Fund Quantum Mechanics Lect & HW Solutions 83

Fund Quantum Mechanics Lect & HW Solutions 83 -...

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5.3. TWO-STATE SYSTEMS 65 and this can be multiplied out, dropping the common factor α 2 and noting that a ψ 1 | ψ 1 A and a ψ 2 | ψ 2 A are one, as a ψ 1 | ψ 2 A ε 2 2 ε + a ψ 1 | ψ 2 A = 0 for which the smallest root can be written as ε = a ψ 1 | ψ 2 A 1 + r 1 − a ψ 1 | ψ 2 A 2 . That is less than a ψ 1 | ψ 2 A , hence ε is small if the overlap is small. The constant α follows from the requirement that the new states must still be normalized, and is found to be α = 1 r 1 2 ε a ψ 1 | ψ 2 A + ε 2 . Note that the denominator is nonzero; the argument of the square root exceeds (1 ε ) 2 . 5.3.2 Solution 2state-b Question: Show that it does not have an eFect on the solution whether or not the basic states ψ 1 and ψ 2 are normalized, like in the previous question, before the state of lowest energy is found. This requires no detailed analysis; just check that the same solution can be described using the nonorthogonal and orthogonal basis states. It is however an important observation for various
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Unformatted text preview: numerical solution procedures: your set of basis functions can be cleaned up and simplied without aFecting the solution you get. Answer: Using the original basis states, the solution, say the ground state of lowest energy, can be written in the form c 1 1 + c 2 2 for some values of the constants c 1 and c 2 . Now the expression for the orthogonalized functions, 1 = ( 1 2 ) 2 = ( 2 1 ) , can for given 1 and 2 be thought of as two equations for 1 and 2 that can be solved. In particular, adding times the second equation to the rst gives 1 = 1 + 2 (1 2 ) . Similarly, adding times the rst equation to the second gives 2 = 2 + 1 (1 2 ) ....
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