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Unformatted text preview: b ' ¢ '
It’s not unitary (because0 not norm preserving).
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This is a “partial NOT” 0operation. 0 1
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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 1test: 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. 0test: 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.
 Fall '13
 BruceMacLennan

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