A1_507A - Phys 507A Solid State Physics I Assignment 1...

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Phys 507A - Solid State Physics I Assignment 1: Electrons in crystals. Due Oct. 2nd 1. Periodic potential in 1d. Consider an electron subject to a 1d periodic potential, U ( x ) = X n = -∞ v ( x - na ) , (1) where v ( x - na ) represents the potential barrier against an electron tunnelling between the ions on opposite sides of the point na . For simplicity we assume that v ( x ) = v ( - x ) (inversion symmetry), and v ( x ) = 0 for | x | ≥ a/ 2. But other than this the potential U ( x ) is quite general. Remarkably, the band structure of the 1d solid can be expressed quite simply in terms of the transmittance and reflectance of an electron hitting the barrier v ( x ). Denote the energy of the incident electron to be = ¯ h 2 K 2 / 2 m . The variable K = 2 m / ¯ h parametrizes the energy of the electron. Consider an electron incident from the left on the potential barrier v ( x ); since v ( x ) = 0 for x a/ 2, in these regions the wave function ψ l ( x ) will have the form ψ l ( x ) = e iKx + r e - iKx , x ≤ - a 2 = t e iKx , x a 2 . (2) Here t and r are the transmission and reflection coefficients, respectively. Their actual dependence on K is given by the form of the barrier v ( x ). However, one can deduce the properties of the band structure of the periodic potential U by appealing only to very general properties of t and r . Because v ( x ) is even, ψ r ( x ) = ψ l ( - x ) is also a solution to the Schr¨ oedinger equation with energy . From Eq. (2) it follows that ψ r ( x ) = t e - iKx , x ≤ - a 2 , = e - iKx + r e iKx , x a 2 . (3) Since ψ l and ψ r are two independent solutions to the single-barrier Schr¨ oedinger equa- tion with the same energy, any other solution with that energy will be a linear combi- nation of these two; in addition, since the crystal potential is identical to v ( x ) in the region | x | ≤ a/ 2, any solution to the crystal Schr¨oedinger equation with energy in the | x | ≤ a/ 2 region must be given by ψ ( x ) = l ( x ) + r ( x ) . (4)
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Now Bloch’s theorem asserts that ψ can be chosen to satisfy ψ ( x + a ) = e ika ψ ( x ) , (5) for a suitable k (different than K !). Differentiating the above equation we also find that ψ 0 ( x + a ) = e ika ψ 0 ( x ) . (6) (a) By imposing the conditions (5) and (6) at x = - a/ 2, and using Eqs. (2)–(4), show that the energy of the Bloch electron is related to its wavevector k by: cos ( ka ) = t 2 - r 2 2 t e iKa + 1 2 t e - iKa , = ¯ h 2 K 2 2 m . (7) Verify that this gives the right answer in the free electron case ( v = 0). We write the complex number t in terms of its magnitude and phase: t = | t | e . (8) The real number δ is known as the phase shift, since it represents the change in phase of the transmitted wave relative to the incident one. Electron conservation requires that the probability of transmission plus the probability of reflection be unity, | t | 2 + | r | 2 = 1 . (9) This, and some other useful information, can be proved as follows. Let φ 1 and φ 2 be
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