Quantum Spectrum Testing Ryan O’DonnellJohn Wright Carnegie Mellon.

Quantum Spectrum Testing

Ryan O’Donnell John Wright

Carnegie Mellon Carnegie Mellon

Picture byJorge Cham

qudit

one of: |1⟩, |2⟩, …, |d⟩ unknownorthonormal

vectors in ℂd

Measuring the qudit…... yields probabilisticinfo about which of

|1⟩, …, |d⟩ it is.

Actual output:

p1 p2 · · · pd

|1⟩ |2⟩ · · · |d⟩

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.)

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

Quantum Questions

#1: Quantum Tomography.

Learn |1⟩, …, |d⟩, p1, …, pd. (More precisely, ρ.)

(Approximately, up to some ϵ, w.h.p.)

Quantum Questions


Learn |1⟩, …, |d⟩, p1, …, pd.

#2: Spectrum Learning.

Learn the multiset {p1, …, pd}.

(Approximately, up to some ϵ, w.h.p.)

Quantum Questions


Learn |1⟩, …, |d⟩, p1, …, pd.



#3: Spectrum Testing.

Determine if {p1, …, pd} satisfies

a certain property. E.g., is

(In the Property Testing model.)

Quantum Questions


Learn |1⟩, …, |d⟩, p1, …, pd.






This one is called the “maximally mixed state”.

Quantum Questions


Learn |1⟩, …, |d⟩, p1, …, pd.





a certain property. E.g.,has ≤ r nonzeros.

This one is called “having rank ≤ r”.

Actual output:

p1 p2 · · · pd

|1⟩ |2⟩ · · · |d⟩

An unknown probability distribution.An unknown set of d orthonormal vectors.

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

#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


[Paninski08]: nec. & sufficient.

Quantum Questions & Answers


Learn |1⟩, …, |d⟩, p1, …, pd.

[Folklore]: necessary.

[FGLE12]: sufficient.

There have been claims that d2 suffices…

Stay tuned…




[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?)


#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.

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.

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

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

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:

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.

|1⟩|3⟩|2⟩|2⟩|1⟩|1⟩|3⟩ ∈ (ℂ3)⊗7

Perhaps d=3, n=7, and result of experiment is

|1⟩|3⟩|2⟩|2⟩|1⟩|1⟩|3⟩ ∈ (ℂ3)⊗7


|1322113⟩

|1⟩|3⟩|2⟩|2⟩|1⟩|1⟩|3⟩ ∈ (ℂ3)⊗7


If you want, you can measure each qudit separately.

However, it’s more effective to do one

giant measurement on (ℂ3)⊗7.

|1322113⟩

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}).

n = 2

Sym2(ℂ3): the symmetric subspace.

An orthonormal basis for it:

dim Sym2(ℂ3) = 6

Proj Sym2(ℂ

3)(

|v⟩ )

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”.)

n = 2


(ℂ3)⊗2 = Sym2(ℂ3) ⊕ Sym2(ℂ3)⊥

with prob. , state is |11⟩,

|| Proj Sym2(ℂ

3)( |11⟩

)||2 = 1

n = 2


(ℂ3)⊗2 = Sym2(ℂ3) ⊕ Sym2(ℂ3)⊥


|| Proj Sym2(ℂ

3)( |12⟩

)||2 =


|| Proj Sym2(ℂ

3)( |11⟩

)||2 = 1

etc. Add it all up…

n = 2


(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.)

n = 2


(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.

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.

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”.)




12345


54423131423144554251





12345

Intuition 2: Since property is symmetric, permuting the d symbols doesn’t matter.

Hence may as well sort the histogram.





12345





1st most freq:2nd most freq:3rd most freq:4th most freq:5th most freq:





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.)

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.

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.

End detour;back to n=2 tests for qubit’s spectrum

n = 2


(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⟩.

n = 2


(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.

n = 2


(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.

?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).

n = 3


(ℂ3)⊗3 = Sym3(ℂ3) ⊕ Alt3(ℂ3) ⊕ S(2,1)(ℂ3)

One can show that this is WLOG.

General n, d


(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.

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].

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…

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


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…


n=20, d = 5 e.g.: w = 54423131423144554251


54423131423144554251

54423131423144554251

54423131423144554251

54423131423144554251

54423131423144554251

Quantum testing: you see λ defined by…


n=20, d = 5 e.g.: w = 54423131423144554251

(Alternatively, you see “RSK”(w).)

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!

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.

Thanks!

Quantum Spectrum Testing Ryan O’DonnellJohn Wright Carnegie Mellon.

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