<p>CHEM 342. Spring 2002. Problem Set #2. Answers. </p><p>Orthogonality  x   2x  1. Prove that the functions 1  Asin  and  2  Asin  are orthogonal  a   a  0  x  a. sinn  mx sinn  mx Hint: sin nx sin mxdx    C  2n  m 2n  m</p><p> a a a *  x   2x  2  x   2x  1 2 dx  Asin Asin dx  A sin sin dx  0 0  a   a  0  a   a  a     2     2         3   sin x sin x sin a sin a 2  a   a  2  a   a   A     A         2     2      2   6    2  2           a   a   0   a   a     sin  sin3  A2     0    2   6           a   a  </p><p>2. Give a mathematical definition for the Kronnecker delta  nm . What is the numerical value of the Kronnecker delta when the two eigenfunctions are orthogonal? What is the numerical value of the Kronnecker delta when n and m are the same eigenfunction (i.e. n = m)? In addition to these two values, can the Kronnecker delta be equal to any other numerical values? *  nm    n  m d</p><p>The eigenfunctions  n and  m are orthogonal when  nm  0 . The other possible value of the Kronnecker delta is 1, and it occurs when n = m. The Kronnecker delta can only equal zero or one; no other values are possible.</p><p>Operators   d    d  3. Find the result of operating with A  y    and B  y    on the function  dy   dy  2 f y  e  y / 2 . Is f(y) an eigenfunction of Aˆ or of Bˆ ?  2 d 2 A f y  ye  y / 2  e  y / 2  dy</p><p> 2 2 A f y  ye  y / 2  e  y / 2  y</p><p> 2 A f y  2ye  y / 2</p><p>1 CHEM 342. Spring 2002. Problem Set #2. Answers. </p><p>This function is not an eigenfunction of Aˆ .</p><p> 2 d 2 B f y  ye  y / 2  e  y / 2  dy</p><p> 2 2 B f y  ye  y / 2  e  y / 2  y</p><p> 2 2 B f y  ye  y / 2  ye  y / 2</p><p> 2 2 B f y  y  ye  y / 2  0e  y / 2  This function is an eigenfunction of Bˆ with eigenvalue of zero.</p><p>4. Find the following commutators for any function f(x). </p><p>(a)   2  dˆ, x   </p><p>(b)   2  d , xˆ  </p><p> 2 ˆ d 2 d  Hint: d  , d  , xˆ  x , and 2 2 . dx dx 2 x  x</p><p>(a)     2  2 2 dˆ, x  f x  dˆ x f x  x dˆf x     2  d 2  df x 2 dˆ, x  f x  x f x  x   dx  dx   2  2  dx  2 df x  df x 2 dˆ, x  f x  f x  x    x   dx dx  dx    2  dˆ, x  f x  2xf x  </p><p>  2  Therefore dˆ, x   2x  </p><p>(b)</p><p>2 CHEM 342. Spring 2002. Problem Set #2. Answers. </p><p>    2  2 2 d , xˆ f x  d xˆf x xˆ d f x      d 2 d 2 f x d 2 , xˆ f x  xf x  x     2    2   dx dx    d  dx df x d 2 f x d 2 , xˆ f x  f x  x  x         2   dx  dx dx  dx    d  df x d 2 f x d 2 , xˆ f x  f x  x  x         2   dx  dx  dx    df x dx df x d 2 f x d 2 f x d 2 , xˆ f x    x  x     2 2   dx dx dx dx dx   2  df x df x d , xˆ f x     dx dx   2   df x d , xˆ f x  2     dx </p><p>  2   d  Therefore d , xˆ  2     dx </p><p> 1  d   d  2 5. Find the result of operating with the operator O   r 2    on the function r 2  dr   dr  r br     Ae . What values must the constants have for to be an eigenfunction of O ?  1  d   d  2 O Ae br  r 2 Ae br  Ae br   2        r  dr   dr  r  br 1  d  2 br  2 br OAe   r  bAe   Ae  r 2  dr   r  1 2 OAe br   bAr 2  be br  2re br  Ae br  r 2 r  br br  br 2e  2 br OAe  bA be    Ae   r  r  2Abe br 2 OAe br  Ab 2e br   Ae br  r r  br br  2 2b 2 OAe  Ae b     r r </p><p>3 CHEM 342. Spring 2002. Problem Set #2. Answers. </p><p> For  to be an eigenfunction of O , the constant b must equal 1, while A can be any real number. In this case, the eigenvalue is 1. </p><p>6. Find the result of operating with the operator  2 on the function x 2  y 2  z 2 . Is it an eigenfunction?  2  2  2  2    x 2 y 2 z 2 2 2 2 2 2 2 2  2 2 2  2 2 2  2 2 2  x  y  z  x  y  z y,z  x  y  z x,z  x  y  z x, y x 2 y 2 z 2     2 x 2  y 2  z 2  2x 2y 2z x y z  2 x 2  y 2  z 2  2  2  2  2 x 2  y 2  z 2  6 This is not an eigenfunction of  2 .</p><p>7. The function   Ax1 x is a well-behaved wave function in the interval 0  x 1 . Calculate the normalization constant (A), and the average value of a series of measurements of x (i.e find the expectation value:  x  ). 1 *dx  1 0 1 A2 x 2 1 x2 dx  1 0 1 A2 x 2 1 2x  x 2 dx  1 0 1 A2 x 2  2x 3  x 4 dx  1 0 13 214 15  A2      1  3 4 5   1  A2    1  30  A  30</p><p>4 CHEM 342. Spring 2002. Problem Set #2. Answers. </p><p>1  xˆ  * xˆdx 0 1  xˆ  x*dx 0 1  xˆ  x30x 2 1 x2 dx 0 1  xˆ  30 x 3  2x 4  x 5 dx 0 14 215 16   xˆ  30     4 5 6  1  xˆ  2</p><p>Expectation Values 8. For the wave function  and the operatorˆ , give an expression that could be used to calculate the average value obtained from repeated measurements (i.e. show an expression for  ˆ  ). * ˆ d  ˆ    *d or  ˆ   * ˆ d , where  is normalized.</p><p>Particle In a Box nx 9. Calculate the value of A so that   Asin is normalized in the region 0  x  a . n a x  1  Hint: sin 2 bxdx    sin2bx C  2  4b </p><p>5 CHEM 342. Spring 2002. Problem Set #2. Answers. </p><p> a *  n  n dx 1 0 2 a  nx  Asin  dx 1 0  a  a nx A2 sin 2 dx  1 0 a a 2  x a  2nx  A   sin   1 2 4n  a  0</p><p>2 a a  A   sin2n sin 0 1 2 4n   a  A2   1  2  2 Therefore, A  a Note: sin2n  0 for n = integer. </p><p>10. For a particle in a one-dimensional box 0  x  a, we used eigenfunctions of the form   Asinkx . Explain why we could not use (a)   Ae kx (b)   Acoskx</p><p>(a) The wave function must be zero for x = a. If k is a real number, then e kx cannot fit this boundary condition. (b) The boundary conditions also require that   0 when x = 0. This cannot be true for the cosine function because cos 0 = 1. Therefore, the allowed eigenfunction must be of the form   Asinkx . </p><p>11. The ground-state wave function for a particle confined to a one-dimensional box of 1/ 2  2   x  length L is     sin  . The box is 10.0 nm long. Calculate the probability that the  L   L  x  1  particle is between 4.95 nm and 5.05 nm. Hint: sin 2 bxdx    sin2bx C  2  4b </p><p>6 CHEM 342. Spring 2002. Problem Set #2. Answers. </p><p>5.05 P  *dx 4.95 5.05  2  2  x  P   sin  dx 4.95  L   L  5.05  2   2  x  P    sin  dx  L 4.95   L  5.05  2  x L  2x  P     sin   L 2 4  L  4.95 5.05  x 1  2x  P    sin   L 2  L  4.95 5.05  4.95 1  25.05 24.95 P   sin  sin  10.0 2  10.0 10.0  P  0.020</p><p>12. What is the ground state energy (i.e. n = 1) for an electron that is confined to a box 34 which is 0.2 nm wide. [Hint: Planck's constant, h,is 6.62610 J s; the mass of an electron, me, is 9.1091031 kg] h 2 n 2 E  8ma 2 2 6.626 1034 J s 12 E    2 89.109 1031 kg0.2 10 9 m E 1.506 1018 J</p><p>Uncertainty 13. The speed of a certain proton is 4.5  105 m/s along the x-axis. If the uncertainty in its momentum along the x-axis is 0.010 %, what is the maximum uncertainty in its location along the x-axis (i.e. x )? </p><p>p x  0.010%p0</p><p>p x  0.00010mv</p><p>7 CHEM 342. Spring 2002. Problem Set #2. Answers. </p><p> h xp  x 4 h x  4p x h x  40.00010mv 6.6261034 J s x    40.000101.67310 27 kg4.5105 m s 1  x  7.0110 10 m</p><p>Tunneling 14. The wave function inside an infinitely long barrier of height V is   Ae kx . Calculate (a) the probability that the particle is inside the barrier; and (b) the average penetration depth of the particle into the barrier (i.e. the expectation value  x  ). Because the barrier is n! infinitely long, this wave function is valid for 0  x   . Hint: x ne ax dx   a n1</p><p>   2kx 2 * 2 2kx 2 e  2  1  A P   dx  A e dx  A    A 0    0 0  2k  2k 2k   0     A2  x  x*dx  xA2e 2kx dx  2 0 0 4k</p><p>8</p>