18.405J/6.841J Advanced Complexity Theory
Lecturer: Dan Spielman
Lecture 13
March 22, 2001
Scribe: Abhinav Kumar
This lecture will be on oracle quantum turing machines.
We wish to compare Quantum complexity classes to ordinary complexity classes.
Fact 1 BQP PSPACE
This is easy to prove, in fact it's on the problem set. A harder proposition is the following.
Fact 2 BQP P#P
Another result says
9A such that NP A 2= BQP A :
So at least in its ability to access an oracle, NP can be more powerful than BQP.
We need to dene the notion of accessing an oracle for a quantum computer. The idea is
to put quantum gates in the circuit representing the oracle. An oracle is viewed as a function
A : f0 1g ! f0 1g. We consider the quantum gate QAn , as shown in the gure.
x 2 f0 1gn
x
y A(x)
y
Gate QAn
To describe the quantum gate QA , assume that A tells us what QA does on basis states jx y >,
x 2 f0 1gn y 2 f0 1g. QA sends this to jx A(x) y >. Extend by linearity to get a unitary
transformation.
Then an oracle quantum turing machine just means a quantum circuit which is allowed to
incorporate some gates of this form.
Let L(A) = fwj9x 2 A jxj = jwjg. Clearly L(A) 2 NP A . We will see that quantum computers
can't decide L(A) very well. Putting in one word of a given length makes very slight dierence to
the functioning of a quantum circuit trying to decide L(A). The rough idea is that if jA \ f0 1gn j =
0 or 1, it doesn't make much of a dierence which of these values it is (to a Quantum circuit). The
proof idea is similar to the proof of P A ( NP A .
Theorem 3 L(A) 2= BQP A
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Proof: Suces to argue about one quantum circuit. The way we elude all Quantum circuits is
the same as in the proof of P A ( NP A . In fact, if a quantum circuit with T oracle gates accepts
L(A) with error probability less than a xed constant, then T = (2n=2 ) (this bound is tight - see
Grover's algorithm). Dene Let the initial state be j0 >.
jT >:= UT QA :::QA U1 j0 >
where Ui 's are unitary matrices, T refers to time steps for the circuit. the language A acts as an
oracle: if x 2 f0 1gn , then A(x) = 0 if x 2 A, 1 if x 2= A. Note that if jA \ f0 1gn j = 0, then QAn ,
is the identity. We will compare two cases: no words against exactly one word. Dene
ji >:= Ui :::U1 j0 >
Our goal is to nd x 2 f0 1gn such that if An = fxg, then jjT ; T jj > const: ) T is huge.
q
Lemma 4 9x 2 f0 1gn such that jjT ; T jj 22Tn2
Before we prove the lemma, we state a corollary:
Claim: If T = o(2n=2 ) then this circuit can't distinguish between jA \ f0 1gn j = 0 or 1.
Now our goal is to nd x so that it's not queried in a meaningful way. Assume w.lo.g. that there
is a distinguished set of wires that are inputs to the QA gate, i.e. assume dedicated input wires for
QA . For a state
j >=
X
c basis state
we dene the query magnitude of x in to be
P
c2Sx
c jc >
jc j2 , where
Sx = f basis states with x on oracle wires g
Def. The query magnitude of x at time i is qix = the query magnitude of x in i .
T
Def. The query magnitude of x, qx = P qix .
i=1
P
Idea: take x of mimimal query magnitude so qx < 2Tn , because qix = 1 for every i.
x
Claim: jjT ; T jj pT 2qx (this ) Lemma 4).
Proof of claim:
Let Vi = Ui QA .
jT > = (VT VT ;1 :::V1 )j0 >
jT > = (UT UT ;1 :::U1 )j0 >
ki = Vi Vi;1 :::Vk+1 Uk :::U1 j0 > ( called the k'th hybrid at time i
so 0T = VT :::V1 j0
kT = VT :::Vk+1 Uk :::U1 j0 >
TT = UT :::U1 j0 >
Compare kk+1 with k+1k+1 .
k = kk , we apply either Vk or Uk . It is easily veried that
jjkk+1 ; k+1k+1 jj 2qkx
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) jjkk+1 ; k+1k+1jj2 2qkx since jjkT ; k+1T jj2 = jjkk+1 ; k+1k+1 jj2
because unitary transformations preserve norms. Therefore
TX
;1
By Cauchy-Schwartz,
k=0
jjkT ; k+1T jj2 jjT ; T jj = jj0T ; TT jj Grover's Algorithm
p
X
k
TX
;1
k=0
2qkx 2qx
p
jjkT ; k+1T jj 2qx T
Search in O( N ) time. Let H be the Hadamard matrix
1
1
1
M = p 1 ;1
2
Also, let us consider the matrix corresponding to the gate NOT ,
QNOT2 = 01 10
We shall use the matrices H and NOT to get a unitary transformation that ips the sign of one bit,
namely U = H NOT H .
1
1
1
0
1
1
1
1
0
U = 2 1 ;1 1 0 1 ;1 = 0 ;1
Using this machinery, we can construct a gate Q0A for an oracle A, which takes jx > to ; jx > if
x 2 A, and leaves jx > unchanged otherwise (note this is a unitary transformation, since it is unitary
on the orthogonal basis vectors, and takes them to orthogonal vectors).
So if A = 01, Q0A j01 >= ;j01 >. Suppose we apply this transform to a uniform linear combination of states
jx0 >= 12 (j00 > +j01 > +j10 > +j11 >)
We get out
1 (j00 > ;j01 > +j10 > +j11 >)
2
Now suppose to the result we apply a Hadamard transform to each bit, then apply the matrix
3
2
1
7
6 ;1
7
6
4
;1 5 (reection about mean)
;1
to it, and again the Hadamard transform to it. We get out just j01 >!
The Hadamard transform acts as a Fourier transform. The general algorithm is based on these lines,
the idea is to iterate, applying these two gates (the diusion transform and Q0A ), when we have more
than 2 inputs.
p
RQ0 RQ0 :::jx > ( 4 2n times)
A
A
0
p
the dominant term will be the word we want to select. This takes time O( N ) since N 2n .
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