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G Tk B 2 ∝ e = p i n i n i > n is the concentration [#/cm3] of electrons > p is the concentration [#/cm3] of holes Every thermally generated electron leaves behind a hole Cite as: Joel Voldman, course materials for 6.777J / 2.372J Design and Fabrication of Microelectromechanical Devices, Spring 2007. MIT OpenCourseW...
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ceptor dopant concentrations ar e NA [#/cm3] (cid:198) p-type material • Boron (3 valence electrons) > Donor dopant concentrations are N D [#/cm3] (cid:198) n-type material • Phosphorous (5 valence electrons) > These introduce new energy levels close to valence or conduction bands (~0.05 eV) > Dopant concentrations...
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/6.777J Spring 2007, Lecture 7E - 10 Main results > Donors or acceptors are fully ionized > Define n0 and p0 as the electron and hole N D N → + + D − e concentrations at equilibrium > n0 and p0 follow a mass-action law N N → + + h A N ≈ D N A ≈ N − A N + D − A n p 0 0 2 n= i - + - + - + − = AN p + = DN n n p= Cite as...
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OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. JV: 2.372J/6.777J Spring 2007, Lecture 7E - 12 Main results > Therefore, the equilibrium majority carrier concentration is determined by the net doping and the minority carrier concentration is determined b...
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′ n τ m Electron-hole pair (EHP) ′ n n t ( ) ~ (0) ′ t − e τ m Cite as: Joel Voldman, course materials for 6.777J / 2.372J Design and Fabrication of Microelectromechanical Devices, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. JV: 2.372J/...
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obey a drift/diffusion flux relation > Drift: carriers move in an electric field > Diffusion: carriers move in a concentration gradient > For p-type material • n is small (cid:206) drift current is small • ∇n can be big (cid:206) diffusion current dominates Electric field [V/cm] drift E n ( q μ= e n diffusion ) D ...
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- - - - - - - - - - - - - - - - > At equilibrium diffusive driving force = electric field in the vicinity of the junction > In order to set up this electric field, the ionized donors and acceptors are relatively depleted of mobile carriers near the junction – the space-charge layer (SCL) or depletion region Oxide...
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.2 in: Senturia, Stephen D. M Boston, MA: Kluwer Academic Publishers, 2001, p. 359. ISBN: 9780792372462. icrosystem Design. Image by MIT OpenCourseWare. p n Equilibrium IN Ec Ev Ec Ev IP Forward bias Image by MIT OpenCourseWare. Adapted from Figu re 6.1 in: Pierret, Robert F. Semiconductor Device Fundamentals. Reading,...
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2007. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. JV: 2.372J/6.777J Spring 2007, Lecture 7E - 20 The Junction-Isolated Diffused Resistor > The structure is a diode, but with two contacts > The diode action prevents currents into the substrate pr...
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the channel region of the substrate can either increase the surface concentration (accumulation) or deplete the surface and eventually invert the surface Source Gate Drain Oxide Channels p-type substrate n+ regions G B D S Body Adapted from Figure 14.10 in: Senturia, Stephen D. Microsystem Design. Boston, MA: Kluwer...
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A m ( D I 1000 800 600 400 200 0 VGS = 4 V triode ID,sat VGS = 3 V saturation VGS = 2 V 0 Image by MIT OpenCourseWare. 1 2 3 4 5 Adapted from Figure 14.13 in: Senturia, Stephen D. Microsystem Design. Boston, MA: Kluwer Academic Publishers, 2001, p. 369. ISBN: 9780792372462. VDS G S e- D e- G S e- VDS>0 V VDS > -V T GS ...
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circuits involving MOSFETS > Can use either full nonlinear characteristics > Or linearized small-signal model • Different models include different components G G 2 K/2(V -V ) Tn GS D S Simple large-signal model, in saturation D CGS g vm gs r0 + - S Simple small-signal model, in saturation Cite as: Joel Voldman, co...
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B _ Top view Adapted from Figures 14.23 and 14.24 in Senturia, Stephen D. Microsystem Design. Boston, MA: Kluwer Academic Publishers, 2001, p. 382. ISBN: 9780792372462. LM158 Image by MIT OpenCourseWare. Cite as: Joel Voldman, course materials for 6.777J / 2.372J Design and Fabrication of Microelectromechanical Devices...
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v − s − R 1 A (0 = = v V − − 0 R 2 − v − ) ⇓ 1 ⎛ ⎜ ⎝ 1 + 1 A + R 2 R 1 ⎞ ⎟ ⎠ ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ A 1 ⇓ → ∞ V o V s R 2 R 1 = − Cite as: Joel Voldman, course materials for 6.777J / 2.372J Design and Fabrication of Microelectromechanical Devices, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/), Massachuset...
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2 R 1 R 2 R 1 Vs + _ V0 R1 R2 Non-inverting Image by MIT OpenCourseWare. Adapted from Figure 14.28 in Senturia, Stephen D. Microsystem Design. Boston, MA: Kluwer Academic Publishers, 2001, p. 388. ISBN: 9780792372462. Cite as: Joel Voldman, course materials for 6.777J / 2.372J Design and Fabrication of Microelectromech...
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J / 2.372J Design and Fabrication of Microelectromechanical Devices, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. JV: 2.372J/6.777J Spring 2007, Lecture 7E - 35 Differentiator C Vs + _ R1 _ + V0 Adapted from Figure 14.31 in Senturia, Ste...
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Turbulent Flow and Transport 3 Concepts in Turbulence 3.1 Comments on laminar flow, its stability, and the transition to turbulent flow. 3.2 Features of turbulent flows (high Reynolds number, "randomness", three− dimensionality of fluctuations, intermittency near free boundaries, etc. 3.3 The energy cascade and...
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, read for example Schlichting. 7th ed. 578−586, or White: 436−440.
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StokesI Theorem Our text states and proves Stokes' Theorem in 12.11, but ituses the scalar form for writing both the line integral and the surface integral involved. In the applications, it is the vector form of the theorem that is most likely to be quoted, since the notations dxAdy =d the like are not in common u...
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( F a T) ds -- JJs ((curl T)-$] d~ . Jc Here the orientation of C is that derived from the counterclockwise orientation of D; and the normal fi to the surface S points in the same direction as 3rJu Y 1gJv . Remark 1. The relation between T and 2 is often described 4 informally as follows: "If youralkaround C in ...
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Proof of the theorem. The proof consists of verifying the following three equations: m e theorem follo& by adding these equations together. We shall in fact verify only the first equation. The others are proved similarly. Alternatively, if one makes the substitutions > 1 j and j + k m.d k -+i m d x--sy and y + z...
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/ ~ v ) du dv . We use the chain rule to compute the integrand. We have where P and its partials are evaluated at ~(u,v), of course. Subtracting, we see that the first and last terms cancel each other. The double integral ( * ) then takes the form Nclw we compute the surfac~ integral of our theorem. Since curl F ...
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t o S, x 2 - 1 z = 0 Let = x t - and let 3, b e t h e unit normal to S 2 , both with positive d component. Evaluate y o d l d S and Jj 1 * a 2 d s . S1 s2 -'Grad 'Curl - - - Div and all that. We study two questions about these operations: I. DO they have natural (i.e., coordinate-free) physical or geo...
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a physical interpretation of curl F, let us -+ , imagine F to be the velocity vector field of a moving fluid. + Let us place a small paddle wheel of radius P in the fluid, with its axis along n. Eventually, the paddle wheel settles -+ down to rotating steadily with angular speed w (considered as positive if cou...
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s i d e r t h e n e x t t w o o p e r a t i o n s , c u r l and d i v . Again, w e c o n s i d e r o n l y f i e l d s t h a t a r e c o n t i n u o u s l y d i f f e r e n t i a b l e i n a r e g i o n U o f R . There a r e analogues of 3 a l l t h e e a r l i e r theorems: I Theorem 5 . - I f F - - - - i s a ...
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h y p o t h e s i s , G = c u r l ? d e f i n e d i n U. W e corn- f o r some +- p u t e : G n d s = S2 11 c u r l F mii d~ = - s 2 . 1 C H mda. - Adding, w e see t h a t , ema ark. The convers'e of Theorem 5 h o l d s a l s o , b u t w e s h a l l n o t a t t e m p t t o prove it. Theorem 6 . - I f 2 = c ...
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to construct a specific such function in the given cases. if 2 Note that -+ = curl ?$ and G = curl H, then -+ + + curl(H-F) = 8 . Hence by Theorem 3, H - 5 = grad 4 in U, for some 4. Theorem 8. -The condition -t d i v G = O &I U does not in general imply - that if - - - - --- is a curl in U. Proof. L ...
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a of s p h e r e ) # 0. 0 erna ark. Suppose w e s a y t h a t a r e g i o n U i n R3 i s "two- simply c o n n e c t e d w i f e v e r y c l o s e d s u r f a c e i n U bounds a s o l i d r e g i o n l y i n g i n U.* The r e g i o n U = R - ( o r i g i n ) 3 i s nott'two- simply connected", f o r example, b u t...
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i s much more one can s a y a b o u t t h e s e matters, b u t one needs t o i n t r o d u c e a b i t of a l g e b r a i c topology i n o r d e r t o do s o . I t i s a b i t l a t e i n t h e s e m e s t e r f o r t h a t ! *The proper mathematical term for t h i s is ffharalogically t r i v i a l in dimension tw...
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8.06 Spring 2016 Lecture Notes 4. Identical particles Aram Harrow Last updated: May 19, 2016 Contents 1 Fermions and Bosons Introduction and two-particle systems . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1 1.2 N particles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ...
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Charged particles in a magnetic field 21 3.1 The Pauli Hamiltonian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 3.2 Landau levels . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24 3.3 The de H...
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·) denotes an ordered list, in which different posi­ tions have different meanings; e.g. in general (rr1, pr1, rr2, pr2) = (rr2, pr2, rr1, pr1). 1 To describe indistinguishable particles, we can use set notation. For example, the sets {a, b, c} and {c, a, b} are equal. We can thus denote the state of N indistinguish...
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, since it may be that swapping two positions could result in an unobservable change, such as multiplying by an overall phase. To be more concrete, consider the case of two indistinguishable particles. Then we should have |ψ((cid:126)r1, (cid:126)r2)| = |ψ((cid:126)r2, (cid:126)r1)|, or equivalently ψ((cid:126)r1, (cid...
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3/2, etc.) are fermions and that particles with integer spin (0, 1, etc.) are bosons. The proof of this involves field theory (or at least the existence of antiparticles) and is beyond the scope of 8.06 (but could conceivable be a term-paper topic). To find a basis for the symmetric and antisymmetric subspaces, we can co...
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105) and |β(cid:105), in which case we get back either the same state (symmetric subspace) or the same state multiplied by -1 (antisymmetric subspace). after normalizing if √ α β β 2 If V is d-dimensional and has basis {|1(cid:105), . . . , |d(cid:105)} then V ⊗ V is d2-dimensional and has basis {|1(cid:105) ⊗ |1(cid:1...
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eyl duality” is the phrase to google to learn more.) (cid:1) = d(d+1) 2 2 In this case, we use spin Example: spin-1/2 particles. The simplest case is when d = 2. notation and describe the single-particle basis with {|+(cid:105), |−(cid:105)}. The resulting basis for Sym2 C2 is {| + +(cid:105), |+−(cid:105)+|−+(cid:105)...
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:105) = |ψ(cid:105) ∀i (cid:54)= j} AntiN V ≡ {|ψ(cid:105) ∈ V ⊗N : F ij|ψ(cid:105) = −|ψ(cid:105) ∀i (cid:54)= j} The corresponding wavefunctions are those satisfying ψ((cid:126)r1, . . . , (cid:126)ri, . . . , (cid:126)rj, . . . , (cid:126)rN ) = ±ψ((cid:126)r1, . . . , (cid:126)rj, . . . , (cid:126)ri, . . . , (cid:...
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In other words if π, ν ∈ SN then (denoted e) and is closed under multiplication and inverse. applying ν then π is another permutation (denoted πν) and there exists a permutation π−1 satisfying ππ−1 = π−1π = e. Additionally F π is a representation meaning that F πν = F πF ν. Verifying these facts is a useful exercise. O...
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1 N ! (cid:88) π∈SN where the second equality used the fact that F π|ψ(cid:105) = |ψ(cid:105) for all |ψ(cid:105) ∈ SymN V . The argument for the antisymmetric projector is similar, but we first need to define sgn(π), which is called the sign of a permutation. It is defined to be 1 if π can be written as a product of an e...
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obeys sgn(νπ) = sgn(ν) sgn(π) for any permutations ν, π. This can be used to prove that Panti is the projector onto the antisymmetric subspace, using an argument similar to the one used for Psym and the symmetric subspace. As a result, we can write a basis for SymN V consisting of the states |ψsym α1,...,α = (cid:105) ...
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:105) = det N      ψα1((cid:126)r1) . . . · · · ψα1((cid:126)rN ) . . . ψαN ((cid:126)r1) · · · ψα N ((cid:126)rN )      . (16) This is called a Slater determinant. For example when N = 2, the wavefunction is of the form ψα1((cid:126)r1)ψα2((cid:126)r2) − ψα2((cid:126)r1)ψα1((cid:126)r2) √ 2 . (17) 1.3 Non-in...
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1(cid:105) ⊗ · · · ⊗ |αN (cid:105), it follows that H|α1(cid:105) ⊗ · · · ⊗ |αN (cid:105) = (Eα1 + . . . + EαN )|α1(cid:105) ⊗ · · · ⊗ |αN (cid:105). (20) N N (cid:105)} is an orthonormal basis of eigenstates of H. The ground state is |0(cid:105)⊗ , Thus {|α1(cid:105) ⊗ · · · ⊗ |α which has energy N E0. The first excite...
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and one 1, but these all refer to exactly the same state. Similarly all the same energies Eα1 + . . . + EαN still exist in the spectrum of symψα1,...,αN (cid:105) H restricted to SymN V , but the degeneracy of up to N ! is now gone. Specifically state | has energy Eα1 + . . . + EαN . Since these are a basis for SymN V w...
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e.g. |ψ0,1,...,N −3,N,N +1(cid:105)). Holes are studied in solid-state physics, and were the way that Dirac originally explained positrons (although this explanation has now been superseded by modern field theory). 0,1,...,N 2,N +1(cid:105)), moving the hole to lower energies (e.g. |ψ 0,1,...,N 2,N (cid:105) which has e...
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n(cid:48) Either way this factorizes as i ranges over 0,1 (for fermions) or over all nonnegativ e integers (for bosons). Pr[n0, n1, . . .] = (cid:89) i≥0 e−βni(Ei−µ) (cid:80) n(cid:48) i e−βn(cid:48) (Ei−µ) i . In other words, each occupation number is an independent random variable. For fermions this results in the Fe...
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could be written either as |n, l, m, s(cid:105) or (dividing into spatial and spin parts) as |n, l, m(cid:105) ⊗ |s(cid:105). More generally, suppose the state space of a single particle is V ⊗ W . Then the state of N distinguishable particles is (V ⊗ W )⊗N ∼= V ⊗N ⊗ W ⊗N . This isomorphism is proved by simply rearrang...
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12:34|α1, α2, α3, α4(cid:105) = |α3, α4, α1, α2(cid:105). (31) (32) What if we would like to understand Anti2(V ⊗ W ) in terms of the symmetric and antisymmetric subspaces of V ⊗2 and W ⊗2? Then it will be convenient to rearrange (31) and write (with some small abuse of notation) Anti2(V ⊗ W ) = {|ψ(cid:105) ∈ V ⊗ V ⊗ ...
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pace of F 3:4. It also contains states in the −1 eigenspace of F 1:2 and the +1 eigenspace of F 3:4, as well as superpositions states in these two spaces. Putting this together we have Anti2(V ⊗ W ) ∼= (Sym2 V ⊗ Anti2 W ) ⊕ (Anti2 V ⊗ Sym2 W ). Similarly the symmetric subspace of two copies of V ⊗ W is Sym2(V ⊗ W ) ∼ (...
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:105), . . . , |N (cid:105) corresponding to wavefunctions ψ1((cid:126)r), . . . , ψN ((cid:126)r). If we had one electron in position |1(cid:105) with spin in state |s1(cid:105), another electron in position |2(cid:105) with spin in state |s2(cid:105), and so on, then the overall state would be 1 √ N ! (cid:88) π∈SN s...
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2((cid:126)r) is nonzero and ψi((cid:126)r) = 0 for i = 2. (Here we use the assumption that the electrons are well separated.) Suppose this field is Bzzˆ in this region and zero elsewhere. This field would correspond to a single-particle Hamiltonian of the form |2(cid:105)(cid:104)2| ⊗ ωzSz 9 (cid:54) for ωz = −µeBz, wh...
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model of N distinguishable spins. Spatial position can be used to distinguish particles, but it does not have to in every case. 2 Degenerate Fermi gas 2.1 Electrons in a box Consider N electrons in a box of size L × L × L with periodic boundary conditions. (Griffiths discusses hard-wall boundary conditions and it is a go...
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the k with the lowest values of k2. These k vectors are contained in (cid:126) (cid:126) (cid:126) 10 a sphere of some radius which we call kF (aka the Fermi wave vector). Since each wavevector can (cid:126) be thought of as taking up “volume” (2π/L)3 in k-space, we obtain the following equation for kF : 4 3 πk3 F · (...
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) (cid:126) k2 2 2m = 2 (cid:90) kF (cid:126) 2m 0 4 ck dk = (cid:126)2 k5 F c 2m 5 = (cid:126)2k2 3 F 2m 5 3 N = N EF . 5 (48) In hindsight we should have guessed that Egs would be some constant times N EF . There are N electrons, each with energy somewhere between 0 and EF depending on their position within the spher...
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EF = O(1)eV . This is (cid:29) kBT for room-temperature T , justifying the assumption that the state is close to the ground state as room temperature, and corresponds to a vF = (cid:112)2EF /m that is (cid:28) c, justifying a non-relativistic approximation. Can we justify neglecting the Coulomb interaction? See the pse...
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be ≈ 1.1 × 10−9 watt - ohm and modeling them K2 as a Fermi gas predicts ≈ 2.44 × 10−8 watt - ohm . In real metals, this number is much closer to the 2 K Fermi gas prediction, which provides some support for the Drude-Sommerfeld theory. σT There are some other aspects of the theory which are clearly too simplistic. The ...
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istic electrons in their ground state. There are a lot of assumptions here. The ground-state assumption is justified because photons will carry away most of the energy of the star. The non-interacting assumption will be discussed on the pset. We consider only electrons and not nuclei because degeneracy pressure scales w...
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3 Edegen(R) = Egs = NeEF (R) 5 (cid:126)2 (cid:18) 3π2Ne 2me V 3 = Ne5 (cid:32) = (cid:126) f 2 5/3 2/3 (cid:19) (cid:18) 3 (cid:19) (cid:18) 10 9π 4 2/3 (cid:19) (cid:33) 5/3 N m R2 e , = λN 5/3 meR2 , (cid:124) (cid:123)(cid:122) ≡λ (cid:125) where again λ is a universal constan t. For f = 0.6 (as is the case for our...
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3 (cid:126)N m c λ e meκ −1/3 N G ∼ (cid:126)c N N 2/3 m2 p = 2 m p m2 Pl N 2/3. In the last step we have introduced the Planck mass mPl which is the unique mass scale associated with the fundamental constants GN , (cid:126), c. Thus the critical value of N at which the non-relativistic assumption breaks down is Ncrit ...
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= 3V k F 23π (cid:19) 3 4πk2dk (cid:126)ck (cid:125) = = κ(cid:48) V (cid:126)c k4 4π2 F N 4/3 V 1/3 where κ(cid:48) ≡ 3 4 (cid:0) 9π 4 (cid:1)1/3 f 4/3(cid:126)c. Now the total energy is Etot = − κN 2 R + κ(cid:48)N 4/3 R Λ ≡ . R (56) The situation is now much simpler than in the non-relativistic case. If Λ < 0 then t...
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by mp. It turns out that neutron stars are stable up to masses of roughly 3.0MSun (although this number is fairly uncertain in part because we don’t know the 15 structure of matter in a neutron star; it may be that the quarks and gluons combine into more exotic nuclear matter). Above 3.0MSun there is no further way to...
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(x + a). Since [Ta, V ] = 0 we have [T , H] = 0 and thus Ta and H can be simultaneously diagonalized. Bloch’s theorem essentially states that we can take eigenstates of H to be also eigenstates of Ta. ˆ ˆ ˆ ˆ ˆ ˆ To see what this means let’s look more carefully at Ta. We have seen this in 8.05 already: (cid:19) (cid:18...
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π/a, it has some momentum-like properties. To see this, first look at the eigenvalue equation satisfied by un,k. (En,k − V (x))eikxun,k(x) = p2 2m eikxun,k(x) = eikx 2 (p + (cid:126)k) 2m un,k(x). Rearranging we have (cid:18) (p + (cid:126)k)2 2m (cid:19) + V | (cid:124) (cid:123)(cid:122) H k (cid:125) By the Hellmann-F...
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(cid:126)a1, (cid:126)a2, (cid:126)a3. The theory of crystallographic groups includes a classification of possible values of (cid:126)a1, (cid:126)a2, (cid:126)a3. The “re- ciprocal lattice vectors” b1, b2, b3 are defined by the relations (cid:126) (cid:126) (cid:126) a a This is equivalent to the matrix equation (cid:12...
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1(cid:105)(cid:104)n| − ∆|n − (cid:105)(cid:104)n|. 1 −∞ This can be rewritten in terms of the translation operator T = (cid:80) n= n |n + 1(cid:105)n as By Bloch’s theorem the eigenstates can be labeled by k, with H = E0I − ∆(T + T †). T |ψk(cid:105) = eika|ψk(cid:105) H|ψk(cid:105) = Ek|ψk(cid:105) From (67) we can c...
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one of these points corresponds to a delocalized state. Band structure. Now suppose each site can support multiple bound states, say with energies E0 < E1 < E2. Then tunneling opens up a band around each energy. If ∆ is small enough then there will be a band gap between these. Kronig-Penney model Another model that can...
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a n = 0 - π a k Figure 3: Band diagram for free electrons. eika for k and integer multiple of 2π . If we let φ := 2πa/L then there is a basis in which we have L 1 . . . 1 eiφ                      T =                      . . . eiφ e2iφ . . . and V = ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗      ...
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k which are vertical lines in Fig. 3. Treat V as a perturbation. For a typical value of k the kinetic energy of these points is well-separated and so V does not significantly mix the free states. However, near ±π/a, the kinetic energy term has a degeneracy, so there the addition of V will lead to a splitting, which will...
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duction band. A small electric field is enough to move this into the conduction band. The resulting material is called an “n-type semiconductor.” On the other hand, aluminum has one few electron than silicon, and so is an “acceptor.” Adding aluminum will increase the number of holes in the valence band and reduce the nu...
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ian. Either way the magnetic field turns out to enter the Hamiltonian 2m ((cid:126)p − q (cid:126)A((cid:126)x))2. We thus obtain the Pauli Hamiltonian replacing the kinetic energy term with 1 (neglecting an additional spin term): c H = (cid:16) (cid:126)p − 1 2m q c (cid:17) 2 (cid:126)A((cid:126)x) + qφ((cid:126)x). (...
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is gauge invariant if (cid:104)ψ(cid:48)|O(cid:48)|ψ(cid:48)(cid:105) = (cid:104)ψ|O|ψ(cid:105).) Remark: This gauge freedom will appear many times in the coming lectures and often leads to seemingly strange results. It is worth remembering that we are already used to a simpler form of f (t(cid:48))dt(cid:48)/(cid:126)...
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:16) 1 m pi − (cid:17) Ai q c =⇒ (cid:126)p = m ˙(cid:126)x + q (cid:126)A, c (79) obtaining a generalized momentum p(cid:126). Next we should remember that A((cid:126)x), φ((cid:126)x) depend on position so that (cid:126) p˙i = − ∂H ∂xi (cid:88) = − j (cid:16) q mc pj − Aj q c (cid:17) ∂Aj ∂xi − q = − ∂φ ∂xi (cid:88) ...
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k (cid:15)ijkBk (cid:88) x˙ j j (cid:18) ∂Aj ∂xi − ∂Ai ∂xj (cid:19) (cid:88) = j,k (cid:15)ijk ˙xjBk = ( ˙(cid:126)x × (cid:126)B)i. 22 (80) (81) (82) We can now rewrite (82) in vector notation as ¨m(cid:126)x = q     −∇(cid:126) φ − (cid:123)(cid:122) (cid:124) (cid:126)E  1 c ˙(cid:126)A    (cid:125) + q ˙ (...
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make things simple, namely (cid:126)A = (−By, 0, 0), φ = 0. (84) (cid:126) Another variant of the Landau gauge is A = (0, Bx, 0). On the pset you will also explore the (cid:126) “symmetric gauge,” defined to be A = (− 1 2 By, 1 Bx, 0). It is nontrivial to show that these result in the same physics, but this too will be ...
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fined l0 to be characteristic length scale of the harmonic oscillator, (cid:126) mωL (cid:113) (cid:126)c . = qB We can now diagonalize H from (85). The eigenstates are labeled by kx and ny, and have energies and wavefunctions given by Ekx,ny = (cid:126)ωL(ny + 1/2) ψ kx,ny = e ikxx 1 4 / − l0 φny (cid:18) − y l0 (cid:1...
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finite region in the plane, say with dimensions L × W . For simplicity we impose periodic boundary conditions. This means that kx = nx, nx ∈ Z. 2π W We also have the constraint that y0 should stay within the sample. (Assume that l0 (cid:28) L, W so we can ignore boundary effects.) Then 0 < y0 < L, or equivalently This im...
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ωL(n + 1/2) = (cid:126)eB 2mec (2n + 1) ≡ µBB(2n + 1) µB = e(cid:126) 2mec . Thus the induced dipole moment is µI = − ∂E B ∂ = −(2n + 1)µB. The minus sign means diamagnetism, corresponding to the fact that the circular orbits caused by a magnetic field will oppose that field. (We neglect here the contributions from spin....
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, according to (90). The number of fully filled LLs is j ≡ (cid:98)ν(cid:99), meaning the largest integer ≤ ν. Thus j ≤ ν < j + 1. The energy of the ground state is then E = j−1 (cid:88) n=0 (cid:124) (cid:123)(cid:122) filled levels j −1 (cid:88) (cid:32) n=0 = N (cid:126) ωL 2 D(cid:126)ωL(n + 1/2) + (N − jD)(cid:126) ...
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uities at integer values of ν. Also observe that for ν < 1 all electrons are in a single LL, so we observe the simple classical prediction of diamagnetism, which we refer to here as “Landau diamagnetism” even though it is the only part of this diagram where the Landau levels do not really play an important role. 26 Fi...
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cid:28) B. To see that this assumption is reasonable, note that in units where (cid:126) = 1, c = 1 an electric field of 1V /cm is equal to 2.4 · 10−4eV2 while a magnetic field of 1 gauss is equal to 6.9 · 10−2eV2. If the velocity is the one induced by the electric field then the magnetic field causes a drift in the positi...
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This has solution d dt (cid:126)v = A(cid:126)v + b = A((cid:126)v + A−1(cid:126)b) (cid:126) d dt ((cid:126)v + A−1(cid:126)b) = A((cid:126)v + A−1b). (cid:126) (cid:126)v(t) + A−1b = eAt((cid:126)v(0) + A−1b), (cid:126) (cid:126) 28 Figure 6: Motion of a charged particle subject to crossed electric and magnetic field...
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is moving (cid:126) (cid:126) at velocity (cid:126)v = vH xˆ. (Suppose for now that we do not know vH .) To leading order in v/c the E, B fields transform as (cid:126) (cid:48) E = E + (cid:126) × (cid:126) B = (E − (cid:126)B(cid:48) = B − × E = (B − (cid:126) (cid:126) v c v c v H c v H c B)yˆ E)zˆ (97a) (97b) (cid:12...
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.) We can also show that jy = −σH Ex. We then conclude that        =   jx jy 0 σH −σH 0    . Ex Ey (This matrix perspective is also useful when computing resistivity which is ρ = σ−1.) (cid:124) (cid:123)(cid:122) σ (cid:125) While the formula we have found is entirely classical, we can interpret in terms o...
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L = (cid:112)(cid:126)c/qB, we then obtain H|kx = p2 y 2m + 1 2 mω2 L(y − y0)2 − qEy y0 ≡ −(cid:96)2 0kx + 1 2 p2 y= 2m p2 1 y + 2m 2 = (cid:18) y2 − 2yy0 + y2 0 − mω2 L (cid:19) 2qE mω2 L y mω2 L (y − y0 − y1)2 − 1 2 mω2 L(2y0y1 + y2 1) y1 ≡ qE mω2 L = qE mωL qB mc = E B c ωL = (100) (101) vH ωL (102) This looks again...
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. We can think of the Landau levels as being “tilted” as follows. (cid:126)E = 0 (cid:126)E (cid:54)= 0 Note that earlier we broke the symmetry between x and y somewhat arbitrarily, and using a different gauge would’ve resulted in wavefunctions that were plane waves in the y direction, or even localized states. This fre...
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− vH , v ωL which is precisely the classical result we would expect from a particle undergoing oscillations with frequency ωL on top of a drift with velocity vH . We can now calculate the group velocity vg = ∂ω of the eigenstates of H. From (103) we ∂k immediately obtain vg = vH , suggesting that each eigenstates moves...
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y1 and 32 so we can write (cid:90) L 0 Sxdy = ωL W ω ≈ L W ωL= W ωL= W (cid:90) L |φn(y − y0 − y1)|2(y − y0) 0 (cid:90) ∞ 0 (cid:90) ∞ 0 y1 |φn(y − y0 − y1)|2(y − y0) |φn(y − y0 − y1)|2 (y − y0 − y1 (cid:125) (cid:123)(cid:122) (cid:124) (cid:124) (cid:125) (cid:123)(cid:122) even odd + y1) We conclude that (cid:126)j...
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LL then we rapidly fill a LL and the conductivity jumps up. This explains the plateaus in the conductivity but not why the conductivity should be an integer multiple of σ0. We will return to this point later after discussing the Aharonov-Bohm effect. 3.5 Aharonov-Bohm Effect To write down the Hamiltonian for a charged par...
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1 = (cid:126)x2. In this case the path becomes a loop and (cid:73) P = (cid:126) · (cid:126) Stokes’ thm (cid:90) = A dl ∇(cid:126) ( (cid:126) × A) · d(cid:126)a = (cid:90) loop surface surface (cid:126)B · d(cid:126)a = Φ. (113) In the last step we have defined Φ to be the magnetic flux through the surface. The last tw...
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particular it only moves through a region where B = 0. (cid:126) However, the vector potential cannot be zero outside the solenoid. Indeed consider a curve C enclosing the solenoid such as the dashed line in Fig. 8(b). The loop integral of A around C is (cid:126) (cid:73) C (cid:126) · (cid:126)A d(cid:96) = πR2B = Φ (...
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a nonzero vector potential but zero magnetic field. We will need a prescription for solving the Schr¨odinger equation in a region where B = 0 but (cid:126)A may be nonzero. To do so, let us fix a point (cid:126)x0, which in our scenario will be the location of the source S. Then given a curve C from (cid:126)x0 to (cid:1...
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)x0→(cid:126)x along C2 (cid:90) (cid:126)x→(cid:126)x0 along C2 (cid:126)A · d(cid:126)(cid:96) (cid:126)A · d(cid:126)(cid:96) (cid:126) A · (cid:126)d(cid:96) (cid:126)x0→(cid:126)x→(cid:126)x0 along C1 then C2 (cid:73) = q (cid:126)c q = Φ (cid:126)c (121a) (121b) (121c) (121d) In this last equation, Φ is the flux e...
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) (cid:126) eig((cid:126)x)∇g = (cid:126) = (cid:126)eig((cid:126)x) q (cid:126)c q (cid:126)Aeig((cid:126)x) c = (cid:126)A((cid:126)x) This means that when p(cid:126) − q (cid:126)A “commutes past” eig it turns into simply p(cid:126). In other words c This implies that (cid:17) (cid:16) q p(cid:126) − (cid:126)A c ei...
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and region B contains point B. This will allow us to use (122) separately in each region. Let gA and gB denote the functions g restricted to regions A and B respectively, and let ψ(0)(S → A → y) and ψ(0)(S → B → y) be the solutions of the free Schr¨odinger equation in those regions. This means that q gA(y) = (cid:126)c...
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2 ∝= cos2 (cid:18) 1 2 (cid:18) kay L − 2π (cid:19)(cid:19) . Φ Φ0 (131) Figure 11: Observed interference pattern when performing the two-slit experiment with a magnetic flux Φ enclosed between the two paths. The resulting interference pattern is depicted in Fig. 11. Observe that the experiment is only sensitive to the ...
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the Thouless charge pump. Suppose we have a 1-d system containing some number of electrons. As 0 ≤ t ≤ T suppose we adiabatically change the Hamiltonian from H(0) to H(T ), and suppose further that H(T ) = H(0). In other words we return to the Hamiltonian that we start with. Then the net flux of electrons from one end o...
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Now suppose we adiabatically increase Φ from 0 to Φ0. By the results of the pset, the final Hamiltonian is the same as the initial Hamiltonian (up to a physically unobservable gauge trans- form), and by the Thouless charge pump argument, this means that an integer number of electrons must have flowed from the inner loop ...
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6.776 High Speed Communication Circuits and Systems Lecture 8 Broadband Amplifiers, Continued Massachusetts Institute of Technology March 1, 2005 Copyright © 2005 by Hae-Seung Lee and Michael H. Perrott The Issue of Velocity Saturation (cid:131) We often assume that MOS current is a quadratic function of Vgs: (cid:1...
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. Perrott 0 0.4 0.6 0.8 1 1.2 (Volts) V gs 1.4 1.6 1.8 2 MIT OCW Example: Gm Versus Voltage for 0.18µDevice Vgs Id M1 W L = 1.8µ 0.18µ 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 ) s t l o V / s p m A i l l i m ( m g H.-S. Lee & M.H. Perrott 0 0.4 0.6 0.8 1 g versus V m gs 1.2 (Volts) V gs 1.4 1.6 1.8 2 MIT OCW Exam...
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directly with W (cid:131) This is independent of bias regime - We can therefore relate gmof devices with different widths given that they have the same current density H.-S. Lee & M.H. Perrott MIT OCW A Numerical Design Procedure for Resistor Amp – Step 1 Vdd αIbias Vin+ M5 R Vo- Ibias M1 M2 2Ibias M6 (cid:131) Two k...
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current density = 115 µA/µm (cid:131) Note that current density reduced as gain increases! - ft effectively decreased MIT OCW 600 700 Example (Continued) (cid:131) Knowledge of the current density allows us to design the amplifier - Recall - Free parameters are W, Ibias, and R (L assumed to be fixed) (cid:131)...
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