6-4 the Time Development of Y Is Given by Equation 6.8 Or

Problem Set # 8 Solutions

6-4 The time development of  is given by Equation 6.8 or

{ -w( ) } Ca {ikx- iw( k) t - a 2 k 2} Y(x, t) =蝌 a( k) ei kx k t dk = e dk , ( p ) -

2 hk with w (k) = for a free particle of mass m. As in Example 6.3, the integral may 2m be reduced to a recognizable form by completing the square in the exponent. Since 骣ht 2 2 2 ih t w (k) t= 琪 k , we group this term together with a 2k 2 by introducing b= a + to 桫2m 2m get

2 2 2 2 骣 ix x ikx-w( k) t - a k = -琪 b k - - . 桫 2b 4b 2

ix Then, changing variables to z=b k - gives 2b

骣Ca-x24b 2-z2 骣 C a - x 2 4 b 2 Y(x, t) =琪 e e = 琪 e . 桫b p - 桫b To interpret this result, we must recognize that  is complex and 2 2 22 2ih t 4 骣 h t separate real and imaginary parts. Thus, b= a + = a + 琪 and the exponent 2m桫 2 m for  is

i t 22a 2 - h 2 xx ( 2m ) x = = + 2 2 2 (imaginary terms) 4b 4 b 2 4 轾a 2 + ht 臌犏 ( 2ma )

Ca -x2轾4a 2 +( t 2 m a )2 Y( ) = { 臌{ h } } x, t1 4 e t 2 . a 4 + h ( ( 2m ) )

We see that apart from a phase factor, Y (x, t) is still a gaussian but with

a 2 1 2 1 4 骣 2 骣ht amplitude diminished by t 2 and a width Dx( t) =a + 琪 where a 4 + h 琪 ( ( 2m ) ) 桫桫2ma a = D x(0) is the initial width. n2 h 2 3h2 6-9 En = , so DE = E2 - E 1 = 8mL2 8mL2 (1240 eV nm c)2 D =( ) = E 32 6.14 MeV 8( 938.28 106 eVc 2)( 10- 5 nm) hc 1240 eV nm l = = = 2.02 10-4 nm D E 6.14 106 eV This is the gamma ray region of the electromagnetic spectrum.

6-21 n = 4 It has n-1=3 nodes and is even meaning antisymmetric about the midpoint of the well.

 x

n  4 0 L  2

Note that the n = 4 wavefunction has three nodes and is antisymmetric about the midpoint of the well.

6.22 The Java applet for this problem is available from our companion website (http://info.brookscole.com/mp3e QMTools Simulations  Problem 6.22). The applet models a proton confined by a square well potential with height U = 26.0 MeV and width L = 10.0 fm (1 fm = 10−15 m = 10–6 nm). These values appear on the Formulas tab of the 2 applet, along with the proton mass value mp = 938.38 MeV/c . On the Coordinate 2D Graphics tab the potential function V( x ) is plotted for 1024 points over the interval [−10 fm, +20 fm]. The listing to the right of the graph includes placeholders for two stationary states of the proton in this well. Follow the applet instructions to display each state in turn and adjust its energy until no discernible mismatch results in the wavefunction. The final form of the applet should resemble the screenshot below: For this potential well, we find the ground and first excited state energies at

E1 = 1.4697 MeV and E2 = 5.8220 MeV (waveforms with zero and one node, respectively).

In a transition from E2 to E1 a photon is emitted with energy

DE = E2 - E 1 =5.8220 MeV - 1.4697 MeV = 4.352 MeV

Using hc=2p h c = 1239.7 MeV fm , we calculate the corresponding wavelength as

l =hc/ D E = 1239.7 MeV� fm / 4.352 MeV) = 284.9 fm 2.849 10-2 nm

2 2 2 For the infinite well model, we use En= ( n h ) /(8 m p L ) to get

2 2 E1=(1239.7) /(8)(938.38)(10) = 2.0472 MeV, and E 2 = 4 E 1 = 8.1888 MeV

Then DE =6.1408 MeV, and l = 2.019 10-2 nm

6.28 A particle within the well is subject to no forces and, hence, moves with uniform speed, spending equal time in all parts of the well. Thus, for such a particle the probability density is uniform. That is, Pc ( x) = constant . The constant is fixed by requiring L 1 the integrated probability to be unity, that is, 1 =Pc ( x) dx = CL or C = . To find x we 0 L weight the possible particle positions according to the probability density Pc to get L 2 L 1 骣x L 2 x= xPc ( x) dx =琪 = . Similarly, x is found by weighting the possible values 0 L 桫20 2 2 of x with Pc :

L 3L 2 2 2 1 骣x L x= x Pc ( x) dx =琪 = . 0 L 桫30 3

The classical and quantum results for x agree exactly; for x2 the quantum L2 prediction is smaller by an amount which, however, goes to zero in the limit of 2(np )2 large quantum numbers n, where classical and quantum results must coincide (correspondence principle).

1 4 2 a 6-32 The probability density for this case is y (x) 2 = C2 e-ax with C = and 0 0 0 (p )

mw 2 a = . For the calculation of the average position x= xy 0 ( x) dx we note that the h - integrand is an odd function, so that the integral over the negative half-axis x < 0 exactly cancels that over the positive half-axis ( x > 0 ), leaving x = 0 . For the calculation of x2 2 2 , however, the integrand x y 0 is symmetric, and the two half-axes contribute equally, giving

1 2 2 1 p 2= 2 2-ax = 2 x2 C0 x e dx 2 C 0 ( )( ) . 0 4a a

1 2 2 1 h 2 2 1 2 h Substituting for C0 and a gives x = = and Dx =( x - x ) = . 2a 2 mw ( 2mw )