = ψ ( x , t + ∆ t ) − ψ ( x , t ) = i ¯ h ∆ t ∂ x 2 2 m LeClair, Patrick (UA) PH253 Lecture 14 February 12, 2020 5 / 29
Time-independent version: h 2 ∂ 2 ψ − ¯ ∂ x 2 + V ψ = E ψ 2 m Time-independent version gives ψ ( t = 0 ) and energy 1 Given potential V ( x , t ) can find ψ 2 Typically: consider static cases, V = V ( x ) 3 What does the wave function tell us? 4 LeClair, Patrick (UA) PH253 Lecture 14 February 12, 2020 6 / 29
Properties of ψ ψ gives probabilities: | ψ ( x , t ) | 2 dx = P ( x , t ) dx 1 Probability particle is in [ x , x + dx ] at time t 2 ∞ � P ( x ) dx = 1. Normalization: particle is somewhere: 3 − ∞ ψ is in general complex: ψ = a + bi or ψ = Ae iB 4 Phase is key for interference of 2 matter waves 5 ψ tot = a ψ 1 + b ψ 2 , but | ψ tot | 2 � = | ψ 1 | 2 + | ψ 1 | 2 6 Single particle or bound state - no interference, ψ can be real. 7 LeClair, Patrick (UA) PH253 Lecture 14 February 12, 2020 7 / 29
Time dependence Let’s say V ( x ) is independent of time (static environment) 1 Then presume wave function is separable into t , x parts 2 I.e., Ψ ( x , t ) = ψ ( x ) ϕ ( t ) 3 V indep. of time required for conservative forces (see ph301/2) 4 Mostly what we will worry about anyway 5 If this is the case, plug into time-dep. Schrödinger 6 One side has only x , the other only t dependence 7 h 2 ∂ 2 Ψ − ¯ h ∂ Ψ ∂ x 2 + V ( x ) Ψ = i ¯ 2 m ∂ t LeClair, Patrick (UA) PH253 Lecture 14 February 12, 2020 8 / 29
Time dependence h 2 ∂ 2 Ψ − ¯ h ∂ Ψ ∂ x 2 + V ( x ) Ψ = i ¯ 2 m ∂ t Ψ ( x , t ) = ψ ( x ) ϕ ( t ) . One side has only x , the other only t . 1 Each side must then be separately equal to the same constant E 2 h ∂ϕ ∂ t = E ϕ i ¯ Now separate & integrate (recall 1/ i = − i ): ∂ϕ ϕ = − iE ϕ = e iEt /¯ h = ⇒ h ∂ t ¯ h ω , ϕ = e − i ω t – simple oscillation; E is energy! With E = ¯ LeClair, Patrick (UA) PH253 Lecture 14 February 12, 2020 9 / 29
Spatial dependence If V independent of time, amplitude oscillates with frequency ω 1 Then Ψ ( x , t ) = ψ ( x ) e − iEt /¯ h 2 Spatial part from time-independent equation - 2nd half of 3 separation h 2 ∂ 2 ψ − ¯ ∂ x 2 + V ψ = E ψ 2 m Can we make this look like something more familiar? LeClair, Patrick (UA) PH253 Lecture 14 February 12, 2020 10 / 29
What is this equation? Do some factoring. Treat ∂ 2 / ∂ x 2 as an operator . � � h 2 h 2 ∂ 2 ψ ∂ 2 − ¯ − ¯ ∂ x 2 + V ψ = ∂ x 2 + V ψ = E ψ 2 m 2 m Looks a little like K + V = E ∂ x . Then p 2 = − ¯ h 2 ∂ 2 h ∂ Let p = − i ¯ ∂ x 2 . . . � p 2 � 2 m + V ψ = E ψ The time-independent equation is just conservation of energy! Must be so: V independent of t requires conservative forces. Classical analogy : p = m d dt , p x = p would give momentum. LeClair, Patrick (UA) PH253 Lecture 14 February 12, 2020 11 / 29
Outline Overview of Schrödinger’s equation 1 Free particles 2 Potential Step 3 LeClair, Patrick (UA) PH253 Lecture 14 February 12, 2020 12 / 29
An electron alone in the universe h 2 ∂ 2 ψ − ¯ ∂ x 2 + V ψ = E ψ 2 m For a free isolated particle, V = 0. Thus, h 2 ∂ 2 ψ ∂ 2 ψ � 2 mE � − ¯ ∂ x 2 = E ψ or ∂ x 2 = − ψ h 2 2 m ¯ We know this equation, it is a = d 2 x dt 2 = − k 2 x 1 Know the solutions are oscillating functions. In general, noting 2 time dependence already found: h � Ae ikx + Be − ikx � ψ ( x ) = e − iEt /¯ LeClair, Patrick (UA) PH253 Lecture 14 February 12, 2020 13 / 29
An electron alone in the universe Ae ikx + Be − ikx � h � ψ ( x ) = e − iEt /¯ Sum left- and right-going sinusoidal waves. What is k ? By analogy: a = d 2 x ∂ 2 ψ � 2 mE � dt 2 = − k 2 x ∂ x 2 = − and ψ h 2 ¯ This implies k 2 = 2 mE /¯ h 2 , or E = ¯ h 2 k 2 /2 m . 1 For a free particle, E = p 2 /2 m , implying | p | = ¯ hk 2 In agreement with de Broglie and classical physics so far 3 h 2 k 2 /2 m Since via Planck E = ¯ h ω = ¯ h ω , implies ¯ 4 hk 2 /2 m Or ω = ¯ 5 LeClair, Patrick (UA) PH253 Lecture 14 February 12, 2020 14 / 29
An electron alone in the universe hk 2 ω = ¯ 2 m Last time: group velocity of wave packet is v group = ∂ω / ∂ k 1 ∂ω / ∂ k = ¯ hk / m = p / m = v 2 Just what we expect for classical particle: p = mv , E = p 2 /2 m 3 Stickier question: where is the particle? 4 LeClair, Patrick (UA) PH253 Lecture 14 February 12, 2020 15 / 29
An electron alone in the universe h � Ae ikx + Be − ikx � ψ ( x ) = e − iEt /¯ Probability it is in [ x , x + dx ] is P ( x ) dx = | ψ ( x ) | 2 dx 1 Probability in an interval [ a , b ] ? 2 b � P ( in [ a , b ]) = P ( x ) dx 3 a ∞ cos 2 x dx . . . not defined � In our case equivalent to 4 − ∞ Plane wave solution is not normalizable , P has no meaning 5 Infinite uncertainty in position, because we know k & p precisely! 6 Makes sense, empty universe with no constraints. Can be 7 anywhere. LeClair, Patrick (UA) PH253 Lecture 14 February 12, 2020 16 / 29
Outline Overview of Schrödinger’s equation 1 Free particles 2 Potential Step 3 LeClair, Patrick (UA) PH253 Lecture 14 February 12, 2020 17 / 29
Slightly less empty universe V ( x ) V 0 I II x 0 Particle of energy E > V o coming from left sees step in potential 1 V ( x ) = 0 for x < 0, V ( x ) = V o for x ≥ 0 2 Write down time-independent Schrödinger equation 3 h 2 ∂ 2 ψ ∂ 2 ψ − ¯ ∂ x 2 + 2 m ∂ x 2 + V ψ = E ψ h 2 ( E − V ) ψ = 0 or 2 m ¯ LeClair, Patrick (UA) PH253 Lecture 14 February 12, 2020 18 / 29
Solution still traveling waves V ( x ) V 0 I II x 0 V is different in the two regions (I, II), solve separately 1 Since E − V 0 > 0 everywhere, same basic solution though. 2 Solution is still traveling waves like free particle 3 h 2 ∂ 2 ψ ∂ 2 ψ − ¯ ∂ x 2 + 2 m ∂ x 2 + V ψ = E ψ h 2 ( E − V ) ψ = 0 or 2 m ¯ LeClair, Patrick (UA) PH253 Lecture 14 February 12, 2020 19 / 29
Region I: free particle V ( x ) V 0 I II x 0 Let k 2 = 2 mE /¯ h 2 , same solution as free particle (ignore e i ω t ) 1 ψ I ( x ) = e ikx + Re − ikx for x < 0 Can choose constant of first term to be 1 1 First term is right-going wave, second is left-going wave. 2 Right-going wave is the reflection of incident wave 3 LeClair, Patrick (UA) PH253 Lecture 14 February 12, 2020 20 / 29
Region II: slightly less free particle V ( x ) V 0 I II x 0 Second region: same! Let q 2 = 2 m ( E − V 0 ) /¯ h 2 . 1 Need two constants now. (ignore e i ω t still) 2 ψ II ( x ) = Te iqx + Ue − ikx for x ≥ 0 First term: transmitted portion. 1 Second term? Wave coming from right – unphysical, so U = 0 2 Overall: like any wave: incident = reflected + transmitted 3 LeClair, Patrick (UA) PH253 Lecture 14 February 12, 2020 21 / 29
Combining the solutions: continuity V ( x ) V 0 I II x 0 ψ I ( x ) = e ikx + Re − ikx ψ II ( x ) = Te iqx Continuity: match solutions at boundary! 1 ψ and its derivatives match at x = 0. So does | ψ | 2 2 ψ I ( 0 ) = 1 + R = ψ II ( 0 ) = T = ⇒ 1 + R = T Total intensity for I and II match at the boundary 1 LeClair, Patrick (UA) PH253 Lecture 14 February 12, 2020 22 / 29
More continuity V ( x ) V 0 I II x 0 ψ I ( x ) = e ikx + Re − ikx ψ II ( x ) = Te iqx Also match ∂ψ / ∂ x at boundary. 1 � � ∂ψ I = ∂ψ II � � � � ∂ x ∂ x � � 0 0 ike ikx + ( − ik ) Re − ikx = iqTe iqx at x = 0 : ik − ikR = iqT k ( 1 − R ) = qT LeClair, Patrick (UA) PH253 Lecture 14 February 12, 2020 23 / 29
Coefficients for transmission & reflection V ( x ) V 0 I II x 0 k ( 1 − R ) = qT Also know 1 + R = T . . . algebra . . . 1 R = k − q 2 k T = k + q k + q e ikx + � � k − q e − ikx x < 0 k + q ψ ( x ) = � � 2 k e iqx x ≥ 0 k + q LeClair, Patrick (UA) PH253 Lecture 14 February 12, 2020 24 / 29
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