quantum well notes

quantum well notes - University of Virginia Spring Semester...

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Unformatted text preview: University of Virginia Spring Semester 2002 Department of Physics Instructor: D. Po cani c PHYS 252: Modern Physics Bound States in a One-dimensional Square Potential Well in Quantum Mechanics (with a brief introduction to symmetry in quantum mechanics) Dinko Po cani c 17 March 2002 These expanded lecture notes include some material that was not covered in class. Thus, they are meant as a reading supplement to the textbook and to any notes taken in class. This paper also provides additional details of the derivation of the ground state energy for a particle in a finite one- dimensional square potential well. c D. Po cani c 2002 2 QM: 1-dim Square Potential Well 1. PARTICLE IN A BOX (INFINITE SQUARE POTENTIAL WELL) Before we turn to the problem of the finite potential well, let us first review the solution for a particle in a hard box, i.e., inside an infinite square potential well, Fig. 1). In the force-free ( V = 0) interior of the box, the time-independent Schr odinger V(x) x-a a V ( x ) = for | x | a for | x | < a Figure 1: Plot of the potential energy of an infinitely deep one- dimensional square well (impene- trable box). equation (TISE) is- h 2 2 m d 2 dx 2 = E . (1) We try a solution of the form: ( x ) = A cos kx + B sin kx . (2) Substitution into Eq. 1 yields- h 2 2 m (- Ak 2 cos kx- Bk 2 sin kx ) = E ( A cos kx + B sin kx ) . (3) This is an identity, i.e., valid for all values of x , if, and only if, h 2 k 2 2 m = E or k = s 2 mE h 2 . (4) Hence ( x ) = A cos s 2 mE h 2 x + B sin s 2 mE h 2 x (5) is a solution of Eq. 1. This wave function does not tell us very much, since none of the above equations places any restriction on E . Actually this was to be expected, since a particle moving freely along the x-axis can have any energy it pleases. Equation 1, however, does not tell the whole story; we have not yet taken into account the influence of the retaining walls. QM: 1-dim Square Potential Well 3 Outside the region- a x a , Eq. 1 has to be replaced with- h 2 2 m d 2 ( x ) dx 2 + V ( x ) = E ( x ) (6) In the limit V this can be satisfied for finite values of E only if = 0 for all | x | a . The infinite potential thus imposes the boundary conditions : (+ ) = 0 = ( a ) and (- ) = 0 = (- a ) . (7) Since * is the probability density for a physical particle, the wavefunction ( x ) has to be continuous everywhere, including x = a . This results in the matching conditions :- A cos ka + B sin ka = (- a ) = 0 and A cos ka + B sin ka = ( a ) = 0 . (8) A trivial solution of Eq. 8 is A = B = 0, which means that ( x ) 0 or that the box is empty . Since cos ka and sin ka cannot simultaneously vanish for any value of ka , there are two very distinct nontrivial solutions; one is B = 0 and hence cos ka = 0 , that is, ( x ) = A cos kx (9) where k = 2 a , 3 2 a , 5 2 a , ... , (2 n + 1)...
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This note was uploaded on 08/12/2008 for the course PHYS 252 taught by Professor Pocanic during the Spring '02 term at UVA.

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quantum well notes - University of Virginia Spring Semester...

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