Unformatted text preview: 4/18/11 PHYS 360 Quantum Mechanics Mon Apr 18, 2011 Lecture 37: Can quantum mechanics explain how a metal works? Can quantum mechanics explain how a metal works? Metals have the unusual property that electrons can ﬂow through the specimen in response to an applied ﬁeld; e.g. connect a wire across a baVery. This means at least one electron per atom is not actually bound to any atom. These act almost like free electrons. However, electrons have extremely strong repulsive interacXons with other electrons, and very strong posiXve interacXons with the nuclei in a solid. How can this lead to nearly “free” electrons? 1 2 HW #10 due today: Ch 5, #5, 7, 9, 13, 14 Lecture Outline: Free electron gas in a box I. Solve 3D parXcle in a box, add parXcles, require Pauli exclusion II. Wave funcXons, energies depend on nx,ny,nz. III. Invent Reciprocal Space (momentum space) IV. Maximum energy is the Fermi Energy V. Isotropic material: Fermi Sphere VI. For ﬁxed density of electrons, determine Total Energy VII. The pressure of this non
interacXng gas is not zero! I. Solve 3D par=cle in a box, add par=cles, require Pauli exclusion 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
interacXng parXcles in a container. What should we include? 1. A three
dimensional box with potenXal inﬁnitely high at the walls. 2. Solve the Schrödinger Eqn with these boundary condiXons (of course). 3. The Pauli Exclusion principle. 3 4 So, we are back to solving for a parXcle in a box: V x , y , z = 0, if 0 < x < lx , 0 < y < l y , and 0 < z < lz ; ∞ , otherwise − ∇ 2ψ = Eψ 2m
2 X ( x ) = Ax sin( kx x ) + Bx cos( kx x ) X (0 ) = X (lx ) = 0 So Bx = 0 and sin( kx lx ) = 0 ( ){ k x lx = nxπ , k y l y = n yπ , k z lz = nz π
nx = 1, 2, 3, ..., n y = 1, 2, 3, ..., nz = 1, 2, 3
The complete, normalized wave funcXon thus is: − 2 d 2 Z = Ez Z 2 m dz 2 since V=0. ψ ( x, y, z ) = X ( x )Y ( y)Z ( z )
− 2 d 2 X = Ex X 2 m dx 2 − 2 d 2Y = EyY 2 m dy 2 ψn n n =
xyz ⎛nπ 8 sin ⎜ x lx l y lz ⎝ lx ⎞ ⎛ n yπ x ⎟ sin ⎜ ⎠ ⎝ ly ⎞ ⎛nπ y ⎟ sin ⎜ z ⎠ ⎝ lz ⎞ z⎟ ⎠
6 E x + E y + Ez = E 5 1 4/18/11 II. Wave func=ons, energies depend on nx,ny,nz. ⎛ n π ⎞ ⎛ n yπ ⎞ ⎛ nz π ⎞ 8 sin ⎜ x x ⎟ sin ⎜ y ⎟ sin ⎜ z⎟ lx l y lz ⎝ lx ⎠ ⎝ l y ⎠ ⎝ lz ⎠ III. Invent Reciprocal Space (momentum space) n n n =
xyz ˆ ˆ k = k x i + k y ˆ + kz k j And the allowed energies are: En n n =
xyz k x = nx π , lx k y = ny π , ly k z = nz π lz 2 2 2 2π 2 ⎛ nx n y nz ⎞ 2 k 2 ⎜ + 2 + ⎟= 2 m ⎝ lx2 l y lz2 ⎠ 2 m It will be useful to label the wave funcXons by their wave vector: ˆ ˆ ˆ k =k i +k j+k k
x y z
7 We now imagine working in “k
space” as opposed to “real” space. 8 IV. Maximum energy is the Fermi Energy 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). The k
space volume per state is: π3 π3 = lx l y lz V Since we label our states with only posiXve values of the integers, the allowed states form one octant of the sphere of radius kF. Assume there are N atoms inside, and each atom contributes q electrons (maybe one or two….). Let’s determine kF. The volume of the octant is equal to the volume per state Xmes the number of states, so: 1 ⎛ 4 3 ⎞ Nq ⎛ π 3 ⎞ ⎜⎟ ⎜ πk ⎟ = 2 ⎝ V ⎠ 8⎝ 3 F⎠
9 Density of states ρ ≡ Nq V
10 k F = 3ρπ 2 ( ) 1/ 3 Let’s determine kF. The volume of the octant is equal to the volume per state Xmes the number of states, so: 1 ⎛ 4 3 ⎞ Nq ⎛ π 3 ⎞ πk = 8⎜ 3 F⎟ 2 ⎜V ⎟ ⎝ ⎠ ⎝⎠ k F = 3ρπ 2 V. For ﬁxed density of electrons, determine Total Energy Now let’s get the total energy Etotal by adding the energies of all occupied levels (in the ground state). Volume of shell = (1/8)(4πk2)dk. The number of electron states in this shell is the density of states Xmes the volume. The energy of these states is ħ2k2/2m. This leads to: Density of states ρ ≡ Nq V ( ) 1/ 3 Now let’s determine the Fermi energy EF: 2 EF = = 3ρπ 2 2m 2m
2 2 kF ( ) 2/ 3 2 ⎛ Nq 2 ⎞ = π⎟ ⎜3 2m ⎝ V ⎠ 2/ 3 dE = 2 k 2 V 2 k dk 2m π 2 Etot =
11 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
12 2 4/18/11 VII. The pressure of this non
interac=ng gas is not zero! Etot =
2 2 5 2 k FV 3π Nq 2V kF 4 k dk = = 2π 2 m ∫0 10π 2 m 10π 2 m P= ( ) 5/ 3 25 3π 2 2 5/3 2 Etot 2 k F = = ρ 3V 3 10π 2 m 5m () 2/ 3 V −2/3 Why is this important? Because it is not zero. It means the “electron gas” inside a metal exerts a pressure, due enXrely to the Pauli Exclusion Principle. (Remember we have ignored all Coulomb interacXons 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. 14 Finally, let’s do something interesXng. 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 → dEtotal
dV dV
25 3π 2 2 5/3 2 Etot 2 k F = = ρ 2 3V 3 10π m 5m The result is: P= () 2/ 3 13 So, what is wrong with our model of a metal as an electron gas in a 3D inﬁnite potenXal well? We are now ﬁrmly in the realm of Condensed MaVer Physics. This tells us: 1. Coulomb aVracXon/repulsion/interacXons are very important BUT “screening” eﬀects and exchange interacXons show that they can nearly cancel out (Landau “Fermi liquid theory”). 2. We MUST include the eﬀect of a weak periodic potenXal. (Why is it weak? See point 1 above.) This leads to Bloch’s Theorem, the heart of condensed maVer physics. 15 Consider a crystal where the atoms are arranged periodically. Looking only at one dimension, the potenXal must saXsfy: V ( x + a) = V ( x )
When this condiXon is put into the Schrödinger Eqn, − 2 d 2ψ + V ( x )ψ = Eψ 2 m dx 2 Bloch proved that it imposes the following constraint on the soluXons: ψ x + a = eiKaψ ( x )
…and ψ x+a ( ) ( ) 2 =ψ x () 2 16 This means we have to change our boundary condiXons. It no longer works to require ψ=0 at all the boundaries. Instead, we employ “periodic boundary condiXons:” ψ x + Na = ψ ( x )
This gives a new requirement for the values of K: K= 2π n , n = 0, ± 1, ± 2, ... Na ( ) ( ) With these tools, we can try more and more accurate models for the actual potenXal, 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. 17 18 3 ...
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 Spring '11
 DURBIN,STEPHEN
 Work, Trigraph, the00, Etot, π lz

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