Quantum Physics I, Lecture Note 10 - MIT OpenCourseWare

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Lecture 10: Solving the Time-Independent Schrödinger EquationB. ZwiebachMarch 14, 2016Contents1 Stationary States12 Solving for Energy Eigenstates33 Free particle on a circle.61Stationary StatesConsider the Schrödinger equation for the wavefunction Ψ(x, t) with the assumption that the potentialenergy V is time independent: Ψ 2 2ˆi HΨ(x,t) V(x)Ψ(x, t) ,(1.1) t2m x2ˆ with the time independent potentialwhere we displayed the form of the Hamiltonian operator HV (x). Stationary states are a very useful class of solutions of this differential equation. The signatureproperty of a stationary state is that the position and the time dependence of the wavefunctionfactorize. Namely,Ψ(x , t) g(t) ψ(x) ,(1.2)for some functions g and ψ. For such a separable solution to exist we need the potential to be timeindependent, as we will see below. The solution Ψ(x, t) is time-dependent but it is called stationarybecause of a property of observables. The expectation value of observables with no explicit timedependence in arbitrary states has time dependence. On a stationary state they do not have timedependence, as we will demonstrate.Let us use the ansatz (1.2) for Ψ in the Schrödinger equation. We then find dg(t)ˆi ψ(x) g(t) Hψ(x),(1.3)dtˆ We can then divide this equation by Ψ(x, t) g(t)ψ(x), givingbecause g(t) can be moved across H.i 1 dg(t)1 ˆ Hψ(x) .g(t) dtψ(x)(1.4)The left side is a function of only t, while the right side is a function of only x (a time dependentpotential would have spoiled this). The only way the two sides can equal each other for all values of tˆ has units of energy.and x is for both sides to be equal to a constant E with units of energy because HWe therefore get two separate equations. The first readsi dg Eg .dt1(1.5)

This is solved byg(t) e iEt/ ,(1.6)and the most general solution is simply a constant times the above right-hand side. From the xdependent side of the equality we getĤψ(x) Eψ(x) .(1.7)ˆ We showed that the eigenvaluesThis equation is an eigenvalue equation for the Hermitian operator H.of Hermitian operators must be real, thus the constant E must be real. The equation above is calledthe time-independent Schrödinger equation. More explicitly it reads 2 d2 V(x)ψ(x) Eψ(x) ,2m dx2(1.8)Note that this equation does not determine the overall normalization of ψ. Therefore we can writethe full solution without loss of generality using the g(t) given above:Stationary state:Ψ(x, t) e iEt/ ψ(x) ,withˆ Eψ .E 2 R and Hψ(1.9)ˆ the full stationary state isNote that not only is ψ(x) an eigenstate of the Hamiltonian operator H,ˆ eigenstatealso an HĤΨ(x, t) EΨ(x, t) ,(1.10)since the time dependent function in Ψ cancels out.We have noted that the energy E must be real. If it was not we would also have trouble normalizingthe stationary state consistently. The normalization condition for Ψ, if E is not real, would giveZZ 1 dx Ψ (x, t)Ψ(x, t) dx eiE t/ e iEt/ ψ (x)ψ(x)ZZ(1.11)2 Im(E)t/ i(E E)t/ dx ψ (x)ψ(x) edx ψ (x)ψ(x). eThe final expression has a time dependence due to the exponential. On the other hand the normalization condition states that this expression must be equal to one. It follows that the exponent mustbe zero, i.e., E is real. Given this, we also see that the normalization condition yieldsZ1 dx ψ (x)ψ(x) 1 .(1.12)1 ˆ on theHow do we interpret the eigenvalue E? Using (1.10) we see that the expectation value of Hstate Ψ is indeed the energyZZZ ˆˆ(1.13)hH iΨ dx Ψ (x, t)HΨ(x, t) dxΨ (x, t)EΨ(x, t) E dxΨ (x, t)Ψ(x, t) E,ˆ the uncertainty Hˆ of the Hamiltonian in a stationarySince the stationary state is an eigenstate of H,state is zero.There are two important observations on stationary states:2

ˆ on a stationary state Ψ is time(1) The expectation value of any time-independent operator Qindependent:ZZ ˆˆ iEt/ ψ(x)dx eiEt/ ψ (x)QehhQiiΨ(x,t) dx Ψ (x, t)QΨ(x, t) ZZ(1.14)iEt/ iEt/ ˆˆ dx eeψ (x)Qψ(x) dx ψ (x)Qψ(x) hhQiiψ(x) ,since the last expectation value is manifestly time independent.(2) The superposition of stationary states with different energies not stationary. This is clear becausea stationary state requires a factorized solution of the Schrödinger equation: if we add twofactorized solutions with different energies they will have different time dependence and theˆ maytotal state cannot be factorized. We now show that that a time-independent observable Qhave a time-dependent expectation values in such a state. Consider a superpositionΨ(x, t) c1 e iE1 t/ ψ1 (x) c2 e iE2 t/ ψ2 (x),(1.15)ˆ eigenstates with energies E1 and E2 , respectively. Consider a Hermitianwhere ψ1 and ψ2 are Hˆoperator Q. With the system in state (1.15), its expectation value isZ 1ˆt)dx Ψ (x, t)QΨ(x,hhQiiΨ 1 Z 1 ˆ 1 (x) c2 e iE2 t/ Qψˆ 2 (x) dx c 1 eiE1 t/ ψ1 (x) c2 eiE2 t/ ψ2 (x) c1 e iE1 t/ Qψ1Z 1 ˆ 1 jc2 j2 ψ2 Qψˆ 2 c 1 c2 ei(E1 E2 )t/ ψ1 Qψˆ 2 c2 c1 e i(E1 E2 )t/ ψ2 Qψˆ 1 dx jc1 j2 ψ1 Qψ1 (1.16)We now see the possible time dependence arising from the cross terms. The first two terms areˆ in the last term wesimple time-independent expectation values. Using the hermitically of Qthen gethhQiiΨ jc1 j2 hhQiiψ1 jc2 j2 hhQiiψ2Z Z1ˆ 2 c1 c 2 e i(E1 E2 )t/ c 1 c2 ei(E1 E2 )t/ dx ψ1 Qψ1 1 ˆ 2 ) dx ψ1 (Qψ(1.17)1 The last two terms are complex conjugates of each other and therefore Z 1ˆ 2 .dx ψ1 QψhhQiiΨ jc1 j2 hhQiiψ1 jc2 j2 hhQiiψ2 2 Re c 1 c2 ei(E1 E2 )t/ (1.18)1 We see that this expectation value is time-dependent if E1 6 E2 and (ψ1 , Qψ2 ) is nonzero. Thefull expectation value hhQiiΨ is real, as it must be for any Hermitian operator.2Solving for Energy EigenstatesWe will now study solutions to the time-independent Schrödinger equationĤψ(x) E ψ(x).3(2.19)

ˆ we are interested in finding the eigenstates ψ and the eigenvalues E,For a given Hamiltonian Hwhich happen to be the corresponding energies. Perhaps the most interesting feature of the aboveequation is that generally the value of E cannot be arbitrary. Just like finite size matrices have a setof eigenvalues, the above, time-independent Schrödinger equation may have a discrete set of possibleenergies. A continuous set of possible energies is also allowed and sometimes important. There areindeed many solutions for any given potential. Assuming for convenience that the eigenstates andtheir energies can be counted we writeψ1 (x) ,E1ψ2 (x) ,.E2.(2.20)Our earlier discussion of Hermitian operators applies here. The energy eigenstates can be organizedto form a complete set of orthonormal functions:Zψi (x)ψj (x) δij .(2.21)Consider the time-independent Schrödinger equation written asd2 ψ2m 2 (E V (x)) ψ .2dx (2.22)The solutions ψ(x) depend on the properties of the potential V (x). It is hard to make general statements about the wavefunction unless we restrict the types of potentials. We will certainly considercontinuous potentials. We also consider potentials that are not continuous but are piece-wise continuous, that is, they have a number of discontinuities. Our potentials can easily fail to be bounded.We allow delta functions in one-dimensional potentials but do not consider powers or derivatives ofdelta functions. We allow for potentials that become plus infinity beyond certain points. These pointsrepresent hard walls.We want to understand general properties of ψ and the behavior of ψ at points where the potentialV (x) may have discontinuities or other singularities. We claim: we must have a continuouswavefunction. If ψ is discontinuous then ψ 0 contains delta-functions and ψ 00 in the above left-handside contains derivatives of delta functions. This would require the right-hand side to have derivativesof delta functions, and those would have to appear in the potential. Since we have declared that ourpotentials contain no derivatives of delta functions we must indeed have a continuous ψ.Consider now four possibilities concerning the potential:(1) V (x) is continuous. In this case the continuity of ψ(x) and (2.22) imply ψ 00 is also continuous.This requires ψ 0 continuous.(2) V (x) has finite discontinuities. In this case ψ 00 has finite discontinuities: it includes the product ofa continuous ψ against a discontinuous V . But then ψ 0 must be continuous, with non-continuousderivative.(3) V (x) contains delta functions. In this case ψ 00 also contains delta functions: it is proportionalto the product of a continuous ψ and a delta function in V . Thus ψ 0 has finite discontinuities.4

(4) V (x) contains a hard wall. A potential that is finite immediately to the left of x a and becomesinfinite for x a is said to have a hard wall at x a. In such a case, the wavefunction willvanish for x a. The slope ψ 0 will be finite as x ! a from the left, and will vanish for x a.0Thus ψ is discontinuous at the wall.In the first two cases ψ 0 is continuous, and in the second two it can have a finite discontinuity. InconclusionBoth ψ and ψ 0 are continuous unless the potential has delta functionsor hard walls in which cases ψ 0 may have finite discontinuities.Let us give an slightly different argument for the continuity of ψ andwith a finite discontinuity, such as the step shown in Fig. 1.dψdx(2.23)in the case of a potentialFigure 1: A potential V (x) with a finite discontinuity at x a.Integrate both sides of (2.22) a to a , and then take ! 0. We find Z a Z a d dψ2mdx 2dx (E V (x))ψ(x) .dx dx a a The left-hand side integrand is a total derivative so we haveZ2m a dψdψ 2dx (V (x) E)ψ(x) . a dx a dx a (2.24)(2.25)By definition, the discontinuity in the derivative of ψ at x a is the limit as ! 0 of the left-handside: dψdψdψ.(2.26) lim a !0 dxdx a dx a Back in (2.25) we then have adψdx 2m lim 2 !0 a Zdx (V (x) E)ψ(x) .(2.27)a The potential V is discontinuous but not infinite around x a, nor is ψ infinite around x a and,of course, E is assumed finite. As the integral range becomes vanishingly small about x a theintegrand remains finite and the integral goes to zero. We thus have dψ 0.(2.28) adx5

There is no discontinuity in dψdx . This gives us one of our boundary conditions.To learn about the continuity of ψ we reconsider the first integral of the differential equation. Theintegration that led to (2.25) now applied to the range from x0 a to x yieldsZdψ(x)dψ2m x(2.29) (E V (x0 ))dx0 . x0dxdx x0Note that the integral on the right is a bounded function of x. We now integrate again from a toa . Since the first term on the right-hand side is a constant we findZZ xdψ2m a dxdx0 (E V (x0 )).(2.30)ψ(a ) ψ(a ) 2 a dx x0x0Taking the ! 0R limit, the first term on the right-hand side clearly vanishes and the second term goesxto zero because x0 dx0 (E V (x0 )) is a bounded function of x. As a result we have a ψ 0 ,(2.31)showing that the wavefunction is continuous at x a. This is our second boundary condition.3Free particle on a circle.Consider now the problem of a particle confined to a circle of circumference L. The coordinate alongthe circle is called x and we can view the circle as the interval x 2 [0, L] with the endpoints identified.It is perhaps clearer mathematically to think of the circle as the full real line x with the identificationx x L,(3.1)which means that two points whose coordinates are related in this way are to be considered the samepoint. If follows that we have the periodicity conditionψ(x L) ψ(x) .(3.2)From this it follows that not only ψ is periodic but all of its derivatives are also periodic.The particle is assumed to be free and therefore V (x) 0. The time-independent Schrödingerequation is then 2 d2 ψ E ψ(x) .(3.3) 2m dx2Before we solve this, let us show that any solution must have E 0. For this multiply the aboveequation by ψ (x) and integrate over the circle x 2 [0 , L). Since ψ is normalized we get 22mZLψ (x)0d2 ψdx Edx2Zψ (x)ψ(x)dx E .(3.4)The integrand on the left hand side can be rewritten as 22mZ0Lhd dψ dψ dψ iψdx E . dxdxdx dx6(3.5)

and the total derivative can be integrated 2 h dψ ψ2mdx x L ψ dψ dxix 0 2 2mZ0Ldψdx2dx E .(3.6)Since ψ(x) and its derivatives are periodic, the contributions from x L and x 0 cancel out and weare left withZ L 2dψ 2E dx 0 ,(3.7)2m 0 dxwhich establishes our claim. We also see that E 0 requires ψ constant (and nonzero!).Having shown that all solutions must have E 0 let us go back to the Schrödinger equation, whichcan be rewritten asd2 ψ2mE 2 ψ.(3.8)dx2 We can then define k via2mE 0.k2 (3.9) Since E 0, the constant k is real. Note that this definition is very natural, since it makesE 2 k 2,2m(3.10)which means that, as usual, p k. Using k 2 the differential equation becomes the familiard2 ψ k 2 ψ .dx2(3.11)We could write the general solution in terms of sines and cosines of kx, but let’s use complex exponentials:ψ(x) eikx .(3.12)This solves the differential equation and, moreover, it is a momentum eigenstate. The periodicitycondition (3.2) requireseik(x L) eikx !eikL 1 !kL 2πn , n 2 Z.(3.13)We see that momentum is quantized because the wavenumber is quantized! The wavenumber hasdiscrete possible values2πnkn , n 2 Z.(3.14)LAll integers positive and negative are allowed and are in fact necessary because they all correspond todifferent values of the momentum pn kn . The solutions to the Schrödinger equation can then beindexed by the integer n:ψn (x) N eikn x ,(3.15)where N is a real normalization constant. Its value is determined fromZ LZ L1 p ,1 ψn (x)ψn (x)dx N 2 dx N 2 L ! N L007(3.16)

so we have11 2πinx eikn x p e L .ψn (x) pLL(3.17)The associated energies are 2 kn2 2 4π 2 n22π 2 2 n2 .(3.18)22m2mLmL2There are infinitely many energy eigenstates. We have degenerate states because En is just afunction of jnj and thus the same for n and n. Indeed ψn and ψ n both have energy En . The onlyp1 , which is a constant wavefunction with zero energy.nondegenerate eigenstate is ψ0 LEn Whenever we find degenerate energy eigenstates we must wonder what makes those states different, given that they have the same energy. To answer this one must find an observable that takesdifferent values on the states. Happily, in our case we know the answer. Our degenerate states can bedistinguished by their momentum: ψn has momentum 2πn L and ψ n has momentum ( 2πn L ).Given two degenerate energy eigenstates, any linear combination of these states is an eigenstatewith the same energy. Indeed ifˆ 1 Eψ1 , Hψˆ 2 Eψ2 ,Hψ(3.19)thenˆˆˆH(aψ1 bψ2 ) aHψ1 bHψ2 aEψ1 bEψ2 E(aψ1 bψ2 ) .(3.20)We can therefore form two linear combinations of the degenerate eigenstates ψn and ψ n to obtainanother description of the energy eigenstates:ψn ψ n cos(kn x) ,ψn ψ n sin(kn x) .(3.21)While these are real energy eigenstates, they are not momentum eigenstates. Only our exponentialsˆ and p̂.are simultaneous eigenstates of both HThe energy eigenstates ψn are automatically orthonormal since they are p̂ eigenstates with nodegeneracies (and as you recall eigenstates of a hermitian operator with different eigenvalues areautomatically orthogonal) :Z LZ1 L 2πi(m n)x Lψn (x)ψm (x)dx edx δmn .(3.22)L 00They are also complete: we can then construct a general wavefunction as a superposition that is infact a Fourier series. For any Ψ(x, 0) that satisfies the periodicity condition, we can writeXΨ(x, 0) an ψn (x),(3.23) n2Zwhere, as you should check, the coefficients an are determined by the integralsZ Lan dx ψn (x) Ψ(x, 0) .(3.24)0The initial state Ψ(x, 0) is then easily evolved in time:XiEn tΨ(x, t) an ψn (x)e .(3.25) n2ZAndrew Turner transcribed Zwiebach’s handwritten notes to create the first LaTeX version of thisdocument.8

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Lecture 10: Solving the Time-Independent Schr odinger Equation B. Zwiebach March 14, 2016 Contents 1 Stationary States 1 2 Solving for Energy Eigenstates 3 3 Free particle on a circle. 6 1 Stationary States Consider the Schr odinger equation for the wavefunction (x;t) with the assumption that the potential energy V is time independent: @ 2 @2

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