quantum spectrum testing ryan o’donnelljohn wright carnegie mellon
TRANSCRIPT
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Quantum Spectrum Testing
Ryan O’Donnell John Wright
Carnegie Mellon Carnegie Mellon
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Picture byJorge Cham
qudit
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one of: |1⟩, |2⟩, …, |d⟩ unknownorthonormal
vectors in ℂd
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Measuring the qudit…... yields probabilisticinfo about which of
|1⟩, …, |d⟩ it is.
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Actual output:
p1 p2 · · · pd
|1⟩ |2⟩ · · · |d⟩
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Actual output:
p1 p2 · · · pd
|1⟩ |2⟩ · · · |d⟩
An unknown probability distribution over an unknown set of d orthonormal vectors.
(Can represent by the “density matrix”
ρ = p1 |1⟩⟨1| + p2 |2⟩⟨2| + · · · + pd |d⟩⟨d|,
a PSD matrix with spectrum {p1, p2, …, pd}.
But we won’t emphasize this notation.)
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Actual output:
p1 p2 · · · pd
|1⟩ |2⟩ · · · |d⟩
An unknown probability distribution over an unknown set of d orthonormal vectors.
It’s expensive, but you can hit the button n times.
d=3, n=7 example:
|1⟩|3⟩|2⟩|2⟩|1⟩|1⟩|3⟩ ∈ (ℂ3)⊗7
with prob. p1 p3 p2 p2 p1 p1 p3
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Quantum Questions
#1: Quantum Tomography.
Learn |1⟩, …, |d⟩, p1, …, pd. (More precisely, ρ.)
(Approximately, up to some ϵ, w.h.p.)
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Quantum Questions
#1: Quantum Tomography.
Learn |1⟩, …, |d⟩, p1, …, pd.
#2: Spectrum Learning.
Learn the multiset {p1, …, pd}.
(Approximately, up to some ϵ, w.h.p.)
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Quantum Questions
#1: Quantum Tomography.
Learn |1⟩, …, |d⟩, p1, …, pd.
#2: Spectrum Learning.
Learn the multiset {p1, …, pd}.
#3: Spectrum Testing.
Determine if {p1, …, pd} satisfies
a certain property. E.g., is
(In the Property Testing model.)
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Quantum Questions
#1: Quantum Tomography.
Learn |1⟩, …, |d⟩, p1, …, pd.
#2: Spectrum Learning.
Learn the multiset {p1, …, pd}.
#3: Spectrum Testing.
Determine if {p1, …, pd} satisfies
a certain property. E.g., is
This one is called the “maximally mixed state”.
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Quantum Questions
#1: Quantum Tomography.
Learn |1⟩, …, |d⟩, p1, …, pd.
#2: Spectrum Learning.
Learn the multiset {p1, …, pd}.
#3: Spectrum Testing.
Determine if {p1, …, pd} satisfies
a certain property. E.g.,has ≤ r nonzeros.
This one is called “having rank ≤ r”.
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Actual output:
p1 p2 · · · pd
|1⟩ |2⟩ · · · |d⟩
An unknown probability distribution.An unknown set of d orthonormal vectors.
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Actual output:
p1 p2 · · · pd
|1⟩ |2⟩ · · · |d⟩
An unknown probability distribution.
If the vectors |1⟩, |2⟩, …, |d⟩ are known,you can measure the outcomes exactly.
Setup becomes equivalent to:
Classical Distribution Learning/Testing
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#1: Learn p1, …, pd (approximately, w.h.p.)
[Folklore]: n = Θ(d) necessary & sufficient.
Classical Distribution Questions
#2: Learn the multiset {p1, …, pd}.
(If you only care about “symmetric” properties.)[RRSS08,Valiants]: still necessary.
#3: Determine if {p1, …, pd} satisfies
a certain property. E.g., is
[Paninski08]: nec. & sufficient.
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Quantum Questions & Answers
#1: Quantum Tomography.
Learn |1⟩, …, |d⟩, p1, …, pd.
[Folklore]: necessary.
[FGLE12]: sufficient.
There have been claims that d2 suffices…
Stay tuned…
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Quantum Questions & Answers
#2: Spectrum Learning.
Learn the multiset {p1, …, pd}.
[ARS88,KW01]: Suggested a natural algorithm.
[HM02,CM06]: Showed it works with .
[O.-Wright’15]: Simpler analysis, gets .
Shows algorithm fails if .
(But maybe another algorithm could work?)
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Quantum Questions & Answers
#3: Spectrum Testing. (Main focus of our work.)
[CHW06]: Distinguishing Unif(d) :=
from Unif(d/2): n = Θ(d) nec. & suff.
[Us]: Distinguishing Unif(d) from Unif(d−Δ): nec. & suff. (for 1 ≤ Δ ≤ d/2).
[Us]: (Analogue of Paninski’s theorem.) Distinguishing Unif(d) from ϵ-far-from-Unif(d): nec. & suff.
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Plan for rest of the talk
Not going to prove any of the theorems.
Just going to try to explain the setup.
Will start by explaining quantum measurement.
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Quantum Measurement in ℂd
Measurer selects an orthogonal
decomposition of ℂd into subspaces,
ℂd = S1 ⊕ S2 ⊕ · · · ⊕ Sm
If the unknown unit state vector is |v⟩,
measurer observes “Sj” with probability
|| Proj Sj( |v⟩
)||2
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ℂ2ℝ2
say this is theunknown |1⟩ say we measure
with coord. axis
decomp S1⊕S2
S1
S2
observe S1 with prob. 1/4
S2 with prob. 3/4
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ℝ2
say this is theunknown |1⟩ say we measure
with coord. axis
decomp S1⊕S2
S1
S2
observe S1 with prob.
(1/3)·(1/4) + (2/3)·(3/4)
|2⟩
p1=1/3:
p2=2/3:
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ℝ2
say this is theunknown |1⟩
Note that if |1⟩, |2⟩are known, we’djust measure with
that decomposition.
S2
S1
|2⟩
p1=1/3:
p2=2/3: Get S1 with prob. p1,
S2 with prob. p2,
Exactly the classical setup.
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|1⟩|3⟩|2⟩|2⟩|1⟩|1⟩|3⟩ ∈ (ℂ3)⊗7
Perhaps d=3, n=7, and result of experiment is
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|1⟩|3⟩|2⟩|2⟩|1⟩|1⟩|3⟩ ∈ (ℂ3)⊗7
Perhaps d=3, n=7, and result of experiment is
|1322113⟩
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|1⟩|3⟩|2⟩|2⟩|1⟩|1⟩|3⟩ ∈ (ℂ3)⊗7
Perhaps d=3, n=7, and result of experiment is
If you want, you can measure each qudit separately.
However, it’s more effective to do one
giant measurement on (ℂ3)⊗7.
|1322113⟩
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n = 2
Fix also d = 3, for concreteness.
Unknown state lies in (ℂ3)⊗2, a 9-dim. space:
span(|11⟩,|12⟩,|13⟩,|21⟩,|22⟩,|23⟩,|31⟩,|32⟩,|33⟩)
There’s a certain subspace of (ℂ3)⊗2 called
Sym2(ℂ3): the symmetric subspace.
One definition: All vectors invariant under S2
(snob’s notation for permutations of {1,2}).
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n = 2
Sym2(ℂ3): the symmetric subspace.
An orthonormal basis for it:
dim Sym2(ℂ3) = 6
Proj Sym2(ℂ
3)(
|v⟩ )
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n = 2
n = 2 idea: Measure w.r.t. the decomposition
(ℂ3)⊗2 = Sym2(ℂ3) ⊕ Sym2(ℂ3)⊥
Problem (?): The basis |1⟩, |2⟩, |3⟩ is unknown, so how does measurer specify it?
No problem: Sym2(ℂ3) has a basis-free definition;
it’s span{v⊗v : v∈ℂ3}.
(This measurement also called the “SWAP test”.)
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n = 2
n = 2 idea: Measure w.r.t. the decomposition
(ℂ3)⊗2 = Sym2(ℂ3) ⊕ Sym2(ℂ3)⊥
with prob. , state is |11⟩,
|| Proj Sym2(ℂ
3)( |11⟩
)||2 = 1
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n = 2
n = 2 idea: Measure w.r.t. the decomposition
(ℂ3)⊗2 = Sym2(ℂ3) ⊕ Sym2(ℂ3)⊥
with prob. , state is |12⟩,
|| Proj Sym2(ℂ
3)( |12⟩
)||2 =
with prob. , state is |11⟩,
|| Proj Sym2(ℂ
3)( |11⟩
)||2 = 1
etc. Add it all up…
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n = 2
n = 2 idea: Measure w.r.t. the decomposition
(ℂ3)⊗2 = Sym2(ℂ3) ⊕ Sym2(ℂ3)⊥
Pr [observing Sym2(ℂ3)]
Minimized iff {p1,p2,p3} = {1,3,1}.
(Using this, can ϵ-test Unif(d) with O(d2/ϵ4) copies.)
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n = 2
n = 2 idea: Measure w.r.t. the decomposition
(ℂ3)⊗2 = Sym2(ℂ3) ⊕ Sym2(ℂ3)⊥
Why just decompose into 6dim + 3dim?
It can’t hurt to further decompose into 91-dimensional subspaces.
But actually… It’s without loss of generality!
To explain why, let’s detour to the classical case.
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Detour:
When testing / learning an Sd-invariant property
of a classical probability distribution p1, …, pd,
i.e., one depending only on {p1, …, pd},
without loss of generality
you can throw away a lot of information you see,
and just remember the “orbit” under Sn × Sd.
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Typical sample when n=20, d=5 might be…
54423131423144554251
Intuition 1: Permuting the n positions doesn’t matter. Hence may as well only retain histogram.
Say we’re trying to test if distribution
p1, …, pd is Unif(d).
(Or any property invariant under relabeling [d]; e.g., “support ≤ r”.)
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Say we’re trying to test if distribution
p1, …, pd is Unif(d).
(Or any property invariant under relabeling [d]; e.g., “support ≤ r”.)
12345
Typical sample when n=20, d=5 might be…
54423131423144554251
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Say we’re trying to test if distribution
p1, …, pd is Unif(d).
(Or any property invariant under relabeling [d]; e.g., “support ≤ r”.)
Typical sample when n=20, d=5 might be…
12345
Intuition 2: Since property is symmetric, permuting the d symbols doesn’t matter.
Hence may as well sort the histogram.
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Say we’re trying to test if distribution
p1, …, pd is Unif(d).
(Or any property invariant under relabeling [d]; e.g., “support ≤ r”.)
Typical sample when n=20, d=5 might be…
12345
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Say we’re trying to test if distribution
p1, …, pd is Unif(d).
(Or any property invariant under relabeling [d]; e.g., “support ≤ r”.)
Typical sample when n=20, d=5 might be…
1st most freq:2nd most freq:3rd most freq:4th most freq:5th most freq:
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Say we’re trying to test if distribution
p1, …, pd is Unif(d).
(Or any property invariant under relabeling [d]; e.g., “support ≤ r”.)
Typical sample when n=20, d=5 might be…
λ1 := 1st most freq:
λ2 := 2nd most freq:
λ3 := 3rd most freq:
λ4 := 4th most freq:
λ5 := 5th most freq:(A sorted histogram is AKA a Young diagram.)
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Claim: An n-sample algorithm for testing an
“Sd-invariant” property of p1, …, pd
(one depending only on {p1, …, pd})
may ignore full sample and just retain the sorted histogram (Young diagram).
Formal proof: If algorithm succeeds with high probability, can check it also succeeds with equally high probability if it first blindly applies a
random perm from Sn to the positions and a random perm from Sd
to the symbols. But then conditioned on the sorted histogramof the sample it sees, the sample is uniformly random among thosewith that sorted histogram. So an algorithm that only sees thesorted histogram can also succeed with equally high probability.
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By the way, for a given sorted histogram /
Young diagram λ = (λ1, λ2, …, λd),
Pr [observing λ]
is a nice symmetric polynomial in p1, …, pd
depending on λ.
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End detour;back to n=2 tests for qubit’s spectrum
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n = 2
n = 2 idea: Measure w.r.t. the decomposition
(ℂ3)⊗2 = Sym2(ℂ3) ⊕ Sym2(ℂ3)⊥
Why is this without loss of generality?
Permutations of the 2 copies shouldn’t matter.
Arbitrary unitary transformations on ℂ3
shouldn’t matter, because algorithm should workfor any unknown orthonormal |1⟩, |2⟩, |3⟩.
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n = 2
n = 2 idea: Measure w.r.t. the decomposition
(ℂ3)⊗2 = Sym2(ℂ3) ⊕ Sym2(ℂ3)⊥
An algorithm may as well apply a random
permutation π ∈ S2 to the 2 copies,
and a random unitary U ∈ U3 to each qudit.
Easy exercise:
If original state |v⟩|w⟩ had length ℓ in Sym2(ℂ3),
randomization’s component in Sym2(ℂ3) will
be a totally (Haar)-random length-ℓ vector.
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n = 2
n = 2 idea: Measure w.r.t. the decomposition
(ℂ3)⊗2 = Sym2(ℂ3) ⊕ Sym2(ℂ3)⊥
∴ WLOG we needn’t further decompose Sym2(ℂ3).
We actually need the same result for Sym2(ℂ3)⊥.
AKA Alt2(ℂ3) = {v∈(ℂ3)⊗2 : πv = sgn(π)v ∀π∈S2}.
It’s an equally easy exercise.
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?n = 3?
Measure w.r.t. the decomposition
? (ℂ3)⊗3 = Sym3(ℂ3) ⊕ Alt3(ℂ3) ?
Unfortunately, that’s not a decomposition.
dim. 27 dim. 10 dim. 1
What’s missing: a 16-dim. space called S(2,1)(ℂ3).
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n = 3
Measure w.r.t. the decomposition
(ℂ3)⊗3 = Sym3(ℂ3) ⊕ Alt3(ℂ3) ⊕ S(2,1)(ℂ3)
One can show that this is WLOG.
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General n, d
Measure w.r.t. the decomposition
(ℂd)⊗n = Symn(ℂd) ⊕ Altn(ℂd) ⊕ ···stuff···
yada yada yada about the groups Sn and Ud,
yada yada yada about representation theory,yada yada yada about “Schur–Weyl duality”,
I will give you the tl;dr.
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General n, d
There’s an explicit subspace decomposition
such that it’s WLOG to measure w.r.t. it.
Interestingly, it’s indexed by Young diagrams:
= {diagrams with n boxes and d rows}
All of this was observed in [CHW06].
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Quantum spectrum testing
You’re testing properties of {p1, …, pd}.
For n copies, WLOG you observe a randomYoung diagram with n boxes and d rows.
For each such λ, Pr [observing λ] is an explicit
symmetric polynomial in p1, …, pd.
We can also describe this distribution on λin a simple quantum-free way…
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The distribution on λ
Pick w ~ [d]n with symbols probs. i.i.d. p1, …, pd.
Classical testing: you just get to see w.
n=20, d = 5 e.g.: w = 54423131423144554251
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The distribution on λ
λ1 = longest incr. subseq
λ1 + λ2 = longest union of2 incr. subseqs
λ1 + λ2 + λ3 = longest union of3 incr. subseqs
· · ·
λ1 + λ2 + λ3 + · · · + λd = n
Quantum testing: you see λ defined by…
Pick w ~ [d]n with symbols probs. i.i.d. p1, …, pd.
n=20, d = 5 e.g.: w = 54423131423144554251
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The distribution on λ
54423131423144554251
54423131423144554251
54423131423144554251
54423131423144554251
54423131423144554251
Quantum testing: you see λ defined by…
Pick w ~ [d]n with symbols probs. i.i.d. p1, …, pd.
n=20, d = 5 e.g.: w = 54423131423144554251
(Alternatively, you see “RSK”(w).)
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Quantum spectrum testing
You’re testing properties of {p1, …, pd}.
For n copies, WLOG you observe a randomYoung diagram λ with n boxes and d rows.
λ encodes L.I.S. information of a random word.
You now know everything you need to know.
Go forth and prove testing upper/lower bounds!
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Our subsequent techniquesKerov’s algebra of polynomials on Young diagrams
Robinson–Schensted–Knuth correspondence
Modified Frobenius coordinates and Maya notation
Fourier analysis over Sn
Gaussian Unitary Ensemble
Littlewood–Richardson structure constants
Okounkov–Olshanski Binomial Formula
Partition 2-quotients
Cyclic sieving phenomenon
I kind of line-by-lined all our proofs at some point.For more details, you might prefer to ask John…
You know the setup, maybe you can simplify it all.
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Thanks!