Quantum Computing
(Fall 2013)
Instructor: Shengyu Zhang.
Lecture 6 Quantum query complexity: Upper bound.
In this lecture, we’ll first finish the proof for the quantum adversary method as a lower bound
of quantum query complexity. Then we show a matching upper bound, by which we prove
the result that the quantum adversary method is in the same order of the quantum query
complexity.
1. Lower bound: Quantum adversary method
Recall that the quantum query complexity 𝑄𝜖 (𝑓) is defined as the minimum number of
queries needed for an 𝜖-error quantum query algorithm to solve a computational problem on
the worst case input. The best lower bound by the quantum adversary method is
‖𝛤‖
𝐴𝑑𝑣(𝑓) =
max
𝛤≠0:𝐻𝑒𝑟𝑚𝑖𝑡𝑖𝑎𝑛 max‖𝛤 ∘ 𝐷𝑖 ‖
𝑖
where 𝐷𝑖 (𝑥, 𝑦) = 1[𝑥𝑖 ≠𝑦𝑖 ]. We’ll first finish the proof of the following theorem.
1
𝑄𝜖 (𝑓) ≥ ( − √𝜖(1 − 𝜖)) 𝐴𝑑𝑣(𝑓)
2
2. Upper bound:
It turns out that the above lower bound is tight. In this section, we’ll show an even strong
result, a quantum query algorithm solving a more general computational task of quantum
state conversion [LMR+11]. Suppose we have a set of pure quantum states {|𝜌𝑥 ⟩: 𝑥 ∈
{0,1}𝑛 } and we’d like to convert them to another set of pure quantum states {|𝜎𝑥 ⟩: 𝑥 ∈
{0,1}𝑛 }. (Usually the Greek letters 𝜌 and 𝜎 are for mixed states, but somehow they are
used in [LMR+11] to denote pure states, and we just respect their notation here.) The
conversion need to be correct for every 𝑥 ∈ {0,1}𝑛 , namely we need to use one universal
algorithm to convert |𝜌𝑥 ⟩ to |𝜎𝑥 ⟩ for all 𝑥 ∈ {0,1}𝑛 simultaneously. In general this is not
doable without the knowledge of 𝑥, as shown by the following basic result.
Exercise. There is a unitary operation 𝑈 that converts the states {|𝑢𝑖 ⟩: 𝑖 = 1, … , 𝑛} to
{|𝑣𝑖 ⟩: 𝑖 = 1, … , 𝑛} if and only if ⟨𝑢𝑖 |𝑢𝑗 ⟩ = ⟨𝑣𝑖 |𝑣𝑗 ⟩ for all 𝑖, 𝑗 ∈ [𝑛].
Since this condition doesn’t hold for general {|𝜌𝑥 ⟩: 𝑥 ∈ {0,1}𝑛 } and {|𝜎𝑥 ⟩: 𝑥 ∈ {0,1}𝑛 }. So
we need to get some knowledge of the index 𝑥. (In the extreme case, consider that we have
the full knowledge of 𝑥. Then we don’t even need the given state |𝜌𝑥 ⟩---we can generate
|𝜎𝑥 ⟩ by ourselves.) We access 𝑥 by queries as usual. The question is what the minimum
number of queries to 𝑥 is needed. We of course allow the algorithm to take ancilla, so the
algorithm takes as input |𝜌𝑥 ⟩|0⟩ and generates a state close to |𝜎𝑥 ⟩|𝑠𝑥 ⟩ for some |𝑠𝑥 ⟩. This
generalizes the function evaluation setting, where all |𝜌𝑥 ⟩ = |0⟩, and target states is
|𝜎𝑥 ⟩|𝑠𝑥 ⟩ = |𝑓(𝑥)⟩|𝑠𝑥 ⟩ for some ancilla |𝑠𝑥 ⟩.
The result is the following:
Theorem. The query complexity of the above state conversion problem is at most
𝑂 ((𝛾2 (𝑀𝜌 − 𝑀𝜎 |𝐷) log(1/𝜖))/𝜖 2 )
where

𝑀𝜌 = [⟨𝜌𝑥 |𝜌𝑦 ⟩]𝑥,𝑦 , 𝑀𝜎 = [⟨𝜎𝑥 |𝜎𝑦 ⟩]𝑥,𝑦 ,

𝐷 = {𝐷1 , … , 𝐷𝑛 } with 𝐷𝑖 = [1[𝑥𝑖 ≠𝑦𝑖 ] ]

𝛾2 (𝐴|𝑍) = min {max max {∑𝑗‖|𝑢𝑥𝑗 ⟩‖ , ∑𝑗‖|𝑣𝑥𝑗 ⟩‖ } : 𝐴𝑥𝑦 = ∑𝑗⟨𝑢𝑥𝑗 |𝑣𝑦𝑗 ⟩(𝑍𝑗 )𝑥𝑦 }
𝑥,𝑦
,
2
2
𝑥
For Boolean function evaluation, note that all |𝜌𝑥 ⟩ = |0⟩ and |𝜎𝑥 ⟩ = |𝑓(𝑥)⟩, thus 𝑀𝜌 = 𝐽
and 𝑀𝜎 = 𝐹 = [1[𝑓(𝑥)=𝑓(𝑦)] ]
𝑥,𝑦
. It turns out that
𝐴𝑑𝑣(𝑓) = 𝛾2 (𝐽 − 𝐹|𝐷)
by SDP duality. Thus the above theorem does implies the tightness of 𝐴𝑑𝑣(𝑓) as a lower
bound of the quantum query complexity.
Let 𝑊 = 𝛾2 (𝑀𝜌 − 𝑀𝜎 |𝐷). An optimum solution |𝑢𝑥𝑗 ⟩, |𝑣𝑥𝑗 ⟩ in the definition of
𝛾2 (𝑀𝜌 − 𝑀𝜎 |𝐷) has the following properties by definition
2
2
∑𝑗‖|𝑢𝑥𝑗 ⟩‖ ≤ 𝑊, ∑𝑗‖|𝑣𝑥𝑗 ⟩‖ ≤ 𝑊, ∀𝑥
(1)
∑𝑗:𝑥𝑗≠𝑦𝑗⟨𝑢𝑥𝑗 |𝑣𝑦𝑗 ⟩ = ⟨𝜌𝑥 |𝜌𝑦 ⟩ − ⟨𝜎𝑥 |𝜎𝑦 ⟩ (= 1[𝑓(𝑥)≠𝑓(𝑦)] for Boolean 𝑓)
(2)
Instead of repeating the proof in the usual way, let me try to extract the line of main ideas.

We actually only need to deal with the case ⟨𝜌𝑥 |𝜎𝑥 ⟩ = 0, since otherwise we can first
attach a |0⟩ to |𝜌𝑥 ⟩, and then transfer it to |1⟩|𝜎𝑥 ⟩, and then discard the |1⟩.

To transfer |0⟩|𝜌𝑥 ⟩ to |1⟩|𝜎𝑥 ⟩, it suffices to keep |𝑡𝑥+ ⟩ unchanged and flipping
|𝑡𝑥− ⟩ to −|𝑡𝑥− ⟩, where
|𝑡𝑥+ ⟩ =
1
√2
(|0⟩|𝜌𝑥 ⟩ + |1⟩|𝜎𝑥 ⟩), and |𝑡𝑥− ⟩ =
(Indeed, |0⟩|𝜌𝑥 ⟩ =
1
√2
(|𝑡𝑥+ ⟩ + |𝑡𝑥− ⟩) →
1
√2
1
√2
(|0⟩|𝜌𝑥 ⟩ − |1⟩|𝜎𝑥 ⟩).
(|𝑡𝑥+ ⟩ − |𝑡𝑥− ⟩) = |1⟩|𝜎𝑥 ⟩.) So it’s enough
to find a 𝑈 s.t. |𝑡𝑥+ ⟩ is close to the 1-eigenspace and |𝑡𝑥− ⟩ is close to the
(−1)-eigenspace.

Actually |𝑡𝑥− ⟩ doesn’t need to be close to (−1)-eigenspace. As long as |𝑡𝑥− ⟩
doesn’t have a large support on eigenspace of 𝑈 corresponding to eigenvalue with
small phases, a tool called Phase Detection can move |𝑡𝑥− ⟩ to close to −|𝑡𝑥− ⟩.
(Namely, as long as 𝑈 moves most of |𝑡𝑥− ⟩’s components, with respect to 𝑈’s
eigenvectors, away in phase, not necessarily to the same far, the Phase Detection
works.) More specifically, ∀𝑈, ∀𝜃 ∗ , there is a circuit 𝑅(𝑈) which, on any
𝑒 𝑖𝜃 -eigenvector |𝛽⟩ of 𝑈 where 𝜃 ≥ 𝜃 ∗ , has
‖𝑅(𝑈)|𝛽⟩|0𝑏 ⟩ + |𝛽⟩|0𝑏 ⟩‖ < 𝛿.
(3)
Here the number of extra qubits needed is 𝑏 = 𝑂(log(1/𝛿) log(1/𝜃 ∗ )), and number
of queries of controlled-𝑈 and controlled-𝑈 † is 𝑂(log(1/𝛿) /𝜃 ∗ ).

The paper finds such a 𝑈, which consists of repeated applications of two reflections.
In what follows, we only consider the Boolean-function evaluation case for simplicity.
Algorithm: Run Phase Detection on unitary 𝑈𝑥 = (2𝛱𝑥 − 𝟏)(2𝛥 − 𝟏) and state
|0⟩|𝜌𝑥 ⟩|0𝑏 ⟩, with precision 𝜃 ∗ = 𝜖 2 /𝑊 and error 𝜖.

Suppose that 𝑊 = 𝛾2 (𝐽 − 𝐹|𝐷), and {|𝑢𝑥𝑗 ⟩, |𝑣𝑥𝑗 ⟩ ∈ ℂ𝑚 : 𝑥, 𝑗} is an optimal solution
in the definition of 𝛾2.

𝛥 is the projection onto span{|𝛹𝑥 ⟩: 𝑥}⊥, where
|𝜓𝑥 ⟩ = (𝜖/√𝑊)|t x− ⟩ − ∑𝑗|𝑗⟩|1 − 𝑥𝑗 ⟩|𝑢𝑥𝑗 ⟩.
Note that here we attached another space 𝐻′ = ℂ𝑛+2+𝑚 to 𝐻 (where |0⟩|𝜌𝑥 ⟩
originally is), s.t. |𝜌𝑥 ⟩ in the specification of the algorithm lives in the 𝐻 part of
𝐻 ⊕ 𝐻′ . So are states |t x± ⟩. But |𝛹𝑥 ⟩ is in the whole space 𝐻 ⊕ 𝐻′, with the first
term in 𝐻 and second in 𝐻′. Note that 𝐻 ⊕ 𝐻′ is the direct sum, not product.

𝛱𝑥 requires that when the first register of 𝐻′ is 𝑗, the second register of 𝐻′ be 𝑥𝑗 .
Query complexity: 𝑂̃(W/ε2 ). Each 2𝛱𝑥 − 𝟏 takes one query, and 2𝛥 − 𝟏 doesn’t need
any query.
Correctness:
(Notation: 𝑃𝜃 is the projection onto the eigenspace of 𝑈 with eigenphase 𝜃, and similarly
for 𝑃≤𝜃 . That is, if 𝑈 = ∑𝑗 𝑒 𝑖𝜃𝑗 |𝜃𝑗 ⟩⟨𝜃𝑗 | is the spectral decomposition of 𝑈, then
𝑃𝜃 = ∑𝑗:𝜃𝑗=𝜃 𝑒 𝑖𝜃𝑗 |𝜃𝑗 ⟩⟨𝜃𝑗 |

and 𝑃≤𝜃 = ∑𝑗:𝜃𝑗 ≤𝜃 𝑒 𝑖𝜃𝑗 |𝜃𝑗 ⟩⟨𝜃𝑗 |. )
|𝑡𝑥+ ⟩ is close to a 1-eigenvector of 𝑈. More precisely, ‖𝑃0 |𝑡𝑥+ ⟩‖2 ≥ 1 − 𝜖 2 .
Fact 1. ‖𝑃0 |𝑡𝑥+ ⟩‖2 ≥ 1 − 𝜖 2 .
[Proof] Actually |𝑡𝑥+ ⟩ is in the subspace 𝛱𝑥 and almost in the subspace 𝛥. For the
former, note that |𝑡𝑥+ ⟩ is in 𝐻, while 𝛱𝑥 only has requirement on 𝐻 ′ . For the latter,
directly bounding ‖Δ⊥ |𝑡𝑥+ ⟩‖ = ‖𝑠𝑝𝑎𝑛{|𝛹𝑥 ⟩: 𝑦}|𝑡𝑥+ ⟩‖ is not that obvious:
⟨𝜓𝑦 |𝑡𝑥+ ⟩ =
𝜖
√𝑊
⟨𝑡𝑦− |𝑡𝑥+ ⟩ =
𝜖
2√𝑊
1[𝑓(𝑥)≠𝑓(𝑦)] , but there are many 𝑦’s with 𝑓(𝑦) =
𝑓(𝑥). Fortunately, one can add a little bit to |𝑡𝑥+ ⟩ to cancel out the annoying factor:
𝜖
𝜖
′ ⟩
′ ⟩
Define |𝑡𝑥+
= |𝑡𝑥+ ⟩ + 2√𝑊 ∑𝑗|𝑗⟩|𝑥𝑗 ⟩|𝑣𝑥𝑗 ⟩, then ⟨𝜓𝑦 |𝑡𝑥+
= 2√𝑊 (1[𝑓(𝑥)≠𝑓(𝑦)] −
′ ⟩
∑𝑗:𝑥𝑗≠𝑦𝑗⟨𝑢𝑥𝑗 |𝑣𝑦𝑗 ⟩) = 0 by Eq.(2). And note that the extra term in |𝑡𝑥+
is of (𝜖/
′ ⟩
2)-small norm because of Eq.(1). (Also note that |𝑡𝑥+
is still in the subspace 𝛱𝑥
because the extra term exactly satisfies the requirement of 𝛱𝑥 .)
□

|𝑡𝑥− ⟩ has a small support on small-eigenvectors of 𝑈.
Fact 2. ‖𝑃≤𝜃 |𝑡𝑥− ⟩‖2 ≤ 𝜃 2 𝑊 2 /4𝜖 2 , ∀θ.
[Proof] We’ll use a spectral gap lemma that is frequently used in one form or another.
Lemma. Suppose 𝛱 and 𝛥 are two projections, and 𝑈 = (2𝛱 − 𝟏)(2𝛥 − 𝟏).
Suppose that {|𝜙𝜃 ⟩} is a complete orthonormal set of eigenvectors of 𝑈, with the
respective eigenvalues 𝑒 𝑖𝜃 . For any vector |𝜔⟩, if Δ|ω⟩ = 0, then for any 𝜃 ∗ ≥ 0,
‖𝑃≤𝜃∗ 𝛱|𝜔⟩‖ ≤
𝜃∗
2
‖|𝜔⟩‖.
We’ll skip the proof of the lemma, and state how to use it to prove Fact 2. The
application is very straightforward by setting |𝜔⟩ =
√𝑊
|𝜓𝑥 ⟩.
𝜖
Note that 𝛥|𝜔⟩ = 0
and |𝑡𝑥− ⟩ = 𝛱𝑥 |𝜔⟩ by definition of 𝛥 and 𝛱𝑥 , respectively. □
Now putting the above two together, we can show the correctness of the algorithm.
‖𝑅(𝑈𝑥 )|0⟩|𝜌𝑥 ⟩ − |1⟩|𝜎𝑥 ⟩‖ = ‖𝑅(𝑈𝑥 )|0⟩|𝜌𝑥 ⟩|0𝑏 ⟩ − |1⟩|𝜎𝑥 ⟩|0𝑏 ⟩‖
=
≤
1
√2
1
√2
‖(𝑅(𝑈𝑥 ) − 1)|𝑡𝑥+ ⟩|0𝑏 ⟩ − (𝑅(𝑈𝑥 ) + 1)|𝑡𝑥− ⟩|0𝑏 ⟩‖
‖(𝑅(𝑈𝑥 ) − 1)|𝑡𝑥+ ⟩|0𝑏 ⟩‖ +
1
√2
‖(𝑅(𝑈𝑥 ) + 1)|𝑡𝑥− ⟩|0𝑏 ⟩‖
We use Fact 1 to bound the item 1, and use Eq.(3) and Fact 2 to bound item 2.
Reference
[LMR+11] Troy Lee, Rajat Mittal, Ben Reichardt, Robert Spalek, Mario Szegedy, Quantum
query complexity of state conversion, FOCS’11.