{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

quantum well notes

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

This preview shows pages 1–4. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 One-dimensional 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: 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 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 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 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)...
View Full Document

{[ snackBarMessage ]}

### Page1 / 13

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

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online