Chapter2_14 - “lowering operator” to the lowest energy...

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PHY4604 R. D. Field Department of Physics Chapter2_14.doc University of Florida The Harmonic Oscillator (3) Energy Eigenvalue Equation: We are looking for solutions of the equation H op |E n > = E n |E n > where |E n > are the “eigenkets” and E n are the allowed energies ( i.e. eigenvalues). The state (a + ) op |E n > = |a + E n > is an “eigenket” with energy ω h + n E since > + >= + = > + >= >= + + + + + + n n n op n n op op op op n op n op op E a E E a E E a H H a E a H E a H | ) ( | ) )( ( | ]) ) ( , [ ) (( | | ) ( h h Thus, > + >= + h n n n op E c E a | | ) ( where c n are constants (that may depend on n ) and similarly > >= h n n n op E d E a | | ) ( Ground State Energy (Lowest Energy State): We know that the norm of the state (a - ) op |E n > = | a - E n > must be positive definite and hence > >=< >=< ≤< + n op n n op op n n n E H E E a a E E a E a | | | ) ( ) ( | | 0 2 1 1 h . Thus, h 2 1 | | >≥ < n op n E H E . This implies that there is a minimum energy state which we will call |E 0 > with H op |E 0 > = E 0 |E 0 > where h 2 1 0 E . The state |E 0 > is the state of lowest energy and E 0 is the ground state energy. Applying the
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Unformatted text preview: “lowering operator” to the lowest energy state gives > − >= − h | | ) ( E d E a op . But there is no state with energy lower than E which implies that d = 0 and | ) ( >= − E a op . Now we can solve for E as follows > >= >= + >= − + 2 1 2 1 | | | ) ) ( ) ( ( | E E E E a a E H op op op h h h . Hence, h 2 1 = E and we normalize so that 1 | >= < E E . Excited States: All the other states are calculated from the ground state using the “raising operator” as follows: > >= + | ) ( ! 1 | E a n E n op n and > + >= n n op E n E H | ) ( | 2 1 h and hf n n E n ) ( ) ( 2 1 2 1 + = + = h Planck’s guess was E n = nhf!...
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This note was uploaded on 05/29/2011 for the course PHY 4064 taught by Professor Fields during the Spring '07 term at University of Florida.

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