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3.052 Nanomechanics of Materials and Biomaterials Thursday 02/22/07 ATOMIC FORCE MICROSCOPY : 3D PLOTS AND 2D SECTION PROFILES Prof. C. Ortiz, MIT-DMSE Two images removed due to copyright restrictions. See Tai and Ortiz, Nano Letters 2006 z (μm) ] Å [ Z 7 1.5 6 5 1.0 4 3 0.5 2 1 0 0 0 0 1 2 2 3 4 x (μm) ...
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-sample separation) non-contact Graph by MIT OCW. Schematic after Thermomicroscopes "Introduction to AFM." Contact (DC and AC or Force Modulation) : ● tip remains in contact with sample ● feedback signal, δc ● potentially destructive due to high lateral (x/y) forces ● high resolution Intermittent Contact (AC, ...
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NH2 X X X X X X X X X X X X X X XXXXXX XXXXXX X X X X X X IV. Chemical Force Microscopy (CFM) Frisbie, et al., 1994 Courtesy of Veeco Instruments. Used with permission. http://www.di.com/AppNotes/ForceVol/FV.array.html Topography & Rough Surface Maps: ness, Electrostati Chemical, Adhesion , Hardness, Elasticity /Visco...
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1996, Sh 66, 141. cantilever = m δt(max) m δt(max) m PROBE TIP SHARPNESS Sheng, et al. J. Microscopy 1999, 196, 1. Image removed due to copyright restrictions. Image removed due to copyright restrictions. 3-D model of sharp probe tip on a protein, from Lieber et al, 2000 (http://cnst.rice.edu) 8 3.052 Na...
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yn J. Miles. Used with permission. Courtesy of Manfred Radmacher. Used with permission. Radmacher, et al., Cardiac Cells http://www.physik3.gwdg.de/~radmacher/ Rat Embryo Fibroblast (*M. Stolz,C. Schoenenberger, M.E. Müller Institute, Biozentrum, Basel Switzerland) Height image of endothelial cells taking in fluid u...
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Shao. Used with permission. 11 3.052 Nanomechanics of Materials and Biomaterials Thursday 02/22/07 HRFS COMBINED WITH AFM : SPATIALLY SPECIFIC SURFACES INTERACTION INFORMATION Prof. C. Ortiz, MIT-DMSE 1 μm1 μm 1 μm 1 μm Grain 2 Grain 2 Grain 2 5 5 5 5 4 4 4 4 3 3 3 3 2 2 2 2 1 1 1 1 Grain 3 ...
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Queueing Systems: Lecture 3 Amedeo R. Odoni October 18, 2006 Announcements • PS #3 due tomorrow by 3 PM • Office hours – Odoni: Wed, 10/18, 2:30-4:30; next week: Tue, 10/24 • Quiz #1: October 25, open book, in class; options: 10-12 or 10:30-12:30 • Old quiz problems and solutions: posted Thu evening along with ...
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(“epochs”) at which all we need to know is the number of customers in the system to determine the probability that at the next epoch there will be 0, 1, 2, …, n customers in the system • Epochs = instants immediately following the completion of a service M/G/1: Transition probabilities for system states at epochs...
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1- ρ and E[B] = 1/(μ – λ) = 1/(1- ρ) • Can also derive closed-form expressions for L, W, Lq and Wq Probability the Server is Busy • Suppose we have been watching the system for a long time, T. ρ, the utilization ratio, is the long-run fraction of time (= the probability) the server is busy; this means, assuming ...
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n ≥ 1 E[T2 | n] = 0, n = 0 • Thus, ⎤ E[T2 ] = ∑ E[T2 | n] ⋅ Pn = ∑ (n −1) ⋅ E[S] ⋅ Pn = E[S] ⋅⎢ ∑ nPn − ∑ Pn ⎥ ⎦⎥ ⎣⎢n≥1 n≥1 n≥1 ⎡ n E[T2 ] = E[S] ⋅ L − E[S] ⋅ ρ Derivation of L and W: M/G/1 [3] • From our “random incidence” result (2.66): 2 σ S E[T | n] = 1 ])2 [ ( S E + ] [ 2 S E ⋅ E[T1 | n] = 0, n =...
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λ2 ⋅σ S + 2λ(1 − ρ) μ Expected values for M/G/1 L = ρ+ 2 ρ 2 + λ2 ⋅σ S 2(1 − ρ) (ρ< 1) W = 2 1 ρ 2 + λ2 ⋅σ S + 2λ(1 − ρ ) μ Wq = 2 ρ 2 + λ2 ⋅σ S 2λ(1 − ρ ) = Lq = ρ 2 + (1 2 2 λ − 2 ⋅σ S ) ρ 2 ) ρ 2 (1 + C S = 2λ(1 − ρ) μ (1 − ρ) ρ 1 ⋅ 2 ) (1 + C S ⋅ 2 Note : CS = σ S S] [ E = μ⋅σ S D...
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Lq 0.58 0.60 1.25 1.31 3.40 3.73 7.81 9.38 Lq (% change) 3.4% 4.8% 9.7% 20.1% Wq (seconds) 69 71 125 130 292 317 625 743 Wq (% change) 2.9% 4% 8.6% 18.9% Can also estimate PHCAP ≅ 40.9 per hour M/G/1 system with non-preemptive priorities: background • • • • • • • r classes of cus...
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Assumptions and Biases • Every design and system has multiple phenomena operating at the same time • This has several consequences – Parts must be designed and tested separately and then tested together – Analytical models and simulations will not be able to encompass all the important things that could happen or m...
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architecture, the way the parts work together, is the key thing • Sometimes no agreement or convergence occurs (car door design) • More design and testing time are devoted to finding and mitigating side effects than in assuring achievement of basic function and performance 8/24/2006 Background © Daniel E Whitney 1997...
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3.032 Mechanical Behavior of Materials Fall 2007 Edge dislocation Dislocation Line b b half-plane Screw b Edge sheared unsheared dislocation line Figure by MIT OpenCourseWare. direction of motion Image source: Callister, W. D., Materials Science and Engineering: An Introduction Lecture 21 (10.29.07) 3.032 Mechanical B...
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3. Transmission electron microscopy diffraction 4. X-ray diffraction Images removed due to copyright restrictions. Please see: Fig. 2.8, 2.13, and 2.14 in Hull, Derek, and Bacon, D. J. Introduction to Dislocations. Boston, MA: Butterworth-Heinemann, 2001. 5. Field ion microscopy diffraction Lecture 21 (10.29.07)
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3'008& 9:;:<& '0=&>=80=&>?&!"#$%&! !"#$%& '()*+),-./(& 0.12.(()2.1& +*&3+*45-)(& 3674(,7& LECTURE 10 MEASUREMENT AND TIMING Charles E. Leiserson !"#$$%&#$'%"()"*+,"-./"01'2#"3,4*56,67" 1 Timing a Code for Sorting #include <stdio.h> #include <time.h> void my_sort(double *A, int n); void fill(double *A, int n); ...
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n); void fill(double *A, int n); struct timespec start, end; int main() { int max = 4 * 1000 * 1000; int min = 500 * 1000; int step = 20 * 1000; double A[max]; for (int n=min; n<max; n+=step){ fill(A, n); Loop over arrays of increasing length. Measure time before sorting. clock_gettime(CLOCK_MONOTONIC, &st...
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is going on? Dynamic Frequency and Voltage Scaling DVFS is a technique to reduce power by adjusting the clock frequency and supply voltage to transistors. • Reduce operating frequency if chip is too hot or otherwise to conserve (especially battery) power. • Reduce voltage if frequency is reduced. Power ∝ C V2 f ...
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can reduce variability, you can compensate for systematic and random measurement errors. © 2008–2018 by the MIT 6.172 Lecturers 9 Sources of Variability • Daemons and background jobs • Interrupts • Code and data alignment • Thread placement • Runtime scheduler ...
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.8% 0.7% 0.6% 0.5% 0.4% 0.3% 0.2% 0.1% 0.0% 0 10 20 30 40 !"#$$%&#$'%"()"*+,"-./"01'2#"3,4*56,67" Performance 12 50 60 70 Rank of Run 80 90 Quiescing the System • Make sure no other jobs are running. • Shut down daemons and cron jobs. • Disconnect the network. • Don’t fiddle with the mouse! • For serial jobs, d...
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"""!"# """"!"!!# !!!!!!!"# !"""""""# Similar: Changing the order in which the *.o files appear on the linker command line can have a larger effect than going between –O2 to –O3. cache and page alignment has changed !"#$$%&#$'%"()"*+,"-./"01'2#"3,4*56,67" 14 LLVM Alignment Switches LLVM tends to cache-align fun...
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OLS FOR MEASURING SOFTWARE PERFORMANCE ! PERFORMANCE MODELING !"#$$%&#$'%"()"*+,"-./"01'2#"3,4*56,67" 17 Ways to Measure a Program • Measure the program externally. • /usr/bin/time • • Instrument the program. • Include timing calls in the program. • E.g., gettimeofday(), clock_gettime(), rdtsc(). • By hand, or with c...
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e-9*(end.tv_nsec - start.tv_nsec); ! On my laptop, clock_gettime(CLOCK_MONOTONIC, …) takes about 83ns. ! That’s about two orders of magnitude faster than a system call. ! clock_gettime(CLOCK_MONOTONIC, …) guarantees never to run backwards. !"#$$%&#$'%"()"*+,"-./"01'2#"3,4*56,67" 20 rdtsc() x86 processors provide a tim...
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Neither is accurate if you don’t obtain enough samples. (gprof samples only 1OO times per second.) © 2008–2018 by the MIT 6.172 Lecturers 23 Hardware Counters • libpfm4 virtualizes all the hardware counters • Modern kernels make it possible...
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A, set A = A!. 5. If A is still not fast enough, go to Step 2. If you can’t measure performance reliably, it is hard to make many small changes that add up. !"#$$%&#$'%"()"*+,"-./"01'2#"3,4*56,67" 27 Problem Suppose that you measure the performance of a deterministic program 100 times on a computer with some inte...
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∙ Speedup of wall clock time © 2008–2018 by the MIT 6.172 Lecturers 30 Summarizing Ratios Trial Program A Program B A/B 1 2 3 4 Mean 9 8 2 10 7.25 3 2 20 2 6.75 3.00 4.00 0.10 5.00 3.03 Conclusion Program ...
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0.00 10 (a) 7.25 2 (a) 6.75 5.00 (g) 1.57 0.20 (g) 0.64 (cid:80) (cid:67)(cid:75) (cid:19) (cid:75) (cid:19) (cid:80) (cid:16) n (cid:67)(cid:19)(cid:67)(cid:20) (cid:67)(cid:80) (cid:2477)(cid:3075) Observation (cid:30) The geometric mean of A/B IS the inverse of the geometric mean of B/A. (cid:30) (ci...
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python Instructions 170889186565542 Time (s) 34864 Cache misses 36615004052 java C gcc -O0 C gcc -O3 2618 1480 430 7509707536406 39322034007 2274589361551 68047140354 278479001783 34049504541 I want to infer how long it takes to run an instruction and how long to take a cache miss. I guess that I c...
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fid out how well the model predicts the other half of the data. How can I tell whether I’m fooling myself? • Triangulate. • Check that different ways of measuring tell a consistent story. • Analogously to a spreadsheet, make sure the sum of the row sums adds up to the sum of the column sums. © 2008–2018 by the MIT 6.1...
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3.024 “Electronic, Optical, and Magnetic Properties of Materials” 1 3.024 Objectives and Approach • How can we understand and predict electrical, optical and magnetic properties? Emphasis on fundamental physical models in lectures • Application to real life situations? Emphasis on real life examples in HW and r...
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http://ocw.mit.edu/fairuse. I V=IR • What determines whether material be a resistor or a diode? Linear, Ohmic V • How would you predict it? • How much voltage or current can material handle? • What happens if we shine light onto a material? I Rectification, Non-linear, Non-Ohmic dark light V 6 Materi...
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can use these waveguides in telecommunication and medicine. 10 Magnets Around Us © R. Nave. All rights reserved. This content is excluded from our Creative Commons license. For more information, see http://ocw.mit.edu/fairuse. From Prof. Beach Can energy dissipation be useful? Neurons expressing heat sensitive pr...
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MIT OpenCourseWare http://ocw.mit.edu 2.161 Signal Processing: Continuous and Discrete Fall 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. Massachusetts Institute of Technology Department of Mechanical Engineering 2.161 Signal Processing - Continuous and Di...
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= 0.05 −→ λ = 4.358 Then 1copyright (cid:4)c D.Rowell 2008 N ≥ log(λ/(cid:2)) log(Ωr/Ωc) = 3.70 7–1 (cid:2) (cid:17) (cid:16) (cid:4) (cid:5) (cid:4) (cid:9) (cid:9) (cid:11) (cid:12) (cid:9) (cid:12) we therefore select N=4. The 4 poles (p1 . . . p4) lie on a circle of radius r = Ωc(cid:2)−1/N = 13.16 and are...
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0 2 Chebyshev Filters The order of a filter required to met a low-pass specification may often be reduced by relaxing the requirement of a monotonically decreasing power gain with frequency, and allowing 7–2 “ripple” to occur in either the pass-band or the stop-band. The Chebyshev filters allow these conditio...
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2xTN −1(x) − TN −2(x), N > 1 with alternate definitions TN (x) = cos(N cos−1(x)) = cosh(N cosh−1(x)) (3) (4) (5) The Chebyshev polynomials have the min-max property: Of all polynomials of degree N with leading coefficient equal to one, the polynomial TN (x)/2N −1 has the smallest magnitude in the interval x amp...
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= � cosh N cosh−1 (Ωr/Ωc) � N ≥ cosh−1 (λ/�) cosh−1 (Ωr/Ωc) The characteristic equation of the power transfer function is � � � � 2 1 + �2TN = 0 or TN s jΩ c s jΩ c j = ± � Now TN (x) = cos(N cos−1(x)), so that � cos N cos−1 If we write cos−1 (s/jΩc) = γ + jα, then � �� s jΩc = ± j � s = Ωc...
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N sinh−1 1 � (11) As in the Butterworth design procedure, we select the left half-plane poles as the poles of the filter frequency response. Design Procedure: 1. Determine the filter order 2. Determine α 3. Determine γn, n = 1 . . . N N ≥ cosh−1 (λ/�) cosh−1 (Ωr/Ωc) α = ± 1 N sinh−1 1 � γn = (2n − 1)...
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1 design. From the previous example we have Ωc = 10 rad/s., Ωr = 20 rad/s., � = 0.3333, λ = 4.358. The required order is N ≥ cosh−1 (λ/�) cosh−1 (Ωr/Ωc) = cosh−1 13.07 cosh−1 2 = 2.47 Therefore take N = 3. Determine α: α = 1 N sinh−1 � � 1 � = 1 3 sinh−1(3) = 0.6061 and sinh α = 0.6438, and cos...
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8 � � p3 = 10 0.6438 × � 1 2 − j1.189 × p4 = 10 −0.6438 × 1 2 − j1.189 × = 3.219 − j10.30 � = −3.219 − j10.30 p5 = 10 (−0.6438 × 0 − j1.189 × 0) = −6.438 p6 = 10 −0.6438 × 1 2 + j1.189 × 3 2 = −3.219 + j10.30 √ � 3 2 √ 3 2 √ � and the gain adjusted transfer function of the resulting Type 1 fil...
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24 may be written in terms of the modified frequency ν H(jν) 2 = | | 2 (ν/Ωc) ˆ�2TN 1 + ˆ�2TN 2 (ν/Ωc) (12) (13) which has a denominator similar to the Type 1 filter, but has a numerator that contains a Chebyshev polynomial, and is of order 2N . We can use a method similar to that used in the Type 1 filter des...
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/2)π N n = 1 . . . N. 2. The poles and zeros are mapped back to the s-plane using s = ΩrΩc/τ and the N left half-plane poles are selected as the poles of the filter. 3. The transfer function is formed and the system gain is adjusted to unity at Ω = 0. Example 2 Repeat the previous Chebyshev Type 1 design example ...
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= 10 1.1024 × 1 2 + j1.4884 × 3 2 = 5.512 + j12.890 τ2 = 10 (1.1024 × 1 + j1.4884 × 0) = 11.024 √ � √ 3 2 τ3 = 10 1.1024 × � 1 2 − j1.488 × 3 2 = 5.512 − j12.890 � × τ4 = 10 −1.1024 � τ5 = 10 −1.1024 × � τ6 = 10 −1.1024 × 1 2 1 2 1 2 − j1.4884 × = −5.512 − j12.890 � − j1.488 × 0 = −11.0...
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and ν2, ν3 = ±j8.666. Mapping these back to the s-plane gives two finite zeros z1, z2 = ±j23.07, z3 = ∞ (the zero at ∞ does not affect the system response) and the unity gain transfer function is H(s) = = −p1p2p3 z1z2 (s − z1)(s − z2) (s − p1)(s − p2)(s − p3) 6.9365(s2 + 532.2) (s + 18.14)(s2 + 11.22s + 203.5) ...
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) (cid:4) (cid:5) (cid:3) (cid:6) (cid:2) (cid:7) (cid:5) (cid:8) (cid:3) (cid:12) (cid:6) (cid:9) (cid:4) (cid:4) (cid:7) (cid:8) (cid:9) (cid:10) (cid:11) (cid:3) (cid:2) (cid:2) (cid:9) (cid:3) (cid:10) (cid:5) (cid:3) (cid:11) (cid:3) (cid:2) (cid:9) (cid:11) (cid:10) (cid:5) (cid:12) (cid:13) 2.3 Comparis...
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Butterworth and the Chebyshev Type 1 filters are all-pole designs and have an asymptotic high-frequency magnitude slope of −20N dB/decade, in this case -80 dB/decade for the Butterworth design and -60 dB/decade for the Chebyshev Type 1 design. The Type 2 Chebyshev design has two finite zeros on the imaginary axis at ...
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byshev Type 1 (N=3) 1 2 3 4 5 Angular frequency (rad/sec) 6 7 8 9 10 Butterworth (N=4) Chebyshev Type 1 (N=3) Chebyshev Type 2 (N=3) 0 20 30 40 50 60 Angular frequency (rad/sec) 70 80 90 100 7–11
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Lecture 9 MOSFET(II) MOSFET I­V CHARACTERISTICS(contd.) Outline 1. The saturation region 2. Backgate characteristics Reading Assignment: Howe and Sodini, Chapter 4, Section 4.4 6.012 Spring 2009 Lecture 9 1 1. The Saturation Region Geometry of problem Regions of operation: • Cut­off: VGS < VT – No inversion ...
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QN, Ey, and V in linear regime as VDS increases: Ohmic drop along channel de-biases inversion layer ⇒ current saturation. 6.012 Spring 2009 Lecture 9 4 The Saturation Region (contd.) What happens when VDS = VGS – VT? Charge control relation at drain: [ QN (L) = −C ox VGS − VDS − VT ]= 0 No inversion layer a...
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�  L − ∆∆∆∆L L L ≈ 6.012 Spring 2009 Lecture 9 7 The Saturation Region (contd.) Experimental finding: ∆L L = λ(VDS − VDSsat ) with Typically, λλλλ∝ 1 L λλλλ= 0.1 µµµµm • V −1 L For L = 1µm, increase of VDS of 1V past VDSsat results in increase in ID of 10%. Improved but approximate model for the...
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⇒ -2φ p - VBS |QB| ↑ ⇒ xdmax↑ • • • Since VGS is constant, Vox unchanged – ⇒ Eox unchanged – ⇒ |QS| = |QG| unchanged |QS| = |QN| + |QB| unchanged, but |QB| ↑ ⇒ |QN| ↓ – ⇒ inversion layer charge is reduced! • VDS + - VGS + - Metal interconnect to gate n+ polysilicon gate n+ source 0 y n+ drain QN(y) Xd(y) VBS ...
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= VFB − 2φφφφp + 1 C ox 2εεεεsqNa (−2φφφφp − VBS) Define backgate effect parameter [units: V1/2]: γγγγ = 1 Cox 2εεεεsqNa VTo = VT (VBS = 0) And: Then: VT (VBS ) = VTo + γγγγ[ −2φφφφp − VBS − −2φφφφp ] 6.012 Spring 2009 Lecture 9 13 Backgate Characteristics (Contd.) Triode Region VDS ~ 0.1V 6.012 Spring 2009 ...
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DIRAC’s BRA AND KET NOTATION B. Zwiebach October 7, 2013 Contents 1 From inner products to bra-kets 2 Operators revisited 2.1 Projection Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Adjoint of a linear operator . . . . . . . . . . . . . . . . . . . . . . . . . . . . ....
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now write v | We have linearity in the second argument ∗, as well as ) u ( = v ( u | ) v ( v | ) ≥ 0 for all v, while v ( v | ) = 0 if and only if v = 0. for complex constants c1 and c2, but antilinearity in the first argument u ( c1v1 + c2v2) | = c1( u v1) | + c2( u v2) | , Two vectors u and v for which ...
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are satisfied. a ( b | ∗ a1b1 + a2b2 . ∗ (1.4) 2. Consider the complex vector space of complex function f (x) functions f (x), g(x) we define C with x ∈ ∈ [0, L]. Given two such f g | ( ) ≡ L 0 f ∗ (x)g(x)dx . The verification of the axioms is again quite straightforward. A set of basis vectors ei} { labe...
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the ket | v | ) represent the vector v. This is a small subtlety with the notation: we think of v V as a vector and also v | ) ∈ V as a vector. It is as if we added some decoration ∈ around the vector v to make it clear | ) by inspection that it is a vector, perhaps like the usual top arrows that are added in ...
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a bra to the left of a ket makes sense: matrix multiplication of a row vector times a column vector gives a number. Indeed, for vectors a = a1 a2  . . .   an        , b = b1 b2  . . .   bn        ∗ ∗ = a1b1 + a2b2 + . . . a bn ) ∗ n we had Now we think of this as a...
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a1, a∗ . . . , a ∗ 2 n = ) (1.15) a ( b | ∗ n ( ) ∗ · jugating the entries. More abstractly the bra labeled by the vector u is defined by its action on u ( | arbitrary vectors as follows v | ) ) → ( As required by the definition, any linear map from V to C defines a bra, and the corresponding ) | : u ( v | ...
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| ( 2 , . . . , α ∗ n 1 , α ∗ α ∗ ( ) 3 (1.19) and the associated vector or ket Note that, by construction = α ) | α1 α2   . . . αn         (1.20) v α | ( This illustrates the point that (i) bras represent dual objects that act on vectors and (ii) bras are (1.21) f (v) =...
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′ ( | | ∈ v ′ ( w | ) = ) → ( v | w ) − ( v ′ | ) = 0 w → ( v ′ v | − ) = 0 . (1.24) In the first step we used complex conjugation and in the second step linearity. Now the vector v must have zero inner product with any vector w, so v ′ v = 0 and v = v . ′ ′ v − We can now reconsider equation (1.3) ...
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of a vector now reads = δij . j i | ( ) As in (1.8) the expansion coefficients are αk = = v | ) αi , i ) | so that X i k ( v | ) = v | ) v i i | )( | ) . X i 4 (1.27) (1.28) (1.29) 2 Operators revisited Let T be an operator in a vector space V . This means that acting on vectors on V it gi...
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.30) (2.31) (2.32) (2.33) (2.34) (2.35) The object a ( Ω is defined to be the bra that acting on the ket | b | ) gives the number a ( Ω | b | . ) We can write operators in terms of bras and kets, written in a suitable order. As an example of an operator consider a bra and a ket a ( | b | . We claim that t...
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= a | ) X n n | an , ) = b | ) n | bn . ) X n (2.39) 5 Assume b | ) is obtained by the action of Ω on a | : ) = Ω a | ) b | ) → Ω n | an = ) n | bn . ) X n X n Acting on both sides of this vector equation with the bra we find m ( | m ( X n Ω | n | an = ) m ( X n n | bn = bm ) We now define th...
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. . . ΩN N      , with Ωij = Ω i | ( j | ) . (2.44) There is one additional claim. The operator itself can be written in terms of the matrix elements and basis bras and kets. We claim that Ω = m | Ωmn( n ) | . X m,n (2.45) We can verify that this is correct by computing the matrix elements using i...
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) we see that in the chosen basis, Pn is represented by a matrix all of whose elements are zero, except for the (n, n) element (Pn)nn which is one: Pn ←→ 0 0 0 0 . . . . . . 0 0 . . . . . . 0 0           . . . 0 . . . 0 . . . . . . . . . 1 . . . . . . . . . 0 . . . 0 . . . 0...
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, m | (2.50) since m ( m | ) one-dimensional subspace of V , the subspace generated by Using the basis vector m | ) with m = n we can define m | . ) Acting on any vector v | ) ∈ Pm,n n | V , this operator gives us a vector in the subspace spanned by ≡ | m m )( )( n | | + . (2.51) and m | ) n | : ) U...
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anticipated this in (1.29), 1 1 | )( ≡ | P1,...,N (2.53) + . . . + N | + N )( . | and we thus write This is the completeness relation for the chosen orthonormal basis. This equation is sometimes called 1 = . i i | )( | X i (2.54) the ‘resolution’ of the identity. Example. For the spin one-half system t...
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X k=1 This is the expected rule for the multiplication of the matrices corresponding to Ω1 and Ω2. 2.2 Adjoint of a linear operator A linear operator Ω on V is defined by its action on the vectors in V . We have noted that Ω can also be viewed as a linear operator on the dual space V ∗ . We defined the linear operator...
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giving a complete list of these maps. The operator v | v ′ Ω† is defined as the one that induces the maps of the corresponding bras. Indeed, ) → | v ′ ) v ( Ωv | ) Ωv ( | → ( | = Ω , v | v ( ) Ω† | = ′ v = ) | v ′ ( = | The first line is just definitions. On the second line, the first equality is obtained ...
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| v | ) for arbitrary vectors a, b, write a bra-ket expression for Ω† . = Ω v | ) a | b ) ( v | ) v → ( Ω† = | v ( b | a ) ( , | Since this equation is valid for any bra we read v ( | Ω† = b | a )( . | 2.3 Hermitian and Unitary Operators A linear operator Ω is said to be hermitian if it is equal to its ad...
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u ( Ω | v | ∗ , ) ∀ u, v . It follows that the expectation value of a Hermitian operator in any state is real Another neat form of the hermiticity condition is derived as follows: v ( Ω | v | ) is real for any hermitian Ω . so that all in all Ωu ( v | ) = u ( Ω† | v | ) = u ( Ω | v | ) = u ( Ωv | ) , He...
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be a unitary operator if U † is an inverse for U , that is, U †U and U U † are both the identity operator: U is a unitary operator: U †U = U U † = 1 (2.74) In finite dimensional vector spaces U †U = 1 implies U U † = 1, but this is not always the case for infinite dimensional vector spaces. A key property of unitary...
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we get a1) | , a2) | , . . . aN ) | , ai| ( aj ) = δij U a1) | , U a2) | , . . . U aN ) | , To show that this is a basis we must prove that βi| U ai) = 0 X i 10 (2.79) (2.80) (2.81) implies βi = 0 for all i. Indeed, the above gives βi| U ai) = X i X i βiU ai) | = U X i βi| ai...
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can be written as N (2.84) (2.85) ai)} {| and bi)} {| (2.86) U = bi)( | ai| . X i=1 ai) | any orthonormal basis and = bi) | . U ai) | Indeed, this is just a rewriting of (2.84), with Exercise: Verify that U in (2.86) satisfies U †U = U U † = 1. Exercise: Prove that aj ) | U ai| ( U bi| ( bj ) | = ...
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, as always in quantum mechanics). Note here that the label on the ket is not a vector! So , ) unless x = 0. For quantum mechanics in three dimensions, ax | x | = a ) for any real a = 1. In particular we have position states xx . Here the label is a vector in a three-dimensional real vector space (our ) | spa...
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position states. We also have a completeness relation 1 = dx x | x )( . | Z This is consistent with our inner product above. Letting the above equation act on equality: (3.89) we find an y | ) The position operator ˆx is defined by its action on the position states. Not surprisingly we let = y | ) Z dx x |...
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≡ ( ψ | ) ∈ C . (3.92) (3.93) (3.94) This is sensible: is a number that depends on the value of x, thus a function of x. We can now do a number of basic computations. First we write any state as a superposition of position eigenstates, by inserting 1 as in the completeness relation x ( ψ | ) ψ | = 1 ψ | ) ) ...
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in (3.98) (3.99) , p | ) p ′ ) , − p )( , | Basis states : p ′ p ( ) | 1 = = δ(p dp p | Z = p p | ) pˆ p | ) Just as for coordinate space we also have R . p ∀ ∈ In order to relate the two bases we need the value of the overlap † pˆ = ˆ p , and p ( p = p ˆ | p ( | . x ( p | . Since we interpret...
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the delta function δ(p p ′ ): − du ei(p−p ′)u = δ(p − 1 2π Z p ′ ) . (3.102) This, then gives the correct value for the overlap justified using the fact that the functions p ( p ′ | , as we claimed. The integral (3.102) can be ) form a complete orthornormal set of functions over the interval x fn(x) 1 ex...
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� ) ) → 1 2π Z ′) dueiu(x−x , (3.106) and back in (3.105) we have justified (3.102). We can now ask: What is p ( ψ | ? We compute ) p ( ψ | ) = Z dx p ( x | x )( ψ | ) = 1 � ~ Z 2π √ dxe−ipx/ ψ(x) ~ � ˜ = ψ(p) , (3.107) which is the Fourier transform of ψ(x), as defined in (6.41) of Chapter 1. Thus the ...
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| (3.110) pˆ . Z p ) ) | pˆ x ( | ψ | ) = ~ � d i dx Z dp x ( p | p )( ψ | ) The completeness sum is now trivial and can be discarded to obtain pˆ x ( | ψ | ) = ~ � d i dx x ( ψ | ) = ~ � d i dx ψ(x) . Exercise. Show that xˆ p ( | ψ | ) = i � ~ d ˜ dp ψ(p) . (3.111) (3.112) (3.113) 14 ...
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1 Introduction 1 INTRODUCTION These notes provide reading material on the Soft-Collinear Effective Theory (SCET). They are intended to cover the material studied in the second half of my effective field theory graduate course at MIT. These latex notes will also appear as part of TASI lecture notes and a review artic...
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namely how they cancel between virtual and real emission diagrams, and how they otherwise signal the presence of nonperturbative physics and the scale ΛQCD as they do for parton distribution functions. Finally it should be remarked that later parts of the notes are still a work in progress (particularly sections ma...
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http://ocw.mit.edu 8.851 Effective Field Theory Spring 2013 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.
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8.701 0. Introduction 0.9 Spin Introduction to Nuclear and Particle Physics Markus Klute - MIT 1 Spin vector, length, and eigenvalues In quantum mechanics, the spin vector S is quantised in terms of its length and its components. Total length is For components along any axis, e.g. z, eigenvalues can be with 2s+1 ...
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18.175: Lecture 9 Borel-Cantelli and strong law Scott Sheffield MIT 18.175 Lecture 9 1 Outline Laws of large numbers: Borel-Cantelli applications Strong law of large numbers 18.175 Lecture 9 2 Outline Laws of large numbers: Borel-Cantelli applications Strong law of large numbers 18.175 Lecture 9 3 Borel-C...
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idea of proof: First, pairwise independence implies that variances add. Conclude (by checking term by term) that VarSn ≤ ESn. Then Chebyshev implies P(|Sn − ESn| > δESn) ≤ Var(Sn)/(δESn)2 → 0, which gives us convergence in probability. Second, take a smart subsequence. Let nk = inf{n : ESn ≥ k 2}. Use Borel Can...
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X 2]2 ≥ 0, so E [X 2]2 ≤ K . The strong law holds for i.i.d. copies of X if and only if it holds for i.i.d. copies of X − µ where µ is a constant. So we may as well assume E [X ] = 0. Key to proof is to bound fourth moments of An. E [A4] = n−4E [S 4] = n−4E [(X1 + X2 + . . . + Xn)4]. Expand (X1 + . . . + Xn)4...
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(Borel Cantelli, expectation of positive r.v. is area between cdf and line y = 1) Claim: S∞ Var(Yk )/k 2 ≤ 4E |X1| < ∞. How to prove it? Observe: Var(Yk ) ≤ E (Y 2) = k 0 since everything positive) 2yP(|X1| > y )dy . Use Fubini (interchange sum/integral, 2yP(|Yk | > y )dy ≤ ∞ 0 k=1 k ∞ t E (Yk 2)/k 2 ≤ k=1...
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n=1 Var(Ym) = E−2 ∞ t Var(Ym) :n m =1 ≥k( ) n m t k(n)−2 . n=1 ∞ t −2E= k(n)−2 =1 n Sum series: S k( ) n t m =1 S n:αn≥m [αn]−2 ≤ 4 n:αn≥m α−2n Combine computations (observe RHS below is finite): ∞ t ≤ 4(1 − α−2)−1m . −2 P(|Tk(n)−ETk (n)| > Ek(n)) ≤ 4(1−α−2)−1E−2 ∞ t E (Y 2 )m m −2 . n=1 m=1 Sinc...
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LECTURE 12 Derived Functors and Explicit Projective Resolutions Let X and Y be complexes of A-modules. Recall that in the last lecture we defined HomA(X, Y ), as well as Homder A (X, Y ) := HomA(P, Y ) for a projective qis −−→ X, i.e., a projective resolution of X. We also defined the Ext- complex P groups Exti A(X, Y ) ...
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−−→ Hom(P1, X2) is a quasi-isomorphism. We are given a map, namely the composition P1 → X1 → X2, which is killed by the differential since it is a map of chain complexes, and therefore defines a cohomology class in H 0Hom(P1, X2). So there is some coho- mology class in H 0Hom(P1, P2) which is a lift of that map through P...
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for Z[H], and then we may regard Z[G] as a direct sum of copies of Z[H]). This implies that Z[G] ⊗Z[H] PH qis −−→ Z[G/H], which is projective as a complex of Z[G]-modules. This is because both Z[G] and PH are bounded, so it will be bounded, and inducing up to Z[G] preserves projective modules as we will still obtain a ...
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→ (cid:88) g, g∈G/H such that the composition corresponds to multiplication by the index [G : H]. This gives an inflation map X hH → X hG such that the composition X hG → X hH → X hG is homotopic to multiplication by [G : H]. More concretely, suppose we had an H-invariant object and a G-invariant ob- ject. Taking coset ...
https://ocw.mit.edu/courses/18-786-number-theory-ii-class-field-theory-spring-2016/4e977d9bd29ad54aee9bd85d49f67416_MIT18_786S16_lec12.pdf
f d = 0. It follows that such maps f are equivalent to G-equivariant maps Z = P 0/dP −1 = Coker(P −1 → P 0) → Ker(X 0 → X 1) = H 0(X) by quasi-isomorphism, which is equivalent to a G-invariant vector in the cohomology (cid:3) H 0(X) (i.e., via the image of 1). We now turn to the problem of constructing explicit project...
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Throughout, our “motto” will be that “all such things come from the bar construction.” Let A be a commmutative ring, and B an A-algebra; the most important case will be A := Z and B := Z[G]. Definition 12.8. For all such A and B, the bar complex BarA(B) is · · · → B ⊗A B ⊗A B b1⊗b2⊗b2(cid:55)→b1b2⊗b3−b1⊗b2b3 with B in ...
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h0(b) := 1 ⊗ b, and h1(b1 ⊗ b2) := 1 ⊗ b1 ⊗ b2. Indeed, we then have (dh(cid:48) +h0d)(b1 ⊗b2) = d(1⊗b1 ⊗b2)+1⊗b1b2 = b1 ⊗b1 −1⊗b1b2 +1⊗b1b2 = b1 ⊗b2, It’s easy to show that defining hn similarly for all n gives a null- as desired. (cid:3) homotopy of the identity. As a reformulation, consider the diagram · · · · · · B ...
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� m (cid:55)→ b1b2 ⊗ m − b1 ⊗ b2m. In words, M is canonically homotopy equivalent to a complex where every term is of the form B ⊗A N , where in this case N stands for B ⊗A · · · ⊗A B ⊗A M , that is, a module induced from some A-module. We now apply this to the case where A := Z, B := Z[G], and M := Z, i.e., Note that ...
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[G] ⊗Z Z[G], since both have a basis by the elements of the product group. This is a great explicit projective resolution of Z for computing group coho- mology! We end up with a complex of the form · · · → 0 → M → Z[G] ⊗ M → Z[G × G] ⊗ M → · · · , with M in degree 0 and G finite. Elements in Z[G] ⊗ M in the kernel of th...
https://ocw.mit.edu/courses/18-786-number-theory-ii-class-field-theory-spring-2016/4e977d9bd29ad54aee9bd85d49f67416_MIT18_786S16_lec12.pdf