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Practice Problem Set 1 Solutions

  1. We are assuming that tex2html_wrap_inline92 and that tex2html_wrap_inline94 . We want to determine the condition under which tex2html_wrap_inline96 :

    eqnarray8

    If the linear combination tex2html_wrap_inline98 is normalized, we must have tex2html_wrap_inline100 .

  2. We want tex2html_wrap_inline102 . According to the uncertainty principle,

    displaymath104

    If we plug in the numbers, we get tex2html_wrap_inline106 . Of course, p = mv so tex2html_wrap_inline110 . For an electron, this gives tex2html_wrap_inline112 . Needless to say, this is a very large error in the speed. For comparison, a free electron at room temperature would have a speed of approximately tex2html_wrap_inline114 so the error imposed by a measurement of position of the given accuracy is greater than typical thermal velocities.

  3. First, write down the Hamiltonian:

    displaymath116

    Schrödinger's equation is therefore

    displaymath118

    If the problem is separable, we should be able to write the wavefunction in the form tex2html_wrap_inline120 . Making this substitution in Schrödinger's equation, we get

    displaymath122

    Divide both sides of the equation by tex2html_wrap_inline124 and rearrange:

    displaymath126

    The first parenthesis involves only terms which depend on x, the second only terms which depend on y. Since they add to a constant (E), it follows that each parenthesis must itself be a constant. We get two separate one-dimensional Schrödinger equations:

    eqnarray52

    with tex2html_wrap_inline134 .

    1. The particle-in-a-box energy equation is

      displaymath136

      The difference in energy between two adjacent energy levels is

      displaymath138

    2. Frequency is proportional to energy so the given frequency is proportional to tex2html_wrap_inline140 (using the above formula with n=1). The next line would have an energy proportional to tex2html_wrap_inline144 which is tex2html_wrap_inline146 bigger than the energy of the first line, so the next line in the spectrum should be found at tex2html_wrap_inline148 .
    3. The energy of a photon is related to its frequency by tex2html_wrap_inline150 so in our case,

      eqnarray76

  4. The expectation value of an observable G is computed by tex2html_wrap_inline154 . All operators corresponding to observable are Hermitian. Applying this property to the expectation value, we get

    displaymath156

    The two inner products are the same, so we are saying that the expectation value must be equal to its complex conjugate. The only way that can be true is if the imaginary part is zero, so the expectation value is real.


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Up: Back to the Chemistry 3730 assignment index

Marc Roussel
Wed Sep 17 15:57:42 MDT 1997