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Lecture38

# Lecture38 - PHYS 360 Quantum Mechanics Wed Apr 20,...

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Unformatted text preview: 4/20/11 PHYS 360 Quantum Mechanics Wed Apr 20, 2011 Lecture 38: Why are some solids metallic, and others insulaEng? As usual, let’s start with a ridiculously oversimpliﬁed model: 1.  Pretend there are no atomic cores. 2.  Pretend the electrons don’t see each other. 3.  Let the electrons behave like a gas of non ­ interacEng parEcles in a container. What should we include? 1.  A three ­dimensional box with potenEal inﬁnitely high at the walls. 2.  Solve the Schrödinger Eqn with these boundary condiEons (of course). 3.  The Pauli Exclusion principle. 1 2 Because of the Pauli Exclusion Principle, only 2 electrons can occupy each state (spin up & spin down). In the ground state, all states below a certain energy E will be ﬁlled. This is called the Fermi energy, EF. This also means all states below a maximum k will be ﬁlled (called the Fermi wave vector kF). Etot = 2 2 5 2 k FV 3π Nq 2V kF 4 k dk = = 2π 2 m ∫0 10π 2 m 10π 2 m ( ) 5/ 3 V −2/3 Finally, let’s do something interesEng. We now know the total energy for a given number of electrons N in a volume V. This is similar to the thermal energy of a gas of atoms. How does the total energy change if the volume changes? From thermodynamics we know that this corresponds to the pressure exerted by the gas: P= dW dE → total dV dV The result is: P= 25 3π 2 2 5/3 2 Etot 2 k F = = ρ 3V 3 10π 2 m 5m () 2/ 3 3 4 P= 25 3π 2 2 5/3 2 Etot 2 k F = = ρ 3V 3 10π 2 m 5m () 2/ 3 Band Structure Why is this important? Because it is not zero. It means the “electron gas” inside a metal exerts a pressure, due enErely to the Pauli Exclusion Principle. (Remember we have ignored all Coulomb interacEons between electrons and nuclei.) This is the pressure that is exerted when you try to compress a solid, but it resists. This is why solids can be strong. The Pauli Exclusion Principle is at the very center of our understanding of materials. 5 1 4/20/11 Consider a crystal where the atoms are arranged periodically. Looking only at one dimension, the potenEal must saEsfy: V ( x + a) = V ( x ) When this condiEon is put into the Schrödinger Eqn, This means we have to change our boundary condiEons. It no longer works to require ψ=0 at all the boundaries. Instead, we employ “periodic boundary condiEons:” ψ x + Na = ψ ( x ) Combine this with ψ x + a = eiKaψ ( x ) ( ) − dψ + V ( x )ψ = Eψ 2 m dx 2 2 2 ( ) to get: Bloch proved that it imposes the following constraint on the soluEons: ψ x + a = eiKaψ ( x ) …and ψ x+a ( ) eiKNa = 1 Hence: K= 2π n , n = 0, ± 1, ± 2, ... Na ( ) ( ) 2 =ψ x () 2 7 With these tools, we can try more and more accurate models for the actual potenEal, and solve for the “band structure” of many, many materials. This tells us exactly why copper is a good metal, why silicon is a semiconductor, and why diamond is a good insulator. 8 The next several slides are on the Kronig ­Penney model of a crystal. It has a liile more detail than necessary for this class…. Let’s determine the “band structure” of the one ­dimensional delta ­funcEon comb (a.k.a. ~Kronig ­Penny model). Put a repulsive (?) delta funcEon of strength α at each site na: V ( x ) = α ∑ δ x − ja j =0 N −1 ( ) Where the potenEal is zero (e.g. 0<x<a) the Schrödinger Eqn is: d 2ψ = − k 2ψ dx 2 with k≡ 2 mE The general soluEon is: ψ I x = A sin( kx ) + B cos( kx ), ψ I 0 = A sin(0) + B cos(0) = ψ II 0 = e− iKa ⎡ A sin k 0 + a + B cos k 0 + a ⎤ ⎣ ⎦ () () ( ) ( ) () (0 < x < a ) Hence: From Bloch’s theorem: ψ x + a = e ψ ( x ) iKa ( ) B = e− iKa ⎡ A sin ka + B cos ka ⎤ ⎣ ⎦ () () …we can also write the general soluEon for an adjacent segment ( ­a<x<0): ψ II x = e− iKa ⎡ A sin k x + a + B cos k x + a ⎤ , − a < x < 0 ⎣ ⎦ We learned back in Ch 2 (eqn 2.125) that the slopes at a delta funcEon are disconEnuous: ⎛ dψ ⎞ dψ I dψ II 2 mα Δ⎜ = − = 2 ψ (0 ) ⎝ dx ⎟ ⎠ dx dx (from Eqn 2.125) () ( ) ( )( ) These two funcEons must be conEnuous at x=0: kA − e− iKa k ⎡ A cos( ka ) − B sin( ka ) ⎤ = ⎣ ⎦ 2 mα B 2 ψ I (0 ) = ψ II (0 ) So…. 2 4/20/11 Re ­arrange the ﬁrst equaEon: cos( Ka ) = cos( ka ) + mα sin( ka ) 2 k A sin( ka ) = ⎡ eiKa − cos( ka ) ⎤ B ⎣ ⎦ Insert into second equaEon and simplify: SubsEtute for the variable k, and simplify constants: z ≡ ka, and β ≡ mα a 2 cos( Ka ) = cos( ka ) + mα sin( ka ) 2 k Which allows one side of the equaEon to be wriien as: f z ≡ cos( z ) + β () sin( z ) z cos( Ka ) = f ( z ) = f ( ka ) cos( Ka ) = f ( z ) = f ( ka ) The allowed values of k are restricted to regions where f(ka) is between +1 and  ­1. Does this picture help you understand energy gaps in solids? No, it does not. The periodic potenEal leads to allowed bands and forbidden gaps. This is another view. What is the x ­axis? The actual band structure of silicon: 18 3 4/20/11 A more physical picture: how two wave funcEons with the same wavelength can have diﬀerent energies, causing an energy gap. 4 ...
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