CH6&amp;7

CH6&amp;7 - 6-20 The Schrödinger equation after...

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Unformatted text preview: 6-20 The Schrödinger equation, after rearrangement, is d2"dx2=2mh2#\$%&’(U x( ))E{ }"x( ). In the well interior, U x( ) =and solutions to this equation are sinkxand coskx, where k2=2mEh2. The waves symmetric about the midpoint of the well (x=) are described by "x( ) =Acoskx"L<x< +LIn the region outside the well, U x( ) =U, and the independent solutions to the wave equation are e±"xwith "2=2mh2#\$%&’(U)E( ). (a) The growing exponentials must be discarded to keep the wave from diverging at infinity. Thus, the waves in the exterior region, which are symmetric about the midpoint of the well are given by "x( ) =Ce#\$xx>Lor x<"L. At x=Lcontinuity of "requires AcoskL=Ce"#L. For the slope to be continuous here, we also must require "AksinkL="Ce"#L. Dividing the two equations gives the desired restriction on the allowed energies: ktankL=". (b) The dependence on E(or k) is made more explicit by noting that k2+"2=2mUh2, which allows the energy condition to be written ktankL=2mUh2"k2#\$%&’(1 2. Multiplying by L, squaring the result, and using tan2"+1=sec2"gives kL( )2sec2kL( ) =2mUL2h2from which the desired form follows immediately, kseckL( ) =2mUh. The ground state is the symmetric waveform having the lowest energy. For electrons in a well of height U=5 eVand width 2L=0.2 nm, we calculate 2mUL2h2=2( )511"103eVc2( )5 eV( )0.1 nm( )2197.3 eV#nmc( )2=1.3127. With this value, the equation for "=kL"cos"=1.312 7( )1 2=1.1457can be solved numerically employing methods of varying sophistication. The simplest of these is trial and error, which gives "=0.799From this, we find k=7.99 nm"1, and an energy E=h2k22m=197.3 eV"nmc( )27.99 nm#1( )22 511\$103eVc2( )=2.432 eV. 6-24 After rearrangement, the Schrödinger equation is d2"dx2=2mh2#\$%&’(U x( ))E{ }"x( )with U x( ) =12m"2x2for the quantum oscillator. Differentiating "x( ) =Cxe#\$x2gives d"dx=#2\$x"x( )+C#\$x2and d2"dx2=#2\$xd"dx#2\$"x( )#2\$x( )Ce#\$x2=2\$x( )2"x( )#6\$"x( ). Therefore, for "x( )to be a solution requires 2"x( )2#6"=2mh2U x( )#E{ } =m\$h%&’()*2x2#2mEh2. Equating coefficients of like terms gives 2"=m#hand 6"=2mEh2. Thus, "=m#2hand E=3"h2m=32h#. The normalization integral is 1="x( )2dx#\$\$%=2C2x2e#2&x2%dxwhere the second step follows from the symmetry of the integrand about x=. Identifying awith 2"in the integral of Problem 6-32 gives 1=2C218"#\$%&’()2"#\$%&’(1 2or C=32"3#\$%&’()1 4. 6-25 At its limits of vibration x= ±Athe classical oscillator has all its energy in potential form: E=1¡m"¡A¡or A=2Em"2#\$%&’(1 2. If the energy is quantized as En=n+12"#\$%&’h(, then the corresponding amplitudes are An=¡n+1( )hm"#\$%&’(1 ¡....
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CH6&amp;7 - 6-20 The Schrödinger equation after...

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