Physics 4617/5617: Quantum Physics Course Lecture Notes
|
|
- Hortense Thornton
- 5 years ago
- Views:
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
Notes on Quantum Mechanics
Notes on Quantum Mechanics Kevin S. Huang Contents 1 The Wave Function 1 1.1 The Schrodinger Equation............................ 1 1. Probability.................................... 1.3 Normalization...................................
More informationQuantum Harmonic Oscillator
Quantum Harmonic Oscillator Chapter 13 P. J. Grandinetti Chem. 4300 Oct 20, 2017 P. J. Grandinetti (Chem. 4300) Quantum Harmonic Oscillator Oct 20, 2017 1 / 26 Kinetic and Potential Energy Operators Harmonic
More information= k, (2) p = h λ. x o = f1/2 o a. +vt (4)
Traveling Functions, Traveling Waves, and the Uncertainty Principle R.M. Suter Department of Physics, Carnegie Mellon University Experimental observations have indicated that all quanta have a wave-like
More informationLecture #8: Quantum Mechanical Harmonic Oscillator
5.61 Fall, 013 Lecture #8 Page 1 Last time Lecture #8: Quantum Mechanical Harmonic Oscillator Classical Mechanical Harmonic Oscillator * V(x) = 1 kx (leading term in power series expansion of most V(x)
More informationUnderstand the basic principles of spectroscopy using selection rules and the energy levels. Derive Hund s Rule from the symmetrization postulate.
CHEM 5314: Advanced Physical Chemistry Overall Goals: Use quantum mechanics to understand that molecules have quantized translational, rotational, vibrational, and electronic energy levels. In a large
More informationPY 351 Modern Physics - Lecture notes, 3
PY 351 Modern Physics - Lecture notes, 3 Copyright by Claudio Rebbi, Boston University, October 2016. These notes cannot be duplicated and distributed without explicit permission of the author. Time dependence
More informationdf(x) = h(x) dx Chemistry 4531 Mathematical Preliminaries Spring 2009 I. A Primer on Differential Equations Order of differential equation
Chemistry 4531 Mathematical Preliminaries Spring 009 I. A Primer on Differential Equations Order of differential equation Linearity of differential equation Partial vs. Ordinary Differential Equations
More informationLecture-XXVI. Time-Independent Schrodinger Equation
Lecture-XXVI Time-Independent Schrodinger Equation Time Independent Schrodinger Equation: The time-dependent Schrodinger equation: Assume that V is independent of time t. In that case the Schrodinger equation
More information5.1 Classical Harmonic Oscillator
Chapter 5 Harmonic Oscillator 5.1 Classical Harmonic Oscillator m l o l Hooke s Law give the force exerting on the mass as: f = k(l l o ) where l o is the equilibrium length of the spring and k is the
More informationPhysics 486 Discussion 5 Piecewise Potentials
Physics 486 Discussion 5 Piecewise Potentials Problem 1 : Infinite Potential Well Checkpoints 1 Consider the infinite well potential V(x) = 0 for 0 < x < 1 elsewhere. (a) First, think classically. Potential
More informationApplied Nuclear Physics (Fall 2006) Lecture 2 (9/11/06) Schrödinger Wave Equation
22.101 Applied Nuclear Physics (Fall 2006) Lecture 2 (9/11/06) Schrödinger Wave Equation References -- R. M. Eisberg, Fundamentals of Modern Physics (Wiley & Sons, New York, 1961). R. L. Liboff, Introductory
More informationChemistry 532 Problem Set 7 Spring 2012 Solutions
Chemistry 53 Problem Set 7 Spring 01 Solutions 1. The study of the time-independent Schrödinger equation for a one-dimensional particle subject to the potential function leads to the differential equation
More informationPhysics 342 Lecture 17. Midterm I Recap. Lecture 17. Physics 342 Quantum Mechanics I
Physics 342 Lecture 17 Midterm I Recap Lecture 17 Physics 342 Quantum Mechanics I Monday, March 1th, 28 17.1 Introduction In the context of the first midterm, there are a few points I d like to make about
More informationLecture 10: Solving the Time-Independent Schrödinger Equation. 1 Stationary States 1. 2 Solving for Energy Eigenstates 3
Contents Lecture 1: Solving the Time-Independent Schrödinger Equation B. Zwiebach March 14, 16 1 Stationary States 1 Solving for Energy Eigenstates 3 3 Free particle on a circle. 6 1 Stationary States
More informationPhysics 443, Solutions to PS 1 1
Physics 443, Solutions to PS. Griffiths.9 For Φ(x, t A exp[ a( mx + it], we need that + h Φ(x, t dx. Using the known result of a Gaussian intergral + exp[ ax ]dx /a, we find that: am A h. ( The Schrödinger
More informationThe Simple Harmonic Oscillator
The Simple Harmonic Oscillator Asaf Pe er 1 November 4, 215 This part of the course is based on Refs [1] [3] 1 Introduction We return now to the study of a 1-d stationary problem: that of the simple harmonic
More informationkg meter ii) Note the dimensions of ρ τ are kg 2 velocity 2 meter = 1 sec 2 We will interpret this velocity in upcoming slides.
II. Generalizing the 1-dimensional wave equation First generalize the notation. i) "q" has meant transverse deflection of the string. Replace q Ψ, where Ψ may indicate other properties of the medium that
More informationOne-dimensional Schrödinger equation
Chapter 1 One-dimensional Schrödinger equation In this chapter we will start from the harmonic oscillator to introduce a general numerical methodology to solve the one-dimensional, time-independent Schrödinger
More informationChem 3502/4502 Physical Chemistry II (Quantum Mechanics) 3 Credits Spring Semester 2006 Christopher J. Cramer. Lecture 7, February 1, 2006
Chem 350/450 Physical Chemistry II (Quantum Mechanics) 3 Credits Spring Semester 006 Christopher J. Cramer ecture 7, February 1, 006 Solved Homework We are given that A is a Hermitian operator such that
More informationThe Schroedinger equation
The Schroedinger equation After Planck, Einstein, Bohr, de Broglie, and many others (but before Born), the time was ripe for a complete theory that could be applied to any problem involving nano-scale
More informationSection 4: Harmonic Oscillator and Free Particles Solutions
Physics 143a: Quantum Mechanics I Section 4: Harmonic Oscillator and Free Particles Solutions Spring 015, Harvard Here is a summary of the most important points from the recent lectures, relevant for either
More informationin terms of the classical frequency, ω = , puts the classical Hamiltonian in the form H = p2 2m + mω2 x 2
One of the most important problems in quantum mechanics is the simple harmonic oscillator, in part because its properties are directly applicable to field theory. The treatment in Dirac notation is particularly
More informationCHAPTER 8 The Quantum Theory of Motion
I. Translational motion. CHAPTER 8 The Quantum Theory of Motion A. Single particle in free space, 1-D. 1. Schrodinger eqn H ψ = Eψ! 2 2m d 2 dx 2 ψ = Eψ ; no boundary conditions 2. General solution: ψ
More informationChem 3502/4502 Physical Chemistry II (Quantum Mechanics) 3 Credits Spring Semester 2006 Christopher J. Cramer. Lecture 9, February 8, 2006
Chem 3502/4502 Physical Chemistry II (Quantum Mechanics) 3 Credits Spring Semester 2006 Christopher J. Cramer Lecture 9, February 8, 2006 The Harmonic Oscillator Consider a diatomic molecule. Such a molecule
More informationDifferential Equations
Electricity and Magnetism I (P331) M. R. Shepherd October 14, 2008 Differential Equations The purpose of this note is to provide some supplementary background on differential equations. The problems discussed
More informationLecture #5: Begin Quantum Mechanics: Free Particle and Particle in a 1D Box
561 Fall 017 Lecture #5 page 1 Last time: Lecture #5: Begin Quantum Mechanics: Free Particle and Particle in a 1D Box 1-D Wave equation u x = 1 u v t * u(x,t): displacements as function of x,t * nd -order:
More informationCHM 532 Notes on Wavefunctions and the Schrödinger Equation
CHM 532 Notes on Wavefunctions and the Schrödinger Equation In class we have discussed a thought experiment 1 that contrasts the behavior of classical particles, classical waves and quantum particles.
More informationThis ODE arises in many physical systems that we shall investigate. + ( + 1)u = 0. (λ + s)x λ + s + ( + 1) a λ. (s + 1)(s + 2) a 0
Legendre equation This ODE arises in many physical systems that we shall investigate We choose We then have Substitution gives ( x 2 ) d 2 u du 2x 2 dx dx + ( + )u u x s a λ x λ a du dx λ a λ (λ + s)x
More informationMATH3383. Quantum Mechanics. Appendix D: Hermite Equation; Orthogonal Polynomials
MATH3383. Quantum Mechanics. Appendix D: Hermite Equation; Orthogonal Polynomials. Hermite Equation In the study of the eigenvalue problem of the Hamiltonian for the quantum harmonic oscillator we have
More information6. Qualitative Solutions of the TISE
6. Qualitative Solutions of the TISE Copyright c 2015 2016, Daniel V. Schroeder Our goal for the next few lessons is to solve the time-independent Schrödinger equation (TISE) for a variety of one-dimensional
More informationLecture 4. 1 de Broglie wavelength and Galilean transformations 1. 2 Phase and Group Velocities 4. 3 Choosing the wavefunction for a free particle 6
Lecture 4 B. Zwiebach February 18, 2016 Contents 1 de Broglie wavelength and Galilean transformations 1 2 Phase and Group Velocities 4 3 Choosing the wavefunction for a free particle 6 1 de Broglie wavelength
More informationPage 404. Lecture 22: Simple Harmonic Oscillator: Energy Basis Date Given: 2008/11/19 Date Revised: 2008/11/19
Page 404 Lecture : Simple Harmonic Oscillator: Energy Basis Date Given: 008/11/19 Date Revised: 008/11/19 Coordinate Basis Section 6. The One-Dimensional Simple Harmonic Oscillator: Coordinate Basis Page
More information4. Complex Oscillations
4. Complex Oscillations The most common use of complex numbers in physics is for analyzing oscillations and waves. We will illustrate this with a simple but crucially important model, the damped harmonic
More informationChem 3502/4502 Physical Chemistry II (Quantum Mechanics) 3 Credits Spring Semester 2006 Christopher J. Cramer. Lecture 8, February 3, 2006 & L "
Chem 352/452 Physical Chemistry II (Quantum Mechanics) 3 Credits Spring Semester 26 Christopher J. Cramer Lecture 8, February 3, 26 Solved Homework (Homework for grading is also due today) Evaluate
More information1 Assignment 1: Nonlinear dynamics (due September
Assignment : Nonlinear dynamics (due September 4, 28). Consider the ordinary differential equation du/dt = cos(u). Sketch the equilibria and indicate by arrows the increase or decrease of the solutions.
More informationPhysics 220. Exam #2. May 23 May 30, 2014
Physics 0 Exam # May 3 May 30, 014 Name Please read and follow these instructions carefully: Read all problems carefully before attempting to solve them. Your work must be legible, with clear organization,
More informationPhysics 217 Problem Set 1 Due: Friday, Aug 29th, 2008
Problem Set 1 Due: Friday, Aug 29th, 2008 Course page: http://www.physics.wustl.edu/~alford/p217/ Review of complex numbers. See appendix K of the textbook. 1. Consider complex numbers z = 1.5 + 0.5i and
More informationBrief review of Quantum Mechanics (QM)
Brief review of Quantum Mechanics (QM) Note: This is a collection of several formulae and facts that we will use throughout the course. It is by no means a complete discussion of QM, nor will I attempt
More information8.04 Spring 2013 March 12, 2013 Problem 1. (10 points) The Probability Current
Prolem Set 5 Solutions 8.04 Spring 03 March, 03 Prolem. (0 points) The Proaility Current We wish to prove that dp a = J(a, t) J(, t). () dt Since P a (t) is the proaility of finding the particle in the
More informationNotes on Fourier Series and Integrals Fourier Series
Notes on Fourier Series and Integrals Fourier Series et f(x) be a piecewise linear function on [, ] (This means that f(x) may possess a finite number of finite discontinuities on the interval). Then f(x)
More informationCHAPTER 36. 1* True or false: Boundary conditions on the wave function lead to energy quantization. True
CHAPTER 36 * True or false: Boundary conditions on the we function lead to energy quantization. True Sketch (a) the we function and (b) the probability distribution for the n 4 state for the finite squarewell
More informationBound and Scattering Solutions for a Delta Potential
Physics 342 Lecture 11 Bound and Scattering Solutions for a Delta Potential Lecture 11 Physics 342 Quantum Mechanics I Wednesday, February 20th, 2008 We understand that free particle solutions are meant
More informationThe Schrödinger Equation in One Dimension
The Schrödinger Equation in One Dimension Introduction We have defined a comple wave function Ψ(, t) for a particle and interpreted it such that Ψ ( r, t d gives the probability that the particle is at
More informationto the potential V to get V + V 0 0Ψ. Let Ψ ( x,t ) =ψ x dx 2
Physics 0 Homework # Spring 017 Due Wednesday, 4/1/17 1. Griffith s 1.8 We start with by adding V 0 to the potential V to get V + V 0. The Schrödinger equation reads: i! dψ dt =! d Ψ m dx + VΨ + V 0Ψ.
More informationProblem Set 5: Solutions
University of Alabama Department of Physics and Astronomy PH 53 / eclair Spring 1 Problem Set 5: Solutions 1. Solve one of the exam problems that you did not choose.. The Thompson model of the atom. Show
More informationChemistry 432 Problem Set 4 Spring 2018 Solutions
Chemistry 4 Problem Set 4 Spring 18 Solutions 1. V I II III a b c A one-dimensional particle of mass m is confined to move under the influence of the potential x a V V (x) = a < x b b x c elsewhere and
More informationUNIVERSITY OF SURREY FACULTY OF ENGINEERING AND PHYSICAL SCIENCES DEPARTMENT OF PHYSICS. BSc and MPhys Undergraduate Programmes in Physics LEVEL HE2
Phys/Level /1/9/Semester, 009-10 (1 handout) UNIVERSITY OF SURREY FACULTY OF ENGINEERING AND PHYSICAL SCIENCES DEPARTMENT OF PHYSICS BSc and MPhys Undergraduate Programmes in Physics LEVEL HE PAPER 1 MATHEMATICAL,
More informationExplanations of animations
Explanations of animations This directory has a number of animations in MPEG4 format showing the time evolutions of various starting wave functions for the particle-in-a-box, the free particle, and the
More informationPhysics 200 Lecture 4. Integration. Lecture 4. Physics 200 Laboratory
Physics 2 Lecture 4 Integration Lecture 4 Physics 2 Laboratory Monday, February 21st, 211 Integration is the flip-side of differentiation in fact, it is often possible to write a differential equation
More informationWaves and the Schroedinger Equation
Waves and the Schroedinger Equation 5 april 010 1 The Wave Equation We have seen from previous discussions that the wave-particle duality of matter requires we describe entities through some wave-form
More informationLecture 3 Dynamics 29
Lecture 3 Dynamics 29 30 LECTURE 3. DYNAMICS 3.1 Introduction Having described the states and the observables of a quantum system, we shall now introduce the rules that determine their time evolution.
More informationdf(x) dx = h(x) Chemistry 4531 Mathematical Preliminaries Spring 2009 I. A Primer on Differential Equations Order of differential equation
Chemistry 4531 Mathematical Preliminaries Spring 009 I. A Primer on Differential Equations Order of differential equation Linearity of differential equation Partial vs. Ordinary Differential Equations
More informationMath Methods for Polymer Physics Lecture 1: Series Representations of Functions
Math Methods for Polymer Physics ecture 1: Series Representations of Functions Series analysis is an essential tool in polymer physics and physical sciences, in general. Though other broadly speaking,
More informationd 1 µ 2 Θ = 0. (4.1) consider first the case of m = 0 where there is no azimuthal dependence on the angle φ.
4 Legendre Functions In order to investigate the solutions of Legendre s differential equation d ( µ ) dθ ] ] + l(l + ) m dµ dµ µ Θ = 0. (4.) consider first the case of m = 0 where there is no azimuthal
More informationREVIEW: The Matching Method Algorithm
Lecture 26: Numerov Algorithm for Solving the Time-Independent Schrödinger Equation 1 REVIEW: The Matching Method Algorithm Need for a more general method The shooting method for solving the time-independent
More informationSuperposition of electromagnetic waves
Superposition of electromagnetic waves February 9, So far we have looked at properties of monochromatic plane waves. A more complete picture is found by looking at superpositions of many frequencies. Many
More informationThe Particle in a Box
Page 324 Lecture 17: Relation of Particle in a Box Eigenstates to Position and Momentum Eigenstates General Considerations on Bound States and Quantization Continuity Equation for Probability Date Given:
More informationPhysics 342 Lecture 23. Radial Separation. Lecture 23. Physics 342 Quantum Mechanics I
Physics 342 Lecture 23 Radial Separation Lecture 23 Physics 342 Quantum Mechanics I Friday, March 26th, 2010 We begin our spherical solutions with the simplest possible case zero potential. Aside from
More informationHigher-order ordinary differential equations
Higher-order ordinary differential equations 1 A linear ODE of general order n has the form a n (x) dn y dx n +a n 1(x) dn 1 y dx n 1 + +a 1(x) dy dx +a 0(x)y = f(x). If f(x) = 0 then the equation is called
More informationPHYS 3313 Section 001 Lecture #20
PHYS 3313 Section 001 ecture #0 Monday, April 10, 017 Dr. Amir Farbin Infinite Square-well Potential Finite Square Well Potential Penetration Depth Degeneracy Simple Harmonic Oscillator 1 Announcements
More informationMAGIC058 & MATH64062: Partial Differential Equations 1
MAGIC58 & MATH6462: Partial Differential Equations Section 6 Fourier transforms 6. The Fourier integral formula We have seen from section 4 that, if a function f(x) satisfies the Dirichlet conditions,
More informationQuantum Physics III (8.06) Spring 2007 FINAL EXAMINATION Monday May 21, 9:00 am You have 3 hours.
Quantum Physics III (8.06) Spring 2007 FINAL EXAMINATION Monday May 21, 9:00 am You have 3 hours. There are 10 problems, totalling 180 points. Do all problems. Answer all problems in the white books provided.
More informationS.E. of H.O., con con t
Quantum Mechanics and Atomic Physics Lecture 11: The Harmonic Oscillator: Part I http://www.physics.rutgers.edu/ugrad/361 Prof. Sean Oh The Classical Harmonic Oscillator Classical mechanics examples Mass
More informationOpinions on quantum mechanics. CHAPTER 6 Quantum Mechanics II. 6.1: The Schrödinger Wave Equation. Normalization and Probability
CHAPTER 6 Quantum Mechanics II 6.1 The Schrödinger Wave Equation 6. Expectation Values 6.3 Infinite Square-Well Potential 6.4 Finite Square-Well Potential 6.5 Three-Dimensional Infinite- 6.6 Simple Harmonic
More informationAn Algebraic Approach to Reflectionless Potentials in One Dimension. Abstract
An Algebraic Approach to Reflectionless Potentials in One Dimension R.L. Jaffe Center for Theoretical Physics, 77 Massachusetts Ave., Cambridge, MA 02139-4307 (Dated: January 31, 2009) Abstract We develop
More informationDamped harmonic motion
Damped harmonic motion March 3, 016 Harmonic motion is studied in the presence of a damping force proportional to the velocity. The complex method is introduced, and the different cases of under-damping,
More informationMassachusetts Institute of Technology Physics Department
Massachusetts Institute of Technology Physics Department Physics 8.32 Fall 2006 Quantum Theory I October 9, 2006 Assignment 6 Due October 20, 2006 Announcements There will be a makeup lecture on Friday,
More informationChapter 16 Waves in One Dimension
Chapter 16 Waves in One Dimension Slide 16-1 Reading Quiz 16.05 f = c Slide 16-2 Reading Quiz 16.06 Slide 16-3 Reading Quiz 16.07 Heavier portion looks like a fixed end, pulse is inverted on reflection.
More informationQuantum Mechanics in One Dimension. Solutions of Selected Problems
Chapter 6 Quantum Mechanics in One Dimension. Solutions of Selected Problems 6.1 Problem 6.13 (In the text book) A proton is confined to moving in a one-dimensional box of width.2 nm. (a) Find the lowest
More informationPHY 396 K. Problem set #5. Due October 9, 2008.
PHY 396 K. Problem set #5. Due October 9, 2008.. First, an exercise in bosonic commutation relations [â α, â β = 0, [â α, â β = 0, [â α, â β = δ αβ. ( (a Calculate the commutators [â αâ β, â γ, [â αâ β,
More informationC/CS/Phys C191 Particle-in-a-box, Spin 10/02/08 Fall 2008 Lecture 11
C/CS/Phys C191 Particle-in-a-box, Spin 10/0/08 Fall 008 Lecture 11 Last time we saw that the time dependent Schr. eqn. can be decomposed into two equations, one in time (t) and one in space (x): space
More informationIntroduction to Quantum Mechanics
Introduction to Quantum Mechanics INEL 5209 - Solid State Devices - Spring 2012 Manuel Toledo January 23, 2012 Manuel Toledo Intro to QM 1/ 26 Outline 1 Review Time dependent Schrödinger equation Momentum
More informationMath 489AB A Very Brief Intro to Fourier Series Fall 2008
Math 489AB A Very Brief Intro to Fourier Series Fall 8 Contents Fourier Series. The coefficients........................................ Convergence......................................... 4.3 Convergence
More informationMathematical Tripos Part IB Michaelmas Term Example Sheet 1. Values of some physical constants are given on the supplementary sheet
Mathematical Tripos Part IB Michaelmas Term 2015 Quantum Mechanics Dr. J.M. Evans Example Sheet 1 Values of some physical constants are given on the supplementary sheet 1. Whenasampleofpotassiumisilluminatedwithlightofwavelength3
More informationPhysics 486 Midterm Exam #1 Spring 2018 Thursday February 22, 9:30 am 10:50 am
Physics 486 Midterm Exam #1 Spring 18 Thursday February, 9: am 1:5 am This is a closed book exam. No use of calculators or any other electronic devices is allowed. Work the problems only in your answer
More informationQuantum Mechanics-I Prof. Dr. S. Lakshmi Bala Department of Physics Indian Institute of Technology, Madras. Lecture - 21 Square-Integrable Functions
Quantum Mechanics-I Prof. Dr. S. Lakshmi Bala Department of Physics Indian Institute of Technology, Madras Lecture - 21 Square-Integrable Functions (Refer Slide Time: 00:06) (Refer Slide Time: 00:14) We
More informationFourier Series and Integrals
Fourier Series and Integrals Fourier Series et f(x) beapiece-wiselinearfunctionon[, ] (Thismeansthatf(x) maypossessa finite number of finite discontinuities on the interval). Then f(x) canbeexpandedina
More informationA Propagating Wave Packet The Group Velocity
Lecture 7 A Propagating Wave Packet The Group Velocity Phys 375 Overview and Motivation: Last time we looked at a solution to the Schrödinger equation (SE) with an initial condition (,) that corresponds
More informationReview of Quantum Mechanics, cont.
Review of Quantum Mechanics, cont. 1 Probabilities In analogy to the development of a wave intensity from a wave amplitude, the probability of a particle with a wave packet amplitude, ψ, between x and
More informationPhysics 137A Quantum Mechanics Fall 2012 Midterm II - Solutions
Physics 37A Quantum Mechanics Fall 0 Midterm II - Solutions These are the solutions to the exam given to Lecture Problem [5 points] Consider a particle with mass m charge q in a simple harmonic oscillator
More informationPhysics 218 Quantum Mechanics I Assignment 6
Physics 218 Quantum Mechanics I Assignment 6 Logan A. Morrison February 17, 2016 Problem 1 A non-relativistic beam of particles each with mass, m, and energy, E, which you can treat as a plane wave, is
More informationAnalogous comments can be made for the regions where E < V, wherein the solution to the Schrödinger equation for constant V is
8. WKB Approximation The WKB approximation, named after Wentzel, Kramers, and Brillouin, is a method for obtaining an approximate solution to a time-independent one-dimensional differential equation, in
More informationThe Sommerfeld Polynomial Method: Harmonic Oscillator Example
Chemistry 460 Fall 2017 Dr. Jean M. Standard October 2, 2017 The Sommerfeld Polynomial Method: Harmonic Oscillator Example Scaling the Harmonic Oscillator Equation Recall the basic definitions of the harmonic
More information08 Fourier Analysis. Utah State University. Charles G. Torre Department of Physics, Utah State University,
Utah State University DigitalCommons@USU Foundations of Wave Phenomena ibrary Digital Monographs 8-04 08 Fourier Analysis Charles G. Torre Department of Physics, Utah State University, Charles.Torre@usu.edu
More informationChapter 6. Q. Suppose we put a delta-function bump in the center of the infinite square well: H = αδ(x a/2) (1)
Tor Kjellsson Stockholm University Chapter 6 6. Q. Suppose we put a delta-function bump in the center of the infinite square well: where α is a constant. H = αδ(x a/ ( a Find the first-order correction
More informationCoordinate systems and vectors in three spatial dimensions
PHYS2796 Introduction to Modern Physics (Spring 2015) Notes on Mathematics Prerequisites Jim Napolitano, Department of Physics, Temple University January 7, 2015 This is a brief summary of material on
More information1 (2n)! (-1)n (θ) 2n
Complex Numbers and Algebra The real numbers are complete for the operations addition, subtraction, multiplication, and division, or more suggestively, for the operations of addition and multiplication
More informationNov : Lecture 18: The Fourier Transform and its Interpretations
3 Nov. 04 2005: Lecture 8: The Fourier Transform and its Interpretations Reading: Kreyszig Sections: 0.5 (pp:547 49), 0.8 (pp:557 63), 0.9 (pp:564 68), 0.0 (pp:569 75) Fourier Transforms Expansion of a
More informationRadiating Dipoles in Quantum Mechanics
Radiating Dipoles in Quantum Mechanics Chapter 14 P. J. Grandinetti Chem. 4300 Oct 27, 2017 P. J. Grandinetti (Chem. 4300) Radiating Dipoles in Quantum Mechanics Oct 27, 2017 1 / 26 P. J. Grandinetti (Chem.
More information1 Infinite-Dimensional Vector Spaces
Theoretical Physics Notes 4: Linear Operators In this installment of the notes, we move from linear operators in a finitedimensional vector space (which can be represented as matrices) to linear operators
More informationWave Mechanics in One Dimension
Wave Mechanics in One Dimension Wave-Particle Duality The wave-like nature of light had been experimentally demonstrated by Thomas Young in 1820, by observing interference through both thin slit diffraction
More informationSecond Quantization Method for Bosons
Second Quantization Method for Bosons Hartree-Fock-based methods cannot describe the effects of the classical image potential (cf. fig. 1) because HF is a mean-field theory. DFF-LDA is not able either
More informationSimple Harmonic Oscillator
Classical harmonic oscillator Linear force acting on a particle (Hooke s law): F =!kx From Newton s law: F = ma = m d x dt =!kx " d x dt + # x = 0, # = k / m Position and momentum solutions oscillate in
More informationContinuum Normalization. good for cooking or inspecting wiggly functions and for crudely evaluating integrals of wiggly integrands.
5.73 Lecture #5 5-1 Last time: Gaussian Wavepackets How to encode x in gk ( )e ikx dk = ψ(x) Continuum Normalization or k in g( x) e ikx = ψ(k) stationary phase: good for cooking or inspecting wiggly functions
More informationLecture 7. 1 Wavepackets and Uncertainty 1. 2 Wavepacket Shape Changes 4. 3 Time evolution of a free wave packet 6. 1 Φ(k)e ikx dk. (1.
Lecture 7 B. Zwiebach February 8, 06 Contents Wavepackets and Uncertainty Wavepacket Shape Changes 4 3 Time evolution of a free wave packet 6 Wavepackets and Uncertainty A wavepacket is a superposition
More informationSOLUTIONS for Homework #2. 1. The given wave function can be normalized to the total probability equal to 1, ψ(x) = Ne λ x.
SOLUTIONS for Homework #. The given wave function can be normalized to the total probability equal to, ψ(x) = Ne λ x. () To get we choose dx ψ(x) = N dx e λx =, () 0 N = λ. (3) This state corresponds to
More informationThe Schrödinger Wave Equation Formulation of Quantum Mechanics
Chapter 5. The Schrödinger Wave Equation Formulation of Quantum Mechanics Notes: Most of the material in this chapter is taken from Thornton and Rex, Chapter 6. 5.1 The Schrödinger Wave Equation There
More informationPhysics 4617/5617: Quantum Physics Course Lecture Notes
Physics 467/567: Quantum Physics Course Lecture Notes Dr. Donald G. Luttermoser East Tennessee State University Edition 5. Abstract These class notes are designed for use of the instructor and students
More informationENGI 9420 Lecture Notes 1 - ODEs Page 1.01
ENGI 940 Lecture Notes - ODEs Page.0. Ordinary Differential Equations An equation involving a function of one independent variable and the derivative(s) of that function is an ordinary differential equation
More informationVibrational motion. Harmonic oscillator ( 諧諧諧 ) - A particle undergoes harmonic motion. Parabolic ( 拋物線 ) (8.21) d 2 (8.23)
Vibrational motion Harmonic oscillator ( 諧諧諧 ) - A particle undergoes harmonic motion F == dv where k Parabolic V = 1 f k / dx = is Schrodinge h m d dx ψ f k f x the force constant x r + ( 拋物線 ) 1 equation
More information