Srednicki Chapter 14
|
|
- Lindsey Dawson
- 5 years ago
- Views:
Transcription
1 Srednc Chapter 4 QFT Problems & Solutons A. George September, Srednc 4.. Derve a generalzaton of Feynman s formula, = Γ ( α ) A αa... A αn n Γ(α df xα n ) (n )! ( x A ) Hnt: start wth: Γ(α) A α = dt t α e At whch defnes the gamma functon. Put an ndex on A, α, and t, and tae the product. Then multply on the rght hand sde by ( = ds δ s ) t Mae the change of varable t = sx, and carry out the ntegral over s. Followng the hnt, we start wth: Γ(α) A α = dt t α e At Puttng an ndex on A, α and t, and tang the product of both sdes, we have: Γ(α ) A α = dt t α e A t α Now we multply on the rght by equaton 4.5: Γ(α ) A α = dt ds t α e A t δ ( s t ) Next we mae the change of varable t = sx : Γ(α ) A α = dx ds (sx ) α e A sx sδ ( s sx )
2 Recall that the delta functon has the property of f(cx) = f(x). Snce s s always postve, c we have: Γ(α ) = ( dx A α ds (sx ) α e A sx δ ) x Now let s add n an dentty: Γ(α ) A α = ( dx ds (sx ) α e A sx (n )! (n )! δ Next, note that all the x terms must be postve (thans to our ntegral), but can t be greater than one (thans to the delta functon). So, we can rewrte the x ntegral: Γ(α ) = ( dx A α ds (sx ) α e A sx (n )! (n )! δ ) x Usng equaton 4.: Γ(α ) A α = df n ds (sx ) α e A sx (n )! x ) Separatng out the s-ntegral: Γ(α ) A α = df n (x ) α (n )! dss α e A sx Ths s qucly gong to get confusng wth the product. To avod ths, let s have the product act on each term separately. Ths, of course, s not always allowed, but t s fne n ths case snce everythng s multpled together. ] [ Γ(α ) A α = df n [ (x ) α (n )! ds s α ] [ ] e A sx These last two terms gve products of exponentals, whch of course s a sum n the exponent. So: [ ] Γ(α ) = df A α n (x ) α ds [ s ] [ α e s A x ] (n )! We can solve ths s-ntegral on Mathematca or the equvalent, the result s: Γ(α ) A α = df n [ (x ) α ] [ ] ( (n )! Γ α Dstrbutng the product on the left hand sde, we have: Γ(α ) A α... A αn n = df n [ (x ) α ] [ ] ( (n )! Γ α A x A x ) α ) α
3 Then: A α... A αn n = Γ ( α ) Γ(α ) (n )! df n (x ) α ( A x) α Srednc 4.. Verfy equaton 4.3. Srednc gave us the openng here when he suggested tang a Gaussan ntegral n cartesan and sphercal coordnates. Let s do cartesan coordnates frst (we ll also use x as our dummy varable): d d xe ( x ) = ( π) d Next let s do t n sphercal coordnates: dω drr d e r = Ω Γ(d/) Equatng these results, we have: Srednc 4.3. (a) Show that: Ω = πd/ Γ(d/) d d qq µ f(q ) = d d qq µ q ν f(q ) = C g µν d d qq f(q ) and evaluate the constant C n terms of d. Hnt: use Lorentz symmetry to argue for the general structure, and evaluate C by contractng wth g µν. The frst equaton s odd, and so ntegrates to zero. The second equaton has two ndces, but cannot depend on any four-vectors. Hence, the general structure s, by Lorentz Symmetry: d d q q µ q ν f(q ) = Ag µν (4.3.) Contractng wth g µν, we fnd: d d qq f(q ) = Ad and so: A = d d d qq f(q ) Pluggng ths nto equaton (4.3.) gves: d d qq µ q ν f(q ) = d gµν d d qq f(q ) 3
4 Comparng ths to equaton 4.53, we fnd that C = /d. (b) Smlarly evaluate d d q q µ q ν q ρ q σ f(q ). As before, Lorentz Symmetry tells us that we must use the g. Ths tme there are three ways to combne the ndces: d d q q µ q ν q ρ q σ f(q ) = Ag µν g ρσ + Bg µρ g νσ + Cg µσ g ρν (4.3.) To evaluate A, B, and C, let s start by contractng wth g µν g ρσ d d q q 4 f(q ) = Ad + Bd + Cd And now let s contract wth g µρ g νσ : d d q q 4 f(q ) = Ad + Bd + Cd And now let s contract wth g µσ g ρν : d d q q 4 f(q ) = Ad + Bd + Cd Equatng the three rght hand sdes, we fnd that A = B = C. Then, we have: d d q q 4 f(q ) = A(d + d ) Hence, we have: A = B = C = d + d d d q q 4 f(q ) Puttng ths nto equaton (4.3.), we have: d d q q µ q ν q ρ q σ f(q ) = gµν g ρσ + g µρ g νσ + g µσ g ρν d + d d d q q 4 f(q ) Srednc 4.4. Compute the values of κ A and κ B. Start wth Srednc 4.39: Π( ) = α Imposng the frst boundary condton, we have: ( ) dxdln(d/m ) + α 6 κ A + κ B m + O(α ) Π( m ) = α m dx( x + x )ln( x + x ) α 6 κ Am + ακ B m = 4
5 Evaluatng the ntegral: ( 3 ) 3π κ A + κ B = (4.4.) Now let s tae the dervatve of equaton 4.39 so that we can mpose the second boundary condton: Π ( ) = α x( x) + m dxx( x) + ln + α m 6 κ A Imposng the second boundary condton: Solvng ths for κ A : Π ( m ) = dxx( x) [ + ln( x + x ) ] + 3 κ A = κ A = 7 3π 3 =.374 Usng ths n equaton (4.4.), we have: κ B = 3π =.9 6 Srednc 4.5. Compute the O(λ) correctons to the propagator n φ 4 theory (see problem 9.), and compute the O(λ) terms n A and B. We essentally have to repeat the entre analyss of ths chapter for the φ 4 theory. Let s start by drawng the O(λ) correctons to the vertex. They are: l x Note that ths second dagram s necessary because t may contrbute at O(λ). Srednc told us (and proved n equatons 4.37, 4.3) that ths dagram s of O(g ) n φ 3 theory. We mght assume that ths dagram s smlarly of O(λ ) n φ 4 theory, but there s no way to now for sure we ll have to assume for the moment that t does correct the propagator at O(λ). We wll have calculated A and B by the end of the problem, so then we ll now whether t contrbuted anythng or not. If not, then no harm s done, we ll just state that the dagram doesn t contrbute and gnore all hgher-order correctons, ncludng those from ths dagram. Our frst step s to wrte down the self-energy. Ths wll be the values of these Feynman dagrams, rememberng to gnore the external propagators. Followng the Feynman rules for φ 4 theory, the frst dagram contrbutes the followng terms to the self-energy: -λ for the vertex /(l + m ɛ) for the nternal propagator of the loop. 5
6 d d l/(π) d to ntegrate over the loop momentum. / for the symmetry factor of the loop because the stated Feynman Rules are for τ (or Π, as the case may be) The second dagram contrbutes only the counterterm vertex factor, whch s the same as n φ 3 theory, snce the counterterm Lagrangan s the same. We also put n the factor of, for the same reason as before. Puttng all ths together, we have: Π( ) = ( ) ( λ) Wrtng ths more neatly, we have: Π( ) = λ It s convenent to wrte ths as: Π( ) = λ d d l (π) d l + m ɛ + ( )( )(A + Bm ) d d l (π) d l + m ɛ + (A + Bm ) d d l (π) d l + m ɛ A Bm The man challenge s to calculate ths ntegral. Srednc presented a bag of trcs, but the only one we need n ths problem s the Wc Rotaton. We defne l = l d and l j = l j. Then l = l, but d d l = d d l. The pont of ths, of course, s that we can get rd of the ɛ n the denomnator, per equaton 4.6. Then, Π( ) = λ d d l (π) d l + m A Bm We now how to solve ths ntegral, but t s worth pausng to consder the dmensonalty. Ths ntegral appears to be well defned only for d = (or lower). However, usng Srednc s trc #3, we could tae two dervatves wth respect to m. Ths would be allowed because we have two degrees of freedom, so we could then ntegrate twce, solve the boundary condtons, and return to the orgnal problem. The advantage of ths s that after dfferentaton, we would have an l 6 n the denomnator, and so the ntegral wll converge so long as d < 6. As t turns out, we get a dmensonless couplng constant for d = 4, so ths theory s renormalzable n the case of nterest. Just as n the text, t s not necessary to actually tae the two dervatves, but t s necessary to now that the theory s renormalzable, and that t s therefore vald to analytcally contnue our results to d = 4. Now we can use equaton 4.7: ) Π( ) = λ Γ ( d (4π) d/ (m ) +d/ A Bm Now t s tme to defne ε = 4 d (4 because we get a dmensonless couplng constant when d = 4). Let s also set λ µ ε λ, such that all mass dmensonalty s projected onto the µ. Ths gves: Π( ) = λ µε Γ ( ) + ε (4π) ε/ (m )(m ) ε/ A Bm (4π) 6
7 whch s: Π( ) = m λ ( (4π) Γ + ε ) ( ) 4π µ ε/ A Bm m Usng equaton 4.6 and 4.33, we smplfy: [ ] Π( ) = m λ [ (4π) ε γ + + ε ( )] 4π µ ln A Bm m Now multply together, and eep only those terms of O() n ε, snce we want to wor n four dmensons. Then: Π( ) = m λ 4π µ (4π) ε γ + + ln A Bm m Next, let s solve equaton 4.35, to fnd that 4π µ = e γ µ. Then: Π( ) = m λ e γ (4π) ε γ + + ln µ A Bm m whch s: and so: Π( ) = m λ (4π) Π( ) = µ ε γ + + ln(eγ ) + ln A Bm m m λ (4π) µ ε + + ln A Bm m By the way, ths s equaton 3.5, so we now that we dd well. The problem s that we need Π( ) to be fnte and ndependent of µ. In terms of, everythng s fne at ths order, so we ll tae A = κ A λ + O(λ ), where κ A s purely numerc. Snce we re gnorng O(λ ) correctons, we have: Π( ) = m λ µ (4π) ε + + ln κ m A λ Bm Now let s mpose the boundary condton Π ( m ) = where the prme represents a dervatve wth respect to. Ths gves κ A λ =. Hence, κ A =, and so A = O(λ ). Ths s because A clearly does not contrbute at lowest order (otherwse the free propagator would have ths counterterm vertex n t), and we ve just shown that t does not contrbute at frst order. The self-energy s now: Π( ) = m λ µ (4π) ε + + ln Bm m Now for B. Ths s nether fnte nor ndependent of µ, so we wll choose: B = λ µ (4π) ε + + ln + λκ m B + O(λ ) 7
8 whch s Now we have: B = λ [ (4π) ε + ( µ ) ] + ln + λκ B + O(λ ) m Π( ) = λκ B m + O(λ ) Imposng the other boundary condton, Π( m ) =, we fnd that κ B =. Then, and B = λ [ (4π) ε + ( µ ) ] + ln + O(λ ) m Π( ) = O(λ ) So the propagator s uncorrected to frst order n λ. Further, we found that the counterterm vertex does contrbute at O(λ). Srednc 4.6. Repeat problem 4.5 for the theory of problem 9.3. Fortunately, ths theory s nearly dentcal to that of the prevous problem, the only dfference s the symmetry factor. Thans to the complex feld n problem 9.3, S =, and so there s an extra factor of n the frst dagram that propagates through the problem. The results are: A = O(λ ) B = λ µ (4π) ε + + ln + O(λ ) m Π( ) = O(λ ) Srednc 4.7. Renormalzaton of the anharmonc oscllator. Consder an anharmonc oscllator, specfed by the Lagrangan L = Z q Z ωω q Z λ λω 3 q 4 We set h = and m = ; λ s then dmensonless. (a) Fnd the Hamltonan H correspondng to L. Wrte t as H = H + H, where H = P + ω Q, and [Q, P] =. We tae the conjugate momentum: Then, the Hamltonan s gven by: P = L q = Z q H = P q L
9 H = Z q + Z ωω q + Z λ λω 3 q 4 whch s, tang Q = q and P of course s the conjugate momentum as defned above, H = Z P + Z ωω Q + Z λ λω 3 Q 4 Then: H = P + ω Q + (Z )P + (Z ω )ω Q + Z λ λω 3 Q 4 where the frst two terms are H and the remander are H. Choosng P to be the conjugate momentum of Q means that the commutaton relaton requred by the problem s automatcally satsfed. (b) Let and be the ground and frst excted states of H, and let Ω and I be the ground and frst excted states of H. We tae all these egenstates to have unt norm. We defne ω to be the exctaton energy of H, ω = E I E Ω. We normalze the poston operator Q by settng I Q Ω = Q = (ω) /. Fnally, to mae thngs mathematcally smpler, we set Z λ equal to one, rather than usng a more physcally motvated defnton. Wrte Z = + A and Z ω = + B, where A = κ A λ + O(λ ) and B = κ B λ + O(λ ). Use Raylegh-Schrödnger perturbaton theory to compute the O(λ) correctons to the unperturbed energy egenvalues and egenstates. Ths s just quantum mechancs. The energy of the unperturbed th state s gven by E = (+)ω. The frst-order correcton to ths s, by Shanar 7..7: E = H whch s: E = (Z )P + (Z ω )ω Q + Z λ λω 3 Q 4 Next we set Z λ =, Z = + A, and Z ω = + B. Then, E = AP + Bω Q + λω 3 Q 4 Splttng ths up, and usng the expanson of A and B, we have: E = κ Aλ P + κ Bλω Q + λω 3 Q 4 Now we quantze the operators n the usual way, see Shanar 7.4. and 7.4.9, whch state that: Q = (a + a) ω ω P = (a a) 9
10 Ths scheme s great because the poston operator s normalzed out of the box to (ω) / (between and ) as Srednc prescrbed. Thus our correcton to the energy egenvalues are: E = ωλ 4 κ A (a a) + ωλ 4 κ B (a + a ) + λω 4 (a + a ) 4 whch s: E = ωλ 4 κ A aa + a a + ωλ 4 κ B aa + a a + λω 4 aaa a + aa aa + aa a a + a aaa Now let s smplfy to the case where = : +a aa a + a a aa E = ωλ 4 κ A + ωλ 4 κ B + 3λω 4 Hence, the ground state energy of the perturbed system s: Now for the case where = : E Ω = ω + λω 4 ( κ A + κ B + 3) + O(λ ) E = 3ωλ 4 κ A + 3ωλ 4 κ B + λω 4 ( ) Hence, the frst excted state energy of the perturbed system s E I = 3ω + 3ωλ 4 ( κ A + κ B + 5) Now for the correctons to the egenstates, gven by Shanar 7..4: Thus: Ω = + n m whch smplfes to: Ω = + n m ω (m+)ω n = n + n m m H m E Em [ κ Aλ m P + ] κ Bλω m Q + λω 3 m Q 4 m λ [ κa m (a a) κ B m (a + a) m (a + a) 4 ] m 4m In the frst two bra-ets, (a ± a) = a a ± aa ± a a + aa. These last two terms annhlate the vacuum, gvng a zero. The second term gves a zero also snce m n. Then: Ω = + n m λ [ κa m a a κ B m a a m (a + a) 4 ] m 4m
11 These frst two terms are only defned f m =. Then, Smplfyng: Ω = + λ [ κa a a κ B a a ] m n m λ 4m m (a + a) 4 m λ(κa + κ B ) Ω = λ 4m m (a + a) 4 m n m Dong the expansons as before, we fnd that the last bra-et gves 4 f m = 4, and 6 f m =. Then, λ(κa + κ B ) Ω = 3λ 6λ 4 4 Ths gves: λ(κa + κ B + 6) Ω = 6λ 4 + O(λ ) We do the same procedure for I, I ll sp the gory detals. The result s: I = λ 6(κ A + κ B + ) 3 λ O(λ ) (c) Fnd the numercal values of κ A and κ B that yeld ω = E I E Ω and I Q Ω = (ω) /. Imposng the frst condton, we have: Ths mples that: Imposng the second condton, we have: [ I Q Ω = ω E I E Ω = ω + ωλ ( κ A + κ B + 6) = ω κ A κ B = 6 (4.7.) λ 6(κ A + κ B + ) ] 3 λ 3 5 ( a + a ) [ ] λ(κa + κ B + 6) 6λ 4 = ω Usng the operators and equaton (4.7.), we have: [ ] I Q Ω = λ 6(κ A + ) 3 λ 3 5 ω 4 [ 6λκA 3 λκ ] A 3λ 6λ
12 Ths gves: I Q Ω = [ λκ ] A ω + O(λ ) Hence, κ A =. Usng equaton (4.7.), we have then that κ B = 6. (d) Now thn of the Lagrangan n equaton 4.54 as specfyng a quantum feld theory n d = dmensons. Compute the O(λ) correcton to the propagator. Fx κ A and κ B by requrng the propagator to have a pole at = ω wth resdue one. Do your results agree wth those of part (c)? Should they? Let s start by wrtng the Lagrangan agan: Now we ll wrte ths n the usual way: L = Z q Z ωω q Z λ λω 3 q 4 L = q ω q + (Z ) q (Z ω )ω q Z λ λω 3 q 4 where the frst two terms are the free feld terms, the second two terms are the counterterms, and the fnal term s the nteracton. Next, let s draw the terms that would correct the propagator at O(λ): l x Ths s a new theory, so we have to use new Feynman rules. Fortunately these are easy to derve from the Lagrangan. For the frst term, we have the vertex factor, the nternal propagator, the symmetry factor, and the ntegral over l. Then, Π ( ) = 4λω 3 dl l + ω ε π where the symmetry factor s snce exchangng the two ends of the propagator s equvalent to exchangng those two vertces, the vertex factor ncludes a factor of 4 snce the Qs can be pared wth the vertces n any of 4! dfferent ways, and the ntegral s one dmensonal snce the problem tells us to wor on one dmenson. By the way, f the factor of 4 seems unusual, that s because most Lagrangans have cleverly-chosen coeffcents to cancel these factors, see for example the Lagrangan of problem 9.. Note also that we have ω rather than m, snce ω has taen the place of mass n our problem. Next we do a Wc rotaton, the result s: Π ( ) = 6λω3 π l + ω dl
13 Ths ntegral s smple to evaluate, usng equaton 4.7 or other methods. We fnd: Π ( ) = 6λω That seems suffcently smplfed! Let s turn to the contrbuton from the second dagram. The only thng we have to worry about s the vertex factor, whch we read off from the counterterm Lagrangan. The result s: Π ( ) = ( (Z ) (Z ω )ω ) Ths should not be unclear, but f t s, t can also be obtaned by comparng our Lagrangan wth equaton 9.9, and mang the necessary substtutons to modfy Feynman Rule #9 on page 77. At any rate, we now use the nformaton n part (b), Z = κ A λ and Z ω = κ B λ. Then, Π ( ) = λ ( κ A κ B ω ) The entre self-energy (at O(λ)) s then: Π( ) = 6λω λ ( κ A κ B ω ) The propagator wll have a pole at = m wth resdue one f the boundary condton 4.7 and 4. are met. Imposng 4., we fnd: Hence κ A =. Imposng 4.7, we have: Π ( ) = λκ A = Π( m ) = 6λω + λκ B ω = whch mples that κ B = 6, n agreement wth the result from part (c). Fnally, we wrte the correcton to the propagator: ( ) = + ω ε + O(λ ) Fnally, we have the queston of whether the agreement between (c) and (d) s expected. Our method of correctng the propagator n part (d) nvolves usng our Feynman Rules, whch we derved from the LSZ formula, whch depends on the Klen-Gordon Equaton. It may seem as though ths s therefore ncompatble wth quantum mechancs. On the other hand, perturbaton theory s not strctly lmted to quantum mechancs: a close loo at the dervaton of perturbaton theory (see for example Grffths 6.) shows that there s no dependence on the Schrödnger equaton or any other tenets that volate feld theory. In fact, the bra-ets that we evaluated n part (b) are dependent only on the defnton of ω and Q, and both of those are defned n a way whch s consstent wth feld theory. Hence, we do expect these to agree. Note: In Srednc s solutons, hs equaton 4.6 has the order of the terms n the denomnator reversed. Ths s easly verfed, see for example Shanar 7..3, Saura 5..44, or Grffths 6.3. Fortunately, ths just results n a negatve sgn that cancels out n the mddle, so hs conclusons are correct, though some terms n hs corrected wave-functons n part (b) are off by a sgn. 3
12. The Hamilton-Jacobi Equation Michael Fowler
1. The Hamlton-Jacob Equaton Mchael Fowler Back to Confguraton Space We ve establshed that the acton, regarded as a functon of ts coordnate endponts and tme, satsfes ( ) ( ) S q, t / t+ H qpt,, = 0, and
More informationThe Feynman path integral
The Feynman path ntegral Aprl 3, 205 Hesenberg and Schrödnger pctures The Schrödnger wave functon places the tme dependence of a physcal system n the state, ψ, t, where the state s a vector n Hlbert space
More informationCausal Diamonds. M. Aghili, L. Bombelli, B. Pilgrim
Causal Damonds M. Aghl, L. Bombell, B. Plgrm Introducton The correcton to volume of a causal nterval due to curvature of spacetme has been done by Myrhem [] and recently by Gbbons & Solodukhn [] and later
More informationLecture 20: Noether s Theorem
Lecture 20: Noether s Theorem In our revew of Newtonan Mechancs, we were remnded that some quanttes (energy, lnear momentum, and angular momentum) are conserved That s, they are constant f no external
More informationC/CS/Phy191 Problem Set 3 Solutions Out: Oct 1, 2008., where ( 00. ), so the overall state of the system is ) ( ( ( ( 00 ± 11 ), Φ ± = 1
C/CS/Phy9 Problem Set 3 Solutons Out: Oct, 8 Suppose you have two qubts n some arbtrary entangled state ψ You apply the teleportaton protocol to each of the qubts separately What s the resultng state obtaned
More information1 GSW Iterative Techniques for y = Ax
1 for y = A I m gong to cheat here. here are a lot of teratve technques that can be used to solve the general case of a set of smultaneous equatons (wrtten n the matr form as y = A), but ths chapter sn
More informationCanonical transformations
Canoncal transformatons November 23, 2014 Recall that we have defned a symplectc transformaton to be any lnear transformaton M A B leavng the symplectc form nvarant, Ω AB M A CM B DΩ CD Coordnate transformatons,
More informationSrednicki Chapter 34
Srednck Chapter 3 QFT Problems & Solutons A. George January 0, 203 Srednck 3.. Verfy that equaton 3.6 follows from equaton 3.. We take Λ = + δω: U + δω ψu + δω = + δωψ[ + δω] x Next we use equaton 3.3,
More informationFrom Biot-Savart Law to Divergence of B (1)
From Bot-Savart Law to Dvergence of B (1) Let s prove that Bot-Savart gves us B (r ) = 0 for an arbtrary current densty. Frst take the dvergence of both sdes of Bot-Savart. The dervatve s wth respect to
More informationA how to guide to second quantization method.
Phys. 67 (Graduate Quantum Mechancs Sprng 2009 Prof. Pu K. Lam. Verson 3 (4/3/2009 A how to gude to second quantzaton method. -> Second quantzaton s a mathematcal notaton desgned to handle dentcal partcle
More informationPHYS 705: Classical Mechanics. Calculus of Variations II
1 PHYS 705: Classcal Mechancs Calculus of Varatons II 2 Calculus of Varatons: Generalzaton (no constrant yet) Suppose now that F depends on several dependent varables : We need to fnd such that has a statonary
More information1 Matrix representations of canonical matrices
1 Matrx representatons of canoncal matrces 2-d rotaton around the orgn: ( ) cos θ sn θ R 0 = sn θ cos θ 3-d rotaton around the x-axs: R x = 1 0 0 0 cos θ sn θ 0 sn θ cos θ 3-d rotaton around the y-axs:
More informationMechanics Physics 151
Mechancs Physcs 5 Lecture 0 Canoncal Transformatons (Chapter 9) What We Dd Last Tme Hamlton s Prncple n the Hamltonan formalsm Dervaton was smple δi δ p H(, p, t) = 0 Adonal end-pont constrants δ t ( )
More informationQuantum Field Theory Homework 5
Quantum Feld Theory Homework 5 Erc Cotner February 19, 15 1) Renormalzaton n φ 4 Theory We take the φ 4 theory n D = 4 spacetme: L = 1 µφ µ φ 1 m φ λ 4! φ4 We wsh to fnd all the dvergent (connected, 1PI
More informationLagrangian Field Theory
Lagrangan Feld Theory Adam Lott PHY 391 Aprl 6, 017 1 Introducton Ths paper s a summary of Chapter of Mandl and Shaw s Quantum Feld Theory [1]. The frst thng to do s to fx the notaton. For the most part,
More informationψ ij has the eigenvalue
Moller Plesset Perturbaton Theory In Moller-Plesset (MP) perturbaton theory one taes the unperturbed Hamltonan for an atom or molecule as the sum of the one partcle Foc operators H F() where the egenfunctons
More information8.6 The Complex Number System
8.6 The Complex Number System Earler n the chapter, we mentoned that we cannot have a negatve under a square root, snce the square of any postve or negatve number s always postve. In ths secton we want
More informationErrors for Linear Systems
Errors for Lnear Systems When we solve a lnear system Ax b we often do not know A and b exactly, but have only approxmatons  and ˆb avalable. Then the best thng we can do s to solve ˆx ˆb exactly whch
More information1 (1 + ( )) = 1 8 ( ) = (c) Carrying out the Taylor expansion, in this case, the series truncates at second order:
68A Solutons to Exercses March 05 (a) Usng a Taylor expanson, and notng that n 0 for all n >, ( + ) ( + ( ) + ) We can t nvert / because there s no Taylor expanson around 0 Lets try to calculate the nverse
More informationEconomics 101. Lecture 4 - Equilibrium and Efficiency
Economcs 0 Lecture 4 - Equlbrum and Effcency Intro As dscussed n the prevous lecture, we wll now move from an envronment where we looed at consumers mang decsons n solaton to analyzng economes full of
More informationSection 8.3 Polar Form of Complex Numbers
80 Chapter 8 Secton 8 Polar Form of Complex Numbers From prevous classes, you may have encountered magnary numbers the square roots of negatve numbers and, more generally, complex numbers whch are the
More informationAndre Schneider P622
Andre Schneder P6 Probem Set #0 March, 00 Srednc 7. Suppose that we have a theory wth Negectng the hgher order terms, show that Souton Knowng β(α and γ m (α we can wrte β(α =b α O(α 3 (. γ m (α =c α O(α
More informationHomework & Solution. Contributors. Prof. Lee, Hyun Min. Particle Physics Winter School. Park, Ye
Homework & Soluton Prof. Lee, Hyun Mn Contrbutors Park, Ye J(yej.park@yonse.ac.kr) Lee, Sung Mook(smlngsm0919@gmal.com) Cheong, Dhong Yeon(dhongyeoncheong@gmal.com) Ban, Ka Young(ban94gy@yonse.ac.kr) Ro,
More informationTHEOREMS OF QUANTUM MECHANICS
THEOREMS OF QUANTUM MECHANICS In order to develop methods to treat many-electron systems (atoms & molecules), many of the theorems of quantum mechancs are useful. Useful Notaton The matrx element A mn
More informationLecture 6/7 (February 10/12, 2014) DIRAC EQUATION. The non-relativistic Schrödinger equation was obtained by noting that the Hamiltonian 2
P470 Lecture 6/7 (February 10/1, 014) DIRAC EQUATION The non-relatvstc Schrödnger equaton was obtaned by notng that the Hamltonan H = P (1) m can be transformed nto an operator form wth the substtutons
More informationOne-sided finite-difference approximations suitable for use with Richardson extrapolation
Journal of Computatonal Physcs 219 (2006) 13 20 Short note One-sded fnte-dfference approxmatons sutable for use wth Rchardson extrapolaton Kumar Rahul, S.N. Bhattacharyya * Department of Mechancal Engneerng,
More informationTransfer Functions. Convenient representation of a linear, dynamic model. A transfer function (TF) relates one input and one output: ( ) system
Transfer Functons Convenent representaton of a lnear, dynamc model. A transfer functon (TF) relates one nput and one output: x t X s y t system Y s The followng termnology s used: x y nput output forcng
More informationAdvanced Quantum Mechanics
Advanced Quantum Mechancs Rajdeep Sensarma! sensarma@theory.tfr.res.n ecture #9 QM of Relatvstc Partcles Recap of ast Class Scalar Felds and orentz nvarant actons Complex Scalar Feld and Charge conjugaton
More informationLecture 12: Discrete Laplacian
Lecture 12: Dscrete Laplacan Scrbe: Tanye Lu Our goal s to come up wth a dscrete verson of Laplacan operator for trangulated surfaces, so that we can use t n practce to solve related problems We are mostly
More informationFoundations of Arithmetic
Foundatons of Arthmetc Notaton We shall denote the sum and product of numbers n the usual notaton as a 2 + a 2 + a 3 + + a = a, a 1 a 2 a 3 a = a The notaton a b means a dvdes b,.e. ac = b where c s an
More informationMore metrics on cartesian products
More metrcs on cartesan products If (X, d ) are metrc spaces for 1 n, then n Secton II4 of the lecture notes we defned three metrcs on X whose underlyng topologes are the product topology The purpose of
More informationON MECHANICS WITH VARIABLE NONCOMMUTATIVITY
ON MECHANICS WITH VARIABLE NONCOMMUTATIVITY CIPRIAN ACATRINEI Natonal Insttute of Nuclear Physcs and Engneerng P.O. Box MG-6, 07725-Bucharest, Romana E-mal: acatrne@theory.npne.ro. Receved March 6, 2008
More informationCME 302: NUMERICAL LINEAR ALGEBRA FALL 2005/06 LECTURE 13
CME 30: NUMERICAL LINEAR ALGEBRA FALL 005/06 LECTURE 13 GENE H GOLUB 1 Iteratve Methods Very large problems (naturally sparse, from applcatons): teratve methods Structured matrces (even sometmes dense,
More informationSalmon: Lectures on partial differential equations. Consider the general linear, second-order PDE in the form. ,x 2
Salmon: Lectures on partal dfferental equatons 5. Classfcaton of second-order equatons There are general methods for classfyng hgher-order partal dfferental equatons. One s very general (applyng even to
More information9 Characteristic classes
THEODORE VORONOV DIFFERENTIAL GEOMETRY. Sprng 2009 [under constructon] 9 Characterstc classes 9.1 The frst Chern class of a lne bundle Consder a complex vector bundle E B of rank p. We shall construct
More information8.323 Relativistic Quantum Field Theory I
MI OpenCourseWare http://ocw.mt.edu 8.323 Relatvstc Quantum Feld heory I Sprng 2008 For nformaton about ctng these materals or our erms of Use, vst: http://ocw.mt.edu/terms. MASSACHUSES INSIUE OF ECHNOLOGY
More information10. Canonical Transformations Michael Fowler
10. Canoncal Transformatons Mchael Fowler Pont Transformatons It s clear that Lagrange s equatons are correct for any reasonable choce of parameters labelng the system confguraton. Let s call our frst
More informationWorkshop: Approximating energies and wave functions Quantum aspects of physical chemistry
Workshop: Approxmatng energes and wave functons Quantum aspects of physcal chemstry http://quantum.bu.edu/pltl/6/6.pdf Last updated Thursday, November 7, 25 7:9:5-5: Copyrght 25 Dan Dll (dan@bu.edu) Department
More informationLectures - Week 4 Matrix norms, Conditioning, Vector Spaces, Linear Independence, Spanning sets and Basis, Null space and Range of a Matrix
Lectures - Week 4 Matrx norms, Condtonng, Vector Spaces, Lnear Independence, Spannng sets and Bass, Null space and Range of a Matrx Matrx Norms Now we turn to assocatng a number to each matrx. We could
More informationHW #6, due Oct Toy Dirac Model, Wick s theorem, LSZ reduction formula. Consider the following quantum mechanics Lagrangian,
HW #6, due Oct 5. Toy Drac Model, Wck s theorem, LSZ reducton formula. Consder the followng quantum mechancs Lagrangan, L ψ(σ 3 t m)ψ, () where σ 3 s a Paul matrx, and ψ s defned by ψ ψ σ 3. ψ s a twocomponent
More informationDifferentiating Gaussian Processes
Dfferentatng Gaussan Processes Andrew McHutchon Aprl 17, 013 1 Frst Order Dervatve of the Posteror Mean The posteror mean of a GP s gven by, f = x, X KX, X 1 y x, X α 1 Only the x, X term depends on the
More informationBernoulli Numbers and Polynomials
Bernoull Numbers and Polynomals T. Muthukumar tmk@tk.ac.n 17 Jun 2014 The sum of frst n natural numbers 1, 2, 3,..., n s n n(n + 1 S 1 (n := m = = n2 2 2 + n 2. Ths formula can be derved by notng that
More informationCHAPTER 14 GENERAL PERTURBATION THEORY
CHAPTER 4 GENERAL PERTURBATION THEORY 4 Introducton A partcle n orbt around a pont mass or a sphercally symmetrc mass dstrbuton s movng n a gravtatonal potental of the form GM / r In ths potental t moves
More informationPhysics 5153 Classical Mechanics. Principle of Virtual Work-1
P. Guterrez 1 Introducton Physcs 5153 Classcal Mechancs Prncple of Vrtual Work The frst varatonal prncple we encounter n mechancs s the prncple of vrtual work. It establshes the equlbrum condton of a mechancal
More informationMath1110 (Spring 2009) Prelim 3 - Solutions
Math 1110 (Sprng 2009) Solutons to Prelm 3 (04/21/2009) 1 Queston 1. (16 ponts) Short answer. Math1110 (Sprng 2009) Prelm 3 - Solutons x a 1 (a) (4 ponts) Please evaluate lm, where a and b are postve numbers.
More informationQuantum Field Theory III
Quantum Feld Theory III Prof. Erck Wenberg February 16, 011 1 Lecture 9 Last tme we showed that f we just look at weak nteractons and currents, strong nteracton has very good SU() SU() chral symmetry,
More information14 The Postulates of Quantum mechanics
14 The Postulates of Quantum mechancs Postulate 1: The state of a system s descrbed completely n terms of a state vector Ψ(r, t), whch s quadratcally ntegrable. Postulate 2: To every physcally observable
More informationSpin-rotation coupling of the angularly accelerated rigid body
Spn-rotaton couplng of the angularly accelerated rgd body Loua Hassan Elzen Basher Khartoum, Sudan. Postal code:11123 E-mal: louaelzen@gmal.com November 1, 2017 All Rghts Reserved. Abstract Ths paper s
More informationLinear Approximation with Regularization and Moving Least Squares
Lnear Approxmaton wth Regularzaton and Movng Least Squares Igor Grešovn May 007 Revson 4.6 (Revson : March 004). 5 4 3 0.5 3 3.5 4 Contents: Lnear Fttng...4. Weghted Least Squares n Functon Approxmaton...
More informationNUMERICAL DIFFERENTIATION
NUMERICAL DIFFERENTIATION 1 Introducton Dfferentaton s a method to compute the rate at whch a dependent output y changes wth respect to the change n the ndependent nput x. Ths rate of change s called the
More informationTHE CHINESE REMAINDER THEOREM. We should thank the Chinese for their wonderful remainder theorem. Glenn Stevens
THE CHINESE REMAINDER THEOREM KEITH CONRAD We should thank the Chnese for ther wonderful remander theorem. Glenn Stevens 1. Introducton The Chnese remander theorem says we can unquely solve any par of
More informationModelli Clamfim Equazione del Calore Lezione ottobre 2014
CLAMFIM Bologna Modell 1 @ Clamfm Equazone del Calore Lezone 17 15 ottobre 2014 professor Danele Rtell danele.rtell@unbo.t 1/24? Convoluton The convoluton of two functons g(t) and f(t) s the functon (g
More informationPHYS 705: Classical Mechanics. Hamilton-Jacobi Equation
1 PHYS 705: Classcal Mechancs Hamlton-Jacob Equaton Hamlton-Jacob Equaton There s also a very elegant relaton between the Hamltonan Formulaton of Mechancs and Quantum Mechancs. To do that, we need to derve
More informationIntegrals and Invariants of Euler-Lagrange Equations
Lecture 16 Integrals and Invarants of Euler-Lagrange Equatons ME 256 at the Indan Insttute of Scence, Bengaluru Varatonal Methods and Structural Optmzaton G. K. Ananthasuresh Professor, Mechancal Engneerng,
More informationOpen string operator quantization
Open strng operator quantzaton Requred readng: Zwebach -4 Suggested readng: Polchnsk 3 Green, Schwarz, & Wtten 3 upto eq 33 The lght-cone strng as a feld theory: Today we wll dscuss the quantzaton of an
More informationMolecular structure: Diatomic molecules in the rigid rotor and harmonic oscillator approximations Notes on Quantum Mechanics
Molecular structure: Datomc molecules n the rgd rotor and harmonc oscllator approxmatons Notes on Quantum Mechancs http://quantum.bu.edu/notes/quantummechancs/molecularstructuredatomc.pdf Last updated
More informationComplex Numbers. x = B B 2 4AC 2A. or x = x = 2 ± 4 4 (1) (5) 2 (1)
Complex Numbers If you have not yet encountered complex numbers, you wll soon do so n the process of solvng quadratc equatons. The general quadratc equaton Ax + Bx + C 0 has solutons x B + B 4AC A For
More informationPhysics 5153 Classical Mechanics. D Alembert s Principle and The Lagrangian-1
P. Guterrez Physcs 5153 Classcal Mechancs D Alembert s Prncple and The Lagrangan 1 Introducton The prncple of vrtual work provdes a method of solvng problems of statc equlbrum wthout havng to consder the
More informationDynamics of a Superconducting Qubit Coupled to an LC Resonator
Dynamcs of a Superconductng Qubt Coupled to an LC Resonator Y Yang Abstract: We nvestgate the dynamcs of a current-based Josephson juncton quantum bt or qubt coupled to an LC resonator. The Hamltonan of
More informationψ = i c i u i c i a i b i u i = i b 0 0 b 0 0
Quantum Mechancs, Advanced Course FMFN/FYSN7 Solutons Sheet Soluton. Lets denote the two operators by  and ˆB, the set of egenstates by { u }, and the egenvalues as  u = a u and ˆB u = b u. Snce the
More informationPhysics 443, Solutions to PS 7
Physcs 443, Solutons to PS 7. Grffths 4.50 The snglet confguraton state s χ ) χ + χ χ χ + ) where that second equaton defnes the abbrevated notaton χ + and χ. S a ) S ) b χ â S )ˆb S ) χ In sphercal coordnates
More informationTensor Analysis. For orthogonal curvilinear coordinates, ˆ ˆ (98) Expanding the derivative, we have, ˆ. h q. . h q h q
For orthogonal curvlnear coordnates, eˆ grad a a= ( aˆ ˆ e). h q (98) Expandng the dervatve, we have, eˆ aˆ ˆ e a= ˆ ˆ a h e + q q 1 aˆ ˆ ˆ a e = ee ˆˆ ˆ + e. h q h q Now expandng eˆ / q (some of the detals
More informationSolutions to Problem Set 6
Solutons to Problem Set 6 Problem 6. (Resdue theory) a) Problem 4.7.7 Boas. n ths problem we wll solve ths ntegral: x sn x x + 4x + 5 dx: To solve ths usng the resdue theorem, we study ths complex ntegral:
More information763622S ADVANCED QUANTUM MECHANICS Solution Set 1 Spring c n a n. c n 2 = 1.
7636S ADVANCED QUANTUM MECHANICS Soluton Set 1 Sprng 013 1 Warm-up Show that the egenvalues of a Hermtan operator  are real and that the egenkets correspondng to dfferent egenvalues are orthogonal (b)
More informationFeynman parameter integrals
Feynman parameter ntegrals We often deal wth products of many propagator factors n loop ntegrals. The trck s to combne many propagators nto a sngle fracton so that the four-momentum ntegraton can be done
More informationMechanics Physics 151
Mechancs Physcs 151 Lecture 3 Lagrange s Equatons (Goldsten Chapter 1) Hamlton s Prncple (Chapter 2) What We Dd Last Tme! Dscussed mult-partcle systems! Internal and external forces! Laws of acton and
More informationTHE VIBRATIONS OF MOLECULES II THE CARBON DIOXIDE MOLECULE Student Instructions
THE VIBRATIONS OF MOLECULES II THE CARBON DIOXIDE MOLECULE Student Instructons by George Hardgrove Chemstry Department St. Olaf College Northfeld, MN 55057 hardgrov@lars.acc.stolaf.edu Copyrght George
More informationHomework Notes Week 7
Homework Notes Week 7 Math 4 Sprng 4 #4 (a Complete the proof n example 5 that s an nner product (the Frobenus nner product on M n n (F In the example propertes (a and (d have already been verfed so we
More informationπ e ax2 dx = x 2 e ax2 dx or x 3 e ax2 dx = 1 x 4 e ax2 dx = 3 π 8a 5/2 (a) We are considering the Maxwell velocity distribution function: 2πτ/m
Homework Solutons Problem In solvng ths problem, we wll need to calculate some moments of the Gaussan dstrbuton. The brute-force method s to ntegrate by parts but there s a nce trck. The followng ntegrals
More informationMechanics Physics 151
Mechancs Physcs 5 Lecture 7 Specal Relatvty (Chapter 7) What We Dd Last Tme Worked on relatvstc knematcs Essental tool for epermental physcs Basc technques are easy: Defne all 4 vectors Calculate c-o-m
More informationLecture Notes 7: The Unruh Effect
Quantum Feld Theory for Leg Spnners 17/1/11 Lecture Notes 7: The Unruh Effect Lecturer: Prakash Panangaden Scrbe: Shane Mansfeld 1 Defnng the Vacuum Recall from the last lecture that choosng a complex
More informationGeorgia Tech PHYS 6124 Mathematical Methods of Physics I
Georga Tech PHYS 624 Mathematcal Methods of Physcs I Instructor: Predrag Cvtanovć Fall semester 202 Homework Set #7 due October 30 202 == show all your work for maxmum credt == put labels ttle legends
More informationEPR Paradox and the Physical Meaning of an Experiment in Quantum Mechanics. Vesselin C. Noninski
EPR Paradox and the Physcal Meanng of an Experment n Quantum Mechancs Vesseln C Nonnsk vesselnnonnsk@verzonnet Abstract It s shown that there s one purely determnstc outcome when measurement s made on
More informationRobert Eisberg Second edition CH 09 Multielectron atoms ground states and x-ray excitations
Quantum Physcs 量 理 Robert Esberg Second edton CH 09 Multelectron atoms ground states and x-ray exctatons 9-01 By gong through the procedure ndcated n the text, develop the tme-ndependent Schroednger equaton
More informationEcon107 Applied Econometrics Topic 3: Classical Model (Studenmund, Chapter 4)
I. Classcal Assumptons Econ7 Appled Econometrcs Topc 3: Classcal Model (Studenmund, Chapter 4) We have defned OLS and studed some algebrac propertes of OLS. In ths topc we wll study statstcal propertes
More informationDigital Signal Processing
Dgtal Sgnal Processng Dscrete-tme System Analyss Manar Mohasen Offce: F8 Emal: manar.subh@ut.ac.r School of IT Engneerng Revew of Precedent Class Contnuous Sgnal The value of the sgnal s avalable over
More informationTrees and Order Conditions
Trees and Order Condtons Constructon of Runge-Kutta order condtons usng Butcher trees and seres. Paul Tranqull 1 1 Computatonal Scence Laboratory CSL) Department of Computer Scence Vrgna Tech. Trees and
More informationDensity matrix. c α (t)φ α (q)
Densty matrx Note: ths s supplementary materal. I strongly recommend that you read t for your own nterest. I beleve t wll help wth understandng the quantum ensembles, but t s not necessary to know t n
More informationAdvanced Circuits Topics - Part 1 by Dr. Colton (Fall 2017)
Advanced rcuts Topcs - Part by Dr. olton (Fall 07) Part : Some thngs you should already know from Physcs 0 and 45 These are all thngs that you should have learned n Physcs 0 and/or 45. Ths secton s organzed
More informationLimited Dependent Variables
Lmted Dependent Varables. What f the left-hand sde varable s not a contnuous thng spread from mnus nfnty to plus nfnty? That s, gven a model = f (, β, ε, where a. s bounded below at zero, such as wages
More information3.1 Expectation of Functions of Several Random Variables. )' be a k-dimensional discrete or continuous random vector, with joint PMF p (, E X E X1 E X
Statstcs 1: Probablty Theory II 37 3 EPECTATION OF SEVERAL RANDOM VARIABLES As n Probablty Theory I, the nterest n most stuatons les not on the actual dstrbuton of a random vector, but rather on a number
More informationLecture 13 APPROXIMATION OF SECOMD ORDER DERIVATIVES
COMPUTATIONAL FLUID DYNAMICS: FDM: Appromaton of Second Order Dervatves Lecture APPROXIMATION OF SECOMD ORDER DERIVATIVES. APPROXIMATION OF SECOND ORDER DERIVATIVES Second order dervatves appear n dffusve
More informationMMA and GCMMA two methods for nonlinear optimization
MMA and GCMMA two methods for nonlnear optmzaton Krster Svanberg Optmzaton and Systems Theory, KTH, Stockholm, Sweden. krlle@math.kth.se Ths note descrbes the algorthms used n the author s 2007 mplementatons
More informationU.C. Berkeley CS294: Beyond Worst-Case Analysis Luca Trevisan September 5, 2017
U.C. Berkeley CS94: Beyond Worst-Case Analyss Handout 4s Luca Trevsan September 5, 07 Summary of Lecture 4 In whch we ntroduce semdefnte programmng and apply t to Max Cut. Semdefnte Programmng Recall that
More informationPhysics 4B. A positive value is obtained, so the current is counterclockwise around the circuit.
Physcs 4B Solutons to Chapter 7 HW Chapter 7: Questons:, 8, 0 Problems:,,, 45, 48,,, 7, 9 Queston 7- (a) no (b) yes (c) all te Queston 7-8 0 μc Queston 7-0, c;, a;, d; 4, b Problem 7- (a) Let be the current
More informationAPPENDIX A Some Linear Algebra
APPENDIX A Some Lnear Algebra The collecton of m, n matrces A.1 Matrces a 1,1,..., a 1,n A = a m,1,..., a m,n wth real elements a,j s denoted by R m,n. If n = 1 then A s called a column vector. Smlarly,
More informationPoisson brackets and canonical transformations
rof O B Wrght Mechancs Notes osson brackets and canoncal transformatons osson Brackets Consder an arbtrary functon f f ( qp t) df f f f q p q p t But q p p where ( qp ) pq q df f f f p q q p t In order
More informationLecture 14: Forces and Stresses
The Nuts and Bolts of Frst-Prncples Smulaton Lecture 14: Forces and Stresses Durham, 6th-13th December 2001 CASTEP Developers Group wth support from the ESF ψ k Network Overvew of Lecture Why bother? Theoretcal
More informationANSWERS. Problem 1. and the moment generating function (mgf) by. defined for any real t. Use this to show that E( U) var( U)
Econ 413 Exam 13 H ANSWERS Settet er nndelt 9 deloppgaver, A,B,C, som alle anbefales å telle lkt for å gøre det ltt lettere å stå. Svar er gtt . Unfortunately, there s a prntng error n the hnt of
More information( ) 2 ( ) ( ) Problem Set 4 Suggested Solutions. Problem 1
Problem Set 4 Suggested Solutons Problem (A) The market demand functon s the soluton to the followng utlty-maxmzaton roblem (UMP): The Lagrangean: ( x, x, x ) = + max U x, x, x x x x st.. x + x + x y x,
More informationLecture 5.8 Flux Vector Splitting
Lecture 5.8 Flux Vector Splttng 1 Flux Vector Splttng The vector E n (5.7.) can be rewrtten as E = AU (5.8.1) (wth A as gven n (5.7.4) or (5.7.6) ) whenever, the equaton of state s of the separable form
More informationLecture 10 Support Vector Machines II
Lecture 10 Support Vector Machnes II 22 February 2016 Taylor B. Arnold Yale Statstcs STAT 365/665 1/28 Notes: Problem 3 s posted and due ths upcomng Frday There was an early bug n the fake-test data; fxed
More informationImplicit Integration Henyey Method
Implct Integraton Henyey Method In realstc stellar evoluton codes nstead of a drect ntegraton usng for example the Runge-Kutta method one employs an teratve mplct technque. Ths s because the structure
More informationClassical Field Theory
Classcal Feld Theory Before we embark on quantzng an nteractng theory, we wll take a dverson nto classcal feld theory and classcal perturbaton theory and see how far we can get. The reader s expected to
More informationNon-interacting Spin-1/2 Particles in Non-commuting External Magnetic Fields
EJTP 6, No. 0 009) 43 56 Electronc Journal of Theoretcal Physcs Non-nteractng Spn-1/ Partcles n Non-commutng External Magnetc Felds Kunle Adegoke Physcs Department, Obafem Awolowo Unversty, Ile-Ife, Ngera
More informationIn this section is given an overview of the common elasticity models.
Secton 4.1 4.1 Elastc Solds In ths secton s gven an overvew of the common elastcty models. 4.1.1 The Lnear Elastc Sold The classcal Lnear Elastc model, or Hooean model, has the followng lnear relatonshp
More informationSome modelling aspects for the Matlab implementation of MMA
Some modellng aspects for the Matlab mplementaton of MMA Krster Svanberg krlle@math.kth.se Optmzaton and Systems Theory Department of Mathematcs KTH, SE 10044 Stockholm September 2004 1. Consdered optmzaton
More informationSection 3.6 Complex Zeros
04 Chapter Secton 6 Comple Zeros When fndng the zeros of polynomals, at some pont you're faced wth the problem Whle there are clearly no real numbers that are solutons to ths equaton, leavng thngs there
More informationOn the correction of the h-index for career length
1 On the correcton of the h-ndex for career length by L. Egghe Unverstet Hasselt (UHasselt), Campus Depenbeek, Agoralaan, B-3590 Depenbeek, Belgum 1 and Unverstet Antwerpen (UA), IBW, Stadscampus, Venusstraat
More informationFrequency dependence of the permittivity
Frequency dependence of the permttvty February 7, 016 In materals, the delectrc constant and permeablty are actually frequency dependent. Ths does not affect our results for sngle frequency modes, but
More information