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0 0 1 0 sum = 1 0 1 sum = 1 0 1 1 1 sum = 3 Decomposition of Error The “ideal” model generates data D. A “learned” model is learned from D. Once learned, model M is fixed. After learning, I and M are conditionally independent given D. D f y I f’ M y^ Decomposition of Error A and B binary (y-hat and y...
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P ( 2 )2 ⎤ ⎥⎦ = Decomposition of Error = 1 2 + 1 2 − ∑ ) B P A P ( ( ) + ∑ P ( AB ) −∑ P ( AB ) + 1 ∑ 2 ( A P 2) − 1 ∑ 2 ( A P (B P 2) − 1 ∑ 2 (B P 2) = 2) + 1 ∑ 2 1 [∑ A P ( 2 2 ) − ∑ AB P ( ) + ∑ B P ( 2 ) ]= − = 1[∑ ) B P A P ( ( ) − ∑ AB P ( )] + 1 [1 − ∑ A P ( 2 2 ...
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MIT OpenCourseWare http://ocw.mit.edu 3.23 Electrical, Optical, and Magnetic Properties of Materials Fall 2007 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. 3.23 Fall 2007 – Lecture 9 BAND STRUCTURE 3.23 Electronic, Optical and Magnetic Properties of Materia...
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⎜ C ⎟ ⎜ ⎟ C ⎜ q G ⎟ + ⎟⎟ ⎜⎜ ⎝ CCq G+ 2 ⎠ ⎝ ⎠ q 3.23 Electronic, Optical and Magnetic Properties of Materials - Nicola Marzari (MIT, Fall 2007) 4 Free electron dispersions, 1-d 3.23 Electronic, Optical and Magnetic Properties of Materials - Nicola Marzari (MIT, Fall 2007) Band Structures: Free Electron Gas...
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V −G + q 2 h 2m − CC −22G ⎛ q G ⎞⎞ ⎛ ⎜ ⎟ ⎜ Cq G ⎟ = E ⎜ ⎟ Cq ⎜ ⎟ ⎜ C + q G ⎟ ⎜ ⎟ ⎝ C + ⎠ q G 2 3.23 Electronic, Optical and Magnetic Properties of Materials - Nicola Marzari (MIT, Fall 2007) Band Edge 3.23 Electronic, Optical and Magnetic Properties of Materials - Nicola Marzari (MIT, Fall 2007) 7 ...
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n k, fn k, r r r r ( ) Ψ n k, The Fermi surface 3.23 Electronic, Optical and Magnetic Properties of Materials - Nicola Marzari (MIT, Fall 2007) Images from the Fermi Surface Database. Used with permission. Please see: http://www.phys.ufl.edu/fermisurface/jpg/K.jpg, http://www.phys.ufl.edu/fermisurface/jpg/Cu.jpg...
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Adjustable Voltage Power Supply +15V 0.1uf 270Ω 1N758 10v 10K V+ V- . Vo 1uf 6.091 IAP 2008 Lecture 4 Appendix p1 555 Block Diagram ready Threshold Control Voltage Trigger VCC 8 6 5 2 5k 5k 5k + Comp A _ + Comp B _ R S Flip Flop Q Inhibit/ Reset 7 3 Discharge Output +15 1 Gnd 4 Reset R C 1k Figure by MIT OpenCourseWar...
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Key Concepts for this section 1: Lorentz force law, Field, Maxwell’s equation 2: Ion Transport, Nernst-Planck equation 3: (Quasi)electrostatics, potential function, 4: Laplace’s equation, Uniqueness 5: Debye layer, electroneutrality Goals of Part II: (1) Understand when and why electromagnetic (E and B) interaction ...
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1 2 2.5 3 1.5 xκ Exact solution Φ = ( ) x 2 RT zF ln + 1 e − κ x tanh − 1 e − κ x tanh ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ ⎤Φ⎛ zF ⎞ 0 ⎟ ⎜ ⎥ 4 RT ⎠ ⎝ ⎥ Φ⎛ zF ⎞ ⎥ 0 ⎟ ⎜ ⎥ 4 RT ⎠ ⎝ ⎦ κ , ⎛ = ⎜ ⎝ 2 2 2 z F c 0 ε RT 1/ 2 ⎞ ⎟ ⎠ Debye-Huckel approximation Φ = Φ ( ) x e κ− x 0 When zF Φ << 0 RT 8 6 4 2 ( ) c x c 0 Φ zF RT 0 = 2 c- (counterion) c+ (...
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RT zF ln ⎛ ⎜ ⎝ c 2 c 1 ⎞ ⎟ ⎠ Nernst Equilibrium potential Diffusion of charged particles -> generate electric field -> stops diffusion of ions
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6.02 Fall 2012 Lecture #14 • Spectral content via the DTFT 6.02 Fall 2012 Lecture 14 Slide #1   Demo: “Deconvolving” Output of Channel with Echo x[n] Channel, h1[.] y[n] z[n] Receiver filter, h2[.] Suppose channel is LTI with h 1[n]=δ[n]+0.8δ[n-1] H1(Ω) = ?? = ∑h1[m]e − jΩm m So: = 1+ 0.8e–jΩ = 1 ...
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012 Lecture 14 Slide #3 DT Fourier Transform (DTFT) for Spectral Representation of General x[n] If we can write h[n] = ∫ H (Ω)e 1 2π<2π> then we can write x[n] = ∫ X(Ω)e 1 2π<2π> jΩn dΩ where H (Ω) = ∑h[m]e − jΩm Any contiguous interval of length 2 m jΩn dΩ where X(Ω) = ∑ x[m]e − jΩm m This Fo...
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π<2π> jΩn dΩ Y (Ω) = H (Ω)X(Ω) Compare with y[n]=(h*x)[n] Again, convolution in time has mapped to multiplication in frequency 6.02 Fall 2012 Lecture 14 Slide #7 Magnitude and Angle Y (Ω) = H (Ω)X(Ω) | Y (Ω) |= |H (Ω) |. | X(Ω) | and < Y (Ω) = < H (Ω)+ < X(Ω) 6.02 Fall 2012 Lecture 14 Slide #...
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Slide #10 Frequency (Hz) 6.02 Fall 2012 © PC Magazine. All rights reserved. This content is excluded from our Creative Commons license. For more information, see http://ocw.mit.edu/fairuse. Lecture 14 Slide #11 Spectral Content of Various Sounds Cymbal Crash Bass Drum Bass Guitar Human Voice Snare Drum Guita...
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] = ∫ X(Ω)e 1 2π<2π> jΩn dΩ ∫ 1⋅ e = = Ω C C 1 2π−Ω sin(Ω πn C / π = Ω Cn) 6.02 Fall 2012 jΩn dΩ , n ≠ 0 , n = 0 DT “sinc” function (extends to ±∞ in time, (cid:3) falls off only as 1/n) Lecture 14 Slide #14 x[n] and X(ΩΩ) 6.02 Fall 2012 Lecture 14 Slide #15 X(Ω) and x[n] 6.02 Fall 201...
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2 Fall 2012 P = 1024, P2 = 1,048,576, P logP ≈ 10,240 Lecture 14 Slide #17 Where do the ΩΩ k live? e.g., for P=6 (even) Ω 3(cid:3) Ω 2(cid:3) Ω 1(cid:3) – exp(jΩ2)(cid:3) exp(jΩ3)(cid:3) = exp(jΩ 3) –1 Ω0(cid:3) 0 j . Ω1(cid:3) Ω2(cid:3) Ω3(cid:3)  exp(jΩ1)(cid:3) exp(jΩ0)(c...
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which is 1/7 of the way from 0 to the positive end of the frequency axis at  (so k approximately 100/7 or 14 in the figure). And that indeed is the neighborhood of where the Fourier coefficients drop off significantly in magnitude. • There are also lower-frequency components corresponding to the fact that the 1 ...
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MIT OpenCourseWare http://ocw.mit.edu 18.727 Topics in Algebraic Geometry: Algebraic Surfaces Spring 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. ALGEBRAIC SURFACES, LECTURE 10 LECTURES: ABHINAV KUMAR Recall that we had left to show that there are no...
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x ∈/ Y . This is because if at x, π : X → X the blowup of X at x, x ∈ Y , then for D ∈ L a curve singular ˜ ˜ then ˜D is a nonsingular rational curve → P1 . After d a fiber of φ = φ ◦ π : X an → P1 is a morphism and further (at most K 2) blowups, we get an X � s.t. φ� : X � one fiber of φ� is a smooth, rational c...
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TX = (Ω1 tangent sheaf. X/k )∨ the 1 2 LECTURES: ABHINAV KUMAR Proof. ∃ a nonsingular elliptic curve D ∈ exact sequence |−K| by the above lemma. The short (1) 0 → OX ((n − 1)D) ⊗ TX → OX (nD) ⊗ TX → OX (nD) ⊗ TX ⊗ OD → 0 gives the long exact sequence in cohomology 2 X (nD) ⊗ X ⊗ OD...
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used that TD = ωD mology gives (5) H 1(O X (nD) ⊗ OD) → H (OX (nD) ⊗ TX ⊗ OD) → H 1(OX (nD) ⊗ N ) 1 D2 > 0, so we get H 1(OX (nD) ⊗ OD) = 0 = H 1(OX (nD) ⊗ N ) as desired, since OX (nD) ⊗ OD and OX (nD) ⊗ N are sheaves of large positive degree on D for � n >> 0. We now return to the proof of the proposition for p...
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= H 2(X, OX ) = 0 is an isomorphism. By Nakayama’s lemma, we get that R2f∗OU = 0, and sim­ ilarly for R1 . Thus, H 1(X �, OX � ) = H 2(X �, OX � ) = 0. See Mumford’s Abelian Varieties or Chapters on Algebraic Surfaces for more details. Now, let K �� be an algebraic closure of K � and Ki of K � inside K ��. Let X �� ...
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fiber of fi is isomorphic to Xi and since Ki ⊃ K is finite, Bi is a DVR and Vi X �� is is an inductive system. By EGA and general nonsense, lim Pic Xi → Pic an isomorphism. −→ Lemma 3. There is a group isomorphism b : Pic Xi following: for L an Li|Xi = L, and we set b([L]) = [Li|X ]. invertible OXi -module, ∃ an in...
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→ X from a rational surface. Corollary 1. In characteristic zero, a unirational surface is rational. Note that this is not true in characteristic > 0, e.g. the Zariski surfaces zp = f (x, y). Proof. Given f : Y X where Y is rational, we have q(Y ) = p2(Y ) = 0. f is separable, so it induces an injective map H 0(X, ...
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Picard group of line bundles up to isomorphism (or divisors up to linear equivalence). Set Pic 0(X) to be those al­ gebraically equivalent to zero: we say that L1 and L2 are algebraically equivalent (i.e. L1 ≈ L2) if ∃ a connected scheme T , points t1, t2 ∈ T , and an invertible ∼= L1 and LX t2 1 ⊗ OX T -module L s...
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ctors Pic X , Pic 0 and Pic 0 X respectively. X are representable by schemes called Pic X This means that we have a natural equivalence Pic X (T ) = Hom k−Sch(T, Pic X ), and T → Pic X corresponds uniquely to a line bundle on X × T up to T ­ Pic X corresponds to a line bundle isomorphism. The identity map Pic X →...
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Mechanical Issues January 4th, 2005 Aaron Sokoloski Agenda „ The Maslab Workshop … Raw Materials … Other Materials … Fasteners … Tools … Safety & Maintenance „ Mechanical issues … Motors … Techniques … Design Principles … Other resources The Maslab Workshop „ Goal: Be able to build a simple robot with the tool...
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Raw Materials „ Prototyping foam (2” blue foam) … Large sheets available … Good for bulky parts … Cuts easily with hot knife … Also can be sculpted with hot knife for interesting / irregular shapes Other materials „ Wooden dowels „ Hollow metal tubing „ Springs „ PVC pipe „ Foam pipe insulation „ Gears „ Oth...
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medium pressure will cut … No metal allowed! Tools „ Hacksaws, wood saw … Cut wood, PVC, cardboard „ Pipe cutter (small red gadget) … Cuts brass tubing – turn and tighten gradually „ Rotary cutting tool … Quick, but inaccurate Tools „ Mitre saw … More accurate wood cuts, any angle … Use clamps for best result „ D...
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motors • Extra high speed and extra high torque motors available • Servos can be modified for larger range of motion Techniques • Many possibilities with wood and bolts Simple Rotating Gripper Techniques: Mounting IR and Servos IR range finder Servomotor Techniques: Metal bending To bend without the brake, mak...
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Lecture 3 Fast Fourier Transform 6.046J Spring 2015 Lecture 3: Divide and Conquer: Fast Fourier Transform • Polynomial Operations vs. Representations • Divide and Conquer Algorithm • Collapsing Samples / Roots of Unity • FFT, IFFT, and Polynomial Multiplication Polynomial operations and representation A poly...
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C(x) = A(x)·B(x) (∀x). Then ck = j=0 bk−j for 0 ≤ k ≤ 2(n−1), because the degree of the resulting polynomial is twice that of A or B. This multiplication is then equivalent to a convolution of the vectors A and reverse(B). The convolution is the inner product of all relative shifts, an operation also useful for sm...
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each xi is distinct. These samples uniquely determine a degree n − 1 polynomial A, according to the Lagrange and Fundamental Theorem of Algebra. Addition and multiplication can be computed by adding and multiplying the yi terms, assuming that the xi’s match. However, evaluation requires interpolation. The runtimes...
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a n−1 ⎤ ⎡ ⎢ ⎥ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ = ⎥ ⎢ ⎢ ⎥ ⎦ ⎣ ⎤ 1 y0 ⎥ y ⎥ ⎥ ⎥ y ⎥ 2 ⎥ ⎥ . . ⎦ . yn−1 where V is the Vandermonde matrix with entries vjk = xk j . Then we can convert between coefficients and samples using the matrix vector product V · A, which is equivalent to evaluation. This takes O(n2). Similarly, we ca...
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3, a5, . . . k ) 3 Lecture 3 Fast Fourier Transform 6.046J Spring 2015 2. Recursively conquer Aeven(y) for y ∈ X 2 and Aodd(y) for y ∈ X 2, where X 2 = {x2 | x ∈ X}. 3. Combine the terms. A(x) = Aeven(x2) ...
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= 1. These points are uniformly spaced around the unit circle in the complex plane (including 1). These points are of the form (cos θ, sin θ) = cos θ +i sin θ = eiθ by Euler’s Formula, for θ = 0, 1 τ, 2 τ, . . . , n−1 τ (where τ = 2π). n n n The nth roots of unity where n = 2£ form a collapsing set, because (e iθ...
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1965. Gauss used the algorithm to determine periodic asteroid orbits, while Cooley and Turkey used it to detect Soviet nuclear tests from offshore readings. A practical implementation of FFT is FFTW, which was described by Frigo and Johnson at MIT. The algorithm is often implemented directly in hardware, for fixed n...
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(e i(j−k)τ /n)m m=0 (e iτ (j−k)/n)n − 1 = eiτ (j−k)/n − 1 = 0 because eiτ = 1. Thus V −1 = n 1 V¯ , because V · V¯ = nI. This claim says that the Inverse Discrete Fourier Transform is equivalent to the Discrete Fourier Transform, but changing xk from e to its complex conjugate e−ikτ /n, and dividing the result...
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vector) represents the phase shift of that signal. For ex­ arg(ak ample, this perspective is particularly useful for audio processing, as used by Adobe Audition, Audacity, etc.: • High-pass filters zero out high frequencies • Low-pass filters zero out low frequencies • Pitch shifts shift the frequency vector • Used ...
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Lecture 1 The Hamiltonian approach to classical mechanics. Analysis of a simple oscillator. Program: 1. Hamiltonian approach to classical mechanics. 2. Vibrations of an electron in a lattice: simple oscillator Questions you should be able to answer by the end of today’s lecture: 1. How to construct a Hamiltonian...
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x  p2 2m  V x  Why is this function important? – Let’s take a look how the Hamiltonian changes with respect to momentum and position:  v  dx dt H p  p m H x    dV x dx       F x m d 2 x dt 2    d  dt  m  dx    dt  dp dt The resulting pair of equations is referred to as ...
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�  pi dt  dpi  H  xi dt    Let’s go back to our initial example. Example I: Particle falling under gravity: The Hamiltonian in this case is: H x, p   p2 2m  V x    2 p 2m  mgx  Then the Hamilton’s equations are:  H dx   p   H  x  dx p   dt m   dp   dt ...
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: Electron vibrations with respect to the lattice ⇒ simple harmonic oscillator In insulating crystalline materials electrons are fairly closely bound to the ions in the lattice. However electrons are not stationary – they have thermal velocity, which allows them to move away from the ions, while the Coulombic attrac...
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d 2 dx p  dt m  K x  l   d 2 x 1 dp  dt 2 m dt   K m x  l 2  x  l  0,  K m The last equation is simply an equation for a simple harmonic oscillator. The solution for this equation is: x  l  Aeit  Beit We can find coefficients A and B from the initial conditions. Let’s say at time...
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w.mit.edu 3.024 Electronic, Optical and Magnetic Properties of Materials Spring 2013 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.
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Key Concepts for section IV (Electrokinetics and Forces) 1: Debye layer, Zeta potential, Electrokinetics 2: Electrophoresis, Electroosmosis 3: Dielectrophoresis 4: Inter-Debye layer force, Van-Der Waals force 5: Coupled systems, Scaling, Dimensionless Numbers Goals of Part IV: (1) Understand electrokinetic phenomena...
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E0 + + + + + + + - - - - - - - E0 - - - - - - + + + + + + - + Positive DEP )σ σ> ( 0 Negative DEP )σ σ< ( 0 Positive / negative DEP, etc. pr pr + - - - + + + + + - - - - + + + - - - + r E Particle moves toward the high field region. r E Particle moves away from the high field region. DEP force is independent of the ...
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cculation Courtesy of J. T. Groves. Used with permission. Source: Figure 2b in Baksh, M. M., M. Jaros, and J. T. Groves. at Membrane Surfaces through Colloid Phase Transitions." (cid:10) "Detection of Molecular Interactions Nature 427 (January 8, 2004): 139-141. Measurement of W in different salt concentration (...
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2.60 2.43 0.067 0.069 0.69 NaCl KCl KNO3 K2SO4 K2Cr2O7 K2 oxalate K3[Fe(CN)6] 43.5 46 60 0.30 0.63 0.69 0.08 Source: “Introduction to Colloid and Surface Chemistry” By Duncan J. Shaw (Butterworth Heinemann) Figure by MIT OCW. Electrostatic interaction within electrolyte solution Interactions and forces in micro / nano...
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Probabilistic models, random variables 6.011, Spring 2018 Lec 12 1 Sample space and events Sample space ° A specific outcome c Collection of outcomes (event) 2 Random variable ° c Real line X(c) 3 Joint pdf © The MathWorks, Inc. All rights reserved. This content is excluded from our Creative Commo...
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MIT OpenCourseWare http://ocw.mit.edu 6.641 Electromagnetic Fields, Forces, and Motion Spring 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. 6.641, Electromagnetic Fields, Forces, and Motion Prof. Markus Zahn Lecture 3: Electroquasistatic and Magnetoquasist...
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su ⎪+εE0 ⎩ z = d z = 0 K 2 b π + π b2 dσsu = 0 ⇒ K = − dt r r b dσsu = − 2 dt b ε 2 dE 0 dt (cid:118)∫ i H ds = C ∂ ( E) i da ⇒ H 2 r ∫ ∂t φ π = π 2 r ε ε S dE 0 dt ⇒ H φ = r dE 0 ε dt 2 (cid:118)∫ i E ds = − µ ∫ ∂ H t ∂ C S i da Courtesy of Hermann A. Haus and James R. Melcher. Used with pe...
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ε 2 ω µb 4 (cid:19) 1 fλ = c = 1 εµ ω 2π λ = c ⇒ ω = 2ε 2 2 c ω µ b 4 π λ ⇒ = 2 π λ2 2 b (cid:19) 1 ⇒ b (cid:19) λ π f=1 MHz in free space ⇒ λ = × 3 10 8 6 10 = 300 m If b (cid:19) 100 m EQS approximation is valid. II. Conditions for Magnetoquasistatic Fields Courtesy of Hermann A. Haus and James ...
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Lecture 3 Page 4 of 12 III. Boundary Conditions 1. Gauss’ Continuity Condition (cid:118)∫ ε0E i da = ∫σsdS ⇒ ε0 (E2n - E1n ) dS = σsdS S S ε (E - E ) = σ ⇒ n i ε (E - E ) = σ 1n 2n 2 0 s s ⎤ 1 ⎦ ⎡ ⎣ 0 2. Continuity of Tangential E (cid:118)∫ E i ds = (E1t - E2t ) dl = 0 ⇒ E1t - E2t = 0 C n× E 1...
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V = 0 i V S n i ⎡ J a - Jb ⎣ ⎤ + ⎦ ∂ ∂t σ s = 0 6.641, Electromagnetic Fields, Forces, and Motion Prof. Markus Zahn Lecture 3 Page 6 of 12 6. Electric Field from a Sheet of Surface Charge a. Electric Field from a Line Charge 6.641, Electromagnetic Fields, Forces, and Motion Prof. Markus Zahn Lectu...
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= y +∞ ∫ x = −∞ dE y = = +∞ σ0y 2πε0 ∫ x2 + y2 dx x = −∞ σ0y 1 2πε0 y tan−1 x +∞ y −∞ 6.641, Electromagnetic Fields, Forces, and Motion Prof. Markus Zahn Lecture 3 Page 9 of 12 ⎧ σ0 ⎪ ⎪2ε0 = ⎨ σ0 ⎪− ⎪ 2ε ⎩ 0 y > 0 y < 0 Checking Boundary condition at y=0 E yy ( = 0 ) − E (y = 0 ) = y + − ...
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I π2 r _ i = − sin φ _ φ i + cos x _ φ i y Thus from 2 symmetrically located line currents dH = x dI π( 2 x2 + y2 2 )1 (− sin φ) 6.641, Electromagnetic Fields, Forces, and Motion Prof. Markus Zahn Lecture 3 Page 11 of 12 = − K dx y 0 2π x2 + y2 H = − x K0 2π y +∞ dx ∫ x2 + y2 x = −∞ = − 0 K...
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MIT Quantum Theory Notes Supplementary Notes for MIT’s Quantum Theory Sequence (cid:13)c R. L. Jaffe 1992-2006 February 8, 2007 Canonical Quantization and Application to the Quantum Mechanics of a Charged Particle in a Magnetic Field (cid:13)c R. L. Jaffe MIT Quantum Theory Notes Contents 1 Introduction 2 3 2 Canonical ...
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of motion . . . . . . . . 15 3.4 Physical Interpretation of Landau Levels . . . . . . . . . . . . 17 . . . . . . . . . 17 3.4.1 The location and size of Landau levels 3.4.2 A more careful look at translation invariance . . . . . . 19 4 The Aharonov Bohm Effect 23 5 Integer Quantum Hall Effect 28 5.1 The ordinary Hall effe...
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a particular point of view, resemble the Hamiltonian formulation of classical mechanics. This similarity has led to a program for guessing the quantum description of systems with classical Hamiltonian formulations. The program is known as “canonical quantization” because it makes use of the “canonical” i.e. Hamiltonian...
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)c R. L. Jaffe MIT Quantum Theory Notes 4 Section 1 describes the method of canonical quantization. We studied this in 8.05. I will not lecture on the material of Section 1; I am providing it here for you to read and review. (cid:13)c R. L. Jaffe, 1998 2 Canonical Quantization 2.1 The canonical method There is a haunting...
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j = = − ∂H ∂xj ∂H pj = ∂pj m ∂V ∂xj (4) Equations (4) look exactly like Hamilton’s equations. When the two lines are combined, we obtain Newton’s second law, mx¨j = −∂V /∂xj . Of course, we have to remember that the content of these equations is very different in quantum mechanics than in classical mechanics: operator m...
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pj, and H. The time development of an arbitrary dynamical (cid:13)c R. L. Jaffe MIT Quantum Theory Notes 6 variable can also be written simply in terms of Poission Brackets. For sim- plicity we consider dynamical variables that do not depend explicitly on the time.1 Then N ˙A ≡ dA dt = ∂ A ∂xj x˙ j + ∂ A ∂pj p˙j (cid:2...
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elegance 1A dynamical variable may or may not depend explicitly upon the time. Any dynamical variable will depend implicitly on the time through the variables xj and pj. Explicit time dependence arises when some agent external to the system varies explicitly with the time. An example is the time dependence of the magne...
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∂s˙ = ms˙; the Hamiltonian is H = p2/2m; and the quantum theory is defined by the operators sˆ, pˆ and H = pˆ2/2m. In short the bead behaves like a free particle on a line. It experiences no forces due to the curving of the wire. 2 mv2 = 1 We can dress up this problem a little by adding gravity. Suppose the wire is plac...
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L = m1~r1 + m2~r2 − V (~r1 − ~r2). ˙ 2 ˙ 2 1 2 1 2 (11) We could quantize this canonically in this form and obtain a two particle Schroedinger equation. Instead let’s make the transformation to relative and center of mass coordinates at the classical level and quantize from there. We define ~r = ~r1 − ~r2 ~R = 1 M (m1~r...
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just like a single particle of mass µ moving in a potential V (~r). These examples may seem overly elementary. If this were all canonical quantization was good for, it would not be necessary for us to spend much time on it. Moreover there are many mistakes to be made by applying the canonical method too naively (as we ...
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de- grees of freedom that classical point particles don’t have. Using guesswork and experimental information, physicists invented operators that describe the behavior of this innately quantum mechanical variable. Nature has forced us to postulate new quantum variables to augment the classical description of a system. T...
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xp (note xp = px) occurred in a Hamiltonian, we could rather confidently replace it by the hermitian form 1 2(xp + px). Sometimes, hermiticity is not enough. General conservation laws, like con- servation of momentum or angular momentum help. If all else fails, it is necessary to leave the ambiguity (parameterized by th...
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we must at least mention a complicated and rich variation of the canonical quantization method that has become an important focus for re- search in recent years. Sometimes the degrees of freedom of complicated systems are not all independent. For example, a particle may be constrained to move on a specified surface (in ...
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the importance of such problems and developed a method to handle quantization under constraints. Other powerful yet practical methods were developed by L. D. Faddeev and V. N. Popov in the 1960’s. Quantum ver- sions of electrodynamics, chromodynamics (the theory of the interactions of quarks and gluons) and gravity all...
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problem quantum mechanically we must first find the classical Hamiltonian that de- scribes the system. Then we will be able to go over to the quantum domain. This requires us to introduce the concept of a vector potential , A, which determines the magnetic field by its curl , ~ ~B = ∇ × ~A, ~ (17) ~ ~ in somewhat the same...
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of the Euler-Lagrange equations, ∂L ∂xj = mx¨j = e c e c x˙ k ∂Ak ∂xj 3 k=1 X 3 xk ˙ k= 1 X (cid:18) ∂Ak ∂xj − ∂Aj ∂ k x (cid:19) (20 ) ~ Notice that A depends on the particle’s position ~x(t) so that when we dif- 3 k=1 x˙ k∂Aj/∂xk by the chain rule. The two ferentiate by t we obtain Aj = ~ terms involving derivatives ...
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1) . (23) ∇~ × ~ A = B eˆ , however other choices such as A = It is easy to verify that 0 3 B0x1eˆ2 would do just as well. They all describe the same magnetic field. In classical mechanics we know that physics depends only on B. The same is true here in the Landau problem, however at this moment we have only the ~ Hamil...
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other components of momentum, however, are not conserved: [pj, H] = 0 (25) 6 (cid:13)c R. L. Jaffe MIT Quantum Theory Notes 16 for j = 1, 2. This comes as a surprise: since the magnetic field is uniform in space we would expect the system to be translation invariant, and therefore to find momentum conserved. On the other...
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, though unusual interpretation. Eigenstates of H and L3 are labeled by the number of quanta of + and − excitation, N ≡ a† a N |n+, n i = n |n , ± ± ± + n i − ± ± − (29) ± quanta carry ±1 unit of angular momentum, but only n - quanta con- tribute to the energy. So the energy eigenstates, known as Landau levels, are − ...
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Hamiltonian, eq. (24), on the other hand, singles out a specific origin of coordinates about which the harmonic oscillator potential is centered. In this section we explore the meaning of the Landau degeneracy and (eventually) explain how the translational invariance so obvious at the classical level, is manifest in the...
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i: (32) hα|x˙ k|αi = = 1 i~ 1 i~ hα|[xk, H]|αi hα|xkE − Exk|αi = 0. (33) so the state centered at (x1, x2) does not wander away with time. On the other hand the components of the momenta p1, and p2 do not vanish in the state |αi, hp1i = hp2i = ~ `0 ~ `0 Im α Re α, (34) but this has no direct physical interpretation bec...
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− − 3.4.2 A more careful look at translation invariance In this section we take a more sophisticated approach to the Landau problem that will clarify both the degeneracy of the Landau levels and the way in which the system manages to respect homogeneity in the xy-plane. We begin with a canonical transformation, trading...
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commutator of P1 and P2 is a c-number, perhaps we can con- struct functions of them which do commute. The simplest possibility is to look at finite translations of the form, T1(b1) ≡ e− T2(b ) ≡ e− 2 P b i 1 1 ~ P b i 2 2 ~ . According to our study of translations, these operators should translate x1 and x2 by b1 and b2...
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nite translations fail to commute only by virtue of this multiplicative factor of unit magnitude. If (and only if) we choose the parameters b1 and b2 so that the phase is a multiple of 2π then the translations commute. This condition is 2mωb1b2/~ = 2πN for an integer N. This defines a rectangle of area b1b2 in the xy-pl...
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) This state is normalized to unity because the operators T are unitary. It has ~= b. What does it look like in terms of the original energy E = 1~ basis |n+, n i? To answer this we must express the translations in terms of the {ak}. Using the Baker-Hausdorf Theorem, a short calculation gives: 2 ωL and h~xi − |~b, 0, 0...
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the Landau problem as towers of nearly 2)~ωL situated in unit cells on a localized energy eigenstates with E = (n + 1 grid labeled by any pair of distances b1 and b2 satisfying (45). − Eigenstates of the finite translations Another, more conventional, ap- proach is to study eigenstates of our maximal set of commuting op...
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3 just like a plane wave, exp i(k1x1 + k2x2), would behave. The difference, of course is that ψ has this simple behavior only for the special translations we have discovered, not for an arbitrary translation, and as a consequence, the “momenta” (k1, k2) are not conserved. States like these arise in many other situations...
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). In the classical domain A and the electrostatic potential, φ can be regarded as merely useful, but inessential, abstractions. ~ ~ ~ ~ In the quantum theory H rather than mx¨ is fundamental, so the possibil- ity exists that physics depends on A. For the case we have studied in detail — motion in a constant magnetic fi...
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85. (cid:13)c R. L. Jaffe MIT Quantum Theory Notes 25 Stokes’ theorem and the defining relation for A, B = ∇ × A, give ~ ~ ~ ~ ~ · ~ dl A = d2S eˆ3 · × ~A ∇~ C I = Z Z Z Z d2S eˆ3 · B = πr2B0, ~ (53) ~ so A cannot vanish everywhere on the circle C. In fact symmetry requires that A point in the azimuthal, φ, direction, s...
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1, follows curve C and ends at a point ~r. Note that ∇~ g(~r, C) = A(~r) ~ e ~c independent of the curve C. Now factor the phase g out of the wavefunction, ψ(~r, t) = exp(ig) χ(~r, t), (58) (55) (56) (57) (cid:13)c R. L. Jaffe MIT Quantum Theory Notes 26 and substitute into eq. (55). The result is that χ obeys the free...
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(~r, t2) and χ2(~r, t2) localized in the vicinity of the point P2. Whatever other (A-independent) relative phase may have accumulated by the time the wave packets have reached P2, there is an A-dependent relative phase, ~ ~ ψ(~r, t2) = exp ~l · ~ d A χ1(~r, t2) + exp ie ~c (cid:20) C1 Z (cid:21) (cid:26) ie ~ c C¯ I (c...
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magnetic flux enclosed by C 0, which is the same as that enclosed by ¯C. Thus it does not matter that the quantum particle cannot follow a sharp trajectory. The Aharonov-Bohm phase is a global property of the motion, not ¯ a property of the particle’s exact path. A similar argument shows that g(C) does not depend on the...
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Griffiths for a description of how the Aharonov-Bohm effect leads to shifts in the energy levels for a “bead on a loop of string” if the string is everywhere in a region with B = 0 but encircles a flux carring solenoid. ~ 6 (cid:13)c R. L. Jaffe MIT Quantum Theory Notes 28 5 Integer Quantum Hall Effect The Hall Effect is an ...
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material in which the electrons propagate. As usual in condensed matter physics, after solving an idealized problem I will have to return to the real world of actual materials and ex- plain why the results of the idealized analysis survive unscathed. Much of my presentation relies heavily on the introductory sections o...
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the local analog of V = IR), in which case ρ0 = 1/σ0. It is also useful to think of the resistivity (and conductivity) as a matrix relating the vector ~j to the vector E. In this simple case, the matrix is diagonal, (64) ~ ~ ~ ρ ρ = 0 0 0 ρ0 . (65) When the magnetic field is turned on, the mobile charges respond to the ...
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elements of the matrix ρ depend on the sign of q. In contrast, the normal resistivity depends only on q2. This is the stuff of undergraduate physics labs. This Hall resistivity describes the behavior of realistic conductors over a wide range of conditions. Surprisingly the behavior of a system of electrons described by ...
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The Hamiltonian is given by, ~ 0 H = 1 2m (−i~ (cid:20) ∂ ∂x − eB y 0 c )2 − 2~ 2 ∂ ∂y2 (cid:21) + eE0y. (70) As usual, it is convenient to introduce some scaled variables. We define the Larmor frequency ωL = eB0/mc as usual, and introduce the natural length ~c/eB0. If we now scale the dimensional scale of the Landau pr...
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η) = eikξϕ(η, k) with ~ω 2 ~ωL 2 = (cid:2) L η + 2αη + (k − η)2 p2 ϕ(η, k) (cid:3) η + (η − k + α)2 + 2αk − α2 ϕ(η, k). p2 (74) This is simply a one-dimensional harmonic oscillator center(cid:3)ed at η = k − α with eigenenergies shifted by ~ωL(αk − α2/2), (cid:2) α2 1 En(k) = ~ωL(n + ) + ~ωL(αk − ) 2 2 (75) The associa...
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B0, passing through the region 0 ≤ x ≤ W and −L/2 ≤ y ≤ L/2. We can relate this degeneracy to the allowed values of k, which labels the degeneracy in eq. (75). From (76) with ~ 6 6 (cid:13)c R. L. Jaffe MIT Quantum Theory Notes 33 α = 0 we see that the state with “momentum” k is localized at η = k or y = k`0. Thus the ...
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: ψ(x = 0, y) = ψ(x = W, y), and use the y-dependence to find the range of k. For α = 0, the eigenstates are centered at η = k − α. This translates into the quantization rule k = (2π`0/W )p with − LW αW + 4π`2 2π`0 0 ≤ p ≤ LW αW + 4π`2 2π`0 0 . (81) The only effect of the electric field is to shift the allowed values of t...
https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2016/3e3e804176b9c697b16abd811b5dfa89_MIT8_06S16_Supplementry.pdf
`0 = eE0L, which is just the change in the classical electrostatic energy of the electron over the length of the conductor. As long as the magnetic field is strong we can assume that the now-smeared-out Landau levels remain well separated from one another. • The other effect of the electric field is to shift the average “...
https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2016/3e3e804176b9c697b16abd811b5dfa89_MIT8_06S16_Supplementry.pdf
(n, k) = − e~k m`0W . (86) where the subscript H on I reminds us that this is the Hall current. Note that the contribution of the Aψ∗ψ term in the current density (85) is odd in y and so vanishes once we integrate over y. ~ Now let us suppose that all states in a given Landau band are filled and calculate the associated...
https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2016/3e3e804176b9c697b16abd811b5dfa89_MIT8_06S16_Supplementry.pdf
2.5 A description of the integer quantum Hall effect A VL B L S VH C 1 mm D I W x y Image by MIT OpenCourseWare, adapted from R. E. Prange and S. M. Girvin, The Quantum Hall Effect, Springer-Verlag, Berlin, 1987. Figure 3: (From R. E. Prange and S. M. Girvin, The Quantum Hall Effect, Springer-Verlag, Berlin, 1987. Now it...
https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2016/3e3e804176b9c697b16abd811b5dfa89_MIT8_06S16_Supplementry.pdf