5.61 F17 Lecture 10: The Time-Dependent Schrödinger Equation

5.61 Fall 2017 Lecture #10 page 1 Lecture #10: The Time-Dependent Schrödinger Equation Last time: 1/2 ⎫ µω ⎤ x̂ = ⎡⎢ x̂ ⎪ ⎥ ⎣  ⎦ ⎬ dimensionless variables −1/2 p̂ = [ µω ] p̂ ⎪⎭ â = 2 −1/2 ( ip̂ + x̂ ) annihilation operator a† = 2 −1/2 ( −ip̂ + x̂ ) creation operator x̂ = 2 −1/2 ( â † + â ) p̂ = 2 −1/2 i ( â † − â ) ⎛  ⎞ x̂ = ⎜ ⎝ 2 µω ⎟⎠ ⎛ µω ⎞ p̂ = ⎜ ⎝ 2 ⎟⎠ 1/2 1/2 ⎛  ⎞ ( â + â ), x̂ = ⎜ ⎝ 2 µω ⎟⎠ n/2 ⎛ µω ⎞ i ( â − â ) p̂ = ⎜ ⎝ 2 ⎟⎠ n/2 n † n † âψ v = [ v ]1/2 ψ v−1 † + â ) n i n ( â † − â ) n , e.g. â 3ψ v = [ v ( v − 1) ( v − 2 )] ψ v− 3 10 1/2 , e.g. a† ψ v = [( v + 10 )…( v + 1)] ψ v+10 1/2 â †ψ v = [ v + 1]1/2 ψ v+1 N̂ = â †â, ( â N̂ψ v = vψ v ψ v = [v!]−1/2 (a † )ψ 0 , ψ 0 is a known Gaussian Operator algebra, to combine terms like â †â â, â â †â, â â â † , is based on ⎡⎣ â, â † ⎤⎦ = 1 . † What is so great about â, â ? It is born with its selection rule and the values of all integrals attached! Z dx † m n v (â ) (â) 6 vf 2 +n v + n m = 4(v | | {z } 6 vi (↠)m (â)n ! vf Suppose you want ∫ vi = m m)(v + n m {z n terms 1) . . . (v m + 1) (v }| m + 1) . . . (v {z m terms 31/2 1)(v)5 } n  p could be â † 2 or â † 3â (in any  pψ ≠ 0 ? Then O dxψ *v+2 O v order). revised 8/7/17 11:25 AM 5.61 Fall 2017 Lecture #10 page 2 Suppose you have p̂ 3 and want the ψ v+ 3 p̂ 3ψ v integral? Only a total of 3 multiplicative â or â † factors are possible in p̂ 3 , thus you only keep the â † 3 term. Recall: semi-classical method based on pclassical(x) = [2µ(E – V(x))]1/2 * node spacing * amplitude envelope of ψ(x) * quantization condition The REAL Schrödinger Equation is the Time Dependent Schrödinger Equation (TDSE). The ordinary time-independent Schrödinger Equation, Ĥ ψ = Eψ , is a special case. Eigenstates do not move, but they encode motion. TDSE: Ĥ Ψ(x,t) = i ∂Ψ ∂t We usually use Ψ for solutions of TDSE and ψ for solutions of the ordinary SE. Suppose we have a complete set of solutions of ordinary SE Ĥ ψ n = En ψ n ( Ĥ is independent of time) , then Ψ n (x,t) = e −iEnt  ψ n (x) satisfies the TDSE. i ∂Ψ ⎛ −i ⎞ = (i) ⎜ ⎟ En e −iEnt  ψ n (x) ⎝ ⎠ ∂t = En e −iEnt  ψ n  this is Ψ n ( x,t ) = En Ψ n (x,t) Ĥ Ψ(x,t) =  Ĥ ψ n e −iEnt  = En ψ n e −iEnt  does not operate on e −iEn t  revised 8/7/17 11:25 AM 5.61 Fall 2017 Lecture #10 page 3 Thus it is evident that, for Ĥ independent of t, if ψn(x) is a solution of the SE, then Ψ n (x,t) = e −iEnt  ψ n (x) is a solution of the TDSE. The TDSE is the truth, the whole truth, and nothing but the truth of Quantum Mechanics. However, we will use the TDSE in 5.61 only occasionally. Here are some very important properties of solutions of the TDSE. 1) The probability density |Ψ*(x,t)Ψ(x,t)|, exhibits motion only if the Ψ(x,t) contains a linear combination of eigenstates that belong to at least two energy eigenvalues, E1 ≠ E2. Ψ(x,t) = c1e −iE1t  ψ 1 + c2 e −iE2t  ψ 2 2 2 2 Ψ * (x,t)Ψ(x,t) = c1 ψ 1 + c2 ψ 2 2 +c1*c2 e −i(E2 −E1 )t  ψ 1*ψ 2 +c1c2*e +i(E2 −E1 )t  ψ 1 ψ *2 The first two terms are independent of t. The second two terms are oscillatory. The second two terms can be re-expressed as 2 Re ⎡⎣ c1*c2 e −i(E2 −E1 )t  ψ 1*ψ 2 ⎤⎦ Re means “the real part.” If C is a complex number ReC = 1 C + C*) ( 2 and in the above Ψ Ψ equation, the fourth term is the complex conjugate of the third term (see McQuarrie, Chapter A) and ∗ ω 12 = E1 E2 . −   For the special case that c1* = c2 = 2 −1/2 and ψ1 and ψ2 are real Ψ * (x,t)Ψ(x,t) = 1 2 1 2 ψ 1 + ψ 2 + ( cosω 12t ) ψ 1ψ 2 .  2 2   regions of static + and – amplitude It is clear that the only way we get motion is from a Ψ(x,t) that contains at least two eigenfunctions of Ĥ that belong to two different energy eigenvalues. The TDSE is the only way we make contact with the familiar world of moving objects. revised 8/7/17 11:25 AM 5.61 Fall 2017 Lecture #10 page 4 There are several easily computed observable dynamical properties that we can use to gain insight into the ways in which Quantum Mechanics encodes motion and to discover what are the Quantum Mechanical laws for motion. 2) If we integrate Ψ*(x,t)Ψ(x,t) over x, we have a simple normalization integral. Since probability is conserved, the normalization integral should not be (and is not) timedependent, no matter how many ψi are present in Ψ. For the two-state superposition state 2 2 ∫ dxΨ * (x,t)Ψ(x,t) = c1 + c2 ! Probability is conserved (and orthogonality simplifies matters). The c1*c2 ψ 1*ψ 2 terms go away because ψ1 is orthogonal to ψ2. The |ψ | and |ψ | terms become 1 because the ψi are normalized to 1. 2 1 3) 2 2 Expectation values of x̂ and p̂ . Here is an example where the integrals xv,v and pv,v ʹ ʹ become important. x̂ t = ∫ dxΨ * (x,t) x̂Ψ(x,t) 2 2 = c1 ∫ dxψ 1* x̂ψ 1 + c2 ∫ dxψ *2 x̂ψ 2 +c1*c2 e −iω 21t ∫ dxψ 1* x̂ψ 2 +c1 c2*e +iω 21t ∫ dxψ 1 x̂ψ *2 = c1 x11 + c2 x 22 + 2 Re ( c1*c2 e −iω 21t x12 )   2 2 stationary xij notation motion Note that, for a harmonic oscillator, x11 = 0, x22 = 0 and x12 ≠ 0 only if v2 = v1 ± 1. For a particle in a box, you can use symmetry to decide whether x11, x22, and x12 are zero. It often helps to shift the box so that it goes from –a/2 < x < a/2 (symmetric) rather than 0 < x < a. For all PIB, x11 = x22 = xcenter of box, but x12 = 0 if both quantum numbers are even or both are odd. WHY? The only case where 〈x〉t is time-dependent for PIB is when Ψ contains at least one even-n and one odd-n eigenstate. We also saw this for the wave equation. What about p̂ t ? It is possible to show (you should show this) that whenever x̂ dependent, so too is p̂ t t is time- and revised 8/7/17 11:25 AM 5.61 Fall 2017 Lecture #10 m d x dt t page 5 = p t, which is one of Newton’s laws. The motion of the center of any wavepacket is governed by Ehrenfest’s Theorem  d r dt t 1  p m and =   ( r, p are 3-D vectors ) t  d p t  = − ∇V ( r ) dt t These two equations express both of Newton’s Laws. 4) Another useful measure of dynamics is the “survival probability”, 2 P(t) = ∫ dxΨ * (x,t)Ψ(x,0) , which provides a measure of how rapidly a time-dependent state departs from (and returns to) the t = 0 form of itself. For the two-component state Ψ(x,t) = c1 ψ 1 e −iE1t  + c2 ψ 2 e −iE2 t  Ψ(x,0) = c1 ψ 1 + c2 ψ 2 2 2 ∫ dxΨ * (x,t)Ψ(x,0) = c1 eiE1t  + c2 eiE2 t  square this to get P(t) 4 4 4 4 2 2 P(t) = c1 + c2 + c1 c2 ⎡⎣ eiω 21t + e −iω 21t ⎤⎦ 2 2 = c1 + c2 + 2 c1 c2 cosω 21t For c1 = c2 1 [1+ cosω 21t ] , 2 which oscillates between 1 and 0. Is it ever possible (at some t) for P(t) < 0? P(t) = 5) Recurrence revised 8/7/17 11:25 AM 5.61 Fall 2017 Lecture #10 page 6 Whenever you have a model system where the energy levels (or energy level differences) are an integer multiple of a common factor, as for (i) particle in a box, En = E1 n2 (ii) harmonic oscillator, Ev+n – Ev = hωn (iii) rigid rotor, EJ = hcBJ(J + 1), thus EJ+1 – EJ = 2hcB[J + 1] you get perfect, 100% recurrences at a series of times that are integer multiples of t grand recurrence = h E1 h ω or or h 2hcB iω t Why? Because, at these special times, every e jk phase factor is +1. Interesting things also occur at tgr/2. Recurrence is a very useful experimental signature and can be a basis for very clever experimental manipulations of dynamics. Next Lecture: we will examine some time evolving Ψ(x,t) “wavepackets” in PIB and HO potentials in order to gain an intuitive understanding of how what is familiar in classical mechanics appears in quantum mechanical systems. revised 8/7/17 11:25 AM