CHAPTER 36. 1* True or false: Boundary conditions on the wave function lead to energy quantization. True

Transcription

1 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 potential. (a) The we function is shown below (b) The probability density is shown below 3 Sketch (a) the we function and (b) the probability distribution for the n 5 state for the finite squarewell potential. (a) The we function is shown below (b) The probability density is shown below

2 4 Show that the expectation value <x> x Ψ dx is zero for both the ground and the first excited states of the harmonic oscillator. The integral xψ dx because the integrand is an odd function of x for the ground state as well as any excited state of the harmonic oscillator. 5* Use the procedure of Example 36- to verify that the energy of the first excited state of the harmonic oscillator is E 3 h _ω. (Note: Rather than solve for a again, use the result a mω/h_ obtained in Example 36-.) dψ The we function is ψ A x e (see Equ. 36-5). Then A e A e and dx d ψ 3 A e A e + 3 a x A e a x ) 4 4 (4 6 A e. We now substitute this into the dx Schrödinger equation. The exponentials and the constant A cancel, so (h _ /m)(4a x 3 6) + /mω x 3 E x. With a /mω/h _, the terms in x 3 cancel, and solving for the energy E we find E 6h _ a/m 3h _ ω/ 3E. 6 Show that the normalization constant A of Equation 36-3 is A (mω/h) /4. Let A (mω/h) /4 (mω/πh _ ) /4. Then ψ ( ) dx (mω/πh _ ) / x / e s exp[( mω/h _ )x ] dx π ds. 7 Find the normalization constant A for the we function of the first excited state of the harmonic oscillator, Equation We require that (a) 3/ y e y A x e dx where a mω/h _. Let y ; then the integral becomes dy π 3 / (a). Consequently, A 3 a π 3 and m ω π A 3 h /4. 8 Find the expectation value <x > x Ψ dx for the ground state of the harmonic oscillator. Use it to show that the erage potential energy equals half the total energy. x ψ dx A x e dx. The integral has already been evaluated in Problem Using that result and

3 A given in Problem 36-6, x h _ /mω. The erage potential energy of the oscillator is /mω x h _ ω/4 E /. 9* Verify that Ψ(x) A xe is the we function corresponding to the first excited state of a harmonic oscillator by substituting it into the time-independent Schrödinger equation and solving for a and E. From Problem 5 we know that the Schrödinger equation for Ψ gives (h _ /m)(4a x 3 6) + /mω x 3 E x. If we now set the coefficients of x 3 and solve for a we find that a /mω/h _, and using this expression and solving for the energy E we find E 6h _ a/m 3h _ ω/ 3E. Find the expectation value <x > x ψ dx for the first excited state of the harmonic oscillator. x 4 ψ x dx A x e ( ) dx. As in Problem 36-7, let y. Then <x 3 π > A a 3a From Problem 36-7, A so x 3a/8 3πmw /h. π Classically, the erage kinetic energy of the harmonic oscillator equals the erage potential energy. We may assume that this is also true for the quantum mechanical harmonic oscillator. Use this condition to determine the expectation value of p for the ground state of the harmonic oscillator. According to the problem statement, p /m k x /, from which p m ω x. From Problem 36-7, x h _ /mω. Thus, p h _ ωm/. We know that for the classical harmonic oscillator, p. It can be shown that for the quantum mechanical harmonic oscillator, <p>. Use the results of Problems 4, 6, and to determine the uncertainty product x p for the ground state of the harmonic oscillator. We associate ( p) and ( x) with the standard deviations as in the statement of Problem Then, ( p) [(p pp p )] and likewise for ( x). Since p x, ( p) p and ( x) x. Now from Problems 36-8 and 36-, ( p) ( x) h _ /4 in the ground state of the harmonic oscillator, so p x h _ /. 3* A free particle of mass m with we number k is treling to the right. At x, the potential jumps from zero to U and remains at this value for positive x. (a) If the total energy is E h _ k /m U, what is the we number k in the region x >? Express your answer in terms of k and in terms of U. (b) Calculate the reflection coefficient R at the potential step. (c) What is the transmission coefficient T? (d) If one million particles with we

4 number k are incident upon the potential step, how many particles are expected to continue along in the positive x direction? How does this compare with the classical prediction? (a) We are given that E h _ k /m U. For x >, h _ k /m + U U. So k k 4mU /h _. k k /. ( k k ) (b) The reflection coefficient is given by Equ So R.94. ( k + k ) mu /h _, whereas (c) T R.97. (d) The number of particles that continue beyond the step is N T ; classically, 6 would continue to move past the step. 4 Suppose that the potential jumps from zero to U at x so that the free particle speeds up instead of slowing down. The we number for the incident particle is again k, and the total energy is U. (a) What is the we number for the particle in the region of positive x? (b) Calculate the reflection coefficient R at x. (c) What is the transmission coefficient T? (d) If one million particles with we number k are incident upon the potential step, how many particles are expected to continue along in the positive x direction? How does this compare with the classical prediction? (a) Proceed as in the preceding problem. Now E h _ k /m U U h _ k /m. Consequently, k (6mU /h _ ) / or k k 3/. (b) Use Equ (c) Use Equ (d) Number of particles transmitted NT R. T.99 NT ; classically 6 particles are transmitted 5 Work Problem 3 for the case in which the energy of the incident particle is.u instead of U. (a) See Problem 36-3; write k and k (b) Use Equ (c) T R (d) Number of particles transmitted NT k (.mu /h _ ) / ; k (.mu /h _ ) / ; k.995k (.. ) (. +. ) R.67 T.39 NT ; classically 6 particles are transmitted 6 A particle of energy E approaches a step barrier of height U. What should be the ratio E/U so that the reflection coefficient is?. Find r k /k using Equ Use the result of Problem k r.76 k + U/E r ; E/U ( r ).3 7* Use Equation 36-9 to calculate the order of magnitude of the probability that a proton will tunnel out of a nucleus in one collision with the nuclear barrier if it has energy 6 MeV below the top of the potential barrier and the barrier thickness is -5 m.

5 . Rewrite α of Equ in units of MeV. T e αa ; m p c 938 MeV α mc (U E) /h _ c; h _ c MeV.m T exp[ 5 ( 938 6) / / ].34 8 A -ev electron is incident on a potential barrier of height 5 ev and width of nm. (a) Use Equation 36-9 to calculate the order of magnitude of the probability that the electron will tunnel through the barrier. (b) Repeat your calculation for a width of. nm. (a). Evaluate α. Estimate T using Equ (b) Repeat with a m α m T 5 8 T 9 A particle is confined to a three-dimensional box that has sides L, L L, and L 3 3L. Give the quantum numbers n, n, n 3 that correspond to the lowest ten quantum states of this box. Use Equ to write E (h /8mL )(n + n /4 + n 3 /9); E (h /88mL )(36n + 9n + n 3 ). The energies in units of h /88mL are listed in the following table. n n n 3 E Give the we functions for the lowest ten quantum states of the particle in Problem 9. The we functions are of the form ψ A sin(n πx/l ) sin (n πy/l ) sin(n 3πz/3L ). * (a) Repeat Problem 9 for the case L L and L 3 4L. (b) What quantum numbers correspond to degenerate energy levels? (a) From Equ. 36-3, E (h /8mL )(n + n /4 + n 3 /6) (h /8mL )(6n + 4n + n 3 ). The table below lists the ten lowest energy levels in units of h /8mL. n n n 3 E

6 (b) There are two degenerate levels, namely, (,,4) and (,,) and (,,6) and (,3,). Give the we functions for the lowest ten quantum states of the particle in Problem. The we functions are of the form ψ A sin(n πx/l ) sin (n πy/l ) sin(n 3πz/4L ). 3 A particle moves in a potential well given by U(x, y, z) for L/ < x < L/, < y < L, and < z < L, and U outside these ranges. (a) Write an expression for the ground-state we function for this particle. (b) How do the allowed energies compare with those for a box hing U for < x < L, rather than for L/ < x < L/? (a) The boundary conditions in the y and z directions are as in Figure 36-. In the x direction, we require that ψ at L/ and at L/. Using the solution given by Equ. 36-9, we see that the boundary conditions are satisfied for A, B if n is an even integer, and are satisfied for A, B if n is an odd integer. The normalization constants A and B are both equal to 3 8 / L. Thus, ψ (x, y, z) B cos(n πx/l) sin(n πy/l) sin(n 3πz/L), n n +; ψ (x, y, z) A sin(n πx/l) sin(n πy/l) sin(n 3πz/L), n n The ground-state we function is ψ A cos(πx/l) sin(πy/l) sin(πz/l). (b) The allowed energies are the same as those for the box with U for < x < L. 4 A particle moves freely in the two-dimensional region defined by x L and y L. (a) Find the we function satisfying Schrödinger s equation. (b) Find the corresponding energies. (c) Find the lowest two states that are degenerate. Give the quantum numbers for this case. (d) Find the lowest three states that he the same energy. Give the quantum numbers for the three states hing the same energy. (a) In Equ. 36-3, set k 3. Thus, ψ(x, y) A sin(nπx/l) sin(mπy/l), where n and m are integers. (b) E n, m (h /8mL )(n + m ) (c) E, E,. These are the two lowest degenerate states. (d) The lowest triply degenerate states are for n, m 7; n 7, m ; n m 5. The energy of that state is E 5h /4mL. 5* What is the next energy level above those found in Problem 4c for a particle in a two-dimensional square box for which the degeneracy is greater than? We need to find the least integral values for n and m such that n + m are the same for more than two choices of n and m. For any pair of values, e.g., n, m, and n, m we he double degeneracy. Therefore, we must find two different sets for which the sum of the squares are the same. For n, m 7 and for n 5, m 5, the sum of the squares equals 5. Consequently, the states n, m 7; n 7, m ; and n 5, m 5 are

7 degenerate (triple degeneracy); the energy of this triply degenerate state is 5h /8mL. The next higher degeneracy is for n 4, m 7; n 7, m 4; n, m 8; and n 8, m. These states are four-fold degenerate. 6 Show that Equation satisfies Equation with U, and find the energy of this state. Substitute Equ into Equ One obtains (h _ /m)[(π /L ) + (4π /L )]ψ, Eψ,. That equation is satisfied if E 5h _ π /ml 5h /8mL. 7 What is the ground-state energy of ten noninteracting bosons in a one-dimensional box of length L? The ten bosons can occupy the same ground state. Consequently, E (h /8mL ) 5h /4mL. 8 What is the ground-state energy of ten noninteracting fermions, such as neutrons, in a one-dimensional box of length L? (Because the quantum number associated with spin can he two values, each spatial state can hold two neutrons.) For fermions, such as neutrons for which the spin quantum number is /, two particles can occupy the same spatial state. Consequently, the lowest total energy for the fermions is E E ( ) 55h /4mL. The integral of two functions over some space interval is somewhat analogous to the dot product of two vectors. If this integral is zero, the functions are said to be orthogonal, which is analogous to two vectors being perpendicular. The following problems illustrate the general principle that any two we functions corresponding to different energy levels in the same potential are orthogonal. 9* Show that the ground-state we function and that of the first excited state of the harmonic oscillator are orthogonal; i.e., show that ψ(x)ψ (x) dx. We need to show that ψ ( x) ψ ( x) dx, where ψ(x) and ψ(x) are given by Equs and 36-5, respectively. Note that ψ(x) is an even function of x and ψ(x) is an odd function of x. It follows that the integral from to must vanish. 3 The we function for the state n of the harmonic oscillator is ψ(x) A ( ) e, where A is the normalization constant for this we function. Show that the we functions for the states n and n of the harmonic oscillator are orthogonal. We note that ψ(x) is an even function of x, whereas ψ(x), given by Equ. 36-5, is an odd function of x. Therefore, ψ ( x) ψ ( x) dx. Q.E.D. 3 For the we functions ψ (x) /L sin( nπx/l) corresponding to a particle in an infinite square well n

8 of potential from to L, show that ψ (x) (x) n ψ m dx, that is, ψn and ψm are orthogonal. L nπx mπx We need to show that sin sin dx. The product of the two sine functions can be rewritten as L L the sum of two cosines, i.e., sin(nπx/l) sin(mπx/l) /{cos[(n m)π x/l] cos[(n + m)π x/l]}. The integral L sin[(n m) π x/l] the first term is ; a similar expression holds for the second term with (n m) replaced by π (n m) (n + m). Since n and m are integers and n m, the sine functions vanish at the two limits x and x L. Therefore, ψn(x)ψ m(x)dx for n m. 3 Consider a particle in a one-dimensional box of length L that is centered at the origin. (a) What are the values of ψ() and ψ()? (b) What are the values of <x> for the states n and n? (c) Evaluate <x > for the states n and n. (See Problem 59 in Chapter 7.) (a) The we functions are ψm (/L) / sin (mπx/l), m n; ψm (/L) / cos (mπx/l), m n +. At x, ψ(), ψ() (/L) /. L / (b) Note that ψm(x) is an even function of x in all cases. Consequently, x xψ m ( x) dx (c) We use the following integrals: x sin () dx x 3 /6 (x 3 /4a /8a 3 )sin () [x cos ()]/4a and x cos () dx x 3 /6 + (x 3 /4a /8a 3 )sin () + [x cos ()]/4a. For m, a π/l, and the first integral between the limits L/ and L/ times the normalization factor (/L) gives x (L /)[ (6/π )]. For m, a π/l and the second integral between the limits L/ and L/ times the normalization factor (/L) gives x (L /)[ (3/π )]. Note that for any value of m, x (L /)[ (6/m π )]. L / 33* Eight identical noninteracting fermions (such as neutrons) are confined to a two-dimensional square box of side length L. Determine the energies of the three lowest states. (See Problem 6.) Each n, m state can accommodate only particles. Therefore, in the ground state of the system of 8 fermions, the four lowest quantum states are occupied. These are (,), (,), (,) and (,). [Note that the states (,) and (,) are distinctly different states since the x and y directions are distinguishable.] The energy of the ground state is E (h /8mL )( ) 5h /ml. The next higher state is achieved by taking one fermion from the (,) state and raising it into the next higherunoccupied state. That state is the (,3) state. The energy difference between the ground state and this state is (h /8mL )( 8) h /4mL. The (3,) is another excited state that is accessible, and it is degenerate with the (,3) state. The three lowest energy levels are therefore E 5h /ml, and two states of energy E E h /4mL. 34 A particle is confined to a two-dimensional box defined by the following boundary conditions: U(x,y) for L/ x L/ and 3L/ y 3L/; and U(x,y) elsewhere. (a) Determine the energies of the lowest three bound states. Are any of these states degenerate? (b) Identify the lowest doubly degenerate bound state by appropriate quantum numbers and determine its energy. (a) The energy levels are the same as for a two-dimensional box of widths L and 3L, i.e., E n, m (h /8mL )(n + m /9) (h /7mL )(9n + m ). The three lowest states are E, 5h /36mL,

9 E, 3h /7mL, and E, 3 h /4mL. None of these states are degenerate. (b) For two states to be degenerate we must he 9(n n ) (m m ). That condition is first satisfied for n, m 3 and n, m 6. The energy of that doubly degenerate state is E, 3 E, 6 5h /8mL. 35 A particle moves in a potential given by U(x) A x. Without attempting to solve the Schrödinger equation, sketch the we function for (a) the ground-state energy of a particle inside this potential and (b) the first excited state for this potential. The we functions for the ground state and first excited state are sketched below. 36 The classical probability distribution function for a particle in a one-dimensional box of length L is P /L. (See Example 7-5). (a) Show that the classical expectation value of x for a particle in a one-dimensional box of length L centered at the origin (Problem 3) is L /. (b) Find the quantum expectation value of x for the nth state of a particle in the one-dimensional box of Problem 3 and show that it approaches the classical limit L / for n >>. L / L (a) The classical expectation value is given by x dx L. L / (b) As shown in Problem 36-3, for any value of the quantum number n, x (L /)[ + (6/n π )]. In the limit n >>, x L /. 37* Show that Equations 36-7 and 36-8 imply that the transmission coefficient for particles of energy E incident on a step barrier U < E is given by 4 kk 4r T ( k + k ) ( + r ) where r k /k. ( k k ) R ( k + k ) and T R ( k + k ( k ) ( k + k ) k ) 4 kk ( k + k ) 4r ( + r ) where r k /k. 38 (a) Show that for the case of a particle of energy E incident on a step barrier U < E, the we numbers k and k are related by

10 k U r k E Use this and the results of Problem 37 to calculate the transmission coefficient T and the reflection coefficient R for the case (b) E.U, (c) E.U, and (d) E.U. (a) In the region U, E h _ k /m. In the region U U, h _ k /m (E U ). It follows that r k k E U E U E (b), (c), (d) Find r and use the result of Problem to determine T and R (b) T.83, R.77; (c) T.976, R.94; (d) T.9993, R Determine the normalization constant A in Problem 3. We require that A ( / ) e dx A ( / ) e dx. Expand the integrand to the 4 three terms 4, a x e e, and e / 4, and integrate term by term using the definite integrals x e dx A b π and n bx 3 5 (n ) π,. b x e dx n n+ n b b a /4 π and A a 8mω π h. Evaluating the three integrals one obtains 4 Consider the time-independent one-dimensional Schrödinger equation when the potential function is symmetric about the origin, i.e., when U(x) U( x). (a) Show that if ψ(x) is a solution of the Schrödinger equation with energy E, then ψ( x) is also a solution with the same energy E, and that, therefore, ψ(x) and ψ( x) can differ by only a multiplicative constant. (b) Write ψ(x) Cψ ( x), and show that C ±. Note that C + means that ψ(x) is an even function of x, and C means that ψ(x) is an odd function of x. (a) Do a spacial inversion, i.e., let x x. The second derivative is an even operator, that is to say, d ψ ( x)/d( x) d ψ (x)/dx. Therefore, if U( x) U(x), the Schrödinger equation for ψ( x) is exactly the same as for ψ(x) and must give the same values for the energy E. Then if ψ( x) does differ from ψ(x), the ratio 4* ψ( x)/ψ (x) cannot be a function of x but must be a constant. So ψ(x) Cψ ( x). (b) The previous result means that replacing the argument of the we function by its negative is equivalent to multiplication by C. Thus, if Cψ ( x) is a good we function and we now replace its argument by its negative, that is, by x, we must multiply by C again. Thus ψ(x) C ψ (x), C, and C ±. In this problem you will derive the ground-state energy of the harmonic oscillator using the precise form of the uncertainty principle, x p h _ /, where x and p are defined to be the standard deviations ( x) [(x x ) ] and ( p) [(p p ) ] (see Equation 8-3). Proceed as follows:. Write the total classical energy in terms of the position x and momentum p using U(x) mω x and K p /m.. Use the result of Equation 8 35 to write ( x ) [( x x ) ] (x ) x and ( p ) [( p p ) ] (p ) p.

11 3. Use the symmetry of the potential energy function to argue that x and p must be zero, so that ( x) (x) and ( p) (p ). 4. Assume that p h _ / x to eliminate (p ) from the erage energy E (p ) /m + mω (x ) and write E as E h _ /8mZ + mω Z, where Z (x ). 5. Set de/dz to find the value of Z for which E is a minimum. 6. Show that the minimum energy is given by (E ) min + h _ ω.. E U + K /mω (x ) + (/m)(p )., 3. ( p) [(p p ) ] [p pp + p ] (p ) since p ; likewise, ( x) (x ). 4. E /mω (x ) + (h _ /8m)/(x ) /mω Z + h _ /8mZ. 5. de /dz /mω h _ /8mZ ; Z h _ /mω. 6. (E ) min /mω h _ /mω + mωh _ /8mω /hω. 4 A particle of mass m near the earth s surface at z can be described by the potential energy U mgz, z > U, z < For some positive value of total energy E, indicate the classically allowed region on a sketch of U(z) versus z. Sketch also the classical kinetic energy versus z. The Schrödinger equation for this problem is quite difficult to solve. Using arguments similar to those in Section 36- about the curvature of the we function as given by the Schrödinger equation, sketch your "guesses" for the shape of the we function for the ground state and the first two excited states. The classically allowed region is for E U(z). In the figure below, this region extends from z to z z m. The kinetic energy is E U(z). In this case, K(z) is a straight line extending from E at z to at z z m. A sketch of the we functions for the lowest three energy states is shown in the third figure.