1.5 The Time-Independent Schrodinger Equation11 How the wave function gives us information on other properties besides the position is d tes of copic exper he y The orld an uite ab pchan s at first reading.As we treat under ding of the will increase It may bother the reader that we wrote down the Schrodinger equation without any attempt to prove its plausibility.By using analogies between geometrical optics and clas- sical mechanics on the one hand,and wave optics and quantum mechanics on the other hand,one can show the plausibility of the Schrodinger equation.Geometrical optics is an approximation to wave optics.valid when the wavelength of the light is much less than the size of the apparatus.(Recall its use in treating lenses and mirrors.)Likewise,classical mechanics is an approximation to wave mechanics,valid when the particle's wavelength is much less than the size of the apparatus.One can make a plausible guess as to how to get om assical me anics base the know on Since med.In any c earguments have been on ation seem plaustb of its dictio withger equa ails of th g o his n in Jan Section 5.3.A refe nce with he author's name ilicd is isred in the ibiogaphy Quantum mechani s provides the law of motion tally,macroscopic objects obey classical mechanics.Hence for quantum mechanics to be a valid theory,it should reduce to classical mechanics as we make the transition from micro- scopic to macroscopic particles.Quantum effects are associated with the de Broglie wave- lengthA=h/me.Sincehis very small,the de Broglie wavelength of macroscopic objects is essentially zero.Thus,in the limit A-0,we expect the time-dependent Schrodinger equation to reduce to Newton's second law.We can prove this to be so(see Prob.7.59) A similar situation holds in the relation between special relativity and classical mechan where c is th I develop will be nonrel com oratceratiomofr Historica y.quantum mecha n quantum mec rg.Bo d tha H rix mechanics)is equivalent to the Schrodinger formulation (called wav In 1926.Dirac and Jordan,working independently.formulated qu ntum mechanies in an abstract version called transformation theory that is a generalization of matrix mechanics and wave mechanics(see Dirac).In 1948,Feynman devised the path integral formulation of quantum mechanics [R.P.Feynman.Rev.Mod.Plrys.20,367(1948);R.P.Feynman and A.R.Hibbs,Quantum Mechanics and Path Integrals,McGraw-Hill,1965]. 1.5 The Time-Independent Schrodinger Equation one of a to che do no n In stead.the simpler time-independent Schrodinger equation is used.We now derive the
1.5 The Time-Independent Schrödinger Equation | 11 How the wave function gives us information on other properties besides the position is discussed in later chapters. The postulates of thermodynamics (the first, second, and third laws of thermodynamics) are stated in terms of macroscopic experience and hence are fairly readily understood. The postulates of quantum mechanics are stated in terms of the microscopic world and appear quite abstract. You should not expect to fully understand the postulates of quantum mechanics at first reading. As we treat various examples, understanding of the postulates will increase. It may bother the reader that we wrote down the Schrödinger equation without any attempt to prove its plausibility. By using analogies between geometrical optics and classical mechanics on the one hand, and wave optics and quantum mechanics on the other hand, one can show the plausibility of the Schrödinger equation. Geometrical optics is an approximation to wave optics, valid when the wavelength of the light is much less than the size of the apparatus. (Recall its use in treating lenses and mirrors.) Likewise, classical mechanics is an approximation to wave mechanics, valid when the particle’s wavelength is much less than the size of the apparatus. One can make a plausible guess as to how to get the proper equation for quantum mechanics from classical mechanics based on the known relation between the equations of geometrical and wave optics. Since many chemists are not particularly familiar with optics, these arguments have been omitted. In any case, such analogies can only make the Schrödinger equation seem plausible. They cannot be used to derive or prove this equation. The Schrödinger equation is a postulate of the theory, to be tested by agreement of its predictions with experiment. (Details of the reasoning that led Schrödinger to his equation are given in Jammer, Section 5.3. A reference with the author’s name italicized is listed in the Bibliography.) Quantum mechanics provides the law of motion for microscopic particles. Experimentally, macroscopic objects obey classical mechanics. Hence for quantum mechanics to be a valid theory, it should reduce to classical mechanics as we make the transition from microscopic to macroscopic particles. Quantum effects are associated with the de Broglie wavelength l = h>mv. Since h is very small, the de Broglie wavelength of macroscopic objects is essentially zero. Thus, in the limit l S 0, we expect the time-dependent Schrödinger equation to reduce to Newton’s second law. We can prove this to be so (see Prob. 7.59). A similar situation holds in the relation between special relativity and classical mechanics. In the limit v>c S 0, where c is the speed of light, special relativity reduces to classical mechanics. The form of quantum mechanics that we will develop will be nonrelativistic. A complete integration of relativity with quantum mechanics has not been achieved. Historically, quantum mechanics was first formulated in 1925 by Heisenberg, Born, and Jordan using matrices, several months before Schrödinger’s 1926 formulation using differential equations. Schrödinger proved that the Heisenberg formulation (called matrix mechanics) is equivalent to the Schrödinger formulation (called wave mechanics). In 1926, Dirac and Jordan, working independently, formulated quantum mechanics in an abstract version called transformation theory that is a generalization of matrix mechanics and wave mechanics (see Dirac). In 1948, Feynman devised the path integral formulation of quantum mechanics [R. P. Feynman, Rev. Mod. Phys., 20, 367 (1948); R. P. Feynman and A. R. Hibbs, Quantum Mechanics and Path Integrals, McGraw-Hill, 1965]. 1.5 The Time-Independent Schrödinger Equation The time-dependent Schrödinger equation (1.13) is formidable looking. Fortunately, many applications of quantum mechanics to chemistry do not use this equation. Instead, the simpler time-independent Schrödinger equation is used. We now derive the
12 Chapter 1 The Schrodinger Equation time-independent from the time-dependent Schrodinger equation for the one-particle. one-dimensional case. We begin by restricting ourselves to the special case where the potential energy V is not a function of time but depends only onx.This will be true if the system experi- ences no time-dependent external forces.The time-dependent Schrodinger equation reads haΨ(x.t) 1.16 i at 2Ψ(.+vx)Ψ(x) 2m2 looki solutions of(1.16)that can be written as the function of (x）=f)(x) a.17 Capital psi is used for the time-dependent wave function and lowercase psi for the factor that depends only on the coordinate x.States corresponding to wave functions of the form (1.17)possess certain properties(to be discussed shortly)that make them of great interest. [Not all solutions of (1.16)have the form (1.17):see Prob.3.51.]Taking partial deriva- tives of (1.17).we have c)_0. =和回 dt 32 2 Substitution into(1.16)gives h0)= v(x)f(t)(x) i dt 玩) de 方1) if(t)dt =-2m(x)dx 1.18) where by to which each side of (1.18 deequal to be tchidof ual must be of t.The left side of (1.18)is inder nden so this function must also he inde of Since the function is independent of both variables,and it must be a constant. We call this constant E. Equating the left side of(1.18)to E,we get df(t) 返d Integrating both sides of this equation with respect to.we have Inf(t)=-iEt/+C where Cis an arbitrary constant of integration.Hence f(t)=ece-E Ae-El where the arbitrary constant A has replaced ec.Since A can be included as a factor in the function)that multiplies f(t)in (1.17).A can be omitted from f(t).Thus f)）=ea
12 Chapter 1 | The Schrödinger Equation time-independent from the time-dependent Schrödinger equation for the one-particle, one-dimensional case. We begin by restricting ourselves to the special case where the potential energy V is not a function of time but depends only on x. This will be true if the system experiences no time-dependent external forces. The time-dependent Schrödinger equation reads - U i 0 1x, t2 0t = - U2 2m 02 1x, t2 0x2 + V1x2 1x, t2 (1.16) We now restrict ourselves to looking for those solutions of (1.16) that can be written as the product of a function of time and a function of x: 1x, t2 = f1t2c1x2 (1.17) Capital psi is used for the time-dependent wave function and lowercase psi for the factor that depends only on the coordinate x. States corresponding to wave functions of the form (1.17) possess certain properties (to be discussed shortly) that make them of great interest. [Not all solutions of (1.16) have the form (1.17); see Prob. 3.51.] Taking partial derivatives of (1.17), we have 0 1x, t2 0t = df1t2 dt c1x2, 02 1x, t2 0x2 = f1t2 d2 c1x2 dx2 Substitution into (1.16) gives - U i df1t2 dt c1x2 = - U2 2m f1t2 d2 c1x2 dx2 + V1x2f1t2c1x2 - U i 1 f1t2 df1t2 dt = - U2 2m 1 c1x2 d2 c1x2 dx 2 + V1x2 (1.18) where we divided by fc. In general, we expect the quantity to which each side of (1.18) is equal to be a certain function of x and t. However, the right side of (1.18) does not depend on t, so the function to which each side of (1.18) is equal must be independent of t. The left side of (1.18) is independent of x, so this function must also be independent of x. Since the function is independent of both variables, x and t, it must be a constant. We call this constant E. Equating the left side of (1.18) to E, we get df1t2 f1t2 = - iE U dt Integrating both sides of this equation with respect to t, we have ln f1t2 = -iEt>U + C where C is an arbitrary constant of integration. Hence f1t2 = eCe-iEt>U = Ae-iEt>U where the arbitrary constant A has replaced eC. Since A can be included as a factor in the function c1x2 that multiplies f1t2 in (1.17), A can be omitted from f1t2. Thus f1t2 = e-iEt>U
1.5 The Time-Independent Schrodinger Equation13 Equating the right side of (1.18)toE.we have 2m dx2 1.19 Equation(1.19)is the time-independent Schrodinger equation for a single particle of mass m moving in one dimension. What is the significance of the constant E?Since E occurs as [E-V(x)]in (1.19). E has the same dimensions as V.so E has the dimensions of energy.In fact,we postulate that E is the energy of the system.(This is a special case of a more general postulate to be discussed in a later chapter.)Thus,for cases where the potential energy is a function ofx only.there exist wave functions of the form 业(xt)=eB(x) (1.20) system wave function n(1)s nplex,but the quantity that is imentally observable is the probability densityx)2.The of th lute alu complex quantity is giv hy the oduct of the with its ever it occurs.(Se Section 17)Thus |Ψ2=业*业 1.2) where the star denotes the complex conjugate.For the wave function(1.20). |(x.)P=【eB()小ebx) =(x)e(x) =ePw*(x)(x)=*(x)地(x) Ψ(x,)2=(x)2 1.22 心eom7.2 assumed that E is a real umber.so上上his tact will be pro stat does f the form(120),the pr obabilit y densi is gi en byΨ(r)Pand )2 and si 6 stat )2 th )is often called the w ave fur ction.althe a stationary state is obtained by multiply ing (x)by rm stationary state should not mislead the reader into thinking that a particle in a stationary state is at rest. What is stationary is the probability density,not the particle itself. We will be concerned mostly with states of constant energy (stationary states)and hence will usually deal with the time-independent Schrodinger equation(1.19).For simplicity we will refer to this equation as"the Schrodinger equation."Note that the Schrodinger equation contains two unknowns:the allowed energies E and the allowed wave functions To solve for two unknowns,we need to impose additional conditions (called bour dary conditions)on besides requiring that it satisfy (1.19).The boundary s dete ce it turns out that only c rtain va 5 Eallow to satisf 00 become clearer when we discuss specific examples in later chapters
1.5 The Time-Independent Schrödinger Equation | 13 Equating the right side of (1.18) to E, we have - U2 2m d2 c1x2 dx2 + V1x2c1x2 = Ec1x2 (1.19) Equation (1.19) is the time-independent Schrödinger equation for a single particle of mass m moving in one dimension. What is the significance of the constant E? Since E occurs as 3E - V1x)4 in (1.19), E has the same dimensions as V, so E has the dimensions of energy. In fact, we postulate that E is the energy of the system. (This is a special case of a more general postulate to be discussed in a later chapter.) Thus, for cases where the potential energy is a function of x only, there exist wave functions of the form 1x, t2 = e-iEt>U c1x2 (1.20) and these wave functions correspond to states of constant energy E. Much of our attention in the next few chapters will be devoted to finding the solutions of (1.19) for various systems. The wave function in (1.20) is complex, but the quantity that is experimentally observable is the probability density 0 1x, t2 0 2 . The square of the absolute value of a complex quantity is given by the product of the quantity with its complex conjugate, the complex conjugate being formed by replacing i with –i wherever it occurs. (See Section 1.7.) Thus 0 0 2 = * (1.21) where the star denotes the complex conjugate. For the wave function (1.20), 0 1x, t2 0 2 = 3e-iEt>U c1x)4*e-iEt>U c1x2 = eiEt>U c*1x2e-iEt>U c1x2 = e0 c*1x2c1x2 = c*1x2c1x2 0 1x, t2 0 2 = 0 c1x2 0 2 (1.22) In deriving (1.22), we assumed that E is a real number, so E = E*. This fact will be proved in Section 7.2. Hence for states of the form (1.20), the probability density is given by 0 1x2 0 2 and does not change with time. Such states are called stationary states. Since the physically significant quantity is 0 1x, t2 0 2 , and since for stationary states 0 1x, t2 0 2 = 0 c1x2 0 2 , the function c1x2 is often called the wave function, although the complete wave function of a stationary state is obtained by multiplying c1x2 by e-iEt>U . The term stationary state should not mislead the reader into thinking that a particle in a stationary state is at rest. What is stationary is the probability density 0 0 2 , not the particle itself. We will be concerned mostly with states of constant energy (stationary states) and hence will usually deal with the time-independent Schrödinger equation (1.19). For simplicity we will refer to this equation as “the Schrödinger equation.” Note that the Schrödinger equation contains two unknowns: the allowed energies E and the allowed wave functions c. To solve for two unknowns, we need to impose additional conditions (called boundary conditions) on c besides requiring that it satisfy (1.19). The boundary conditions determine the allowed energies, since it turns out that only certain values of E allow c to satisfy the boundary conditions. This will become clearer when we discuss specific examples in later chapters
14 Chapter 1 The Schrodinger Equation 1.6 Probability Probability plays a fundamental role in quantum mechanics.This section reviews the mathematics of probability. There has been much controversy about the proper definition of probability.One defi- nition is the following:If an experiment has n equally probable outcomes,m of which are favorable to the occurrence of a certain event A,then the probability that A occurs is m/n. Note that this definition is circular,since it specifies equally probable outcomes when eoually probsble outcomesc An alternative efinit ility of A occurring is then defined as Thus,if we toss a coin repeatedly.the fraction of heads will approach 1/2 as we increase the number of tosses. ing a heart when a card is om a sta card ck cont ning 13 hearts.Th are 5 card comes.Ther prob ere are e out 3/52 The pro ty Io ime ty of two re exam ptacethcnmlCcadahertd 52. and fo ssibilities the nd d We hay 52.5 possible outcomes.Since there are 13 hea ther are 13.12 differ s to draw two hearts.The desired probability is (1312)/(52.51)=1/17.This calculation illustrates the theorem:The probability that two events A and B both occur is the probability that A occurs,multiplied by the conditional probability that B then occurs,calculated with the as- sumption that A occurred.Thus,if A is the probability of drawing a heart on the first draw. the probability of A is 13/52.The probability of drawing a heart on the second draw.given that the first draw yielded a heart,is 12/51 since there remain 12 hearts in the deck.The probability of drawing two hearts is then (13/52)(12/51)=1/17.as found previously. In quantum mechanics we must deal with probabilities involving a continuous vari ate.It does not make much sense to talk about the prob d ar a particular po nt su s on tne x ax r any fin Lof the and x+d element ofe al to th th of the axis.Hence the probability that the ticle will he found be en x andx dx is equal to g(x)dt.where g(x)is some function that tells how the probability varies over the axis.The functiong(x)is called the probability density,since it is a probability per unit length.Since probabilities are real,nonnegative numbers.g(x)must be a real function that is everywhere nonnegative.The wave function can take on negative and complex values and is s not a probability density.Quantum mechanics postulates that the probability density is[Eq.(1.15). What is the probability that the particle lies in son me finite region of space a≤x≤by To find this probability,we sum up the probabilitiesdr of finding the particle in all
14 Chapter 1 | The Schrödinger Equation 1.6 Probability Probability plays a fundamental role in quantum mechanics. This section reviews the mathematics of probability. There has been much controversy about the proper definition of probability. One definition is the following: If an experiment has n equally probable outcomes, m of which are favorable to the occurrence of a certain event A, then the probability that A occurs is m>n. Note that this definition is circular, since it specifies equally probable outcomes when probability is what we are trying to define. It is simply assumed that we can recognize equally probable outcomes. An alternative definition is based on actually performing the experiment many times. Suppose that we perform the experiment N times and that in M of these trials the event A occurs. The probability of A occurring is then defined as lim NS M N Thus, if we toss a coin repeatedly, the fraction of heads will approach 1>2 as we increase the number of tosses. For example, suppose we ask for the probability of drawing a heart when a card is picked at random from a standard 52-card deck containing 13 hearts. There are 52 cards and hence 52 equally probable outcomes. There are 13 hearts and hence 13 favorable outcomes. Therefore, m>n = 13>52 = 1>4. The probability for drawing a heart is 1>4. Sometimes we ask for the probability of two related events both occurring. For example, we may ask for the probability of drawing two hearts from a 52-card deck, assuming we do not replace the first card after it is drawn. There are 52 possible outcomes of the first draw, and for each of these possibilities there are 51 possible second draws. We have 52 # 51 possible outcomes. Since there are 13 hearts, there are 13 # 12 different ways to draw two hearts. The desired probability is 113 # 122> 152 # 512 = 1>17. This calculation illustrates the theorem: The probability that two events A and B both occur is the probability that A occurs, multiplied by the conditional probability that B then occurs, calculated with the assumption that A occurred. Thus, if A is the probability of drawing a heart on the first draw, the probability of A is 13>52. The probability of drawing a heart on the second draw, given that the first draw yielded a heart, is 12 >51 since there remain 12 hearts in the deck. The probability of drawing two hearts is then 113>522112>512 = 1>17, as found previously. In quantum mechanics we must deal with probabilities involving a continuous variable, for example, the x coordinate. It does not make much sense to talk about the probability of a particle being found at a particular point such as x = 0.5000c, since there are an infinite number of points on the x axis, and for any finite number of measurements we make, the probability of getting exactly 0.5000c is vanishingly small. Instead we talk of the probability of finding the particle in a tiny interval of the x axis lying between x and x + dx, dx being an infinitesimal element of length. This probability will naturally be proportional to the length of the interval, dx, and will vary for different regions of the x axis. Hence the probability that the particle will be found between x and x + dx is equal to g1x2 dx, where g1x2 is some function that tells how the probability varies over the x axis. The function g1x2 is called the probability density, since it is a probability per unit length. Since probabilities are real, nonnegative numbers, g1x2 must be a real function that is everywhere nonnegative. The wave function can take on negative and complex values and is not a probability density. Quantum mechanics postulates that the probability density is 0 0 2 [Eq. (1.15)]. What is the probability that the particle lies in some finite region of space a . x . b? To find this probability, we sum up the probabilities 0 0 2 dx of finding the particle in all
1.6 Probability 15 the infinitesimal regions lying between a and b.This is just the definition of the definite ntegral ΨPd=Pr(a≤xsb) 1.23 where Pr denotes a probability.A probability of I represents certainty.Since it is certain that the particle is somewhere on the x axis,we have the requirement Ψ2d=1 1.24 When satisfies(1.24).it is said to be normalized.For a stationary state.= anddx 1. EXAMPLE Aone-particle,one-dimensional system has=aat=0.where a=1.0000 nm.Att=0.the particle's position is measured.(a)Find the probability that the measured value lies betweenx=1.5000 nm and x=1.5001 nm.(b)Find the probability that the measured value is betweenx =0 and x =2 nm.(c)Verify that is normalized 0.225313nm to e 'nm 2dx a-le-2a dx (1nm)-le-2(1.5 am)/(1 nm)(0.0001 nm) =4.979×10-6 (See also Prob.1.14.) (b)Use of Eq.(1.23)and x=x for x 0 gives 2nm Pr(0≤x≤2nm)= yd=∫2 。 e2咖dk =-e-2咖18m=-(e4-1)=0.490 (e)Use of∫fx)d=9efx)d+ffx)d,ld=-x forx≤0，andx=x for x 2 0,gives r=ae2咖+a e咖ah =al(传ae2a1ex)+a1(-支ae2al6)=支+支=1 EXERCISE Eo -1.000nt time of a po find the rem hat th ticle is found betw (Answer:0.0000258.)
1.6 Probability | 15 the infinitesimal regions lying between a and b. This is just the definition of the definite integral L b a 0 0 2 dx = Pr1a . x . b2 (1.23) where Pr denotes a probability. A probability of 1 represents certainty. Since it is certain that the particle is somewhere on the x axis, we have the requirement L - 0 0 2 dx = 1 (1.24) When satisfies (1.24), it is said to be normalized. For a stationary state, 0 0 2 = 0 c0 2 and 1 - 0 c0 2 dx = 1. Ex a m p l e A one-particle, one-dimensional system has = a-1>2 e- 0 x 0>a at t = 0, where a = 1.0000 nm. At t = 0, the particle’s position is measured. (a) Find the probability that the measured value lies between x = 1.5000 nm and x = 1.5001 nm. (b) Find the probability that the measured value is between x = 0 and x = 2 nm. (c) Verify that is normalized. (a) In this tiny interval, x changes by only 0.0001 nm, and goes from e-1.5000 nm-1>2 = 0.22313 nm-1>2 to e-1.5001 nm-1>2 = 0.22311 nm-1>2 , so is nearly constant in this interval, and it is a very good approximation to consider this interval as infinitesimal. The desired probability is given by (1.15) as 0 0 2 dx = a-1 e-20 x 0>a dx = 11 nm2-1 e-211.5 nm2>11 nm2 10.0001 nm2 = 4.979 * 10-6 (See also Prob. 1.14.) (b) Use of Eq. (1.23) and 0 x 0 = x for x Ú 0 gives Pr10 . x . 2 nm2 = L 2 nm 0 0 0 2 dx = a-1 L 2 nm 0 e-2x>a dx = -1 2 e-2x>a 2 nm 0 = -1 2 1e-4 - 12 = 0.4908 (c) Use of 1 - f1x2 dx = 1 0 - f1x2 dx + 1 0 f1x2 dx, 0 x 0 = -x for x . 0, and 0 x = x for x Ú 0, gives L - 0 0 2 dx = a-1 L 0 - e2x>a dx + a-1 L 0 e-2x>a dx = a-111 2 ae2x>a 0 - 2 + a-11- 1 2 ae-2x>a 0 2 = 1 2 + 1 2 = 1 Exercise For a system whose state function at the time of a position measurement is = 132a3>p21>4 xe-ax2 , where a = 1.0000 nm-2 , find the probability that the particle is found between x = 1.2000 nm and 1.2001 nm. Treat the interval as infinitesimal. (Answer: 0.0000258.)