An algebraic approach to phase retrieval

Transcription

1 University of Michigan joint with Aldo Conca, Dan Edidin, and Milena Hering.

2 I learned about frame theory from... AIM Workshop: Frame theory intersects geometry Frame theory intersects geometry July 29 to August 2, 2013 at the American Institute of Mathematics, Palo Alto, California organized by Bernhard Bodmann, Gitta Kutyniok, and Tim Roemer This workshop, sponsored by AIM and the NSF, will be devoted to outstanding problems that are in the intersection of frame theory and geometry. Frames are families of vectors in Hilbert spaces which provide stable expansions. These families are more general than orthonormal bases because they can incorporate linear dependencies. Applications in engineering, quantum theory Cynthia and Vinzant pure mathematics An algebraic have lead approach to several to phase design retrieval

3 Frames and intensity measurements A frame is a collection of vectors Φ = {φ 1,..., φ n } spanning C d. (a redundant basis )

4 Frames and intensity measurements A frame is a collection of vectors Φ = {φ 1,..., φ n } spanning C d. (a redundant basis ) A frame defines intensity measurements of a signal x C d : φ k, x 2 = φ k xx φ k for k = 1,..., n.

5 Frames and intensity measurements A frame is a collection of vectors Φ = {φ 1,..., φ n } spanning C d. (a redundant basis ) A frame defines intensity measurements of a signal x C d : φ k, x 2 = φ k xx φ k for k = 1,..., n. Phase Retrieval: Recover x from its measurements φ k, x 2.

6 Frames and intensity measurements A frame is a collection of vectors Φ = {φ 1,..., φ n } spanning C d. (a redundant basis ) A frame defines intensity measurements of a signal x C d : φ k, x 2 = φ k xx φ k for k = 1,..., n. Phase Retrieval: Recover x from its measurements φ k, x 2. Some Questions: How do we recover the signal x?

7 Frames and intensity measurements A frame is a collection of vectors Φ = {φ 1,..., φ n } spanning C d. (a redundant basis ) A frame defines intensity measurements of a signal x C d : φ k, x 2 = φ k xx φ k for k = 1,..., n. Phase Retrieval: Recover x from its measurements φ k, x 2. Some Questions: How do we recover the signal x? When is recovery of signals in C d possible?

8 Frames and intensity measurements A frame is a collection of vectors Φ = {φ 1,..., φ n } spanning C d. (a redundant basis ) A frame defines intensity measurements of a signal x C d : φ k, x 2 = φ k xx φ k for k = 1,..., n. Phase Retrieval: Recover x from its measurements φ k, x 2. Some Questions: How do we recover the signal x? When is recovery of signals in C d possible? When is recovery of signals in C d stable?

9 Frames and intensity measurements A frame is a collection of vectors Φ = {φ 1,..., φ n } spanning C d. (a redundant basis ) A frame defines intensity measurements of a signal x C d : φ k, x 2 = φ k xx φ k for k = 1,..., n. Phase Retrieval: Recover x from its measurements φ k, x 2. Some Questions: How do we recover the signal x? When is recovery of signals in C d possible? When is recovery of signals in C d stable?

10 Motivation and Applications In practice the signal is some structure that is too small (DNA, crystals) or far away (astronomical phenomena) or obscured (medical images) to observe directly. Figure 1: A typical setup for structured illuminations in diffraction imaging using a phase mask. (picture from Candés-Eldar-Strohmer-Voroninski 2013)

11 Motivation and Applications In practice the signal is some structure that is too small (DNA, crystals) or far away (astronomical phenomena) or obscured (medical images) to observe directly. If some measurements are possible, then one hopes to reconstruct this structure. Figure 1: A typical setup for structured illuminations in diffraction imaging using a phase mask. (picture from Candés-Eldar-Strohmer-Voroninski 2013)

12 Motivation and Applications In practice the signal is some structure that is too small (DNA, crystals) or far away (astronomical phenomena) or obscured (medical images) to observe directly. If some measurements are possible, then one hopes to reconstruct this structure. Figure 1: A typical setup for structured illuminations in diffraction imaging using a phase mask. (picture from Candés-Eldar-Strohmer-Voroninski 2013) Here our signal x lies in a finitedimensional space (C d ), and its measurements are modeled by φ k, x 2 for φ k C d.

13 Phase Retrieval: recovering a vector from its measurements When do the frame measurements φ k, x 2 determine x C d?

14 Phase Retrieval: recovering a vector from its measurements When do the frame measurements φ k, x 2 determine x C d? (Never: φ k, x 2 invariant under x e iθ x)

15 Phase Retrieval: recovering a vector from its measurements When do the frame measurements φ k, x 2 determine x C d? (Never: φ k, x 2 invariant under x e iθ x) The frame measurements define a map M Φ : (C d /S 1 ) R n by x ( x, φ k 2) k

16 Phase Retrieval: recovering a vector from its measurements When do the frame measurements φ k, x 2 determine x C d? (Never: φ k, x 2 invariant under x e iθ x) The frame measurements define a map M Φ : M Φ : (C d /S 1 ) R n by x ( x, φ k 2) k or { rank-1 Hermitian d d matrices } R n by X (trace(x A k )) k. where X = xx, A k = φ k φ k

17 Phase Retrieval: recovering a vector from its measurements When do the frame measurements φ k, x 2 determine x C d? (Never: φ k, x 2 invariant under x e iθ x) The frame measurements define a map M Φ : M Φ : (C d /S 1 ) R n by x ( x, φ k 2) k or { rank-1 Hermitian d d matrices } R n by X (trace(x A k )) k. Better question: When is the map M Φ injective? where X = xx, A k = φ k φ k

18 Question: How many measurements? We need n 4d measurements to recover vectors in C d.

19 Question: How many measurements? We need n 4d measurements to recover vectors in C d. (Balan-Casazza-Edidin, 2006): For n 4d 2, M Φ is injective for generic Φ C d n.

20 Question: How many measurements? We need n 4d measurements to recover vectors in C d. (Balan-Casazza-Edidin, 2006): For n 4d 2, M Φ is injective for generic Φ C d n. (Heinosaari-Mazzarella-Wolf, 2011): For n < 4d 2α 3, M Φ is not injective, where α = # of 1 s in binary expansion of d 1.

21 Question: How many measurements? We need n 4d measurements to recover vectors in C d. (Balan-Casazza-Edidin, 2006): For n 4d 2, M Φ is injective for generic Φ C d n. (Heinosaari-Mazzarella-Wolf, 2011): For n < 4d 2α 3, M Φ is not injective, where α = # of 1 s in binary expansion of d 1. Conjecture (Bandeira-Cahill-Mixon-Nelson, 2013) (a) If n < 4d 4, then M Φ is not injective. (b) If n 4d 4, then M Φ is injective for generic Φ.

22 Question: How many measurements? We need n 4d measurements to recover vectors in C d. (Balan-Casazza-Edidin, 2006): For n 4d 2, M Φ is injective for generic Φ C d n. (Heinosaari-Mazzarella-Wolf, 2011): For n < 4d 2α 3, M Φ is not injective, where α = # of 1 s in binary expansion of d 1. Conjecture (Bandeira-Cahill-Mixon-Nelson, 2013) (a) If n < 4d 4, then M Φ is not injective. (b) If n 4d 4, then M Φ is injective for generic Φ. We prove (b) by writing injectivity as an algebraic condition.

23 A nice reformulation of non-injectivity Observation (Bandeira et al., among others): M Φ is non-injective a nonzero matrix Q C d d Herm with rank(q) 2 and φ k Qφ k = 0 for each 1 k n.

24 A nice reformulation of non-injectivity Observation (Bandeira et al., among others): M Φ is non-injective a nonzero matrix Q C d d Herm with Why? rank(q) 2 and φ k Qφ k = 0 for each 1 k n.

25 A nice reformulation of non-injectivity Observation (Bandeira et al., among others): M Φ is non-injective a nonzero matrix Q C d d Herm with Why? rank(q) 2 and φ k Qφ k = 0 for each 1 k n. M Φ (x) = M Φ (y) φ k xx φ k = φ k yy φ k for 1 k n

26 A nice reformulation of non-injectivity Observation (Bandeira et al., among others): M Φ is non-injective a nonzero matrix Q C d d Herm with Why? rank(q) 2 and φ k Qφ k = 0 for each 1 k n. M Φ (x) = M Φ (y) φ k xx φ k = φ k yy φ k for 1 k n φ k (xx yy )φ k = 0 for 1 k n

27 A nice reformulation of non-injectivity Observation (Bandeira et al., among others): M Φ is non-injective a nonzero matrix Q C d d Herm with Why? rank(q) 2 and φ k Qφ k = 0 for each 1 k n. M Φ (x) = M Φ (y) φ k xx φ k = φ k yy φ k for 1 k n φ k (xx yy )φ }{{} k = 0 for 1 k n rank 2

28 A nice reformulation of non-injectivity Observation (Bandeira et al., among others): M Φ is non-injective a nonzero matrix Q C d d Herm with Why? rank(q) 2 and φ k Qφ k = 0 for each 1 k n. M Φ (x) = M Φ (y) φ k xx φ k = φ k yy φ k for 1 k n φ k (xx yy )φ }{{} k = 0 for 1 k n rank 2 More algebraic question: When does (span R {φ 1 φ 1,..., φ nφ n}) intersect the rank-2 locus of C d d Herm?

29 Getting (Real) Algebraic Consider the incidence set { } (Φ, Q) P(C d n ) P(C d d Herm ) : rank(q) 2 and φ k Qφ k = 0 k.

30 Getting (Real) Algebraic Consider the incidence set { } (Φ, Q) P(C d n ) P(C d d Herm ) : rank(q) 2 and φ k Qφ k = 0 k. Φ C d n U + iv where U, V R d n

31 Getting (Real) Algebraic Consider the incidence set { } (Φ, Q) P(C d n ) P(C d d Herm ) : rank(q) 2 and φ k Qφ k = 0 k. Φ C d n U + iv where U, V R d n Q C d d Herm X + iy where X R d d sym, Y R d d skew

32 Getting (Real) Algebraic Consider the incidence set { } (Φ, Q) P(C d n ) P(C d d Herm ) : rank(q) 2 and φ k Qφ k = 0 k. Φ C d n U + iv where U, V R d n Q C d d Herm X + iy where X R d d sym, Y R d d skew incidence set real projective variety in P((R d n ) 2 ) P(R d d sym R d d skew )

33 Getting (Real) Algebraic Consider the incidence set { } (Φ, Q) P(C d n ) P(C d d Herm ) : rank(q) 2 and φ k Qφ k = 0 k. Φ C d n U + iv where U, V R d n Q C d d Herm X + iy where X R d d sym, Y R d d skew incidence set real projective variety in P((R d n ) 2 ) P(R d d sym R d d skew ) Consequence: The bad frames, {Φ : M Φ is non-injective}, are the projection of a real (projective) variety. ( a closed semialgebraic subset of P((R d n ) 2 ))

34 Getting (Complex) Algebraic Let B d,n be the set of (U, V, Q) in P(C d n C d n ) P(C d d ), where U = (u 1,..., u n ) and V = (v 1,..., v n ), satisfying rank(q) 2 and (u k iv k ) T Q(u k + iv k ) = 0 for all 1 k n.

35 Getting (Complex) Algebraic Let B d,n be the set of (U, V, Q) in P(C d n C d n ) P(C d d ), where U = (u 1,..., u n ) and V = (v 1,..., v n ), satisfying rank(q) 2 and (u k iv k ) T Q(u k + iv k ) = 0 for all 1 k n. Theorem (CEH-) The projective variety B d,n has dimension 2dn + 4d 6 n

36 Getting (Complex) Algebraic Let B d,n be the set of (U, V, Q) in P(C d n C d n ) P(C d d ), where U = (u 1,..., u n ) and V = (v 1,..., v n ), satisfying rank(q) 2 and (u k iv k ) T Q(u k + iv k ) = 0 for all 1 k n. Theorem (CEH-) The projective variety B d,n has dimension 2dn + 4d 6 n = 2dn 1 + 4d 4 1 n. dim( P((C (d n) ) 2 ) ) dim( {rk-2 in P(C d d )} ) constraints

37 Getting (Complex) Algebraic Let B d,n be the set of (U, V, Q) in P(C d n C d n ) P(C d d ), where U = (u 1,..., u n ) and V = (v 1,..., v n ), satisfying rank(q) 2 and (u k iv k ) T Q(u k + iv k ) = 0 for all 1 k n. Theorem (CEH-) The projective variety B d,n has dimension 2dn + 4d 6 n = 2dn 1 + 4d 4 1 n. dim( P((C (d n) ) 2 ) ) dim( {rk-2 in P(C d d )} ) constraints Sketch of proof: The preimage π2 1 (Q) of any matrix Q is the product of n quadratic hypersurfaces.

38 The set of bad frames is small. Theorem (CEH-) The projective variety B d,n has dimension 2dn + 4d 6 n.

39 The set of bad frames is small. Theorem (CEH-) The projective variety B d,n has dimension 2dn + 4d 6 n. As a consequence, for n 4d 4, dim ( π 1 (B d,n ) ) 2dn 2 and codim ( π 1 (B d,n ) ) 1.

40 The set of bad frames is small. Theorem (CEH-) The projective variety B d,n has dimension 2dn + 4d 6 n. As a consequence, for n 4d 4, dim ( π 1 (B d,n ) ) 2dn 2 and codim ( π 1 (B d,n ) ) 1. {Φ : M Φ is non-injective} π 1 (B d,n ) a hypersurface in (C d n ) 2

41 The set of bad frames is small. Theorem (CEH-) The projective variety B d,n has dimension 2dn + 4d 6 n. As a consequence, for n 4d 4, dim ( π 1 (B d,n ) ) 2dn 2 and codim ( π 1 (B d,n ) ) 1. {Φ : M Φ is non-injective} π 1 (B d,n ) a hypersurface in (C d n ) 2 Corollary For n 4d 4, M Φ is injective for generic Φ C d n = (R d n ) 2. There is a Zariski-open set of frames Φ for which M Φ is injective.

42 Example: d = 2, n = 4d 4 = 4 A 2 2 Hertmitian matrix Q defines the real quadratic polynomial q(a, b, c, d) = ( a ic b id )( )( ) x 11 x 12 + iy 12 a + ic x 12 iy 12 x 22 b + id

43 Example: d = 2, n = 4d 4 = 4 A 2 2 Hertmitian matrix Q defines the real quadratic polynomial q(a, b, c, d) = ( a ic b id )( )( ) x 11 x 12 + iy 12 a + ic x 12 iy 12 x 22 b + id = x 11 (a 2 + c 2 ) + x 22 (b 2 + d 2 ) + 2x 12 (ab + cd) + 2y 12 (bc ad).

44 Example: d = 2, n = 4d 4 = 4 A 2 2 Hertmitian matrix Q defines the real quadratic polynomial q(a, b, c, d) = ( a ic b id )( )( ) x 11 x 12 + iy 12 a + ic x 12 iy 12 x 22 b + id = x 11 (a 2 + c 2 ) + x 22 (b 2 + d 2 ) + 2x 12 (ab + cd) + 2y 12 (bc ad). Since any Q has rank 2, the frame ( ) a1 + ic Φ = 1 a 2 + ic 2 a 3 + ic 3 a 4 + ic 4 b 1 + id 1 b 2 + id 2 b 3 + id 3 b 4 + id 4 defines injective measurements M Φ whenever

45 Example: d = 2, n = 4d 4 = 4 A 2 2 Hertmitian matrix Q defines the real quadratic polynomial q(a, b, c, d) = ( a ic b id )( )( ) x 11 x 12 + iy 12 a + ic x 12 iy 12 x 22 b + id = x 11 (a 2 + c 2 ) + x 22 (b 2 + d 2 ) + 2x 12 (ab + cd) + 2y 12 (bc ad). Since any Q has rank 2, the frame ( ) a1 + ic Φ = 1 a 2 + ic 2 a 3 + ic 3 a 4 + ic 4 b 1 + id 1 b 2 + id 2 b 3 + id 3 b 4 + id 4 defines injective measurements M Φ whenever a1 2 + c2 1 b1 2 + d 1 2 a 1 b 1 + c 1 d 1 b 1 c 1 a 1 d 1 a2 det 2 + c2 2 b2 2 + d 2 2 a 2 b 2 + c 2 d 2 b 2 c 2 a 2 d 2 a3 2 + c2 3 b3 2 + d 3 2 a 3 b 3 + c 3 d 3 b 3 c 3 a 3 d 0. 3 a4 2 + c2 4 b4 2 + d 4 2 a 4 b 4 + c 4 d 4 b 4 c 4 a 4 d 4

46 Example: d = 3 The rank-two matrices in C 3 3 form an 8-dimensional hypersurface defined by the 3 3 determinant det(q).

47 Example: d = 3 The rank-two matrices in C 3 3 form an 8-dimensional hypersurface defined by the 3 3 determinant det(q). 4d 4 = 8 measurements: For φ 1,..., φ 8 C 3, we expect codim(span{φ 1 φ 1,..., φ 8 φ 8 }) = 1.

48 Example: d = 3 The rank-two matrices in C 3 3 form an 8-dimensional hypersurface defined by the 3 3 determinant det(q). 4d 4 = 8 measurements: For φ 1,..., φ 8 C 3, we expect codim(span{φ 1 φ 1,..., φ 8 φ 8 }) = 1. {Q : φ k Qφ k = 0} = just one point in P(C 3 3 )

49 Example: d = 3 The rank-two matrices in C 3 3 form an 8-dimensional hypersurface defined by the 3 3 determinant det(q). 4d 4 = 8 measurements: For φ 1,..., φ 8 C 3, we expect codim(span{φ 1 φ 1,..., φ 8 φ 8 }) = 1. {Q : φ k Qφ k = 0} = just one point in P(C 3 3 ) expect {Q : φ k Qφ k = 0} V (det(q)) =.

50 Example: d = 3 The rank-two matrices in C 3 3 form an 8-dimensional hypersurface defined by the 3 3 determinant det(q). 4d 4 = 8 measurements: For φ 1,..., φ 8 C 3, we expect codim(span{φ 1 φ 1,..., φ 8 φ 8 }) = 1. {Q : φ k Qφ k = 0} = just one point in P(C 3 3 ) expect {Q : φ k Qφ k = 0} V (det(q)) =. injective

51 Example: d = 3 The rank-two matrices in C 3 3 form an 8-dimensional hypersurface defined by the 3 3 determinant det(q). 4d 4 = 8 measurements: For φ 1,..., φ 8 C 3, we expect codim(span{φ 1 φ 1,..., φ 8 φ 8 }) = 1. {Q : φ k Qφ k = 0} = just one point in P(C 3 3 ) expect {Q : φ k Qφ k = 0} V (det(q)) =. 4d 5 = 7 measurements: injective For φ 1,..., φ 7 C 3, we expect codim(span{φ 1 φ 1,..., φ 7 φ 7 }) = 2.

52 Example: d = 3 The rank-two matrices in C 3 3 form an 8-dimensional hypersurface defined by the 3 3 determinant det(q). 4d 4 = 8 measurements: For φ 1,..., φ 8 C 3, we expect codim(span{φ 1 φ 1,..., φ 8 φ 8 }) = 1. {Q : φ k Qφ k = 0} = just one point in P(C 3 3 ) expect {Q : φ k Qφ k = 0} V (det(q)) =. 4d 5 = 7 measurements: injective For φ 1,..., φ 7 C 3, we expect codim(span{φ 1 φ 1,..., φ 7 φ 7 }) = 2. {Q : φ k Qφ k = 0} = a line in P(C 3 3 )

53 Example: d = 3 The rank-two matrices in C 3 3 form an 8-dimensional hypersurface defined by the 3 3 determinant det(q). 4d 4 = 8 measurements: For φ 1,..., φ 8 C 3, we expect codim(span{φ 1 φ 1,..., φ 8 φ 8 }) = 1. {Q : φ k Qφ k = 0} = just one point in P(C 3 3 ) expect {Q : φ k Qφ k = 0} V (det(q)) =. 4d 5 = 7 measurements: injective For φ 1,..., φ 7 C 3, we expect codim(span{φ 1 φ 1,..., φ 7 φ 7 }) = 2. {Q : φ k Qφ k = 0} = a line in P(C 3 3 ) {Q : φ k Qφ k = 0} V (det(q)) = finitely many points in P(C 3 3 ).

54 Example: d = 3 The rank-two matrices in C 3 3 form an 8-dimensional hypersurface defined by the 3 3 determinant det(q). 4d 4 = 8 measurements: For φ 1,..., φ 8 C 3, we expect codim(span{φ 1 φ 1,..., φ 8 φ 8 }) = 1. {Q : φ k Qφ k = 0} = just one point in P(C 3 3 ) expect {Q : φ k Qφ k = 0} V (det(q)) =. 4d 5 = 7 measurements: injective For φ 1,..., φ 7 C 3, we expect codim(span{φ 1 φ 1,..., φ 7 φ 7 }) = 2. {Q : φ k Qφ k = 0} = a line in P(C 3 3 ) {Q : φ k Qφ k = 0} V (det(q)) = finitely many points in P(C 3 3 ). finitely many = 3 = deg(det(q))

55 Example: d = 3 The rank-two matrices in C 3 3 form an 8-dimensional hypersurface defined by the 3 3 determinant det(q). 4d 4 = 8 measurements: For φ 1,..., φ 8 C 3, we expect codim(span{φ 1 φ 1,..., φ 8 φ 8 }) = 1. {Q : φ k Qφ k = 0} = just one point in P(C 3 3 ) expect {Q : φ k Qφ k = 0} V (det(q)) =. 4d 5 = 7 measurements: injective For φ 1,..., φ 7 C 3, we expect codim(span{φ 1 φ 1,..., φ 7 φ 7 }) = 2. {Q : φ k Qφ k = 0} = a line in P(C 3 3 ) {Q : φ k Qφ k = 0} V (det(q)) = finitely many points in P(C 3 3 ). finitely many = 3 = deg(det(q)) Since 3 is odd, at least one must be Hermitian.

56 Example: d = 3 The rank-two matrices in C 3 3 form an 8-dimensional hypersurface defined by the 3 3 determinant det(q). 4d 4 = 8 measurements: For φ 1,..., φ 8 C 3, we expect codim(span{φ 1 φ 1,..., φ 8 φ 8 }) = 1. {Q : φ k Qφ k = 0} = just one point in P(C 3 3 ) expect {Q : φ k Qφ k = 0} V (det(q)) =. 4d 5 = 7 measurements: injective For φ 1,..., φ 7 C 3, we expect codim(span{φ 1 φ 1,..., φ 7 φ 7 }) = 2. {Q : φ k Qφ k = 0} = a line in P(C 3 3 ) {Q : φ k Qφ k = 0} V (det(q)) = finitely many points in P(C 3 3 ). finitely many = 3 = deg(det(q)) Since 3 is odd, at least one must be Hermitian. non-injective

57 Fewer measurements? Conjecture For n 4d 5 and every Φ C d n, M Φ is not injective.

58 Fewer measurements? Conjecture For n 4d 5 and every Φ C d n, M Φ is not injective. Smallest open question: (d = 4, n = 4d 5 = 11) Given vectors φ 1,..., φ 11 C 4, does there always exist a Hermitian rank-two matrix Q C 4 4 Herm for which φ k Qφ k = 0?

59 Fewer measurements? Conjecture For n 4d 5 and every Φ C d n, M Φ is not injective. Smallest open question: (d = 4, n = 4d 5 = 11) Given vectors φ 1,..., φ 11 C 4, does there always exist a Hermitian rank-two matrix Q C 4 4 Herm for which φ k Qφ k = 0? dim({rk-2 in C 4 4 }) = 12 and deg({rk-2 in C 4 4 }) = 20.

60 Fewer measurements? Conjecture For n 4d 5 and every Φ C d n, M Φ is not injective. Smallest open question: (d = 4, n = 4d 5 = 11) Given vectors φ 1,..., φ 11 C 4, does there always exist a Hermitian rank-two matrix Q C 4 4 Herm for which φ k Qφ k = 0? dim({rk-2 in C 4 4 }) = 12 and deg({rk-2 in C 4 4 }) = 20. We expect 20 rank-two matrices Q P(C d d ) with φ k Qφ k = 0. Must there be a Hermitian one?

61 Final Thoughts Frame theory and phase retrieval bring together many areas of mathematics and produce interesting algebraic questions. Tools from algebraic geometry can be used to tackle problems that look very non-algebraic.

62 Final Thoughts Frame theory and phase retrieval bring together many areas of mathematics and produce interesting algebraic questions. Tools from algebraic geometry can be used to tackle problems that look very non-algebraic. SOME REFERENCES R. Balan, P. Casazza, D. Edidin. On signal reconstruction without phase. Appl. Comput. Harmon. Anal., 20 (2006) A. Bandeira, J. Cahill, D. Mixon, and A. Nelson. Saving phase: Injectivity and stability for phase retrieval. arxiv: E. J. Candés, Y. Eldar, T. Strohmer, V. Voroninski. Phase retrieval via matrix completion. SIAM J. Imaging Sci., 6(1), (2013) T. Heinosaari, L. Mazzarella, M. Wolf. Quantum tomography under prior information. Comm. Math. Phys. 318(2) (2013), Thanks!