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Unformatted text preview: University of Virginia Spring Semester 2002 Department of Physics Instructor: D. Poˇ cani´ c PHYS 252: Modern Physics Bound States in a Onedimensional Square Potential Well in Quantum Mechanics (with a brief introduction to symmetry in quantum mechanics) Dinko Poˇ 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. Poˇ cani´ c 2002 2 QM: 1dim 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 forcefree ( V = 0) interior of the box, the timeindependent Schr¨ odinger V(x) xa 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 ψ dx 2 = Eψ . (1) We try a solution of the form: ψ ( 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 ψ ( 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 xaxis 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: 1dim Square Potential Well 3 Outside the region a ≤ x ≤ a , Eq. 1 has to be replaced with ¯ h 2 2 m d 2 ψ ( x ) dx 2 + V ψ ( x ) = Eψ ( x ) (6) In the limit V → ∞ this can be satisfied for finite values of E only if ψ = 0 for all  x  ≥ a . The infinite potential thus imposes the boundary conditions : ψ (+ ∞ ) = 0 = ψ ( a ) and ψ (∞ ) = 0 = ψ ( a ) . (7) Since ψ * ψ is the probability density for a physical particle, the wavefunction ψ ( x ) has to be continuous everywhere, including x = ± a . This results in the matching conditions : A cos ka + B sin ka = ψ ( a ) = 0 and A cos ka + B sin ka = ψ ( a ) = 0 . (8) A trivial solution of Eq. 8 is A = B = 0, which means that ψ ( 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, ψ ( x ) = A cos kx (9) where k = π 2 a , 3 π 2 a , 5 π 2 a , ... , (2 n + 1)...
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 Spring '02
 POCANIC
 Physics, mechanics, odd parity, Even Parity, square potential, 1dim Square Potential

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