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Quantum Computing (Fall 2013) Instructor: Shengyu Zhang. Lecture 6 Quantum query complexity: Upper bound. In this lecture, weβll 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 ππ (π) is defined as the minimum number of queries needed for an π-error quantum query algorithm to solve a computational problem on the worst case input. The best lower bound by the quantum adversary method is βπ€β π΄ππ£(π) = max π€β 0:π»πππππ‘πππ maxβπ€ β π·π β π where π·π (π₯, π¦) = 1[π₯π β π¦π ]. Weβll first finish the proof of the following theorem. 1 ππ (π) β₯ ( β βπ(1 β π)) π΄ππ£(π) 2 2. Upper bound: It turns out that the above lower bound is tight. In this section, weβll 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 {|ππ₯ βͺ: π₯ β {0,1}π } and weβd like to convert them to another set of pure quantum states {|ππ₯ βͺ: π₯ β {0,1}π }. (Usually the Greek letters π and π 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 π₯ β {0,1}π , namely we need to use one universal algorithm to convert |ππ₯ βͺ to |ππ₯ βͺ for all π₯ β {0,1}π simultaneously. In general this is not doable without the knowledge of π₯, as shown by the following basic result. Exercise. There is a unitary operation π that converts the states {|π’π βͺ: π = 1, β¦ , π} to {|π£π βͺ: π = 1, β¦ , π} if and only if β©π’π |π’π βͺ = β©π£π |π£π βͺ for all π, π β [π]. Since this condition doesnβt hold for general {|ππ₯ βͺ: π₯ β {0,1}π } and {|ππ₯ βͺ: π₯ β {0,1}π }. So we need to get some knowledge of the index π₯. (In the extreme case, consider that we have the full knowledge of π₯. Then we donβt even need the given state |ππ₯ βͺ---we can generate |ππ₯ βͺ by ourselves.) We access π₯ by queries as usual. The question is what the minimum number of queries to π₯ is needed. We of course allow the algorithm to take ancilla, so the algorithm takes as input |ππ₯ βͺ|0βͺ and generates a state close to |ππ₯ βͺ|π π₯ βͺ for some |π π₯ βͺ. This generalizes the function evaluation setting, where all |ππ₯ βͺ = |0βͺ, and target states is |ππ₯ βͺ|π π₯ βͺ = |π(π₯)βͺ|π π₯ βͺ for some ancilla |π π₯ βͺ. The result is the following: Theorem. The query complexity of the above state conversion problem is at most π ((πΎ2 (ππ β ππ |π·) log(1/π))/π 2 ) where ο· ππ = [β©ππ₯ |ππ¦ βͺ]π₯,π¦ , ππ = [β©ππ₯ |ππ¦ βͺ]π₯,π¦ , ο· π· = {π·1 , β¦ , π·π } with π·π = [1[π₯π β π¦π ] ] ο· πΎ2 (π΄|π) = min {max max {βπβ|π’π₯π βͺβ , βπβ|π£π₯π βͺβ } : π΄π₯π¦ = βπβ©π’π₯π |π£π¦π βͺ(ππ )π₯π¦ } π₯,π¦ , 2 2 π₯ For Boolean function evaluation, note that all |ππ₯ βͺ = |0βͺ and |ππ₯ βͺ = |π(π₯)βͺ, thus ππ = π½ and ππ = πΉ = [1[π(π₯)=π(π¦)] ] π₯,π¦ . It turns out that π΄ππ£(π) = πΎ2 (π½ β πΉ|π·) by SDP duality. Thus the above theorem does implies the tightness of π΄ππ£(π) as a lower bound of the quantum query complexity. Let π = πΎ2 (ππ β ππ |π·). An optimum solution |π’π₯π βͺ, |π£π₯π βͺ in the definition of πΎ2 (ππ β ππ |π·) has the following properties by definition 2 2 βπβ|π’π₯π βͺβ β€ π, βπβ|π£π₯π βͺβ β€ π, βπ₯ (1) βπ:π₯πβ π¦πβ©π’π₯π |π£π¦π βͺ = β©ππ₯ |ππ¦ βͺ β β©ππ₯ |ππ¦ βͺ (= 1[π(π₯)β π(π¦)] for Boolean π) (2) Instead of repeating the proof in the usual way, let me try to extract the line of main ideas. ο· We actually only need to deal with the case β©ππ₯ |ππ₯ βͺ = 0, since otherwise we can first attach a |0βͺ to |ππ₯ βͺ, and then transfer it to |1βͺ|ππ₯ βͺ, and then discard the |1βͺ. ο· To transfer |0βͺ|ππ₯ βͺ to |1βͺ|ππ₯ βͺ, it suffices to keep |π‘π₯+ βͺ unchanged and flipping |π‘π₯β βͺ to β|π‘π₯β βͺ, where |π‘π₯+ βͺ = 1 β2 (|0βͺ|ππ₯ βͺ + |1βͺ|ππ₯ βͺ), and |π‘π₯β βͺ = (Indeed, |0βͺ|ππ₯ βͺ = 1 β2 (|π‘π₯+ βͺ + |π‘π₯β βͺ) β 1 β2 1 β2 (|0βͺ|ππ₯ βͺ β |1βͺ|ππ₯ βͺ). (|π‘π₯+ βͺ β |π‘π₯β βͺ) = |1βͺ|ππ₯ βͺ.) So itβs enough to find a π s.t. |π‘π₯+ βͺ is close to the 1-eigenspace and |π‘π₯β βͺ is close to the (β1)-eigenspace. ο· Actually |π‘π₯β βͺ doesnβt need to be close to (β1)-eigenspace. As long as |π‘π₯β βͺ doesnβt have a large support on eigenspace of π corresponding to eigenvalue with small phases, a tool called Phase Detection can move |π‘π₯β βͺ to close to β|π‘π₯β βͺ. (Namely, as long as π moves most of |π‘π₯β βͺβs components, with respect to πβs eigenvectors, away in phase, not necessarily to the same far, the Phase Detection works.) More specifically, βπ, βπ β , there is a circuit π (π) which, on any π ππ -eigenvector |π½βͺ of π where π β₯ π β , has βπ (π)|π½βͺ|0π βͺ + |π½βͺ|0π βͺβ < πΏ. (3) Here the number of extra qubits needed is π = π(log(1/πΏ) log(1/π β )), and number of queries of controlled-π and controlled-π β is π(log(1/πΏ) /π β ). ο· The paper finds such a π, 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 ππ₯ = (2π±π₯ β π)(2π₯ β π) and state |0βͺ|ππ₯ βͺ|0π βͺ, with precision π β = π 2 /π and error π. ο· Suppose that π = πΎ2 (π½ β πΉ|π·), and {|π’π₯π βͺ, |π£π₯π βͺ β βπ : π₯, π} is an optimal solution in the definition of πΎ2. ο· π₯ is the projection onto span{|πΉπ₯ βͺ: π₯}β₯, where |ππ₯ βͺ = (π/βπ)|t xβ βͺ β βπ|πβͺ|1 β π₯π βͺ|π’π₯π βͺ. Note that here we attached another space π»β² = βπ+2+π to π» (where |0βͺ|ππ₯ βͺ originally is), s.t. |ππ₯ βͺ in the specification of the algorithm lives in the π» part of π» β π»β² . So are states |t x± βͺ. But |πΉπ₯ βͺ is in the whole space π» β π»β², with the first term in π» and second in π»β². Note that π» β π»β² is the direct sum, not product. ο· π±π₯ requires that when the first register of π»β² is π, the second register of π»β² be π₯π . Query complexity: πΜ(W/Ξ΅2 ). Each 2π±π₯ β π takes one query, and 2π₯ β π doesnβt need any query. Correctness: (Notation: ππ is the projection onto the eigenspace of π with eigenphase π, and similarly for πβ€π . That is, if π = βπ π πππ |ππ βͺβ©ππ | is the spectral decomposition of π, then ππ = βπ:ππ=π π πππ |ππ βͺβ©ππ | ο· and πβ€π = βπ:ππ β€π π πππ |ππ βͺβ©ππ |. ) |π‘π₯+ βͺ is close to a 1-eigenvector of π. More precisely, βπ0 |π‘π₯+ βͺβ2 β₯ 1 β π 2 . Fact 1. βπ0 |π‘π₯+ βͺβ2 β₯ 1 β π 2 . [Proof] Actually |π‘π₯+ βͺ is in the subspace π±π₯ and almost in the subspace π₯. For the former, note that |π‘π₯+ βͺ is in π», while π±π₯ only has requirement on π» β² . For the latter, directly bounding βΞβ₯ |π‘π₯+ βͺβ = βπ πππ{|πΉπ₯ βͺ: π¦}|π‘π₯+ βͺβ is not that obvious: β©ππ¦ |π‘π₯+ βͺ = π βπ β©π‘π¦β |π‘π₯+ βͺ = π 2βπ 1[π(π₯)β π(π¦)] , but there are many π¦βs with π(π¦) = π(π₯). Fortunately, one can add a little bit to |π‘π₯+ βͺ to cancel out the annoying factor: π π β² βͺ β² βͺ Define |π‘π₯+ = |π‘π₯+ βͺ + 2βπ βπ|πβͺ|π₯π βͺ|π£π₯π βͺ, then β©ππ¦ |π‘π₯+ = 2βπ (1[π(π₯)β π(π¦)] β β² βͺ βπ:π₯πβ π¦πβ©π’π₯π |π£π¦π βͺ) = 0 by Eq.(2). And note that the extra term in |π‘π₯+ is of (π/ β² βͺ 2)-small norm because of Eq.(1). (Also note that |π‘π₯+ is still in the subspace π±π₯ because the extra term exactly satisfies the requirement of π±π₯ .) β‘ ο· |π‘π₯β βͺ has a small support on small-eigenvectors of π. Fact 2. βπβ€π |π‘π₯β βͺβ2 β€ π 2 π 2 /4π 2 , βΞΈ. [Proof] Weβll use a spectral gap lemma that is frequently used in one form or another. Lemma. Suppose π± and π₯ are two projections, and π = (2π± β π)(2π₯ β π). Suppose that {|ππ βͺ} is a complete orthonormal set of eigenvectors of π, with the respective eigenvalues π ππ . For any vector |πβͺ, if Ξ|Οβͺ = 0, then for any π β β₯ 0, βπβ€πβ π±|πβͺβ β€ πβ 2 β|πβͺβ. Weβll skip the proof of the lemma, and state how to use it to prove Fact 2. The application is very straightforward by setting |πβͺ = βπ |ππ₯ βͺ. π Note that π₯|πβͺ = 0 and |π‘π₯β βͺ = π±π₯ |πβͺ by definition of π₯ and π±π₯ , respectively. β‘ Now putting the above two together, we can show the correctness of the algorithm. βπ (ππ₯ )|0βͺ|ππ₯ βͺ β |1βͺ|ππ₯ βͺβ = βπ (ππ₯ )|0βͺ|ππ₯ βͺ|0π βͺ β |1βͺ|ππ₯ βͺ|0π βͺβ = β€ 1 β2 1 β2 β(π (ππ₯ ) β 1)|π‘π₯+ βͺ|0π βͺ β (π (ππ₯ ) + 1)|π‘π₯β βͺ|0π βͺβ β(π (ππ₯ ) β 1)|π‘π₯+ βͺ|0π βͺβ + 1 β2 β(π (ππ₯ ) + 1)|π‘π₯β βͺ|0π βͺβ 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, FOCSβ11.