0 0 0 0 0 0 1 0 this is a partial not 0operation 0 1

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Unformatted text preview: b ' ¢ ' It’s not unitary (because0 not norm preserving). 0 0 0 0 0 0 1 0 This is a “partial NOT” 0operation. 0 1 0 0 1 1 1 1 1 1 0 0 1 1 0 1 0 1 0 1 0 0 1 1 1 1 1 1 0 0 1 1 0 1 0 l 1 0 180 CHAPTER III. QUANTUM COMPUTATION ¶8. Creation operation: Its conjugate it the creation operation ✓ ◆ 00 ⇤ a= = | 1i 0| . h 10 The creator transforms |0i to |1i, but leaves |1i alone, returning 0. It matches |0i and resets it to |1i. Note that a⇤ is the adjoint (conjugate transpose) of a. This is the other half of NOT. ¶9. Number operation or 1-test: Consider11 ✓ ◆ 00 ⇤ Na = a a = = |1i 1| . h 01 This has the e↵ect of returning |1i for input |1i, but 0 for |0i: Na = a⇤ a = |1ih0| |0ih1| = |1ih0 | 0ih1| = |1i 1|. h Thus it’s a test for |1i. (This is a partial identity operation.) ¶10. 0-test: Similarly,12 1 ⇤ Na = aa = ✓ 10 00 ◆ = | 0i 0| . h (Feynman writes this 1 Na because he writes 1 = I .) This has the e↵ect of returning |0i for input |0i, but 0 for |1i. This is test for |0i. (This is the rest of the identity operation.) ¶11. Universality: The two operations a and a⇤ are su cient for creating all 2 ⇥ 2 matrices, and therefore all transformations on a single qubit. Note that ✓ ◆ wx = waa⇤ + xa + ya⇤ + za⇤ a. yz 11 This matrix is not the same as that given in F82 and F85, since Feynman uses the basis |1i = (1, 0)T , |0i = (0, 1)T . 12 This matrix is di↵erent from that given in F82 and F85, as explained in the previous footnote. F. been s tudied i n n ormal c omputers. B ut u ntil w e f ind a s pecific i mplement ation f or t his c omputer, I d o n ot k now h ow t o p roceed t o a nalyze t hese effects. H owever, i t a ppears t hat t hey w ould b e v ery i mportant, i n p ractice. T his c omputer s eems t o b e v ery d elicate a nd t hese i mperfections m ay p roduce c onsiderable h avoc. T he t ime n eeded t o m ake a s tep o f c alculation depends o n t he s trength o r t he e nergy o f t he i nteractions i n t he t erms of t he H amiltonian. I f e ach o f t he t erms i n t he H amiltonian is s upposed t o b e o f t he o rder o f 0.1 e lectron v olts, t hen i t ppears that t he t ime f or the c ursor UNIVERSAL aQUANTUM COMPUTERS to m ake e ach s tep, if d one i n a b allistic f ashion, is o f t he o rder 6 x 10 -15 sec. T his does n ot r epresent <o 181 C ...... p~ q I F c = t GO p T O q A ND P UT c = O I F c = O GO p T O r A ND P UT c = I I H : q * c p + r *c*p I F c = I GO r T O p A ND P UT c : O IF c = O GO q T O p A ND P UT c = I + p *c*q + p * c r F ig. 7. S witch. Figure III.40: Switch element. 0/1 annotations on the wires show the c values. [fig. from F85] ¶12. Negation: F writes Aa for the negation operation applied to a. Obviously, Aa = a + a⇤ (it annihilates |1i and creates from |0i) and 1 = aa⇤ + a⇤ a (it passes |0i and passes |1i. Prove that Aa Aa = 1 (exercise). ¶13. CNOT: F writes Aa,b for the CNOT operation applied to lines a and b. Aa,b = a⇤ a(b + b⇤ ) + aa⇤ . Notice that this is a tensor product on the regis...
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This document was uploaded on 03/14/2014 for the course COSC 494/594 at University of Tennessee.

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