CHAPTER 2

TIME-INDEPENDENTSCHRODINGER EQUATION

2.1 STATIONARY STATES

In Chapter 1 we talked a lot about the wave function, and how you use it tocalculate various quantities of interest. The time has come to stop procrastinating,and confront what is, logically, the prior question: How do you get Y (x . t ) in the`b,'st place? We need to solve the Schr6dinger equation,

[2.1]

for a speci6ed potential 1 V (x . t ) . In this chapter (and most of this book) I shallassume that V is fl.Icfependenf of t . In that case the Schr6dinger equation can besolved by the Inethod of separation of variables (the physicist's first line of attackon any partial differential equation): We look for solutions that are simple products,

LP(x . T) = (X) j (f ). [2.2]

where (losner"case) is a function of x alone, and j is a function of t alone` Onits face, this is an absurd restriction, and we cannot hope to get more than a tiny

lit is tiresome to keep saying .'potential energy function... so most people just call V the..potential`. even though this invites occasional confusion with e/ectrlc potential. which is actmtllypotential energy per mt it charge.

24

Section 2.1: Stationa States 25

subset of all solutions in this way. But hang on, because the solutions we do obtaintum out to be of great interest. Moreover (as is typically the case with separationof variables) we will be able at the end to patch together the separable solutionsin such a way as to construct the most general solution.

For separable solutions we have

(ordinary derivatives, now), and the Schr6dinger equation reads

Or, dividing through by j :

Idj h2 I d2

th-y dt- - `2` d-x2 + V [2.3]

Now, the left side is a function of t alone, and the right side is a function ofx alone.2 The only way this can possibly be true is if both sides are in factcovzstant -otherwise, by varying t , I could change the left side without touchingthe right side, and the two would no longer be equal. (That's a subtle but crucialargument, so if it's new to you, be sure to pause and think it through,) For reasonsthat will appear in a moment, we shall call the separation constant E. Then

ih _1_ dj, ,, E,j dt ~

dj IE

dt- " ` h- j '

or

and

'or

[2.4]

h2 1 d2

-2-m dx2- + V" E'

[2.5]

Separation of variables has turned a partial differential equation into two ov.dl-`navy differential equations (Equations 2.4 and 2.5). The first of these (Equation 2,4)

2Note that this would nol be true if V were a function of / as well as .\-.

26 Cbapter 2 Time-Independent Schr6dinger Equation

is easy to solve Oust multiply through by dt and integrate); the general solution isC exp(-i Et|h), but we might as well absorb the constant C into (since the quantityof interest is the product j ). Then

j(t) ,_ e-lEt/h. [2.6]

The second (Equation 2.5) is called the time-independent Schr8dinger equation;we can go no further with it until the potential V(x) is specified.

The rest of this chapter will be devoted to solving the time-independentSchr6dinger equation, for a variety of simple potentials. But before I get tothat you have every right to ask: What 's so great about separable solutions?After all, most solutions to the (time dependent) Schr6dinger equation do nottake the form (x )j Ct ) . I offer three answers-two of them physical, and onemathematical:

1. They are stationary states. Although the wave function itself,

F (x , t ) ,_ f (x)e-i El /h ,

does (obviously) depend on t , the probabtl-ity density,

{2,7]

IF(x, t)|2 _- F*F -, *e+iE'/he-iEI/h ,, |(X)|2, [2.8]

does not -the time-dependence cancels out.3 The same thing happens in calculat-ing the expectation value of any dynamical variable; Equation 1.36 reduces to

[2.9]

Every expectation value ts constant in time; we might as well drop the factor j (t)altogether, and simply use in place of Y. (Indeed, it is common to refer to as"tfte wave function," but this is sloppy language that can be dangerous, and it isimportant to remember that the true wave function always carries that exponentialtime dependent factor.) In particular, (x) is constant, and hence (Equation 1.33)(p} - O. Nothing ever happens in a stationary state.

2. They are states of dejinite total energy. In classical mechanics, the totalenergy (kinetic plus potential) is called the Hamiltonian:

[210]H (x , p) - r + v (x ) .` ' ' 2m ` '

3For normalizable solutions, E must be real (see Problem 2.1 (a))

Section 2.1: Stationary States "27

The corresponding Hamiltonian operator, obtained by the canonical substitutionp (h/i)(3|Jx), is therefore4

^ k2 a2H - - 2 Bx- 2 + V(x)` [2.11]

Thus the time-independent Schr6dinger equation (Equation 2.5) can be written

{2.12]

and the expectation value of the total energy is

(Notice that the normalization of F entails the normalization of .) Moreover,

and hence

So the variance of H is

s2H -_ (H 2) , (H )2 -, E 2 - E2 -, O. [2,14]

But remember, if s -, O, then every member of the sample must share the samevalue (the distribution has zero spread). Conclusion.` A separable solution has theproperty that eve,.y measurement of the total energy is certain to rerun? the valueE . (That's why I chose that letter for the separation constant)

3. The general solution is a !Wear combination of separable solutions. Aswe're about to discover, the time-independent Schr6dinger equation (Equation 2.5)yields an infinite collection of solutions (I(x), 2(X), 3(X), . . . ), each withits associated value of the separation constant (El , E2, E3, . . . ); thus there is adifferent wave function for each allowed energy:

Fl (x , t ) _- tll (x)e-i E I 'Ir' , .F2 (x , t ) -` 2 (X)e-I E2'Ih , . . .

Now (as you can easily check for yourself) the (time-dependent) Schr6dingerequation (Equation 2.1) has the property that any linear combinations of solutions

4Whenevcr confusion might arise. I'll put a '.fiat" (") on the operator, to distinguish it from thedynamical variable it represents.

5A {mear combmanon of the functions fl (z) ~ f2 (z) . . is an expression of the form

.f (Z) - cl f I (z) + c2 f2 (z) + . ̀ -

where ct , c2 , . . . are any (complex) constants.

28 Chaj>ter 2 Time-Independent Scbr6dinger Equa,tion

is itself a solution. Once we have found the separable solutions, then, we canimmediately construct a much more general solution, of the form

F (X. t) `- S Cu (x)e-i E"' Ir' .ll-l

[2.15]

It so happens that every solution to the (time-dependent) Schr6dinger equationcan be written in this form-it is simply a matter of 6nding the right constants(cl , C2 . . . ) SO as to fit the initial conditions for the problem at hand. You'll seein the following sections how all this works out in practice, and in Chapter 3 we'llput it into more elegant language, but the main point is this: Once you've solvedthe time-independent Schr6dinger equation, you're essentially done; getting fromthere to the general solution of the time-dependent Schr6dinger equation is, inprinciple, simple and straightforward

A lot has happened in the last four pages, so let me recapitulate, from asomewhat different perspective. Here's the generic problem: You're given a (time-independent) potential V(x), and the starting wave function F(x, 0); your job isto find the wave function, F(x, t), for any subsequent time t . To do this you mustsolve the Clime-dependent) Schr6dinger equation (Equation 2.1). The strategy6 isfirst to solve the time-nzdependent Schr6dinger equation (Equation 2.5); this yields,in general, an infinite set of solutions ( Lll (x ), L/12(X), 3 (x ), . . . ), each with its ownassociated energy (El , E2 , E3, . . ). To fit Y(x, 0) you write down the generallinear combination of these solutions:

qJ (x , 0) - Cu fl(X):

n.-l

[2.16]

the miracle is that you can a/ways~ match the specified initial state by appropriatechoice of the constants cl , c2, 03 , . . . To construct Y(x , t ) you simply tack ontoeach term its characteristic time dependence, exp(-i E,, t / h) :

[2.171

[2.18]

The separable solutions themselves,

Fn (x , t ) - tln (x)e-i En' /h ,

6Occasionally you can solve the time-dependent Schrgdinger equation without recourse Lo sep-aration of variables-see, for instance Problems 2.49 and 2.50. But such cases al.e extremely rare.

Section 2.1: Stationary States 29

are stationary states, in the sense that all probabilities and expectation values areindependent of time, but this property is emphatically not shared by the generalsolution (Equation 2.17); the energies are different, for different stationary states,and the exponentials do not cancel, when you calculate | F |2.

Example 2.1 Suppose a particle starts out in a linear combination of just ttvostationary states:

LP (X , 0) -` CI 1 (X ) + c2 t1V2 (X ) .

(To keep things simple Gil assume that the constants cu and the states fl (x) m`ereal.) What is the wave function F (x , t ) at subsequent times? Find the probabilitydensity, and describe its motion.

Solution: The first part is easy:

F (.- , t ) ,` CI I (x )e-i E I ' /h + c22 (x )e-i E2'Ih .

where El and E2 are the energies associated with I and t12. It follows that

| F CA- , t )12 = (CI I el E I ' Ih + c22ei E2lh ) (cl I e-i E I t /h + c22e-i E/h )

CI used Euler's formula, exp i q = cos q + i sin q, to simplify the result.) Evidentlythe probability density oscillates sinusoidally, at an angular frequency ( E2 - E I ) /h;this is certainly not a stationary state. But notice that it took a linear combivlationof states (with different energies) to produce motion.7

*Problem 2.1 Prove the following three theorems:

For normalizable solutions, the separation constant E must be real. Ht.nt.,Write E (in Equation 2,7) as Eo + ir (with Eo and G real), and show thatif Equation 1.20 is to hold for all t, P must be zero.

The time-independent wave function tl(x) can always be taken to be real(unlike Y(x , t ), which is necessarily complex). This doesn't mean that everysolution to the time independent Schr6dinger equation is real; what it saysis that if you've got one that is not, it can always be expressed as a linearcombination of solutions (with the same energy) that are. So you might aswell stick to 's that are real. Hint'. If tl(x) satisfies Equation 2.5, for agiven E, so too does its complex conjugate, and hence also the real linearcombinations ( + *) and i ( - * ) .

(a)

Cb)

7This is nicely illustrated by an applet at the Web site http://thorin.adnc.com/~topquark/quantum/deepwellmain.html.

30 Chapter 2 Time'.fndependent Schr6dinger Equation

Cc) If V(x) is an even fnnc6on (that is, V(-x) = V(x)) then (x) can alwaysbe taken to be either even or odd. Hint.` If (x) satisfies Equation 2.5, fora given E, so too does (-x), and hence also the even and odd linearcombinations (x) (-x).

*Problem 2.2 Show that E must exceed the minimum value of V(x), for everynormalizable solution to the time-independent Schdinger equation. What is theclassical analog to this statement? Hint: Rewrite Equation 2.5 in the form

if E < Vmin , then and its second derivative always have the same sign -arguethat such a function cannot be normalized.

THE INFIMTE SQUARE WELL_______________________________________2.2

Suppose

[2.19]

(Figure 2.I). A particle in this potential is completely free, except at the two ends(x = O and x = ct ), where an infinite force prevents it from escaping. A classicalmodel would be a cart on a frictionless horizontal air track, with perfectly elasticbumpers-it just keeps bouncing back and forth forever. (This potential is artifi-cial, of course, but I urge you to treat it with respect. Despite its simplicity-orrather, precisely because of its simplicity-it serves as a wonderfully accessi-ble test case for all the fancy machinery that comes later. We'll refer back to itfrequently.)

FIGURE 2.1: The infinite square well poten-tial (Equation 2.19).

Section 2.2: Tbe Innite Square Well 31

Outside the well, (x) = 0 (the probability of finding the particle there iszero). Inside the well, where V = 0, the time-independent Schr6dinger equation(Equation 2.5) reads

/t2 d2

`z dx2- " E' [2'20]

or

[2.21]

(By writing it in this way, I have tacitly assumed that E 0; we know fromProblem 2.2 that E < 0 won't work.) Equation 2.21 is the classical simple har-monic osciflatol equation; the general solution is

(x ) -` A sinkx + B cos kx, [2.22]

where A and B are arbitrary constants. Typically, these constants are fixed by theboundary conditions of the problem. What are the appropriate boundary con-ditions for (x)? Ordinarily, both and d|dx are continuous, but where thepotential goes to infinity only the first of these applies. CPR prove these boundaryconditions, and account for the exception when V -` oc, in Section 2.5; for now Ihope you will trust me.)

Continuity of (x) requires that

(0) - (a) - 0, [2.23]

so as to join onto the solution outside the well. What does this tell us about A andB? Well,

(0) - A sin 0 + B cos 0 = B ,

so B `- 0, and hence(x) = A sinkx [2.24]

Then (a) = A sinka, so either A `- 0 (in which case we're left with the triv-ial-non-normalizable-solution (x) = 0), or else sinka = 0, which meansthat

ka - 0, P, -+ 2P , 3P , . . . [2.25]

But k - 0 is no good (again, that would imply (x) = 0), and the negativesolutions give nothing new, since sin(`q) -` ` sin(q) and we can absorb theminus sign into A. So the disti,zct solutions are

HPkn - - . with n -` I, 2, 3, . . . [2.26]

32 Cbapter 2 Time-Judependent Schr6dinger Equation

yl(x) 4

FIGURE 2.2: Thefirst three stationary states of the infinite squarewell (Equation 2.28).

Curiously, the boundary condition at x = a does not determine the constantA, but rather the constant k, and hence the possible values of E:

[2.27]

In radical contrast to the classical case, a quantum particle in the infinite squarewell cannot have just any old energy-it has to be one of these special aHowedvalues.8 To find A, we normalize :

This only determines the magnitude of A, but it is simplest to pick the positive realroot: A = a (the phase of A carries no physical signi6cance anyway). Insidethe well, then, the solutions are

[2.28]

As promised, the time"-independent Schr6dinger equation has delivered aninfinite set of solutions (one for each positive integer n). The first few of these areplotted in Figure 2.2. They look just like the standing waves on a string of length a;Ikl , which carries the lowest energy, is called the ground state, the others, whoseenergies increase in proportion to 772, are called exdted states. As a collection, thefunctions n(x) have some interesting and important properties:

1. They are alternately even and odd, with respect to the center of the well:I is even, 2 is odd, 3 is even, and so on.9

SNolice that the quantization of enelgy emerged as a rather technical consequence of the bound-aly conditions on solutions lo the time-independent Schr6dInger equation.

9To make this symmetry more apparent. some authors center the well at the origin (running itfrom -a to +a). The even functions are then cosines. and the odd ones are sines. See Problem 2.36.

Section 2.2: The Inm'te Square Well 33

2. As you go up in energy, each successive state has one more node (zero-crossing): I has none (the end points don't count), 2 has one, 3 has two, andso on.

3. They are mutually orthogonal, in the sense that

[2.29]

whenever m n. Proof:

__!sin[(m - n)P ] _ sin[(m + n)P]}_,O.

P I (m - n) (m + n) I

Note that this argument does not work if m = n. (Can you spot the point at whichit fails?) In that case nolmalization tells us that the integral is I . In fact, we cancombine orthogonality and normalization into a single statement:1o

[2.30]

[2.31]

where Soto (the so-called Kronecker delta) is defined in the usual way,

& _ j 0, if m n;umn - I I if ,n _- n .

We say that the 's are orthonormal4. They are complete, in the sense that any other function, f (x), can be

expressed as a linear combination of them:

00Y\ |nP \>ell sin(-x) [2.32]

� . ,a In-I

loin this case file s are real. so the * on m is unnecessary. but for future purposes It`s a goodidea to get in the habit of putting it there.

34 Chapter 2 Time-fndependent Schr6dinger Equation

I'm not about to prove the completeness of the functions sin (nnx/a), but if you'vestudied advanced calculus you will recognize that Equation 2.32 is nothing but theFourier series for f (x), and the fact that "any" function can be expanded in thisway is sometimes called Dirichlet's theorem.ll

The coefficients cvz can be evaluated-for a given f (x)-by a method I callFourier's trick, which beautifully exploits the orthononnalify of {n}; Multiplyboth sides of Equation 2.32 by tn (x)* , and integrate.

(Notice how the Kronecker delta kills every tenn in the sum except the one forwhich n = m.) Thus the nth coefficient in the expansion of f (x) 1812

cvz = ) tbvn (x )* f (.,-) dx. [2.34]

These four properties are extremely powerful, and they are not peculiar to theinfinite square well. The first is true whenever the potential itself is a symmetricfunction; the second is universal, regardless of the shape of the potential.13 Orthog-onality is also quite general-G ll show you the proof in Chapter 3, Completenessholds for all the potentials you are likely to encounter, but the proofs tend to benasty and laborious; Gm afraid most physicists simply assume completeness, andhope for the best.

The stationary states (Equation 2.18) of the infinite square well are evidently

[2.35]

I claimed (Equation 2.17) that the most general solution to the (time-dependent)Schr6dinger equation is a linear combination of stationary states:

' 1 See, for example, Mary Boas, Mathematical Methods z`vz the Physical Sciences. 2d ed. (NewYork: John Wiley. 1983) p. 313: f (x) can even have a finite number of finite discontinuities.

I2It doesn't matter whether you use vzz or ,z as the -dummy index" here (as long as you areconsistent on the two sides of the equation. of course): whatever Letter you use. it just stands for �.anypositive integer.`.

l3See. for example, John L. Powell and Bemd Crascnlann, Qzzavzrmn Mechanics (Addison-Wesley, Reading. MA, 1961 ). p. 126.

Section 2.2.' Tbe Inm'te Square Well 35

(If you doubt that this is a solution, by all means check it!) It remains only forme to demonstrate that I can fit any prescribed initial wave function, Y(x 0), byappropriate choice of the coefficients cn:

The completeness of the 's (confirmed in this case by Dirichlet's theorem) gum-antees that I can always express Y(x . 0) in this way, and their orthononnalitylicenses the use of Fourier's trick to determine the actual coefficients:

[2.37]

That does it: Given the initial wave function, F (x . O), we first compute theexpansion coefficients C/t , using Equation 2.37, and then plug these into Equation 2.36to obtain F (x . t ). Armed with the wave function, we are in a position to compute anydynamical quantities of interest, using the procedures in Chapter I. And this sameritual applies to any potential-the only things that change are the functional folmof the s and the equation for the allowed energies.

Example 2.2 A particle in the inBnite square well has the initial wave function

Y (x . 0) = Ax (ct - x) (O _< x a).

for some constant A (see Figure 2.3). Outside the well, of course, Y = O. Findq(x, t).

Ame

FIGURE 2.3: The starting wave function in Example 2.2.

36 Cbapter 2 Time-Independent Scbro..dinger Equation

Solution: First we need to determine A, by normalizing F (X . O):

so

A |30rt- --&1 Va5'

The nth coefficient is (Equation 2.37)

CH

__ j O, _ if tr is even.

| 8v[i-5/ Cup )3. if n is odd.

Thus (Equation 2.36):

Loosely speaking, en tells you the "amount of ti that is contained in Y."Some people like to say that |cvl 12 is the "1:)robability of Duding the particle in thenth stationary state," but this is bad language; the particle is in the state F , notYti , and, anyhow, in the laboratory you don't ..find a particle to be in a particularstate"-you measut`e some obsen'able, and what you get is a number. As we.IIsee in Chapter 3, what |cu 12 tells you is the probabili that a measurement of the

Section 2.2.' The Inm.te Square Well 37

energy wou\d yield the value En (a competent measurement will always return oneof the "allowed" values-hence the name-and |cu|2 is the probability of gettingthe particular value En ).

Of course, the sum of these probabilities should be 1,

[2.38]

Indeed, this follows from the normalization of F (the Ct, ' s are independent of time,so I'm going to do the proof for t = O; if this bothers you, you can easily generalizethe argument to arbitrary t).

00 00 co

-SSC,,,*Cudmu " S |cu|2'

vi- I m-I ti -I

(Again, the Kronecker delta picks out the term m = n in the summation over m.)Moreover, the expectation value of the energy must be

[2.39]

and this too can be checked directly: The time-independent Schr6dinger equation(Equation 2.12) says

H ,/ = Et, ,/ [2.40]

SO

(H ) `, / F* H F dx -,J (Sc,,,)* H (Scu ) dx

" SS cm* cu Eu | dx `- S lcnl2 En.

Notice that the probability of getting a particular energy is independent of time, andso, ct fortiori, is the expectation value of H . This is a manifestation of conservationof energy in quantum mechanics.

38 Chaptor 2 Time-f?2dependent Schr6dinger Equation

Example 2.3 In Example 2.2 the starting wave function (Figure 2.3) closely re-sembles the ground state I (Figure 2.2). This suggests that |cl 12 should dominate,and in fact -

Icl|2 =(8p/-5) `0.998555 . . .

The rest of the coef6cients make up the difference:14

|85\2 oc 1S |c,,|- = ( -') S = 1.?,=1 \ | VI -1.3.5 ...

The expectation value of the energy, in this example, is

oc |8v/j[-5 \2 ,72 p2tz2 480h2 oo 1 5h2(H) = S | ~ ) "2,,za'2L -` p 4'o 2 S. - 2 '

,/=1.3.5.... I / ,/=1.3.5....

As one might expect, it is very close to E I = P2h2/2ma2 - slightly larger, becauseof the admixture of excited states.

Problem 2.3 Show that there is no acceptable solution to the {time"independent)Schr6dinger equation for the in6nite square well with E = 0 or E < O. (This is aspecial case of the general theorem in Problem 2.2, but this time do it by explicitlysolving the Schr6dinger equation, and showing that you cannot meet the boundaryconditions.)

*Problem 2.4 Calculate (x), (X2), (p}, {p2}, sx, and sp, for the nth stationary stateof the in6nite square well. Check that the uncertainty principle is satisfied. Whichstate comes closest to the uncertainty limit?

*Problem 2.5 A particle in the infinite square well has as its initial wave functionan even mixture of the first two stationary states:

t4You can look up the series

I I I p6

+ + + ' . - - 96

and

I } 1 Jr+

I-:f + + 5 + ` ̀ ̀ = 9

in math tables. under ..Sums of Reciprocal Powers.. or "Riemann Zeta Function."

Section 2.2: The Inme Square Well 39

(a) Normalize F (x , O). (That is, find A. This is very easy, if you exploit theorthonormality of tltl and 11t2. Recall that, having nonnalized Y at t = 0,you can rest assured that it stays normalized-if you doubt this, check itexplicitly after doing part (b).)

Cb) Find F (x , t) and | F (x , t )12. Express the latter as a sinusoidal function oftime, as in Example 2.1. To simplify the result, let w P2h/2ma2 .

(c) Compute (x). Notice that it oscillates in time. What is the angular frequencyof the osclllation? What is the amplitude of the oscillation? (If your amplitudeis greater than a/2, go directly to jail.)

Cd) Compute (p). (As Peter Lorre would say, "Do it ze kveek nay, Johnny!")

Ce) If you measured the energy of this particle, what values might you get, andwhat is the probability of getting each of them? Find the expectation valueof H . How does it compare with El and E2?

Problem 2.6 Although the overall phase constant of the wave function is of nophysical signi6cance (it cancels out whenever you calculate a measurable quantity),the re/ative phase of the coefficients in Equation 2.17 does matter. For example,suppose we change the relative phase of tltl and t1t2 in Problem 2.5:

where F is some constant. Find F (x , t), | F (x , t ) 12, and (x) , and compare yourresults with what you got before. Study the special cases & = P/2 and & = P.(For a graphical exploration of this problem see the applet in footnote 7.)

*Problem 2.7 A particle in the in6nite square well has the initial wave functionl5

(a) Sketch F(x. O), and determine the constant A.

Cb) Find F (x t ) .

\5There is no restriction in principle on the shape of the starting wave function. as longas it is normalizable. In pm-Ocular. (x . 0) need not have a continuous derivative-in fact, itdoesn't even have to be a cottlivmotts function. However. if you try lo calculate (H) usingf F(x . 0)* H F(x . 0) dx in such a case, you may encounter technical difficulties~ because the secondderivative of F (x . O) is ill-defined. It works in Problem 2.9 because the discontinuities occur at the endpoints. where the wave function is zero anyway~ In Problem 2.48 youII see bow to manage cases likeProblem 2.7.

40 Cbapter 2 Time-Independent Scbr6dinger Equation

(c) What is the probability that a measurement of the energy would yield thevalue El?

Cd) Find the expectation value of the energy.

Problem 2.8 A particle of mass m in the infinite square well (of width a) startsout in the left half of the well, and is (at r = O) equally likely to be found at anypoint in that region.

(a) What is its initial wave function, F(x, O)? (Assume it is real. Don't forgetto normalize it.)

Cb) What is the probability that a measurement of the energy would yield thevalue p2h2/2ma2?

Problem 2.9 For the wave function in Example 2.2, find the expectation value ofH , at time t = O, the "old fashioned" way:

( H ) - ] F (x . O)*H F (x, O) dx .

Compare the result obtained in Example 2.3. Note.' Because (H) is independent oftime, there is no loss of generality in using r = O.

2.3 THE HARMONIC OSCILLATOR______________________________________

The paradigm for a classical harmonic oscillator is a mass tn attached to a springof force constant k. The motion is governed by Hooke's law,

d2xF -, -kx -, m

dt2

(ignoring friction), and the solution is

x(r) - A sin(wr) + B cos(wt),

where/k

w- \|-Vm

[2.41]

[2.42]

is the (alIgular) frequency of oscillation. The potential energy is

V(x) = 2-1 kx2;

its graph is a parabola.

Section 2.3: Tbe Harmom"c Oscillator 41

FIGURE 2.4: Parabolic approximation (dashed curve) to an arbitrary potential, inthe neighborhood of a local minimum.

Of course, there's no such thing as a pe,fect harlnonic oscillator-if youstretch it too far the spring is going to break, and typically Hooke's law failslong before that point is reached. But practically any potential is approximatelyparabolic, in the neighborhood of a local minimum (Figure 2.4). Formally, if weexpand V(x) in a Taylor series about the minimum:

V(x) = V(xO) + V'(xo)(x - xO) + ~V"(xo)(x - xO)2 + ' . ' ,

subtract V(xO) (you can add a constant to V(x) with impunity, since that doesn'tchange the force), recognize that V'(xO) = O (since xo is a minimum), and drop thehigher-order terms (which are negligible as long as (x - xO) stays small), we get

V(x) @ V"(XO)(x - xO)2,2

which describes simple harmonic oscillation (about the point xO), with an effectivespring constant k = V"(xO).16 That's why the simple harmonic oscillator is soimportant; Virtually any oscillatory motion is approximately simple harmonic, aslong as the amplitude is small.

The quantum problem is to solve the Schr6dinger equation for the potential

V(x) = mW2X2 [2.43]2

(it is customary to eliminate the spring constant in favor of the classical frequency,using Equation 2.41). As we have seen, it suffices to solve the time-independentSchr6dinger equation:

h2 d2 I - 2/ dx + mm-x = E' [2'44]

loNote that V"(XO) O. since by assumption xO is a mivtimum. Only in the rare case V"(xO) = 0is the oscillation not even approximately simple harmonic.

42 Chapter 2 Time-independent Schro"dinger Equation

In the literature you will find two entirely different approaches to this problem.The first is a Sn.aightforward "brute force" solution to the differential equation,using the power series method; it has the virtue that the same strategy can beapplied to many other potentials (in fact, we'll use it in Chapter 4 to treat theCoulomb potential). The second is a diabolicaHy clever algebraic technique, usingso-called ladder operators. Gll show you the algebraic method 6rst, because it isquicker and simpler (and a lot more fun);17 if you want to skip the power seriesmethod for now, that' s fine, but you should certainly plan to study it at somestage.

2.3.1 Algebraic Method

To begin with, let's rewrite Equation 2.44 in a more suggestive form:

[2.45]

idea is to

[2.46]

and oper-this does

[2.47]

where p - (h|i)d|dx is, of course, the momentum operator. The basicfactor the Hamiltonian,

H = [ p2 + (InWX)2].

If these were numbers, it would be easy:

it 2 + V2 `- (i u + v)(-iu + v) .

Here, however, it' s not quite so simple, because p and x are operators,-ators do not, in general, commute (xp is not the same as px). .Still,motivate us to examine the quantities

04_ -__ 1_

- L /?...hInto

(the factor in front is just there to make the final results look nicer).Well, what is the product a-a+?

1a-ct+ -` 2 2t--nw (iP + mwx)(-iP + "'wx)

17-We 'II encounter some of the same strategies in file theoly of angular momentum (Chapter 4).and the technique generalizes to a broad class of potentials in super-symmetric quantum mechanics(sce. for example. Richard W~ Robinett. Qtlanttmt Mechavtics. (Oxford U,P.. New York. 1997)~ Section14.4),

Section 2.3: Tbe Harmonic Oscillator 43

As anticipated, there's an extra term, involving (xp - px). We call this the com-mutator of x and p; it is a measure of how badly they fail to commute. In general,the commutator of operators A and B (written with square brackets) is

[A, B] - AB - BA.

In this notation,

[2.48]

[2.49]u_a+ _- I [p2 + (mwx)2] - [x. p].' 2kmw " ` ' A 2h ̀ r A

We need to figure out the colnmutator of x and p, Warning.' Operators arenotoriously slippery to work with in the abstract, and you are bound to makemistakes unless you give them a "test function," f (x), to act on. At the end youcan throw away the test function, and you'll be left with an equation involving theoperators alone. In the present case we have:

[x. P] f (x) -[x~d-dx (f ) - ~ d-dx (xf )]-~(x~-'-\-~- f)- ik f (.,-).

[2.50]Dropping the test function, which has served its purpose,

[2.511

This lovely and ubiquitous result is known as the canonical commutation rela-tion.18

With this, Equation 2.49 becomes

[2.52]

[2.53]

or

H - hw(a-a+ - -~)Evidently the Hamiltonian does nor factor perfectly-there' s that extra -I/2 on theright. Notice that the ordering of a+ and a- is important here; the same argument.with a+ on the left, yields

1 Ia+a- - H - 2-'

In particular,[a- , a+] - 1.

[2.54]

[2.55]

}Sin a deep sense all of the mysteries of quantum mechanics can be traced to the fact that positionand momentum do not commute. Indeed some authors take the canonical commutation relation as anario'fl of the theory. and use it to derine p - (h|i)d|dx.

44 Chapter 2 Time-Independent Scbr6dInger Equation

So the Hamiltonian cml equally well be written

[2.56]

In terms of a , then, the Schr6dinger equationl9 for the harmonic oscillator takesthe form

kw(a +_ ~) = [2.57]

(in equations like this you read the upper signs all the way across, or else the lowersigns).

Now, here comes the crucial step: I claim that saries the Schr6dingerequation with e,lergy E, (that is: H = E ), then a+ saris.|ies the Schr6dingerequation with energy ( E + hw): H (a+ ) - ( E + kw)(a+ ) . Proof:

H (a+F) - kw(a+a- + - (a+) - kw(a+a-a+ + -~a+)

- hwa+(a-a+ + -:)- a+[kw(a+a- + I + ] - a+(H + kw) - a+(E + kw) - (E + kw)(a+) .

(I used Equation 2.55 to replace a-a+ by a+a- + I, in the second line. Noticethat whereas the ordering of a+ and a- does matter, the ordering of a andany constants-such as k, w, and E-does not; an operator commutes with anyconstant.)

By the same token, a- is a solution with energy ( E - kw):

Here, then, is a wonderful machine for generating new solutions, with higher andlower energies-if we coLlld just find one solution, to get started! We call aladder operators, because they allow us to climb up and down in energy; a+ isthe raising operator, and a- the lowering operator. The "ladder" of states isillustrated in Figure 2.5.

19I'm gelling tired of writing "tinIc-independent Schr6dinger equation,.. so when its clear fromthe context which one I mean. III just call it the ..Schr6dinger equation."

Section 2.3: The Harmonic OsaIlator 45

FIGURE 2.5: The "ladder" of states for the harmonic oscillator.

But wait! What if I apply the lowering operator repeatedly? Eventually Gmgoing to reach a state with energy less than zero, which (according to the generaltheorem in Problem 2.2) does not exist! At some point the machine must fail.How can that happen? We know that a- is a new solution to the Schr6dingerequation, but there is no guarantee that it will be norma/izable-it might be zero,or its square-integral might be infinite. In practice it is the former: There occurs a"lowest rung" (call it O) such that

a- o ̀ , 0.

We can use this to determine o(x):

[2.58]

46 Cbalter 2 Time-Independent Scbr6dinger Equation

or

This differential equation is easy to solve:

constant,

so

We might as well normalize it right away:

1 = |A|2 {00 e-",wx2/h dx ,, |A|2 \/p-kJ -!:>o Y mw

so A2 -` mw|P.,[-k, and hence

[2.59]

To detennine the energy of this state we plug it into the Schr6dinger equation (inthe form of Equation 2.57), hw(a+a- + I/2)o = Eoo, and exploit the fact thatct-o = O:

Eo - -1 kw. [2.60]< 2 ` -

With our foot now securely planted on the bottom rung (the ground state of thequantum oscillator), we simply apply the raising operator (repeatedly) to generatethe excited states,20 increasing the energy by kw with each step:

[2.611

where An is the normalization constant. By applying the raising operator (repeat-edly) to o, then, we can (in principle) construct a1l2l the stationary states of

2010 the case of the harmonic oscillator it is customary. for some leason, to depall from the usualplacticc. and number the slates starting with fl = 0, instead of n = I. Obviously, the lower limit on thesum in a formula such as Equation 2.17 should be altered accordingly.

21 Note that we obtain a[[ the (normalizable) solutions by this procedure. For if there were someother solution we could generate from it a second ladder. by lepeated application of the raising andlowering operators. But the bottom rung of this new ladder would have to satisfy Equation 2.58. andsince that leads inexorably lo Equation 2.59. the bottom rungs would be the same, and hence the twoladders would in fact be identical.

Section 2.3.' Tbe Harmom~c Oscillator 47

the halrnonic oscillator. Meanwhile, without ever doing that explicitly, we havedetermined the allowed energies.

Example 2.4 Find the 6rst excited state of the harmonic oscillator.

Solution: Using Equation 2.61,

[2.62]I mw \ 1I4 [2mw mt,) "2

- Al ) \| h--xe-A

We can nornlalize it "by hand":

so, as it happens, A I -, 1.I wouldn't want to calculate 50 this way (applying the raising operator fifty

times!), but never mind: In principle Equation 2.61 does the job-except for thenormalization.

You can even get the normalization algebraically, but it takes some fancyfootwork, so watch closely. We know that a:l:: ,r is proportional to n ::J::: I ,

[2.63]

but what are the proportionality factors, c,r and dn? First note that for "any" 22

functions f (x ) and g (x ) ,

[2.64]

(In the language of linear algebra, a is the hermitian conjugate of a::I:: .)

Proof:

2201. course, the integrals must e and this means that f (x ) and g (x ) must go to zero at:::J:: oc.

48 Chapter 2 Time-Independent Scbr6dinger Equation

and integration by parts takes f f *(dg|dx) dx to - f (d f|dx)*g dx (the boundaryterms vanish, for the reason indicated in footnote 22), so

QEDIn particular,

But (invoking Equations 2.57 and 2.61)

a+a-n " nfl , a-a+ n `- (n + 1)n . [2.65]

But since n and fl ::1:: I are normalized, it follows that |cn 12 = n+ I and |dn 12 -, n,

and hence Cl+ = n +_ _I. [2.66]

and so on� Clearly

I I I, -,- - ,~ \n-,_ \ [2.67]

which is to say that the normalization factor in Equation 2.61 is An = 1/71. (inparticular, A I = 1, confirming our result in Example 2.4).

Section 2.3.' The Hamlonic Oscillator 49

As in the case of the infinite square well, the stationary states of the harnlonicoscillator are orthogonal;

.~/ ,,, dx = 8mn . [2.68]

J -

Thl. s can be proved using Equation 2.65, and Equation 2.64 twice-first movinga+ and then moving a-:

Unless m - n, then, f t n dx must be zero. Orthononnality means that wecan again use Fourier's trick (Equation 2.34) to evaluate the coefficients, when weexpand F(x, O) as a linear combination of stationary states (Equation 2.16), and|cnl2 is again the probability that a measurement of the energy would yield thevalue En .

Example 2.5 Find the expectation value of the potential energy in the nth stateof the harmonic oscillator.

There's a beautiful device for evaluating integrals of this kind (involving powersof x or p); Use the definition (Equation 2.47) to express x and p in terms of theraising and lowering operators:

[2.69]

In this example we are interested in x2:

x2 - 2"k-Iw [(a+ )2 + (a+c,- ) + (a-a+) + (a- )2].

So

(V) = k 4w ) [Ca +) 2 + (a+a_ ) + (a_ a + ) + (a_ ) 2] d x.

50 Cbapter 2 Time-Independent Scbr6dl.n'ger Equat,`on

But (a+)2n is (apart from normalization) n+2, which is orthogonal to n, andthe same goes for (a-)21l!ln, which is proportional to n-2. So those terms dropout, and we can use Equation 2.65 to evaluate the remaining two:

As it happens, the expectation value of the potential energy is exactly half thetotal (the other half, of course, is kinetic). fhis is a peculiarity of the harmonicoscillator. as we'll see later on.

*Problem 2.10

(a) Construct 1//2 Cx).

(b) Sketch o, Ft , and 2.

(c) Check the orthogonality of o, t , and 2, by explicit integration. Hm.t: Ifyou exploit the even-mess and odd-mess of the functions, there is really onlyone integral left to do.

*Problem 2.11

(a) Compute (x), (p). (X2), and (p2), for the states /IO (Equation 2.59) and ]//t(Equation 2.62), by explicit integration. Comment.. In this and other problemsinvolving the harmonic oscillator it simpli6es matters if you introduce thevariable mw|v[-hx and the constant a (mw|ph)II4.

Cb) Check the uncertainty principle for these states.

(c) Compute (T) (the average kinetic energy) and (V) (the average potentialenergy) for these states. (No new integration allowed!) Is their sum what youwould expect?

*Problem 2.12 Find (x), (p) , (x2) , (p2), and (T ), for the nth stationary state of thehartnonic oscillator, using the method of Example 2.5. Check that the uncertaintyprinciple is satisfied

Problem 2.13 A particle in the harmonic oscillator potential starts out in the state

F (x . 0) = A [3/10(x ) + 41|II (x )].

(a) Find A.

(b) Construct F (x . t ) and | F (x . t ) |2.

Section 2.3: The Harmonic Oscillator 51

(c) Find (x) and (p). Don't get too excited if they oscillate at the classicalfrequency; what would it have been had I specified 2 (x ), instead of I (x )?Check that Ehrenfest' s theorem (Equation 1.38) holds for this wave function.

(d) If you measured the energy of this particle, what values might you get, andwith what probabilities?

Problem 2.14 A particle is in the ground state of the harmonic oscillator withclassical frequency w, when suddenly the spring constant quadruples, so w' = 2w,without initially changing the wave function (of course, \I' will now evolve differ-ently, because the Hamiltonian has changed). What is the probability that a mea-surement of the energy would still return the value hw/2? What is the probabilityof getting kw? [Answer: O.943.]

2.3.2 Analytic Method

We return now to the Schr6dinger equation for the harmonic oscillator,

- 2~ + 2-1 mW2 x2 |/! - E [2.70]

and solve it directly, by the series method. Things look a little cleaner if weintroduce the dimensionless variable

[2.71]

[2.72]

[2.73]

in terms of the Schr6dinger equation reads

Sr 2E^-___..- .kW

Our problem is to solve Equation 2.72, and in the process obtain the "allowed"values of K (and hence of E).

To begin with, note that at veIy large (which is to say, at very large x), 2

completely dominates over the constallt K, so in this regime

[2.74]

which has the approximate solution (check it!)

() Ae2/2 + Be+!2. [2.75]

52 Chapter 2 Time-"Independent Scbr6dinger Equation

The B term is clearly not normalizable (it blows up as |x| ); the physicallyacceptable solutions, then, have the asymptotic form

[2.76]

[2.771

This suggests that we "peel off" the exponential part,

in hopes that what remains, h(), has a simpler functional form than () itself.23Differentiating Equation 2.77,

and

so the Schr6dinger equation (Equation 2.72) becomes

[2.78]

I propose to look for solutions to Equation 2.78 in the form of power seriesin :24`

coh () - ao + al + a2 2 + . - - = S aJ' J . [2.79]

J=o

Differentiating the series term by tenn,

and

23Note that although we invoked some approximations to motivctte Equation 2.77 what fol-lows is e"mct. Tile device of snipping off the asymptotic behavior is the standaz'd first step inthe power series method for solving differential equations-see for example. Boas (footnote II ).Chapter 12.

24This is known as the Frobeuius method for solving a differential equation. According toTaylors theorem. atty reasonably well'behaved function can be explessed as a power series, soEquation 2.79 ordinarily involves no loss of generality. For conditions on the applicability of themethod. see Boas (footnote 11) or George B. Arfken alld HansJurgen Weber, Mathematical Metlze>dsji,r Plzysz.cists Sill ed.. Academic Press. Orlando (2000). Section 8.5.

Section 2.3: The Harmom'c Oscillator 53

Putting these into Equation 2.78, we find

It follows (from the uniqueness of power series expansions25) that the coefficientof each power of must vanish,

( j +I)(J' + 2)aJ+2 - 2 jaj + (K - I)uJ - 0,

and hence that(2 j + 1 - K)

ai+2 -` .ai ., ` (J + I)(J +2)

[2.81]

This recursion formula is entirely equivalent to the Schr6dinger equation.Starting with aO, it generates all the even-numbered coefficients:

(1 ` K) (5 - K) (5 - K)(I - K)a2 `- 2--aO' u4 `- 12--a2 " 2aO. ' - ' '

and starting with ul , it generates the odd coefficients:

(3 - K) (7 - K) (7 - K)(3 - K)u3 -` 6--ul . a5 -` 20-- 03 - 12uI ' ' ' '

We write the complete solution as

[2.82]

where

hcvcn ( ) - aO + a2 2 + u44 + . . .

is an even function of , built on aO, and

is ml odd function. built on al . Thus Equation 2.81 deternlines h() in tenns oftwo arbitrary constants (aO and al )-which is just what we would expect, for asecond-order differential equation.

However. not all the solutions so obtained are nonnalizable. For at very largej. the recursion fomlula becomes (approximately)

2aJ'+2 ~~ -:OJ' ,

w , -

See. for example. Aliken CIootnote 24). Section 5.7.