{VERSION 6 0 "IBM INTEL NT" "6.0" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 1 }{CSTYLE "2D Math" -1 2 "Times" 0 1 0 0 0 0 0 0 2 0 0 0 0 0 0 1 }{CSTYLE "2D Input" 2 19 "" 0 1 255 0 0 1 0 0 0 0 0 0 0 0 0 1 } {CSTYLE "" -1 256 "" 1 16 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 257 "" 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 258 "" 1 12 0 0 0 0 0 2 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 259 "" 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 260 "" 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 } {CSTYLE "" -1 261 "" 1 12 0 0 0 0 0 2 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 262 "" 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 263 "" 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 }{PSTYLE "Normal" -1 0 1 {CSTYLE "" -1 -1 "Tim es" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 } {PSTYLE "Heading 1" -1 3 1 {CSTYLE "" -1 -1 "Times" 1 18 0 0 0 1 2 1 2 2 2 2 1 1 1 1 }1 1 0 0 8 4 1 0 1 0 2 2 0 1 }} {SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT 256 43 "Quantum Computation and Q uantum Information" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 73 "Michael A. Nielsen, Isaac L. Chuang (Cambridge, 2000) ISB N 0 521 63235 8." }}{PARA 0 "" 0 "" {TEXT -1 128 "Maple implementation by Marko Horbatsch (Computational Physics using Maple; www.yorku.ca/m arko) August 2002, modified Oct. 2003." }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 0 "" 0 "" {TEXT -1 208 "We implement the matrix representati on for a few quantum logic gates, such as the conditional NOT, derive \+ the swap, the Hadamard gate, the formation of entangled states and wor k through quantum teleportation." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "with(LinearAlgebra):" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 129 "We intr oduce a basis for a 2-qubit; the first is the control qubit, the secon d is the target qubit; they form an entangled state." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "s00:=Vector([1,0,0,0]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 71 "s01:=Vector([0,1,0,0]): s10:=Vector([0,0, 1,0]): s11:=Vector([0,0,0,1]):" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 130 "The conditional NOT gate of quantum computing leaves the control \+ qubit as is, and adds the target and control bit in binary logic." }} {PARA 0 "" 0 "" {TEXT -1 56 "The following picture is taken from the t ext (Fig. 1.6):" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {OLE 1 26114 1 "[xm]Br=WfoRrB:::wk;nyyI;G:;:j::>:B>N:F:nyyyyy]::yyyyyy :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: :fyyyyyI@nYnyyyYE:G:I:K:M:O:Q:S:UJ:n;v;;JBB:]:_:a:c:e:g:i:k:m:o:q:s:u: w:y:;[:F>N>V>^>f>n>v>>?F?N?V?^?f?n?v?>`:B:wAwAyA:::::::::::::::::::::: :::::::::::::::::::::::::::::NDYmq^H;C:ELq^H_mvJ::::::::gj:;H:<:T><::[hoyU@oW_AX:A:;b:>Z:j:vCSmlF@[KaFFcmnnHEM:>:::::::oJ;@jyyy yyI:;Z::::::Jd:B:j:vC S=>:::::::::J?N:ry?:yayA:<::::::?J:^<>:F:;JyKyKR<:TN>Jb@pkhpjXpl>;:::JtaMSAA<::oXlOGZOiaobwqXygshye rOuyHrhTjZMluqMdik_\\kmqu@Ej_=pX=fV?bRiuHy]EWgNWkCG:D:<::JHhxranaxYcQY 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?LvVdNrQXyyTMHVqhLYqWwZ::::::::::::::::::::::::::::::::::::NZ:vc:^g:ff:>D_E :EE:aE:[C:[C:>Z::::::::kB:?R:Z:vyyyyyAZ:>Z::::::j;<:SS:;B:<:=ja <:>Z:>:::::::::J?:<::::::wyyyyyB:::::::::::::yay=J:B:::::::: :::::::::::jysy:>:<::::::::3:" }{TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 58 "U_cnot:=Matrix([[1,0,0,0],[0,1,0,0],[0,0,0,1],[0 ,0,1,0]]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 55 "U_cnot . s00, U_cnot . s01, U_cnot . s10, U_cnot . s11;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 74 "In order to verify the truthtable we need to compare two \+ vectors. Maple's " }{TEXT 19 5 "evalb" }{TEXT -1 119 " is unreliable w hen it performs this comparison, the results change between versions. \+ Thus we define our own procedure " }{TEXT 19 5 "Evalb" }{TEXT -1 32 ", which compares equations with " }{TEXT 19 6 "Vector" }{TEXT -1 46 " o bjects on both sides component by component:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 88 "Evalb:=proc(VectorEqn::`=`) local L,R,dL,dR,i,re s; L:=lhs(VectorEqn): R:=rhs(VectorEqn):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 101 "if type(L,Vector) and type(R,Vector) then dL:=op(1,L ): dR:=op(1,R): if dR<>dL then RETURN(false) fi: " }}{PARA 0 "> " 0 " " {MPLTEXT 1 0 103 "for i from 1 to dL do: res:=simplify(R[i]-L[i]): i f res <> 0 then RETURN(false): fi: od: RETURN(true): " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "else evalb(VectorEqn): fi: end:" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 25 "Now apply it to our case:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "Evalb(U_cnot . s00 = s00);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "Evalb(U_cnot . s01 = s01);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "Evalb(U_cnot . s10 = s11) ;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "Evalb(U_cnot . s11 = s 10);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 211 "We see that: when the co ntrol qubit (qubit 1) is zero, the target qubit is left unchanged, and when qubit 1 has value 1, then qubit 2 is flipped. This represents th e conditional NOT. We have verified eq. (1.18)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 141 "Now one should verify us ing binary arithmetic that given the notation |A> and |B> for the cont rol and target qubits that |B> goes into |A+B>." }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 63 "true and true, true and false, false and true, false and false;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "true o r true, true or false, false or true, false or false;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 66 "This does not implement 0 + 0 = 0, 0 + 1 = 1 + 0 = 1, 1 + 1 = 0 !" }}{PARA 0 "" 0 "" {TEXT -1 61 "But the XOR \+ between boolean states does do the desired thing:" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 14 "true xor true," }{TEXT -1 1 " " }{MPLTEXT 1 0 15 "true xor false," }{TEXT -1 1 " " }{MPLTEXT 1 0 15 "false xor tru e," }{TEXT -1 1 " " }{MPLTEXT 1 0 16 "false xor false;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 261 "Note: the linearity of matrix multiplica tion will guarantee that we only need to know the action for the trans formation matrix on basis states. The expansion coefficients, which me asure how much probability amplitude we have in either state factor ou t linearly." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 109 "Now let us understand how three conditional NOTs can form a sw ap, as explained in eq. (1.20), and figure 1.7." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {OLE 1 17410 1 "[xm]Br=WfoRrB:::wk;nyy I;G:;:j::>:B>N:F:nyyyyy]::yyyyyy:::::::::::::::::::::::::::::::::::::: :::::::::::::::::::::::::::::::::fyyyyyA>;F;N;V;^;f;;JA jA>:[Z:F=F=N=V=^=f=n=v=nYnYvY::::::::::::::::::::::::::: ::::::::::::::::::::::::::::::::::::_lqvGcMJ:::::::JEf:yyy xI^:^;AJ:>:;H:<:T><::[yWIV@oW_AX:A:;b:>Z:j:vCSmlF@[KaFFcmnnHEM:>:::::: :oJ;@jyyyyyI:;Z::::::Jd:B:j:vCS=>:::::::::J?N:ry?:yayA:<::::::?J:^<>:F:;JyKyKR<:TN>Jb@pkhpjXpl>;:::JtaMSAA<::::::::::=Z< Z::::::::::jysyA:C:=Z:vYxY;J:j=X:ET;NW@C:US:f:F[< Z:f:FZ:QB:F^?N;]tRGUF[uGWcT[uS=KK^=VY[j=:sg :B:=J;Dlc`qsLqlp@OV:B:yayAmmq:McJKxWCJ:f?=j;B:;:::Ja@Na`N:>Z::::J `X=?B:]tR:>Z=OaVgg[?hfgfr^eKWbgncCgbxG`VGd=ofk:B:;r:;bArJo;:p<:G=Zoq@x YfyYj?\\DobCoB=b:=R:SS:S[yIv@nl??icoFpn`N?>;IR:_j;>:=B:=J:@JGf:B:?A:=J;>Z;F@?Z:^s:R:S;=B:@JPZ;_fiO f_?^J?gi_hiWhbwgj?^qn_jbF[CF_cH[CHocF_CB]cGosFccH[cGZ;>bZX:C::;:GR:Aj;>:=J=R:o:;Z:>Z;n\\;V=Z:JI>?Z:N:;b: =B:C:>:Gc:?J:>CF:Z:>CKL:xyp_[;_fikiHn`NZyqWWRJCuU >Z^?j:B:=jg_JSBer>aB@]B=SB=SZK^[FB=: :::=R=QrWb=[B?Wb=WB?cb=WJALkBlJ@^;]R>SB=l:]B=^;:::ZyAr:F\\BBBBZ;NZ>fZ@vZBF[DV[;R:E:=Z:J: j:Pj;ljS@Zat^ax_hp?it_ixPofpojPpnpprPqvpy`AwdawhAxlaxpAytay xKb:?R:?B;Cr:CR;vZ@fZoF:?R>@lRDK<@nb@prDlCPJS@rBMkJAU[FrLGtjY`^Nf[Qh

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First we have to find the matrix that will do that! Let's ma ke a guess, and transpose the two non-zero 2-by-2 blocks that form U_c not." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "U_rcnot:=Matrix([[0 ,1,0,0],[1,0,0,0],[0,0,1,0],[0,0,0,1]]);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 55 "Let us verify the action of this transformation matrix:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "U_rcnot . s00, U_rcnot . \+ s01, U_rcnot . s10, U_rcnot . s11;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 111 "The question is: did we flip qubit 1 depending on qubit 2 bein g 0 or 1, and leave qubit 2 alone? Let's compare:" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 19 "s00, s01, s10, s11;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 27 "What did we generate above?" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "s01,s00,s10,s11;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 155 "We kept qubit 1, while flipping qubit 2 when qubit 1 was |0>, rather than |1>. So this matrix was not what we wanted! How should we find the right matrix?" }}{PARA 0 "" 0 "" {TEXT -1 21 "We want the ou tcomes:" }}{PARA 0 "" 0 "" {TEXT -1 57 "|00> -> |00> , |10> -> |10> , \+ |01> -> |11> , |11> -> |01>" }}{PARA 0 "" 0 "" {TEXT -1 219 "We could \+ set up a system of equations to determine the matrix (cannon approach \+ to swat a fly). It makes more sense to write the source and target spi nors side-by-side, and to determine the required coupling coefficient. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "s00,s00;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 69 "The first row better have a 1 in its firs t entry (and be [1,0,0,0] ?)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "s10,s10;" }}}{EXCHG {PARA 0 " " 0 "" {TEXT -1 75 "The third row better be [0,0,1,0], or at least hav e a 1 in the third entry." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "s01,s11;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 206 "The second row (pr oducing the action for |01>) should have a zero in the second column. \+ The fourth row has to produce the 1 in the fourth column, while it act s on s01. So it needs a 1 in the second column." }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 8 "s11,s01;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 57 "U_try:=Matrix([[1,0,0,0],[0,0,0,1],[0,0,1,0],[0,1,0,0]]);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "Evalb(U_try . s00 = s00);" } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "Evalb(U_try . s01 = s11); " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "Evalb(U_try . s10 = s10 );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "Evalb(U_try . s11 = s 01);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 91 "OK, so apparently this do es what we want! Now let's compute the action of swapping circuit:" }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "U_swap:=U_cnot . U_try . U_ cnot;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 82 "Indeed, this is the matr ix listed on page XXV for the swap! What does the swap do?" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "U_swap . s00,s00;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "U_swap . s01, s10;" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 18 "U_swap . s10, s01;" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 18 "U_swap . s11, s11;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 132 "Note that swapping is not the same as inverting both qub its. This would agree on the 2nd and 3rd case, but not the first and f ourth." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 257 10 "Exercise1:" }}{PARA 0 "" 0 "" {TEXT -1 120 "Find the matrix which \+ inverts both qubits. Construct this matrix carefully by comparing sour ce and target qubit spinors." }}}{SECT 1 {PARA 3 "" 0 "" {TEXT 258 22 "Solution to Exercise 1" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "s0 0,s11;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 37 "first column in fourth \+ row needs a 1." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "s10,s01;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 36 "third column in second row need s a 1" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "s01,s10;" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 36 "second column in third row needs a 1" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "s11,s00;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 36 "fourth column in first row needs a 1" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 57 "U_inv:=Matrix([[0,0,0,1],[0, 0,1,0],[0,1,0,0],[1,0,0,0]]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "U_inv . s00, s11;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 " U_inv . s10, s01;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "U_inv \+ . s01, s10;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "U_inv . s11 \+ , s00;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 116 "The Hadamard gate matrix can be used to \+ create so-called Bell states, or entangled states as explained in Fig. 1.12." }}{PARA 0 "" 0 "" {TEXT -1 104 "First one applies the 2-by-2 H adamard gate to the control qubit, and then carries out a conditional \+ NOT." }}{PARA 0 "" 0 "" {TEXT -1 104 "Entangled states are such 2-qubi t states which are not factorizable as a product of single-qubit state s." }}{PARA 0 "" 0 "" {TEXT -1 18 "Here is Fig. 1.12:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {OLE 1 21505 1 "[xm]Br=WfoRrB:::wk; nyyI;G:;:j::>:B>N:F:nyyyyy]::yyyyyy::::::::::::::::::::::::::::::::::: ::::::::::::::::::::::::::::::::::::fyyyyyA?nYnyyyYE:G:I:K:M:O:Q:S:UJ: n;v;;JBB:]:_:a:c:e:g:i:k:m:o:q:s:u:w:y:;[:F>N>V>^>f>n>v>nYnYvY:::::::: :::::::::::::::::::::::::::::::::::::::::::::::::JcvGYMt>^:fBWMtNHm=;: 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vslyIDW@yJMTMZaVsEyx`qDMW;yQeYmsMmwuQ@qmALYDZdpi>oadX`t YDMgxaxP=vyap_Hs>YOHu[n^ZAr]@jY_ZHvy:F_c@k>> \\@gbjvayaTUt^;VFuyuMYNKH:YSZSfKKbwCioAd@MrxstBIc;_BLsYKMTHGHuyEVeXI;v ymH:;B:;:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: ::::::::J;`:>`:J:<:::::::>]:NZ;VZ:>:=B:;B:yyyyyy ;B:;B::::::VZ:JQ@J:Z:vYxI:;Z::::::JyyyyyI<:: ::::::::::jysy:>:<:::::::::::::::::::vYxI:;Z::::::::6:" }{TEXT -1 0 " " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "H2:=Matrix([[1,1],[1,-1 ]])/sqrt(2);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 273 "In order to be a ble to work this out with our 2-qubit states, we need to embed the Had amard matrix so that it will act on qubit 1 in the 2-qubit state in sp inor representation. We want to mix the components associated with qub it 1 irrespective of what's going on in qubit2." }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 16 "s00,s10,s01,s11;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "s00+s10;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 " s01+s11;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 121 "We really have to co nstruct the Hadamard matrix acting on qubit 1. Naive embedding doesn't work! What is it that we want?" }}{PARA 0 "" 0 "" {TEXT -1 139 "H2 |0 0> = s2 (|00> + |10>) , H2 |10> = s2 (|00> - |10>) , H2 |01> = s2 (|0 1> + |11>) , H2 |11> = s2 (|01> - |11>) ; where s2 = 1/sqrt(2)." }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "s00, s00 + s10;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "s01, s01 + s11;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "s10, s00 - s10;" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 15 "s11, s01 - s11;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "sq2:=1/sqrt(2):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 74 "H2q1:=Matrix([[sq2,0,sq2,0],[0,sq2,0,sq2],[sq2,0,-sq2 ,0],[0,sq2,0,-sq2]]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "H2 q1 . s00, s00 + s10;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "H2q 1 . s01, s01 + s11;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "H2q1 . s10, s00 - s10;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "H2q1 \+ . s11, s01 - s11;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 167 "OK, so it w orked. Now we can complete the action, by following up on H2q1 by a cN OT gate, which is not decomposable, and therefore leads to an entangle d 2-qubit state:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "B:=U_cn ot . H2q1;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 32 "Figure 1.12 can now be verified:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "B . s00 , \+ s00 + s11;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "B . s01 , s01 + s10;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "B . s10 , s00 - \+ s11;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "B . s11 , s01 - s10 ;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 260 11 "Exercise 2:" }}{PARA 0 "" 0 "" {TEXT -1 134 "Construct the matrix for the Hadamard gate for qubi t2, and then compute the matrix H4 which performs the Hadamard gate on both qubits." }}}{SECT 1 {PARA 3 "" 0 "" {TEXT 261 22 "Solution to Ex ercise 2" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 33 "First, we have to stat e the goal:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "s00, s00 + s01;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "s10, s10 + s11;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "s01, s00 - s01;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "s11, s10 - s11;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 10 "Therefore:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 74 "H2q2:=Mat rix([[sq2,sq2,0,0],[sq2,-sq2,0,0],[0,0,sq2,sq2],[0,0,sq2,-sq2]]);" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 13 "Let's verify:" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 22 "H2q2 . s00, s00 + s01;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "H2q2 . s10, s10 + s11;" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 22 "H2q2 . s01, s00 - s01;" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 22 "H2q2 . s11, s10 - s11;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 32 "OK, now take the matrix product:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "2* H2q1 . H2q2;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 557 "We multiplied by 2 to have a clear view of the matrix. W ithout the factor of 2 we have computed the Hadamard matrix W4, which \+ represents the application of a Hadamard gate to both qubits. It is ob tained as an outer product of two Hadamard gates W2 for each qubit (th e order did not matter). An outer product represents a factorizable op eration. The important thing to produce the Bell states was to apply U _cnot, the conditional NOT gate, which is represented by a matrix that cannot be factorized into two matrices acting on qubit 1 and qubit 2 \+ separately." }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 259 21 "Quantum Teleportation" }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 587 "According to Fig. 1.13 an entangled pair shared between two remote communication partners A \+ and B (A=Alice, B=Bob) can be used to transfer a quantum state from on e end to the other by means of a few gates, and a classical communicat ion channel by which the outcome of a measurement done at A can be tra nsmitted to B. The unknown state |psi> = a|0> + b|1> is being transfer red from A to B, i.e., while B will be able to reconstruct it, A will \+ wind up with a destroyed state. This has to be so to be compliant with the no-cloning theorem, which does not permit to duplicate quantum st ates." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {OLE 1 27138 1 "[xm]Br=WfoRrB:::wk;nyyI;G:;:j::>:B>N:F:nyyyyy]::yyyyyy::::::::::::: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::fyyyyyY@nYny yyYE:G:I:K:M:O:Q:S:UJ:n;v;;JBB:]:_:a:c:e:g:i:k:m:o:q:s:u:w:y:;[:F>N>V> ^>f>n>v>>?F?N?V?^?f?n?v?>`:B:];_;wAwAyA::::::::::::::::::::::::::::::: :::::::::::::::::::_lqvGcMJ:::::::JEf:yyyxI^:^;A:;JZ=Z:Z@[ ::JBImDaKWQs[Aj;J:D:;B:F:YLpfF]KRnC=MtFGgml>:;::::::JGNZ;vyyyyy=J:B::: :::^:V^V[:j:VBYmp>HYLkNG>::::::::N:c:;:=J:nYnY^:f:n:v::;`:Z@?;>d[WfeGfagFK::::^xsNpk[::JUyOJDMuIl@tJ?Mtd`RDtW:VuAj \\P:SA=j>>Z=nHN:s:;j<>Z:V[l;B:@:GKHjw?:sg:B:=J;Dlc`qsLqlp@OV:B:yayA mmq:McJKxWCJ:f?=j;B:;:::Ja@Na`N:>Z:Nf^?qyHqknbV^ieggeWeughiXy`y]VK;ZYZ=OA:RE:nF:frUky]wy=hJNc@ Gc@S:?R:=BE=BEvy[I[GgbrGg`k`p^dJ:\\ZAFZLr:Z;B:=b@P:>:;bR=r@P:@:dp: 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Z::::::::kB:?R:AB:;j:Z:B:F:YD:J:, and a target qubit line on which is encoded one half of the ent angled 2-qubit state. B has one line on which it has encoded the other qubit of the entangled state. The entire system, therefore, is a 3-qu bit state, and the objective is to transfer the information from qubit 1 (Alice) to qubit 3 (Bob)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 322 "Alice will apply a cNOT followed by a Hadamard gate on q1, and then measure probabilities on q1 and q2. These probab ilities will be transmitted classically to Bob, who will then know whi ch gate to apply to q3, such that it will contain |psi> as originally \+ stored on q1. Our aim is to calculate the evolution of the states." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 145 "To be ab le to apply the cNOT and Hadamard gates we need to set up A's 2-qubit \+ state. The overall state is a product of |psi> = a|0> + b |1>, and" } }{PARA 0 "" 0 "" {TEXT -1 69 "1/sqrt(2) (|00> + |11>). There are eight (2^3) possible basis states." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 263 "s000:=Vector([1,0,0,0,0,0,0,0]): s001:=Vector([0,1,0,0,0,0,0,0] ): s010:=Vector([0,0,1,0,0,0,0,0]): s011:=Vector([0,0,0,1,0,0,0,0]): s 100:=Vector([0,0,0,0,1,0,0,0]): s101:=Vector([0,0,0,0,0,1,0,0]): s110: =Vector([0,0,0,0,0,0,1,0]): s111:=Vector([0,0,0,0,0,0,0,1]):" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 163 "Now we need to define the cNOT ga te for q1 and q2. We need an 8-by-8 matrix that serves as an embedding of the 4-by-4 cNOT gate, and which ignores the state of q3." }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "s000,s001;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 43 "Both these states are to be left unchanged." }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "#UcNOT:=Matrix([[1,0,],[0,1 ,],[],[],[],[],[],[]]):" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 39 "Also t he next two should be left as is:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "s010,s011;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 74 "#UcNOT:=Matrix([[1,0,0,0,],[0,1,0,0,],[0,0,1,0,],[0,0,0,1,],[],[], [],[]]):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "s100,s110,s101, s111;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 78 "In the above two pairs: \+ the first spinor is to be transformed into the second." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 156 "#UcNOT:=Matrix([[1,0,0,0,0,0,0,0], [0,1,0,0,0,0,0,0],[0,0,1,0,0,0,0,0],[0,0,0,1,0,0,0,0],[0,0,0,0,0,0,],[ 0,0,0,0,0,0,],[0,0,0,0,1,0,0,0],[0,0,0,0,0,1,0,0]]):" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 41 "and finally, the remaining pair of pairs:" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "s110,s100,s111,s101;" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 161 "UcNOT:=Matrix([[1,0,0,0,0,0 ,0,0],[0,1,0,0,0,0,0,0],[0,0,1,0,0,0,0,0],[0,0,0,1,0,0,0,0],[0,0,0,0,0 ,0,1,0],[0,0,0,0,0,0,0,1],[0,0,0,0,1,0,0,0],[0,0,0,0,0,1,0,0]]);" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 13 "Check it out:" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 54 "Evalb(UcNOT . s000 = s000) ,Evalb(UcNOT . s0 01= s001);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 55 "Evalb(UcNOT . s010 = s010) ,Evalb(UcNOT . s011 = s011);" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 56 "Evalb(UcNOT . s100 = s110) , Evalb(UcNOT . s101 = s 111);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 56 "Evalb(UcNOT . s110 = s100) , Evalb(UcNOT . s111 = s101);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 61 "Now construct the Hadamard gate matrix representation for q1." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 4 "sq2;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "s000, s000 + s100, s001, s001 + s10 1;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 114 "We will start with a matri x filled by zeroes, and will replace the non-zero entries according to the requirements." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "Hq1:= Matrix(8,8):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "Hq1[1,1]:=s q2: Hq1[5,1]:=sq2: Hq1[2,2]:=sq2: Hq1[6,2]:=sq2:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "s010, s010 + s110, s011, s011 + s111;" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "Hq1[3,3]:=sq2: Hq1[7,3]:=sq2 : Hq1[4,4]:=sq2: Hq1[8,4]:=sq2:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "s100, s000 - s100 , s101, s001 - s101;" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 61 "Hq1[1,5]:=sq2: Hq1[5,5]:=-sq2: Hq1[2,6]:=sq2: \+ Hq1[6,6]:=-sq2:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "s110, s0 10 - s110 , s111, s011 - s111;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 61 "Hq1[3,7]:=sq2: Hq1[7,7]:=-sq2: Hq1[4,8]:=sq2: Hq1[8,8]:=-sq2: " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 4 "Hq1;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "Determinant(Hq1);" }}}{EXCHG {PARA 0 "" 0 " " {TEXT -1 117 "Let's check it out... we copy the lines used for the m atrix construction, and turn them into equations to be checked:" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 81 "Evalb(sqrt(2)*Hq1 . s000 = s 000 + s100), Evalb(sqrt(2)*Hq1 . s001 = s001 + s101);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 81 "Evalb(sqrt(2)*Hq1 . s010 = s010 + s 110), Evalb(sqrt(2)*Hq1 . s011 = s011 + s111);" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 82 "Evalb(sqrt(2)*Hq1 . s100 = s000 - s100) , Eval b(sqrt(2)*Hq1 . s101 = s001 - s101);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 82 "Evalb(sqrt(2)*Hq1 . s110 = s010 - s110) , Evalb(sqrt( 2)*Hq1 . s111 = s011 - s111);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 95 " Now we are able to put the teleportation gate together: Equation 1.29 \+ states the initial |psi>." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "psi:=sq2*(a*(s000 + s011) + b*(s100 + s111));" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 19 "psi:=simplify(psi);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "psi1:= UcNOT . psi;" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 18 "psi2:= Hq1 . psi1;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 572 "What is the significance of psi2 as compared to the init ial psi, i.e., what was the effect of applying the gates? Somehow, the information (values of a,b) got transferred all over the place. Note, Alice never knew what her amplitudes a and b were, and the claim is, \+ that if she measures the probability content of q1 and q2, and transmi ts it classically to Bob, then Bob can infer what simple gate to apply in order to re-surrect the superposition a|0> + b|1> in 'his' qubit, i.e., in q3 (without knowing the values of a,b). How can we find out \+ whether this indeed works?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 482 "The claim is, that if Alice measures a |00>, the n Bob simply has a|0> + b|1> in q3. For Alice observing |01>, he gets b|0> + a|1>, for Alice measuring |10> he gets a|0> - b|1>, and when \+ she sees |11>, he obtains -b|0> + a|1>. This means that upon receivin g information from Alice on the classical channel as to what she measu red on q1/q2, Bob can apply a corresponding transformation (in the lat ter three cases) in order to rotate the state into a|0> + b|1>, thus \+ receiving psi." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 122 "How do we obtain the collapse of psi2 to a single-qubit \+ wavefunction, i.e., what do we do to simulate Alice's measurement?" }} {PARA 0 "" 0 "" {TEXT -1 158 "It is a projection onto the 2-qubit stat e that should do the trick. Two degrees of freedom would be contracted away, with a remaining two-spinor wavefunction." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 89 "One way to convince onese lf, is to realize which superposition the state psi2 represents:" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 86 "phi2:=simplify((a*s000+b*s00 1) + (a*s011+b*s010) + (a*s100-b*s101) + (a*s111-b*s110));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "Evalb(phi2=2*psi2);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 205 "The four states above are grouped in suc h a way that they correspond to Alice's four possible measurement outc omes, namely |00> (in which case qubit 3 is in the original state of q 1), |01>, |10>, and |11>." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 " " 0 "" {TEXT -1 81 "This calculation then carries out the matrix repre sentation of eqs. 1.28 to 1.36." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}} {EXCHG {PARA 0 "" 0 "" {TEXT 262 45 "Classical computing with qubits: \+ Toffoli gate" }}{PARA 0 "" 0 "" {TEXT -1 259 "The idea is to demonstra te the a classical NAND gate can be implemented on a quantum computer. This will be very inefficient, but is interesting from a fundamental \+ point of view: a quantum computer, in principle, can be built to simul ate a classical computer." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 " " 0 "" {OLE 1 27648 1 "[xm]Br=WfoRrB:::wk;nyyI;G:;:j::>:B>N:F:nyyyyy]: :yyyyyy::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: ::::::::fyyyyyi?nYn?f:n:v:>;F;N;V;^;f;;JAjA>:[Z:F=F=N=V= ^=f=n=v=>>`:B:];_;a;c;wAyA::::::::: ::::::::::::::::::::::::::::::::::::::::NDYmq^H;C:ELq^H_mvJ::::::::gj< vyyuyJ:;H:<:T><::[tPwV@oW_AX:A:;v=B:F:YLpfF]KRnC=MtFGgml>:;::::: :JGNZ;vyyyyy=J:B::::::^:vvJK:j:VBYmp>HYLkNG>::::::::N:F:;JyKyK:i:k:m:oJ:V=^=f=n=v=>>::::cYH _WVB::FBUKR>@=lZ>@[K[FA_kVfBkkxjcnG=MrVHKmq^H=ZiHj:F;HJXX:?jJT::Q b:B:E:Sb:akd^=Vy<^xy]:JBAZ:R:B:cTRKuUUURXQZB:?w;ORRgrJKxWyrJKxW;CjOxW> @CZ:f_;jD:Z:V=K;;[:>BZ:>:@: <:::jyK?dJ?lxGxyyyyy;:NlU^qLN[DN[D:lHvY::_Fi_b;IR:_j;>Z:F:=J:@JGf:B:?A:=J;>Z;F`:N::sN:`jN`t> 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A NAND gate can be obtained by \+ using the first two qubits as input, setting the third input to 1, and using the q3 output as the result. Another application is by using it as a FANOUT, i.e., a bit readout gate (by setting q1=1, q3=0 on input , and obtaining q3=q2=a for q2=a input." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 263 11 "Exercise 3:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 99 "Go through the truth tabl e of the Toffoli gate, and verify the NAND gate behaviour described ab ove." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{MARK "122 3 0 " 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }