Physics 7440, Solutions to Problem Set # 8

Transcription

1 Physics 7440, Solutios to Problem Set # 8. Ashcroft & Mermi. For both parts of this problem, the costat offset of the eergy, ad also the locatio of the miimum at k 0, have o effect. Therefore we work with the simpler form for the eergy ǫk k im ij k j. a For simplicity of otatio, we defie the matrix W by W ij Mij. So, the, M W. W ad M are both real, symmetric matrices. Moreover they are positive defiite sice otherwise k 0 would ot be a miimum, which implies amog other thigs that both the determiat ad the diagoal elemets are positive. I geeral we have W W xx W xy W x W xy W yy W y. W x W y W The -axis is the axis alog which we apply the magetic field ad is fixed. However, we ca choose the x ad y axes ay way we like so log as they are perpedicular to the axis, ad it is possible to choose them so that W xy 0. This is because choosig the x ad y axes this way correspods to makig a orthogoal trasformatio to diagoalie the upper-left block of W, which we ca certaily do. So with this choice of axes we have a somewhat simpler form W W xx 0 W x 0 W yy W y. 3 W x W y W To proceed, we ote that the electro orbits for a magetic field alog the -directio are give by the itersectio of costat eergy surfaces with surfaces of costat k. We will show that such orbits form ellipses, fid their area as a fuctio of eergy ǫ ad k, ad use this to get the cyclotro effective mass. The surface of costat eergy ǫ is specified by the equatio ǫ ǫk [ Wxx kx +W yy ky +W k ] +W x k k x +W y k k y. 4 If we also treat k as a costat here, the the resultig equatio for k x ad k y specifies the orbit at k with eergy ǫ. We rewrite this as follows: ǫ W k W xxk x +W yyk y +W xk k x +W y k k y. 5 By completig the square twice o the right-had side, we have ǫ W k W xx k x + W xk +Wyy k y + W yk W x k W y k. 6 W xx W yy W xx W yy earragig terms, we have ǫ [ W x k +k + W y k ] W W xx k x + W xk +Wyy k y + W yk. 7 W xx W yy W xx W yy Now, we recogie this as the equatio specifyig a ellipse for the variables k x ad k y. We are oly iterested i the area of this ellipse, so we ca drop the costat terms iside the paretheses o the right-had side, sice these just shift the ceter of the ellipse. So we work with the equatio ǫ [ W x k +k + W y k ] W W xx kx +W yy k W xx W y. 8 yy

2 If, by divig out by the left-had side, we put this equatio ito the form the the area of the ellipse is Aǫ,k πab. After some algebra we fid Aǫ,k k x /a +k y /b, 9 π [ ǫ W x k Wxx W yy +k + W y k ] W. 0 W xx W yy Usig this, ad Ashcroft & Mermi Eq..44, we get the simple result m A π ǫ. Wxx W yy Now we just have to express this i the desired form. Usig stadard formulas for the iverse of a 3 3 matrix, ad applyig them to W, we have so that M W W xxw yy detw W xxw yy detm, ad therefore W xx W yy M detm, 3 m detm M. 4 b The specific heat depeds o the bad structure oly through the desity of states at the Fermi eergy, accordig to c v π 3 k BTDǫ F. 5 We eed to calculate Dǫ F for the give dispersio. To simplify the calculatio, we are free to choose axes to that M, ad hece M is diagoal, ad The desity of states is ǫk ix,y, k i m i. 6 d 3 k Dǫ 4π 3δ[ ǫ / k ] i/m i 7 i d 3 q m x m y m 4π 3δ[ ǫ /q ] 8 d 3 q detm 4π 3δ[ ǫ /q ]. 9 I the secod lie above, we have chaged variables to q i k i / m i. The q-itegral above is exactly Dǫ F as evaluated for free electros i Ashcroft & Mermi chapter, with the mass set to uity m. Therefore lookig at A & M Eq..6, we have We wat to write this i the form Dǫ F detm π ǫf. 0 Dǫ F m m ǫ F π, where m is the specific heat effective mass. Equatig the above two equatios, we see that m detm /3.

3 3. Ashcroft & Mermi.4 a We imagie applyig a electric field E, which is felt i the same way by every bad. Therefore the curret desity of the th bad is j ρ E. 3 The total curret desity is the j j ρ ]E. 4 The we have for the electric field i terms of the curret desity: E ρ Sice by defiitio E ρ j, we have the desired result, ρ ] j 5 ρ ]. 6 b I this case ad similarly for ρ. We have ad ρ ρ ρ ρ H, 7 H ρ ρ + H ρ + H ρ H, 8 H ρ ρ H. 9 H ρ Addig these ad takig the iverse, we have after some algebra ρ [ρ +ρ ] ρ +ρ +H + Writig this as we have the desired results: ad ρ ρ + H +ρ ρ + H Hρ + H Hρ + H Hρ + H + Hρ + H ρ ρ + H +ρ ρ + H ρ H ρ H ρ, 3 ρ ρ ρ ρ +ρ +ρ +ρ H ρ +ρ +H +, 3 ρ + ρ + + H ρ +ρ +H + 33 c For simplicity, assume bad is a electro bad all occupied orbits closed, ad bad is a hole bad all empty orbits closed. Bad has a desity e of electros ad bad a desity h of holes. We focus o this case because it allows us to cosider the case of a compesated metal eff e h 0. We could easily repeat the aalysis for electro bads ad hole bads. 30

4 I the high field limit, / e ec, ad / h ec. Therefore, i the high field limit, + ec [ + ] e h e h ec e h eff ec e h Because this is oero, oly the terms proportioal to H i the umerator ad deomiator of will cotribute i the high-field limit, ad we have + 35 ec e h e ec h ec eff eff ec. 36 Now we cosider the case of a compesated metal eff 0, ad work out the high-field behavior of both ρ ad. Now + 0, so the Hall coefficiet is field-idepedet: ρ + ρ ρ +ρ. 37 Notice that this ca o loger be simply related to the desities i the two bads. Ideed, eve the sig of is completely determied by which of ρ ad ρ is larger, which has othig to do with carrier desities. As for the resistivity, the H -term i the deomiator drops out because + 0, but this is ot the case for the term i the umerator. The H -term i the umerator the domiates i the large-field limit, ad we have which icreases proportioal to H for large fields. 3. Ashcroft & Mermi.6 ρ ρ +ρ H ρ +ρ, 38 Let T be the operator that traslates a wavefuctio by the Bravais lattice vector. Actig o a geeral wavefuctioφr, wehavet φr φr+. Moreover,othe positiooperator ˆr, wehavet ˆrT ˆr+. The latter statemet ca be verified by cosiderig However, we ca also calculate this by writig T ˆrφr T [rφr] r +φr T ˆrφr [T ˆrT ]T φr [T ˆrT ]φr The two expressios are oly cosistet provided T ˆrT ˆr +. Now let s cosider the problem at had. We re told that ψr, 0 is a liear combiatio of Bloch wavefuctios, all with the same crystal mometum k. This meas that We wat to show that T ψr,0 e ik ψr,0. 4 T ψr,t e ik ψr,t, 4 where k k eet/. This is what we wat to show, because this statemet is true if ad oly if ψr,t is a liear combiatio of Bloch wavefuctios, all with crystal mometum k. To show this cosider the followig expressio: T ψr,t T e iĥt/ ψr,0 [T e iĥt/ T ]T ψr,0 e ik [T e iĥt/ T ]ψr,0. 43 We eed to compute the term i square brackets. First, ote that T ĤT m +Uˆr ++ee ˆr + 44 m +Uˆr+eE ˆr +ee 45 Ĥ +ee. 46

5 I the secod lie we used the fact that Ur is a periodic potetial. Next, I claim that, for ay operator Â, T eât exp [T ÂT. ] 47 This ca be proved by usig the Taylor expasio of the matrix expoetial: [ T eât T!Â] T 48! exp! T Â T 49 [T ÂT ] 50 T ÂT. 5 The crucial step is the d to last lie. It is easiest to illustrate by cosiderig a particular term, for example Combiig these results, we have Goig back to Eq. 43, we have as desired. T Â T TÂT TÂT [TÂT ]. 5 T e iĥt/ T e iĥt/ iete / e iete / e iĥt/. 53 T ψr,t e ik e iete / e iĥt/ ψr,0 e ik ψr,t, 54 5