Physics 4617/5617: Quantum Physics Course Lecture Notes

Transcription

1 Physics 4617/5617: Quantum Physics Course Lecture Notes Dr. Donald G. Luttermoser East Tennessee State University Edition 5.1

2 Abstract These class notes are designed for use of the instructor and students of the course Physics 4617/5617: Quantum Physics. This edition was last modified for the Fall 2006 semester.

3 III. The Time-Independent Schrödinger Equation A. Stationary States. 1. How does one determine the wave function Ψ(x, t) from first principles? a) It is solved through a knowledge of V (i.e., the potential energy function) in the Schrödinger equation. b) In this class, we will assume that the potential V is independent of t = then the Schrödinger equation can be solved by the method of separation of variables: Ψ(x, t) =ψ(x) f(t), (III-1) where ψ (lowercase) is a function of x alone and f is a function of t alone. c) For separable solutions we have Ψ t = ψ df dt, 2 Ψ x 2 = d2 ψ dx 2 f (III-2) (ordinary derivatives now) and the Schrödinger equation (Eq. II-1) becomes i hψ df dt = h2 2m d) Dividing through by ψf gives d 2 ψ dx 2 f + Vψf. (III-3) i h 1 f df dt = h2 2m 1 ψ d 2 ψ dx 2 + V. (III-4) e) The left side is a function of t alone, and the right side, a function of x alone = the only way this can be true is III 1

4 if both sides of the equation are constant (which we will set equal to E). Then or and or h2 2m h2 2m i h 1 f df dt = E, df dt = iē h f, 1 ψ d 2 ψ dx 2 + V = E, d 2 ψ + Vψ= Eψ. dx2 (III-8) (III-5) (III-6) (III-7) 2. Separation of variables has turned a partial differential equation into two ordinary differential equations (Eqs. III-6 & III-8). a) Eq. (III-6) is easy to solve by just rearranging terms and integrating: f(t) =Ce iet/ h. (III-9) Typically the integration constant C is absorbed into the solution for ψ, hence we state the solution to Eq. (III-6) as f(t) =e iet/ h. (III-10) b) Meanwhile, Eq. (III-8) is referred to as the time-independent Schrödinger equation (in one dimension). To solve it, we must specify the potential V (x). 3. When is a separable solution to the Schrödinger equation valid? a) Stationary states: i) These are states which even though the wave function depends upon time: Ψ(x, t) =ψ(x) e iet/ h, (III-11) III 2

5 the probability density does not: Ψ(x, t) 2 =Ψ Ψ=ψ e +iet/ h ψe iet/ h = ψ(x) 2. (III-12) ii) In a stationary state, x is constant and p = 0 (from Eq. II-45) = nothing ever happens in a stationary state. b) States of Definite Total Energy: i) In classical physics, the total energy (kinetic plus potential) is called the Hamiltonian: H(x, p) = p2 2m + V (x). (III-13) ii) The corresponding Hamiltonian operator is defined by 2 Ĥ = h2 2m x + V (x), (III-14) 2 where the hat (i.e., ˆ) means this is an operator. iii) Thus the time-independent Schrödinger equation (Eq. III-8) can be written as Ĥψ = Eψ. (III-15) iv) The expectation value of the total energy is H = + ψ Ĥψdx = E + ψ 2 dx = E. (III-16) v) Moreover, Ĥ 2 ψ = Ĥ(Ĥψ)=Ĥ(Eψ)=E(Ĥψ)=E2 ψ, (III-17) III 3

6 and hence H 2 = + ψ Ĥ 2 ψdx = E 2 + ψ 2 dx = E 2. (III-18) vi) So the standard deviation in H is given by σ 2 H = H 2 H 2 = E 2 E 2 =0. (III-19) vii) If σ = 0, then every member of the sample must share the same value = the distribution has zero spread. viii) Hence, a separable solution has the property that every measurement of the total energy is certain to return the value E. c) The general solution of the time-independent Schrödinger equation is a linear combination of separable solutions. i) The time-independent Schrödinger equation yields an infinite collection of solutions (ψ 1 (x),ψ 2 (x),ψ 3 (x),...), each with its associated value of the separation constant (E 1,E 2,E 3,...)= thus there is a different wave function for each allowed energy: Ψ 1 (x, t) =ψ 1 (x)e ie 1t/ h, Ψ 2 (x, t) =ψ 2 (x)e ie 2t/ h,... (III-20) ii) Hence the general solution of the wave function for these cases is Ψ(x, t) = n=1 c n ψ n (x) e ie nt/ h. (III-21) iii) Every solution to the time-independent Schrödinger III 4

7 B. Infinite Square Well. equation can be written in this form it is simply a matter of finding the right constants (c 1,c 2,...). 1. Assume we have a potential of the following form: V (x) = as shown in Figure III-1. 0, if 0 x a,, otherwise (III-22) V(x) a x Figure III 1: The infinite square well potential (Eq. III-22). a) A particle in this potential is completely free, except at the two ends (x = 0 and x = a), where an infinite force prevents it from escaping. b) Outside the well, ψ(x) = 0 = the probability of finding the particle there is zero. c) Inside the well, V = 0, and the time-independent Schrödinger equation becomes or d 2 ψ dx 2 = k2 ψ, h2 2m d 2 ψ dx 2 = Eψ, where k III 5 2mE h (III-23). (III-24)

8 d) Eq. (III-24) is just the (classical) simple harmonic oscillator equation with the general solution of ψ(x) =A sin kx + B cos kx, (III-25) where A and B are constants that are fixed by the boundary conditions of the problem: i) Both ψ and dψ/dx are continuous. ii) Continuity of ψ(x) requires that ψ(0) = ψ(a) = 0, (III-26) so as to join the solution outside the well, hence, so ψ(0) = A sin 0 + B cos 0 = B =0, ψ(x) = A sin kx. (III-27) (III-28) iii) At the other boundary, ψ(a) = A sin ka = 0. Since ψ must be normalizable, A 0, so ka =0, ±π,±2π,±3π,... (III-29) iv) But k = 0 is no good since that would give ψ(x) = 0, and the negative solutions give nothing new, since sin( θ) = sin(θ) and we can absorb the minus sign into A. So the distinct solutions are k n = nπ, with n =1, 2, 3,... (III-30) a v) Note that the boundary condition at x = a does not determine A, but rather the constant k, and hence the possible values of E: E n = h2 kn 2 2m = n2 π 2 h 2 2ma 2. (III-31) III 6

9 Exercise: Prove Eq. (III-25) is the solution to Eq. (III-24). Exercise: Prove Eq. (III-31). e) Unlike the classical case, a quantum particle in the infinite square well cannot have just any old energy it can only have the allowed values of Eq. (III-31) = allowed states. f) As usual, we find the amplitude A from the normalization criterion: 1 = a 0 ψ ψdx a = 0 A 2 sin 2 (kx) dx = A 2 a 1 [1 cos(2kx)] dx 0 2 = A 2 [ a 2 dx a ] cos(2kx) dx 0 0 = A 2 2 = A 2 2 = A 2 2 [ x a 0 1 2k sin(2kx) a 0 { a 1 [ ( nπ sin 2 2k a ( a 1 ) 2k sin 2nπ ] ) a sin 0 ]} A = = A 2a 2 2 a (III-32) (we are only worried about the magnitude of A here), so our final wave function becomes ψ n (x) = 2 ( ) nπ a sin a x. (III-33) 2. As can be seen, for an infinite square well, there are an infinite amount of solutions to the time-independent Schrödinger equa- III 7

10 tion. Figure III-2 shows the first 3 of these stationary states as can be seen, these are just standing waves. a) The lowest energy state, ψ 1 is called the ground state. b) The higher energy states scale as n 2 and are called excited states. c) These states alternate from even to odd, with respect to the center of the well. (ψ 1 is even, ψ 2 is odd, ψ 3 is even, and so on.) d) As you go up in energy, each successive state has one more node (zero crossing). ψ 1 has none (the end points don t count), ψ 2 has 1, ψ 3 has 2, etc. e) They are mutually orthogonal (i.e., the wave functions are said to be orthonormal), in the sense that ψ m (x) ψ n(x) dx = δ mn, (III-34) where δ mn is the Kronecker delta: δ mn = 0, if m n; 1, if m = n. (III-35) f) They are complete, in the sense that any other function, f(x), can be expressed as a linear combination of them: f(x) = n=1 c n ψ n (x) = 2 a n=1 ( ) nπ c n sin a x. (III-36) Proof of this completion criterion requires knowledge of Fourier series (see below). Exercise: Prove Eq. (III-34) with the wave function in Eq. (III-33). 3. The time-dependent form of the solution to this Schrödinger III 8

11 Ψ 1 (x) Ψ 2 (x) Ψ 3 (x) a a x x a x Figure III 2: The first 3 stationary states of the infinite square well. equation is thus Ψ(x, t) = c n n=1 2 a sin ( nπ a x ) e i(n2 π 2 h/2ma 2 )t. a) Then our initial condition equation becomes Ψ(x, 0) = n=1 c n ψ n (x). (III-37) (III-38) b) By making use of the orthonormality of the solutions, the initial condition equations then give us the mechanism for finding the coefficients of the series: c n = 2 a a ( ) nπ sin 0 a x Ψ(x, 0) dx. (III-39) Example III 1. function A particle in an infinite square well has the initial wave Ψ(x, 0) = Ax(a x). (a) (b) Normalize Ψ(x, 0). Graph it. Which stationary state does it most closely resemble? On that basis, estimate the expectation value of the energy. Compute x, p and H, att =0.(Note : This time you cannot get p by differentiating x, because you only know x at one instant in time.) How does H compare with your estimate in (a)? III 9

12 Solution (a): Ψ(x, 0) = Ax(a x) Ψ (x, 0) = Ax(a x) a a 1 = 0 Ψ Ψ dx = A 2 x(a x)x(a x) dx 0 = A 2 a ( 0 x2 a 2 2ax + x 2) dx = A 2 a ( a 2 x 2 2ax 3 + x 4) dx 0 A = = A 2 [ 1 3 a2 x ax x5 ] a 0 = A 2 [ 1 3 a5 1 2 a a5 ] = A2 [ 10a 5 15a 5 +6a 5] = A a5 30 a. 5 To graph this function, note that at the endpoints (i.e., x = 0 and x = a), Ψ(x, 0) = 0. Also, expanding out the integrand gives ax x 2, which clearly indicates that this wave function has a parabolic shape. We can find the extrema of this parabola by taking the derivative and setting it equal to 0: a 2x = 0, or x = a/2. The second derivative tells us the orientation of the parabola: 2, so the curve is concave downward as shown on the graph on the next page. As one can see from an inspection of this curve (see figure on next page), this wave function resembles the harmonic oscillator solution in the ground state (n = 1) since it resembles a sine function (note that we will be investigating simple harmonic oscillators in section III.E). As such, we can guess the expectation value of the energy will follow Eq. (III-31): E = π2 h 2 h2 =4.93 2ma2 ma 2. III 10

13 Ψ(x,0) The wave function Ψ(x, 0) = x(a x). x a Solution (b): a x = 0 Ψ xψ dx = A 2 a [ x(a x)x 2 (a x) ] dx 0 = A 2 a [ x 3 (a x) 2] dx 0 = A 2 a [ x 3 (a 2 2ax + x 2 ) ] dx 0 = A 2 a ( a 2 x 3 2ax 4 + x 5 ) ) dx 0 = A 2 [ 1 4 a2 x ax x6 ] a 0 = A 2 [ 1 4 a6 2 5 a a6 ] = 30 ( 15a 6 24a 6 +10a 6) 60a 5 = 1 = a 2a 5a6 2. Which is just what you would expect from the appearance of the graph. p = a 0 Ψ h i x Ψ dx = A2 h i a 0 III 11 { x(a x) x [x(a x)] } dx

14 a = A 2 h [x(a x)(a 2x)] dx i 0 = A 2 h a [ x(a 2 3ax +2x 2 ) ] dx i 0 = A 2 h a [ a 2 x 3ax 2 +2x 3] dx i 0 = A 2 h [ 1 i 2 a2 x 2 ax ] a 4 x4 0 = A 2 h ( a 4 2a 4 + a 4) =0. 2i The particle initially has an expectation of being at rest. Now, so 2 x H = 2 Ψ= 2 a 0 Ψ h2 2m 2 x + V 2 Ψ dx. x2[x(a x)] = (a 2x) = 2, x H = h2 a 2m x dx + V a 2 0 Ψ Ψ dx = h2 a m A2 x(a x) dx + V 0 = h2 [ 1 m A2 2 ax2 1 ] a 3 x3 + V 0 Ψ 2 Ψ ( 1 2 a3 1 3 a3 ) = h2 m A2 + V = h2 ( 6m A2 3a 3 2a 3) + V = h2 6m 30 a 5 a3 + V = 5 h2 ma 2 + V. From this, it is clear that E =5 h2 ma, 2 0 III 12

15 which is close to the solution found for the harmonic oscillator. C. Fourier Analysis. 1. As we continue on with our work with the Schrödinger equation, we will often encounter wave functions that take the form Ψ(x, t) = 1 A(k) 2π ei(kx ωt) dk. (III-40) a) In the wave function, x and t represent their usual meanings of the position and time independent variables, respectively. The new variables introduced here in the integral are: i) The wave number, k, which is inversely related to the wavelength, λ, of a wave through k 2π λ. Using Eq. (I-77) we can show that the wave number is related to the particle/wave s linear momentum via p = hk. ii) The angular frequency, ω, which is directly related to the frequency, ν, of a wave through ω 2πν. Using Eq. (I-77) we can show that the angular frequency is related to the particle/wave s energy via E = hω. iii) Finally note that one can write a dispersion relation for a wave with these new variables. Using III 13

16 Eq. (I-75) for a non-relativistic free particle (i.e., V = 0), the total energy of the wave is E = p2 2m. Making use of our relations above, this energymomentum equation becomes the following dispersion relation: ω(k) = hk2 2m. b) At t = 0, this equation takes on a form that may be familiar: Ψ(x, 0) = 1 A(k) 2π eikx dk. (III-41) c) Eq. (III-41) reveals that the amplitude function A(k) is the Fourier transform of the wave function Ψ(x, t) at t =0= the amplitude function is related to the wave function at t = 0 by a Fourier integral. 2. Fourier analysis the generation and deconstruction of Fourier series and integrals are the mathematical methods that underlies the construction of wave packets by superposition. a) Mathematicians commonly use Fourier analysis to rip functions apart, representing them as sums or integrals of simple component functions, each which is characterized by a single frequency. b) This method can be applied to any function f(x) that is piecewise continuous i.e., that has at most a finite number of finite discontinuities. As we have already noted, wave functions must be continuous, so as such, satisfies this condition and so are prime candidates for Fourier analysis. III 14

17 c) Whether we represent f(x) via a Fourier series or Fourier integral depends on whether or not this function is periodic = any function that repeats itself is said to be periodic. d) More precisely, if there exists a finite number L such that f(x + L) =f(x), then f(x) isperiodic with period L. e) We can write any function that is periodic (or that is defined on a finite interval) as a Fourier series. f) However if f(x) is non-periodic or is defined on the infinite interval from to +, we must use a Fourier integral. 3. Fourier Series. Fourier series are not mere mathematical devices; they can be generated in the laboratory (or telescope) = a spectrometer decomposes an electromagnetic wave into spectral lines, each with a different frequency and amplitude (intensity). Thus, a spectrometer decomposes a periodic function in a fashion analogous to the Fourier series. a) Suppose we want to write a periodic, piecewise continuous function f(x) as a series of simple functions. Let L denote the period of f(x), and choose as the origin of coordinates the midpoint of the interval defined by this period L/2 x L/2. b) If we let a n and b n denote (real) expansion coefficients, we can write the Fourier series of this function as f(x) =a 0 + [ ( a n cos 2πn x ) ( + b n sin 2πn x )]. n=1 L L (III-42) III 15

18 c) We calculate the coefficients in Eq. (III-42) from the function f(x) as a 0 = 1 L/2 f(x) dx, L (III-43) L/2 a n = 2 L b n = 2 L L/2 L/2 f(x) cos (2πn x L L/2 L/2 f(x) sin (2πn x L ) ) dx (n =1, 2,...) (III-44) dx (n =1, 2,...). (III-45) d) Notice that the summation in Eq. (III-42) contains an infinite number of terms. In practice we retain only a finite number of terms = this approximation is called truncation. i) Truncation is viable only if the sum converges to whatever accuracy we want before we chop it off. ii) Truncation is not as extreme an act as it may seem. If f(x) is normalizable, then the expansion coefficients in Eq. (III-42) decrease in magnitude with increasing n, i.e., a n 0 and b n 0asn. iii) Under these conditions, which are satisfied by physically admissible wave functions, the sum in Eq. (III-42) can be truncated at some finite maximum value n max of the index n. (Trial and error is typically needed to determine the value of n max that is required for the desired accuracy.) iv) If f(x) is particularly simple, all but a small, finite number of coefficients may be zero. One should III 16

19 always check for zero coefficients first before evaluating the integrals in Eqs. (III-43, 44, 45). 4. The Power of Parity. One should pay attention as to whether one is integrating an odd or an even function. Trigonometric functions have the well-known parity properties: sin( x) = sin x (odd) (III-46) cos( x) = + cos x (even). (III-47) As such, if f(x) is even or odd, then half of the expansion coefficients in its Fourier series are zero. a) If f(x) isodd [f( x) = f(x)], then a n =0 (n =0, 1, 2, 3,...) f(x) = n=1 ( b n sin 2πn x ) L. (III-48) b) If f(x) iseven [f( x) =+f(x)], then b n =0 (n =1, 2, 3,...) f(x) = n=0 a n cos ( 2πn x ) L. (III-49) c) If f(x) is either an even or and odd function, it is then said to have definite parity. 5. The Complex Fourier Series: If f(x) does not have a definite parity, we can expand it in a complex Fourier series. a) To derive this variant on the Fourier series in Eq. (III- 42), we just combine the coefficients a n and b n so as to introduce the complex exponential function e i2πnx/l : f(x) = n= c n e i2πnx/l. (III-50) III 17

20 b) Note carefully that in the complex Fourier series in Eq. (III-50) the summation runs from to. The expansion coefficients c n for the complex Fourier series are L/2 c n = 1 L f(x) L/2 e i2πnx/l dx. (III-51) Exercise: Derive Eqs. (III-50) and (III-51) and thereby determine the relationship of the coefficients c n of the complex Fourier series of a function to the coefficients a n and b n of the corresponding real series. 6. Fourier Integrals: Any normalizable function can be expanded in an infinite number of sine and cosine functions that have infinitesimally differing arguments. Such an expansion is called a Fourier integral. a) A function f(x) can be represented by a Fourier integral provided the integral f(x) 2 dx exists = all wave functions satisfy this condition for they are normalizable. b) The Fourier integral has the form f(x) = 1 2π g(k)eikx dk, which is the inverse Fourier transform. (III-52) c) The function g(k) plays the role analogous to that of the expansion coefficients c n in the complex series (Eq. III-50). The relationship of g(k) tof(x) is more clearly exposed by the inverse of Eq. (III-52), g(k) = 1 2π f(x)e ikx dx, (III-53) which is the famed Fourier transform equation. In mathematical parlance, f(x) and g(k) are said to be Fourier transforms of one another. III 18

21 d) More precisely, g(k) is the Fourier transform of f(x), and f(x) is the inverse Fourier transform of g(k). e) When convenient, we will use the shorthand notation A(k) =F [Ψ(x, 0)] and Ψ(x, 0) = F 1 [A(k)] (III-54) to represent Eqs. (III-53) and (III-52), respectively. f) Many useful relationships follow from the intimate relationship between f(x) and g(k). For our purposes, the most important is the Bessel-Parseval relationship: f(x) 2 dx = D. The Gaussian Wave Packet. g(k) 2 dk. (III-55) 1. In Eqs. (III-40) and (III-41), we stated that there will often be time when the wave function will be expressed in terms of an integral of an amplitude function = calculating the wave function is the inverse Fourier transform of the amplitude function. 2. Hence, to calculate the amplitude function, one merely takes the Fourier transform of the wave function: A(k) =F[Ψ(x, 0)] = 1 Ψ(x, 2π 0)e ikx dx. (III-56) 3. The Bessel-Parseval relationship (Eq. III-55) guarantees that the Fourier transform of a normalized function is normalized: A(k) 2 dk = Ψ(x, 0) 2 dx =1. (III-57) Example III 2. The Amplitude Function for a Gaussian. The Gaussian function is a wave packet with a well-defined center and a single peak. Its amplitude function has the same properties. Initially (i.e., t =0), III 19

22 the Gaussian function contains one parameter, a real number L that governs its width. Show how the amplitude function of a Gaussian scales with L and prove that both the Gaussian wave function and its amplitude function obey normalization rules. Solution: The most general form of such a function has a center at x 0 and corresponds to an amplitude function that is centered at k 0: ( ) 1 1/4 Ψ(x, 0) = e ik x e [(x x )/(2L)] 2. 2πL 2 For simplicity though, assume the Gaussian is centered at x = 0 with an amplitude function centered at k =0,i.e., ( ) 1 1/4 Ψ(x, 0) = e x2 /(4L 2). 2πL 2 Ψ(x,0) L Ψ(x,0) for L = 0.5 Ψ(x,0) for L = 1.0 Ψ(x,0) 2 for L = x Two Gaussian wave packets of the form in Example III-2; parameter L for these functions takes on the values L =0.5 and L =1.0 (solid curves). The corresponding probability density for L = 1.0 is shown as a dashed curve. In the figure above, you ll find two such wave packets (with different values of L). Each exhibits the characteristic shape of a Gaussian function III 20

23 (i.e., a Bell-shaped curve): Each has a single peak and decreases rather sharply as x increases from zero. But because the Gaussian function decays exponentially, it never actually equals zero for finite x. (It is, nonetheless, normalizable.) You ll also find in this figure above that the probability density Ψ(x, 0) 2 for a Gaussian. This figure illustrates one of the special properties of a Gaussian wave function = its probability is also a Gaussian function, one that has the same center but is narrower than the state function from which it is calculated. To find the amplitude function for this Gaussian, make sure that both the state function and the amplitude function are normalized as requested in the problem. We begin by substituting the state function listed above into Eq. (III-56) for the amplitude function: A(k) = 1 ( ) 1 1/4 [ 2π 2πL exp 1 ] ikx 2 4L 2x2 We note that a Table of Integrals give dx. π βx /(4α) dx = e αx2 α eβ2 (α >0). Let α =1/(4L 2 ) and β = ik, then we see that A(k) =( 2 π L2 ) 1/4 e k2 L 2. Comparing the mathematical form of the amplitude function and the initial wave function, we see that the Fourier transform of a Gaussian function of variable x is a Gaussian of variable k. Now, check for normalization Ψ Ψ dx = = = ( 1 2πL 2 ( 1 2πL 2 ( 1 2πL 2 ) 1/2 /(2L 2) dx e x2 ) 1/2 2 0 ) 1/2 2 π 1 2/ 2L e x2 /(2L 2) dx III 21

24 = = ( 1 2πL 2 2πL2 2πL 2 ) 1/2 2πL 1/2 = 1=1. QED A(k) A(k) dk = = = = = ( ) 2 1/2 π L2 L 2 dk e 2k2 ( ) 2 1/2 π L2 2 e 2k2 L 2 dk 0 ( ) 2 1/2 1 π L2 2 2 π 2L ( ) 2 1/2 π π L2 2πL2 2πL 2 1/2 2L 2 = 1=1. QED E. The Quantum Simple Harmonic Oscillator. 1. From classical mechanics, the equation of motion of a mass connected to a spring is govern by Hooke s Law: F = kx = m d2 x dt, 2 where k is the spring constant. 2. The solution to Hooke s Law gives a simple harmonic oscillation in position: x(t) =A sin(ωt)+b cos(ωt), III 22

25 where A and B are constants determined from either initial conditions or boundary conditions, and ω = k m (III-58) is the (angular) frequency of oscillation. The potential energy is V (x) = 1 2 kx2. (III-59) 3. As can be seen by the above equation, the potential for Hooke s law is parabolic. Practically any potential can be approximated by a parabola, if one takes the limits small enough in the neighborhood of a local minimum. a) One can expand any function about a local minimum in a Taylor series. In the case of the potential: V (x) =V (x )+ 1 1! V (x )(x x )+ 1 2! V (x )(x x ) 2 + (III-60) where x is the location of the local minimum. b) We can set V (x ) = 0 since the value of this constant is irrelevant to the force. Also V (x ) = 0 since x is a minimum. If we drop terms higher than the second derivative, we get V (x) = 1 2 V (x )(x x ) 2, (III-61) which describes simple harmonic oscillation about the point x, with an effective spring constant k = V (x ). c) It should be noted from this Taylor series approximation that any oscillatory motion is approximately simple harmonic, as long as the amplitude is small. III 23

26 4. The quantum problem is to solve the Schrödinger equation for the potential V (x) = 1 2 mω2 x 2. (III-62) a) With this potential, the time-independent Schrödinger s equation becomes h2 d 2 ψ 2m dx mω2 x 2 ψ = Eψ. (III-63) b) There are 2 ways to solve this equation: The power series expansion method and the ladder operator method. This second method is primarily just clever algebra and we will go through that solution first. 5. The Ladder Operator (Algebraic) Method. a) Let s rewrite Eq. (III-63) as )2 1 ( h d +(mωx) 2 ψ = Eψ. 2m i dx (III-64) b) We must factor the term in the square brackets. We can do this by realizing that if these terms were numbers, we could factor them as u 2 + v 2 =(u iv)(u + iv). However in our case, u and v are operators and operators do not, in general, commute (i.e., uv vu). c) Let us introduce the a-operators as a ± 1 ) ( h d 2m i dx ± imωx. (III-65) Let s check the various products of these operators: i) First, (a a + )f(x) = 1 2m ( h i d dx imωx )( h i III 24 ) d dx + imωx f(x)

27 ii) = 1 ) ( h d )( h 2m i dx imωx df i dx + imωxf = 1 h 2d2 f 2m dx + hmω d 2 dx (xf) hmωx df ] dx +(mωx)2 f = 1 h 2d2 f df + hmωx 2m dx2 dx + hmωf hmωx df ] dx +(mωx)2 f = 1 )2 ( h d +(mωx) 2 + hmω f(x) 2m i dx a a + = 1 )2 ( h d +(mωx) (III-66) 2m i dx 2 hω Pulling the (1/2) hω term to the other side of the equation, we get the following for Schrödinger s equation: (a a + 2 hω 1 ) ψ = Eψ. (III-67) iii) From a similar treatment, one can easily show that a + a = 1 )2 ( h d +(mωx) 2 1, (III-68) 2m i dx 2 hω and the Schrödinger equation becomes (a + a + 2 hω 1 ) ψ = Eψ. (III-69) iv) Also note that a a + a + a = hω. (III-70) d) If ψ satisfies the Schrödinger equation with energy E, then a + ψ satisfies the Schrödinger equation with energy (E + III 25

28 hω): (a + a hω ) (a + ψ) = ( a + a a hωa 1 ) + ψ ( = a + a a hω 1 ) [( ψ = a + a a + 2 hω 1 ) ] ψ + hωψ = a + (Eψ + hωψ) = (E + hω)(a + ψ). QED(III-71) e) By the same token, a ψ is a solution with energy (E hω): (a a + 2 hω 1 ) ( (a ψ) = a a + a 2 hωa 1 ) ψ ( = a a + a 2 hω 1 ) [( ψ = a a + a + 2 hω 1 ) ] ψ hωψ = a (Eψ hωψ) = (E hω)(a ψ). QED(III-72) f) We call the a-operators the ladder operators: a + is called the raising operator, and a, the lowering operator = if we can find one solution with energy E, we can find all other solutions by using these operators. g) There exist a lowest energy state for the quantum harmonic oscillator which satisfies a ψ =0, (III-73) where ψ is referred to as the ground state. i) As a result, we can write the lowering operator for this ground state as ) 1 ( h dψ 2m i dx imωxψ =0, (III-74) or dψ dx = mω h xψ. (III-75) ii) The solution to this differential equation is trivial: dψ = mω ψ h xdx III 26

29 ln ψ = mω 2 h x2 + constant ψ = A e (mω/2 h) x2. (III-76) iii) Plugging this into Schrödinger equation, (a + a + (1/2) hω)ψ = E ψ and noting Eq. (III-73), we see that E = 1 2 hω. (III-77) Note that so hω = h 2π ω = 2πν h = h 2π, 2πν = hν. h) From this ground state, we can easily calculate the excited states with the raising operator: ( ψ n (x) =A n (a + ) n e (mω/2 h) x2, with E n = n + 1 ) hω. 2 (III-78) i) For instance, the first excited state is ψ 1 (mω/2 h) x2 = A 1 a + e 1 ( h = A 1 2m i = A 1 [ h 2m i ) d dx + imωx (mω/2 h) x2 e ( mω ) ] h x e (mω/2 h) x2 + imωxe (mω/2 h) x2. (III-79) ii) This simplifies to ψ 1 =(ia 1 ω 2m)xe (mω/2 h) x2. (III-80) Note that since ψ must be real, the amplitude A 1 must be imaginary (see Example III-3 below). III 27

30 Example III 3. The raising and lowing operators: (a) The raising and lowering operators generate new solutions to the Schrödinger equation, but these new solutions are not correctly normalized. Thus a + ψ n is proportional to ψ n+1, and a ψ n is proportional to ψ n 1, but we would like to know the precise proportionality constants. Use integration by parts and the Schrödinger equation (Eqs. III-67 and III-69) to show that a +ψ n 2 dx = (n +1) hω a ψ n 2 dx = n hω, and hence (with i s to keep the wave function real) (III-81) (III-82) a + ψ n = i (n + 1) hω ψ n+1 (III-83) a ψ n = i n hω ψ n 1. (III-84) (b) Use Eq. (III-83) to show that the normalization constant A n in Eq. (III-78) is A n = ( ) mω 1/4 ( i) n π h. n!( hω) n (III-85) Solution (a): a +ψ n 2 dx = (a +ψ n ) (a + ψ n ) dx = 1 ) ] [( h d 2m i dx + imωx ψ n ) ] [( h d i dx + imωx ψ n dx = 1 ( h dψ ) n 2m i dx imωxψ n ) ( h dψ n i dx + imωxψ n dx III 28

31 At this point, note that = 1 [ h 2dψ n 2m dx hmωxψ n dψ n dx hmωxdψ n dψ n dx +(mωx)2 ψ n ψ n d dx (ψ n ψ n)=ψn dψ n dx + dψ n dx ψ n. As such, we can simplify the integral above as a +ψ n 2 dx = 1 2m [ h 2dψ n dψ n dx dx +(mωx)2 ψnψ n ] dx ψ n ] dx. hmωx d dx (ψ n ψ n) dx = 1 [ 2m h2dψ n dψ n dx dx dx + (mωx)2 ψn ψ n dx hmωx d ] dx (ψ n ψ n) dx. (A) The first integral in Eq. (A) is found by using integration by parts, let u = dψ n dv = dψ dx n du = d2 ψ n dx v = ψ dx 2 n, then h2dψ n dx dψ n dx dx = h2dψ n dx = h 2 dψ n = h 2 ψ n dψ n dx dx dx dψ n dψ n dx ψ n d 2 ψ n dx 2 = h 2 d 2 ψ n ψ n dx, (A1) dx2 where ψn (dψ n/dx) 0 since the wave function must be normalizable. The second integral in Eq. (A) can be written as dx (mωx)2 ψ n ψ n dx = ψ n (mωx)2 ψ n dx. (A2) III 29

32 The third and final integral in Eq. (A) in once again solved with integral by parts, let u = x dv = d(ψn ψ n) du = dx v = ψn ψ n, then hmω x d dx (ψ n ψ n) dx = hmω xd(ψ n ψ n) = hmω [xψ n ψ n ψ n ψ n dx = hmω, (A3) which results once again since xψn ψ n 0 in order for the wave function to be normalizable. Now plugging Eqs. (A1), (A2), and (A3) back into Eq. (A) gives a +ψ n 2 dx = 1 h 2 ψ 2m nd 2 ψ n dx + 2 ψ n (mωx)2 ψ n dx 1 2m ( hmω), or with use of the Schrödinger equation (Eq. III-64), we get a +ψ n 2 dx = 1 h 2 ψ 2m nd 2 ψ n dx 2 + ψ n(mωx) 2 ψ n dx = hω ψ n h2 2m d 2 = ψ ne n ψ n dx + 2 hω 1 = E n ψ n ψ n dx + 2 hω 1 dx mω2 x 2 ψ n dx hω = E n + 2 hω 1 =(n ) hω hω = (n +1) hω. QED Note that we could have solved this much more simply by using operator algebra and noting that a + = a (see next section): (a +ψ n ) (a + ψ n ) dx = III 30 (ψ n a )(a + ψ n ) dx ]

33 = ψ na a + ψ n dx = ψ n(e n + 2 hω) 1 ψ n dx =(E n + 2 hω) 1 ψ n ψ n dx = E n + 2 hω 1 =(n ) hω hω =(n +1) hω. QED Now for the proof of the second integral relation (a ψ n ) (a ψ n ) dx = 1 ) ] [( h d 2m i dx imωx ψ n ) ] [( h d i dx imωx ψ n dx = 1 ( h dψ ) n 2m i dx + imωxψ n ) ( h dψ n i dx imωxψ n dx = 1 [ h 2dψ n dψ n 2m dx dx + hmωxdψ n dx ψ n ] + hmωxψn dψ n dx = 1 [ h 2dψ n 2m dx + hmωx d dx (ψ n ψ n) dx +(mωx)2 ψn ψ n dψ n dx +(mωx)2 ψn ψ n ] dx. (B) The first two integrals in Eq. (B) are identical to those in Eq. (A) and the third is just the negative of the third integral in Eq. (A). As such, Eq. (B) reduces to a ψ n 2 dx = 1 2m h 2 ψ nd 2 ψ n dx 2 + ψ n(mωx) 2 ψ n dx = = 1 2 hω ψ n h2 2m d 2 ψ ne n ψ n dx 1 2 hω dx mω2 x 2 ψ n dx 1 2 hω III 31

34 = E n ψ nψ n dx 2 hω 1 = E n + 2 hω 1 =(n ) hω 2 2 hω = n hω. QED We next have to prove the solutions of the raising and lowering operators. Assume that a + ψ n = cψ n+1, for some constant c. With ψ n and ψ n+1 normalized, a +ψ n 2 dx = c 2 ψ n+1 2 dx = c 2 =(n +1) hω, so c = (n + 1) hω. Note however, to achieve consistency with the integrals above, c = i (n + 1) hω, so a + ψ n = i (n + 1) hω ψ n+1. Similarly, a ψ n = bψ n 1, for some constant b, so a ψ n 2 dx = b 2 ψ n 1 2 dx = b 2 = n hω, so b = n hω. Once again, to achieve consistency, b = i n hω, so a ψ n = i n hω ψ n 1. Solution (b): From Eqs. (III-76) and (III-78), ψ = A e (mω/2 h)x2, e (mω/2 h)x2 = ψ A ψ n = A n (a + ) n e (mω/2 h)x2 = A n A (a + ) n ψ = A n (a + ) n 1 (a + ψ ) A }{{} i hωψ 1 = A n A (i hω)(a + ) n 2 (a + ψ 1 ) }{{} i 2 hωψ 2 = = A n (i hω)(i 2 hω)(i 3 hω) (i n hω)ψ n A = A n i n n!( hω)n ψ n, A ( i) n A n = A. n!( hω) n III 32

35 Normalizing ψ gives or so 1= A 2 / h dx = A e mωx2 2 π h mω A n = A = ( ) mω 1/4, π h ( ) mω 1/4 ( i) n π h. n!( hω) n 6. The Power Series Expansion (Analytic) Method. a) Let mω dξ mω ξ x and h dx = h in the Schrödinger equation (III-86) h2 d 2 ψ 2m dx mω2 x 2 ψ = Eψ. Then using the chain rule, d 2 ψ = d dx 2 dx = d dξ ( ) dψ = d ( ) dψ dξ dx dξ dx dx = d ( dψ dξ dξ ( ) dψ mω mω dξ h h = mω h plugging this back into Eq. (III-87) gives h2 mω d 2 ψ 2m h dξ + 1 ( h 2 2 mω2 mω hω 2 (III-87) ) dξ dξ dx dx d 2 ψ dξ 2, (III-88) ) ξ 2 ψ = Eψ d 2 ψ dξ 2 + hω 2 ξ2 ψ = Eψ d 2 ψ dξ 2 ξ2 ψ = 2E hω ψ III 33

36 ( ξ 2 2E hω ) ψ d 2 ψ dξ 2 = d 2 ψ dξ 2 = ( ξ 2 K ) ψ, (III-89) where K 2E hω. Finally, we can rewrite Eq. (III-89) as d2 dξ 2 ξ2 + K ψ(ξ) =0, which is called Weber s equation. (III-90) (III-91) b) To solve this DE, let s first look at the asymptotic limit of very large ξ (hence x) such that ξ 2 K, then d 2 ψ dξ 2 ξ2 ψ, which has the approximate solution ψ(ξ) Ae ξ2 /2 + Be +ξ2 /2. (III-92) (III-93) i) The B term is clearly not normalizable since ψ blows up as ξ. ii) The realistic solution thus has the asymptotic form ψ(ξ) Ae ξ2 /2, at large ξ. (III-94) c) Since we know the asymptotic form of the solution, we can re-examine Eq. (III-89 or III-91) and assume the solution to this equation has the functional form ψ(ξ) =Ah(ξ) e ξ2 /2, (III-95) where h is some function of ξ that contains the constant energy term K and the coefficient A is determined from III 34

37 the normalization condition. Since we insist that the wave function for a real wave be normalizable, then ( h(ξ) e ξ2 /2 ) 0. (III-96) lim ξ ± i) Note that h(ξ) must depend upon the energy term K since the asymptotic form Ae ξ2 /2 does not. ii) Also note that since the asymptotic form is an even function, the h(ξ) portion of the wave function will dictate the parity of the complete wave function = if h(ξ) is an even function, then the wave function will have even parity, and if h(ξ) is an odd function, then the wave function will have odd parity. d) We now need to come up with the functional form of h(ξ). i) The first derivative of Eq. (III-95) is ( ) dψ dh dξ = A dξ ξh e ξ2 /2. (III-97) ii) The second derivative is thus d 2 ψ dξ 2 = A d2 h dh 2ξ dξ2 dξ +(ξ2 1)h e ξ2 /2. (III-98) iii) Using these derivatives in the Schrödinger equation (e.g., Eq. III-91) gives d 2 h dh 2ξ +(K 1)h =0. dξ2 dξ (III-99) e) Now, let s assume that h(ξ) can be expressed as a power series h(ξ) =a 0 + a 1 ξ + a 2 ξ 2 + = a j ξ j. (III-100) III 35 j=0

38 (Note that the a parameters here are series coefficients and not the raising and lowering operators we were discussing earlier.) i) The first derivative of this series is dh dξ = a 1 +2a 2 ξ +3a 3 ξ 2 + = j=0 ja j ξ j 1. (III-101) ii) The second derivative is d 2 h dξ 2 = 2a a 3 ξ +3 4a 4 ξ 2 + = j=0 (j + 1)(j +2)a j+2 ξ j. (III-102) iii) Putting these into Eq. (III-99), we find j=0 [(j + 1)(j +2)a j+2 2ja j +(K 1)a j ] ξ j =0. (III-103) iv) It follows from the uniqueness of the power series expansion that the coefficient of each power of ξ must vanish, (j + 1)(j +2)a j+2 2ja j +(K 1)a j =0, and hence that a j+2 = (2j +1 K) (j + 1)(j +2) a j. (III-104) v) This recursion formula is entirely equivalent to the Schrödinger equation itself. Given a 0 we can in principle derive a 2,a 4,a 6,..., and given a 1 it generates a 3,a 5,a 7,... III 36

39 f) Let us therefore write h(ξ) =h e (ξ)+h o (ξ), (III-105) where h e (ξ) are the even terms of the series expansion (i.e., aneven function): h e (ξ) a 0 + a 2 ξ 2 + a 4 ξ 4 + (III-106) and h o (ξ) are the odd terms of the series expansion (i.e., an odd function): h o (ξ) a 1 ξ + a 3 ξ 3 + a 5 ξ 5 + (III-107) g) Our next job is to figure out the functional form of the a coefficients. Essentially, we will guess at a form for a j and test it to see if it is consistent with Eqs. (III-91) and (III-99). But first, let s note the series expansions of exponential functions. i) For the exponential function, we have ii) e ±ξ =1±ξ + 1 2! ξ2 ± 1 3! ξ3 + = b j ξ j, (III-108) where j=0 b j =(±1) j 1 j!. (III-109) If we now substitute ξ 2 for ξ in the above series expansion, we get e ξ2 = 1+ξ ! (ξ2 ) ! (ξ2 ) ! (ξ2 ) 4 + = 1+ξ ξ ξ ξ8 + where = j =0 (even) b j ξ j, b j = 1 (j/2)!. III 37 (III-110) (III-111)

40 iii) We next ask if the series expansion above converges. To test this, we use the d Alembert ratio test which takes the limit of the ratio of two successive terms as the counter goes to infinity: b j+2 ξ j+2 b j ξ j = 1 [(j+2)/2]! ξj+2 1 = (j/2)! ξj 1 (j/2) + 1 ξ2 = 2 j +2 ξ2. Now as taking the limit of this ratio as j,we get lim j 2 2 j +2 ξ2 lim j j ξ2 0, thus this series converges. (III-112) h) We are now faced with a wave function solution to the Schrödinger equation of the form ψ(ξ) = Ah(ξ) e ξ2 /2 = A j=0 a j ξ j e ξ2 /2 (III-113) = (infinite series in ξ) (decaying Gaussian). i) For this wave function to be physically admissible, it must go to zero as ξ. But does the infinite sum produce a finite number? Let s assume that our wave function has even parity. Then we will only look at h e (ξ) in this case. (We could make a similar argument for the h o (ξ) odd parity solution.) ii) For very large j, we see from Eq. (III-104) that the recursion formula takes on the approximate form of a j+2 2 j a j. (III-114) iii) But what is a j when j is large? Let us take the ratio of two successive terms and run the d Alembert III 38

41 ratio test. Using Eq. (III-104), this ratio is a j+2 ξ j+2 a j ξ j = 2j +1 K (j + 2)(j +1) ξ2. Now as taking the limit of this ratio as j,we get lim j 2j +1 K 2j 2 (j + 2)(j +1) ξ2 lim j j 2 ξ2 = lim j j ξ2, (III-115) which is precisely the same limit for large j as e ξ2! From this coincidence, we can make use of Eq. (III- 111) to write that a j C (j/2)! (for even j), for some constant C at large j. (III-116) iv) This yields (at large ξ, where the higher powers dominate) h(ξ) C 1 (j/2)! ξj C 1 k! ξ2k Ce ξ2. (III-117) v) If h goes like e ξ2, then ψ goes like e ξ2 /2 (see Eq. III-113), which is precisely the asymptotic behavior we don t want! vi) There is only one way out of this dilemma = the power series must terminate! There must occur some highest j (call it n = j max ) such that a n+2 =0. i) We see from Eq. (III-104) that this will occur when K =2n +1, (III-118) III 39

42 or using Eq. (II-90), we see that the requirement that for h(ξ) to be a finite polynomial, the energies of a simple harmonic oscillator are restricted to the form 2E = 2n +1 hω E n = hω 2 = (2n +1) ( n ) hω n =0, 1, 2,... (III-119) Hence we recover, by a completely different method, the fundamental quantization condition found in Eq. (III-78). j) As a result of this, we see that for any energy not in the form of Eq. (III-119), the wave function of the simple harmonic oscillator blows up at large ξ and the wave function is no longer un-normalizable, hence unphysical. From this we see that the boundary conditions demand energy quantization. k) The recursion formula now reads 2(n j) a j+2 = (j + 1)(j +2) a j (j n). (III-120) i) If n =0(i.e., the ground state), there is only one term in the series (we must pick a 1 = 0 to kill h o, and j = 0 in Eq. (III-120) yields a 2 = 0): and hence h 0 (ξ) =a 0, ψ 0 (ξ) =A 0 a 0 e ξ2 /2, which reproduces Eq. (III-76). (III-121) (III-122) ii) For n = 1 we pick a 0 = 0, and Eq. (III-120) with j = 1 yields a 3 = 0, so h 1 (ξ) =a 1 ξ, III 40 (III-123)

43 Table III 1: The first few Hermite polynomials, H n (x). H 0 = 1 H 1 = 2x H 2 = 4x 2 2 H 3 = 8x 3 12x H 4 = 16x 4 48x H 5 = 32x 5 160x x and hence ψ 1 (ξ) =A 1 a 1 ξe ξ2 /2 (III-124) confirming Eq. (III-80). iii) For n =2,j = 0 yields a 2 = 2a 0, and j =2 gives a 4 = 0, so h 2 (ξ) =a 0 (1 2ξ 2 ) (III-125) and ψ 2 (ξ) =A 2 a 0 (1 2ξ 2 ) e ξ2 /2, (III-126) and so on. l) h n (ξ) is thus a polynomial of degree n in ξ, involving even powers only, if n is an even integer, and odd powers only, if n is an odd integer = Hermite polynomials (designated by H n (ξ)) share these same characteristics (see Table III-1). Note that in Hermite polynomials, the coefficient of the term with the maximum power is arbitrarily set to 2 n. m) Normalization. i) Finally, we need to figure out the functional form of the normalization constant A n. We need to convert back to the x label (from ξ, see Eq. III-86). To III 41

44 simplify matters, let mω0 β = h for the ground state. so ξ = βx (III-127) ii) For the n = 0 ground state, normalization gives 1 = = ψ ψdx A2 0 e β2 x 2 dx = A 2 π 0 β 2 A 0 = β2 1/4 π. (III-128) iii) We can continue this for some of the higher-order wave functions (I leave it to the students to do this on their own). We can compare these results to derive the following recursion relationship for the normalization constant: A n = 1 2n n! A 0 = 1 2n n! β2 1/4 π. (III-129) iv) With this, the normalized stationary states for the simple harmonic oscillator are ψ n (ξ) = β2 1/4 π 1 2n n! H n(ξ)e ξ2 /2, which is identical to Eq. (III-78). (III-130) v) Finally, we can write the complete wave equation by including the time dependence (i.e., Eqs. III-10, III 42

45 III-119, III-129, and III-130) as Ψ n (x, t) = β2 1/4 π 1 2n n! where β = mω 0 / h (see Eq. III-127). H n (βx) e β2 x 2 /2 e i(2n+1)ω 0t/2, (III-131) Example III 4. In the ground state of the harmonic oscillator, what is the probability (correct to 3 significant digits) of finding the particle outside the classically allowed region? Hint: Look in a math table under Normal Distribution or Error Function. Solution: For the ground state, the wave function is ( ) mω 1/4 ψ 0 = e ξ2 /2, π h so the probability is mω mω P =2 e ξ2 dx =2 h e ξ2 dξ. π h x 0 π h mω ξ 0 1 The classically allowed region extends out to: 2 mω2 x 2 0 = E 0 = x 0 = h/mω, soξ 0 = 1. Therefore, P = 2 e ξ2 dξ = 2[1 F ( 2)]=0.157, π 1 where F (z) is the notation used in the CRC Tables. 1 2 hω, or Example III 5. In this problem we explore some of the more useful theorems (stated without proof) involving Hermite polynomials. (a) The Rodrigues formula states that ( d H n (ξ) =( 1) n e ξ2 dξ Use it to derive H 3 and H 4. )n e ξ2. III 43

46 (b) The following recursion relation gives you H n+1 in terms of the 2 preceding Hermite polynomials: H n+1 (ξ) =2ξH n (ξ) 2nH n 1 (ξ). Use it, together with your answer to (a), to obtain H 5 and H 6. (c) If you differentiate an n-th order polynomial, you get a polynomial of order (n 1). For the Hermite polynomials, in fact: dh n dξ =2nH n 1(ξ). Check this by differentiating H 5 and H 6. (d) H n (ξ) is the n-th z-derivative, at z =0, of the generating function exp( z 2 +2zξ); or, to put it another way, it is the coefficient of z n /n! in the Taylor series expansion for the function: e z2 +2zξ = n=0 Use this to rederive H 0, H 1, and H 2. z n n! H n(ξ). Solution (a): d ) = 2ξe dξ (e ξ2 ξ2 ( d )2 e ξ2 = d )=( 2+4ξ 2 )e dξ dξ ( 2ξe ξ2 ξ2 ( d )3 e ξ2 = d dξ dξ [( 2+4ξ2 )e ξ2 ]=[8ξ +( 2+4ξ 2 )( 2ξ)]e ξ2 = (12ξ 8ξ 3 )e ξ2 ( d )4 e ξ2 = d dξ dξ [(12ξ 8ξ3 )e ξ2 ] = [12 24ξ 2 + (12ξ 8ξ 3 )( 2ξ)]e ξ2 = (12 48 ξ 2 +16ξ 4 )e ξ2 H 3 (ξ) = e ξ2 ( d dξ )3 e ξ2 = 12ξ +8ξ 3 III 44

47 H 4 (ξ) = e ξ2 ( d dξ )4 e ξ2 = ξ 2 +16ξ 4. Solution (b): H 5 (ξ) = 2ξH 4 8H 3 =2ξ(12 48 ξ 2 +16ξ 4 ) 8( 12ξ +8ξ 3 ) = 120ξ 160ξ 3 +32ξ 5 H 6 (ξ) = 2ξH 5 10H 4 = 2ξ(120ξ 160ξ 3 +32ξ 5 ) 10(12 48 ξ 2 +16ξ 4 ) = ξ 2 480ξ 4 +64ξ 6. Solution (c): dh 5 dξ dh 6 dξ = ξ ξ 4 = 10(12 48ξ 2 +16ξ 4 ) = (2)(5)H 4 = 1440ξ 1920ξ ξ 5 = 12(120ξ 160ξ 3 +32ξ 5 ) = (2)(6)H 5. Solution (d): d +2zξ )=( 2z +2ξ)e dz (e z2 z2 +2zξ ; putting in z = 0 gives H 0 (ξ) =2ξ. ( d )2 (e z2 +2zξ )= d dz dz [( 2z +2zξ ]=[ 2 +( 2z +2ξ) 2 ]e +2ξ)e z2 z2 +2zξ ; putting in z = 0 gives H 1 (ξ) = 2+4ξ 2. ( d )3 (e z2 +2zξ ) = d dz dz {[ 2+( 2z +2ξ)2 ]e z2 +2zξ } = {2( 2z +2ξ)( 2) + [ 2+( 2z +2ξ) 2 ]( 2z +2ξ)}e z2 +2zξ ; III 45

48 putting in z = 0 gives H 2 (ξ) = 8ξ +( 2+4ξ 2 )(2ξ) = 12ξ +8ξ 3. F. The Free Particle 1. A free particle is simply one where V (x) = 0 for all x. The time-independent Schrödinger equation (TISE) becomes h2 d 2 ψ 2m dx = Eψ, 2 (III-132) or d 2 ψ 2mE dx = 2 k2 ψ, where k. h (III-133) a) Instead of using sines and cosines, we will express the solution in exponential notation: ψ(x) =Ae ikx + Be ikx. (III-134) b) Since there are no boundary conditions, k (hence E) can take on any positive value = a continuum of possible energies. c) Taking on the standard time dependence gives hk ik(x Ψ(x, t) =Ae 2m t) hk ik(x+ + Be 2m t). (III-135) 2. Any function of x and t that depends upon these variables in the special combination (x ± vt) (for some constant v) represents a wave of fixed profile, traveling in the x-direction, at speed v. a) As such, the first term in Eq. (III-135) represents a wave traveling to the right. b) The second term, a wave traveling to the left. III 46

49 c) Since the only difference between the 2 terms in Eq. (III- 135) is the sign of k, we can write hk ik(x Ψ k (x, t) =Ae 2m t), (III-136) and let k run negative (as well as positive) to cover the case of waves traveling to the left: 2mE k>0 = traveling to the right, k ±, with h k<0 = traveling to the left. (III-137) 3. The speed of the wave is v quantum = h k 2m = E 2m. (III-138) a) On the other hand, the classical speed of a free particle with energy E is given by E =(1/2)mv 2 (pure kinetic since V = 0), so v classical = 2E m =2v quantum. (III-139) b) The quantum mechanical wave function travels a half the speed of the particle it is suppose to represent! c) This wave function is not normalizable! = in other words, there is no such thing as a free particle with a definite energy. 4. The general solution to the TISE is still a linear combination of separable solutions (only this time, it s an integral over the continuous variable k, instead of a sum over the discrete index n): Ψ(x, t) = 1 2π + hk 2 φ(k)ei(kx 2m t) dk. (III-140) III 47

50 a) Now this wave function can be normalized [for appropriate φ(k)], but it necessarily carries a range of k s, and hence a range of energies and speeds. b) As such, this is called a wave packet. c) The initial wave function is thus Ψ(x, 0) = 1 2π + φ(k)eikx dk. (III-141) 5. This is a classic problem of Fourier analysis (see III.C); the answer is provided by Plancherel s theorem: f(x) = 1 2π + F (k)eikx dk F (k) = 1 2π + f(x)e ikx dx. a) F (k) is called the Fourier transform of f(x). b) f(x) is called the inverse Fourier transform of F (k). (III-142) c) The necessary and sufficient condition on f(x) is that + f(x) 2 dx be finite. 6. The solution to the generic, free-particle quantum problem is thus φ(k) = 1 + Ψ(x, 2π 0)e ikx dx. (III-143) a) A wave packet is a sinusoidal function whose amplitude is modulated by φ (Figure III-3); it consists of ripples contained within an envelope. b) What corresponds to the particle velocity is not the speed of the individual ripples (the so-called phase velocity), but rather the speed of the envelope (the group velocity) III 48

51 Ψ v g v p x Figure III 3: A wave packet. The envelope travels at the group velocity; the ripples travel at the phase velocity. which, depending upon the nature of the waves, can be greater than, less than, or equal to the velocity of the ripples that go to make it up. c) For a free particle in quantum mechanics, the group velocity is twice the phase velocity as shown in Eq. (III-139). 7. But what are the respective group and phase velocities of the wave packet with the general form Ψ(x, t) = 1 + φ(k) 2π ei(kx ωt) dk? (III-144) a) Here the dispersion relation, that is, the formula for ω as a function of k, isω =( hk 2 /2m). b) The amplitude function φ(k) insures that the integrand in Eq. (III-144) is negligible except in the vicinity of k (the position in k-space of the peak value of φ(k)). As such, III 49

52 we can Taylor expand: ω(k) = ω + ω (k k ), (III-145) where ω k. is the derivative of ω with respect to k at point c) Let s k k, then Ψ(x, t) = 1 + φ(k + s) e i[(k +s)x (ω +ω s)t] ds. 2π (III-146) d) At t =0, Ψ(x, 0) = 1 + φ(k + s) e i(k +s)x ds (III-147) 2π and at later times Ψ(x, t) = 1 e i( ω t+k ω t) + φ(k +s) e i(k +s)(x ω t) ds. 2π (III-148) e) Except for the shift from x to (x ω t), this last integral is the same as the one for Ψ(x, 0), thus Ψ(x, t) = e i( ω t+k ω t) Ψ(x ω t, 0). (III-149) f) Apart from the phase factor in front (which won t affect Ψ 2 in any event), the wave packet moves along at speed v group = dω = hk dk k m, (III-150) which is contrasted with the ordinary phase velocity v phase = ω k = hk 2m. (III-151) g) Hence, the group velocity is twice as great as the phase velocity: v classical = v group =2v phase. (III-152) III 50

53 Example III 6. A free particle has the initial wave function Ψ(x, 0) = Ae ax2, where A and a are constants (a is real and positive). (a) Normalize Ψ(x, 0). (b) Show that Ψ(x, t) can be written in the form ( 2a Ψ(x, t) = π ) 1/4 e ax2 /[1+(2i hat/m)] 1+(2i hat/m). Hint : Integrals of the form + +bx) dx e (ax2 can be handled by completing the square. Let y a [x +(b/2a)], and note that (ax 2 + bx) y 2 (b 2 /4a). (c) Find Ψ(x, t) 2. Express your answer in terms of the quantity w a/[1 + (2 hat/m)2. Sketch Ψ 2 (as a function of x) att =0, and again for some very large t. Qualitatively, what happens to Ψ 2 as time goes on? (d) Find x, p, x 2, p 2, σ x, and σ p. Hint : You will need to show that p 2 = a h 2, but it may take some algebra to reduce it to this simple form. (e) Does the uncertainty principle hold? At what time t does the system come closest to the uncertainty limit? Solution (a): or 1= A 2 + π dx = A 2 e 2ax2 2a A = ( ) 2a 1/4. π III 51

54 Solution (b): Using Eq. (III-143), we can write φ(k) = 1 + A e ikx dx = 1 + A +ikx) dx. 2π e ax2 2π e (ax2 Using the substitution variables in the question (i.e., y from the last page and let b = ik), the integral above as the functional form So, + e (ax2 +bx) dx = + e y2 +(b 2 /4a) 1 a dy = 1 a e b2 /4a + e y2 dy = φ(k) = 1 ( ) 2a 1/4 π /4a 2π π a e k2 1 = /4a. (2πa) 1/4 e k2 Now using φ(k) in Eq. (III-140), we can solve for Ψ, π a eb2 /4a. Ψ(x, t) = /4a 2π (2πa) 1/4 e e k2 i(kx hk2t/2m) dk = 1 2π 1 (2πa) 1/4 = 1 2π 1 (2πa) 1/4 + e k 2 4a +ixk i hk2 t 2m dk + e [k2 ( 1 4a + i ht 2m ) ixk] dk. (A) At this point, let ( 1 β = 4a + i ht ) and γ = ix. 2m Then making use of Eq. (A) above, the integral in the equation above becomes + ( 1 e [k2 4a + i ht 2m ) ixk] + dk = +γk) dk e (βk2 III 52

55 = π β eγ2 /4β = π 1 4a + i ht 2m e x2 /4( 1 4a + i ht 2m ). Substituting this integral solution into the equation for Ψ above, we get 1 π Ψ(x, t) = e x2 /4( 1 4a + i ht 2m ) 2π(2πa) 1/4 = ( 2a π 1 4a + i ht 2m ) 1/4 e ax2 /(1+2i hat/m) 1+2i hat/m. Solution (c): Let θ = 2 hat/m, then Ψ 2 = 2a π The exponent is ax2 (1 + iθ) ax2 (1 iθ) and Ψ 2 = 2a π Or, with w a, we get 1+θ 2 e ax2 /(1+iθ) e ax2 /(1 iθ) (1 + iθ)(1 iθ). = ax2(1 iθ +1+iθ) (1 + iθ)(1 iθ) = 2ax2 1+θ 2, e 2ax2 /(1+θ 2 ) (1 + θ2 ) Ψ 2 = 2 x 2. π we 2w2. Ψ 2 Ψ 2 x t = 0 t >> 0 x As t increases, the graph of Ψ 2 flattens out and broadens, as can be III 53