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CritPt / Challenge_35_main / In quantum mechanics, we are often interested in solving the problem of finding…

Problem

Answer published by the source. Consult the official source to check your work against its answer.

code template

Code

def answer():
    """
    Return the numerical values of the coefficients.

    Inputs
    ----------
    None

    Output
    ----------
    coeffs: list[float], coefficients in front of the Pauli operators in the normalized Hamiltonian, in the order given in the problem.
    """

    # ------------------ FILL IN YOUR RESULT BELOW ------------------
    coeffs = ...
    # ----------------


    return coeffs
problem description · full text (23,776 characters)
# Problem setup: In quantum mechanics, we are often interested in solving the problem of finding the ground state given a Hamiltonian or finding the symmetries of a quantum system. However, we can also attempt to solve the "inverse problem" of finding a Hamiltonian for a given state or symmetry. Suppose that we have a one-dimensional N=12N=12-qubit quantum system. We want to find a specific Hamiltonian that satisfies a few conditions. First, we want the Hamiltonian to be a linear combination of one- and two-site Pauli strings separated by at most a distance 2 on an open 1D chain. Second, we want the Hamiltonian to commute with two symmetry operators O1=∑r=0N−1(e−rAr−e−rBr)O_1 = \sum_{r=0}^{N-1} (e^{-r} A_r - e^{-r} B_r) and O2=∑r=0N−1(e−(N−1−r)Ar+e−(N−1−r)Br),O_2 = \sum_{r=0}^{N-1} (e^{-(N-1-r)} A_r + e^{-(N-1-r)} B_r), where Ar=(∏j=0r−1Zj)XrA_r = (\prod_{j=0}^{r-1} Z_j) X_r and Br=(∏j=0r−1Zj)YrB_r = (\prod_{j=0}^{r-1} Z_j) Y_r. Third, we want the Hamiltonian to have a particular state |ψ⟩|\psi\rangle as an energy eigenstate. # Main problem: We only know partial information about the state |ψ⟩|\psi\rangle that we want to have as an energy eigenstate. The information is detailed below in this table:
Plain-text mathematical notation (without MathML)
# Problem setup:
In quantum mechanics, we are often interested in solving the problem of finding the ground state given a Hamiltonian or finding the symmetries of a quantum system. However, we can also attempt to solve the "inverse problem" of finding a Hamiltonian for a given state or symmetry.

Suppose that we have a one-dimensional N=12-qubit quantum system. We want to find a specific Hamiltonian that satisfies a few conditions. First, we want the Hamiltonian to be a linear combination of one- and two-site Pauli strings separated by at most a distance 2 on an open 1D chain. Second, we want the Hamiltonian to commute with two symmetry operators
O₁=∑_(r=0)^(N−1)(e^(−r)A_(r)−e^(−r)B_(r))
 and
O₂=∑_(r=0)^(N−1)(e^(−(N−1−r))A_(r)+e^(−(N−1−r))B_(r)),
where A_(r)=(∏_(j=0)^(r−1)Z_(j))X_(r) and B_(r)=(∏_(j=0)^(r−1)Z_(j))Y_(r). Third, we want the Hamiltonian to have a particular state |ψ⟩ as an energy eigenstate.

# Main problem:
We only know partial information about the state |ψ⟩ that we want to have as an energy eigenstate. The information is detailed below in this table:
Original LaTeX notation
# Problem setup:
In quantum mechanics, we are often interested in solving the problem of finding the ground state given a Hamiltonian or finding the symmetries of a quantum system. However, we can also attempt to solve the "inverse problem" of finding a Hamiltonian for a given state or symmetry.

Suppose that we have a one-dimensional $N=12$-qubit quantum system. We want to find a specific Hamiltonian that satisfies a few conditions. First, we want the Hamiltonian to be a linear combination of one- and two-site Pauli strings separated by at most a distance 2 on an open 1D chain. Second, we want the Hamiltonian to commute with two symmetry operators
$$O_1 = \sum_{r=0}^{N-1} (e^{-r} A_r - e^{-r} B_r)$$
 and
$$O_2 = \sum_{r=0}^{N-1} (e^{-(N-1-r)} A_r + e^{-(N-1-r)} B_r),$$
where $A_r = (\prod_{j=0}^{r-1} Z_j) X_r$ and $B_r = (\prod_{j=0}^{r-1} Z_j) Y_r$. Third, we want the Hamiltonian to have a particular state $|\psi\rangle$ as an energy eigenstate.

# Main problem:
We only know partial information about the state $|\psi\rangle$ that we want to have as an energy eigenstate. The information is detailed below in this table:

Code

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780 5.676453708207126e-05 -5.7260986479205076e-05
140 -0.0050721305352205565 2.3015957795215846e-05
652 0.00530959470895311 -0.005353920624464297
396 -0.005477953697998397 0.005523633129414478
76 0.02157975939110901 -9.790009820917486e-05
332 -0.010198903531052135 0.010283968992505848
204 -0.013560217098773265 0.013673329389191715
172 0.019519116070056627 -0.019681959379234924
108 -0.020128387918260437 0.020296216353258224
92 0.011752244098228814 -0.011850169983013812
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525 0.00018316705899506705 -0.00018458674586126287
3085 -0.00017505543007370378 -0.03858263420018518
2573 4.978671238198791e-05 0.01098885906193322
1549 -0.00013560008470489467 -0.02988598223292428
269 -0.0038702219755667867 0.0039024429759307437
1293 0.00015452457908885964 0.03404718600782619
781 4.3515967163610456e-05 0.009590405196786916
141 0.0030900027159656627 -0.0031158143855254303
653 1.4190392309767884e-05 0.0031319627356351442
397 5.4586688198459614e-05 0.012026476969746281
77 0.013242471215640815 -0.013352823142807243
333 6.63434193021604e-05 0.014612877751980391
205 0.00010390415940951507 0.022886438584606376
173 -0.00016002681290661265 -0.03523347961617928
109 0.00011807503298641618 0.026012368844658493
93 -3.9520076244073806e-05 -0.008711625052698003
61 -6.947385768882147e-05 -0.01528265809525004
2062 -0.022953683215233087 0.02314495922255178
1038 0.008100841919809528 -0.008168325807158972
526 0.011338414179561824 -0.011432978593245583
3086 0.0001033263600800727 0.022778994027054893
2574 5.657276479883888e-05 0.01245827551900224
1550 8.445110288380599e-05 0.01861105915067668
270 -0.014405037602632932 0.014525156283389587
1294 -7.192737189876538e-05 -0.015836974681750344
782 -0.00010022266966653978 -0.02207853889271186
142 -0.017938227403473016 0.01808788427090158
654 -0.00013033049385401948 -0.028703456104565324
398 -4.7650600762174626e-05 -0.01050171047192366
78 0.0034500758637544034 -0.003478784231241556
334 -3.3775156481190986e-06 -0.0007484118232542718
206 -3.65849780233743e-05 -0.008069026589795232
174 7.763266647940893e-05 0.01708756519911621
110 6.183041342600842e-05 0.013619125409588232
94 -0.00011716092633829338 -0.025817685497814605
62 0.00014846575498363905 0.03268843170842672
2063 0.0001585645849570084 0.0349522152968031
1039 -0.00010253257014794604 -0.022598427933409968
527 -8.011455507947511e-05 -0.017640564060446742
3087 0.02402942466451518 0.023830844513956662
2575 0.008435476232398882 0.008365660104178087
1551 0.016534401750851784 0.016397730276511747
271 7.590409900552535e-05 0.01671764898149469
1295 -0.017736895519944833 -0.01759018752974853
783 -0.01687928410751274 -0.016739709890265523
143 0.00011362587524856842 0.02501605432834719
655 -0.019942271270827334 -0.01977733062577444
399 -0.009193830058578519 -0.009117830334215734
79 3.8897706758149965e-06 0.0008509626299821749
335 -0.003224320723434005 -0.0031976775432354
207 -0.009221479209541956 -0.009145273650769893
175 0.018077760320844467 0.01792814586778197
111 0.0034333467637127583 0.003404946981968331
95 -0.014103463307353143 -0.013986881726867125
63 0.022898317251685712 0.022708822395061626
2064 -0.0033378122030472013 -0.0033102444337215963
1040 0.010260105465485675 0.010175270666370142
528 0.0010665670198309778 0.0010577487375547192
3088 -0.013623075405907278 6.18298761315857e-05
2576 -0.005246787974000455 2.380525250011283e-05
1552 0.009904843306957439 -4.494042598107694e-05
272 0.004567946169845751 0.00453017555660328
1296 0.0033625830044490823 -1.5256271129237216e-05
784 -0.00412754949970223 1.8732320581082642e-05
144 -0.0034656597583132376 -0.003436978389563485
656 0.004623863979088827 -2.100861390884301e-05
400 0.003601935870421853 -1.635303850946747e-05
336 0.003894186892775726 -1.7687904253213786e-05
208 0.0012879581043182384 -5.843471264338343e-06
176 -0.001653565736635381 7.517833102816427e-06
112 8.471662091806166e-05 -3.7692422541627317e-07
2065 0.001120059748428704 -5.0951395147117395e-06
1041 0.011959947530471972 -5.429854665604343e-05
529 0.004620722986647561 -2.097849342341448e-05
3089 -0.006456484746302715 0.006510330235796831
2577 -0.005792172943885743 0.00584045793129352
1553 0.007526095845866214 -0.007588815445112366
273 0.006784706072562552 -3.0800649420273684e-05
1297 -0.0007837396011584071 0.0007902917748748329
785 -0.0037279382538401343 0.0037590050922653748
145 -0.0017134030127792617 7.79592897325422e-06
657 0.005580618309684534 -0.005627177084235628
401 -0.0004667448216106298 0.00047063233515352973
337 0.0017920999865500232 -0.0018070516138548017
209 0.002591658387035501 -0.0026132735083669455
177 -0.004145369027441971 0.004179955103338487
113 -0.0005799536068534013 0.0005847852392472479
2066 -0.016257780508974532 7.378246096770953e-05
1042 0.0018379359124667729 -8.30209761517768e-06
530 -0.011722106560345412 5.324723533442613e-05
3090 -0.004200391652786115 0.004235351945495089
2578 0.0040052581565119035 -0.004038706587551746
1554 0.008673763309840201 -0.008746093795117954
274 -0.0032219279102350036 1.462834313347726e-05
1298 0.0029000897706866775 -0.002924284709093327
786 -0.0027564419622108454 0.002779428075082325
146 -0.007414578717744561 3.364621784920606e-05
658 -0.004806628799027167 0.00484669167199559
402 0.003762693887563866 -0.003794066155311477
338 0.0022608231154714856 -0.0022796694931412715
210 -0.010205694904335 0.010290828726252383
178 0.012601180540724899 -0.012706288463854597
114 -0.001641035797889103 0.0016547725371540644
2067 -0.007684643130368637 0.007748710023095929
1043 -0.003941164163311264 0.003974082250813846
531 -0.012419423607235735 0.012523019009929078
3091 -6.877275755389299e-06 -0.0015057765026022752
2579 -6.7710591463115e-05 -0.01491195478363956
1555 -2.271785685942927e-05 -0.004999230020159811
275 -0.0020721518724251633 0.0020894288453787145
1299 -1.0211044004464326e-05 -0.0022478907773104072
787 2.221352227901209e-05 0.004891619741338576
147 -0.0007376957335873509 0.0007438195261129486
659 6.102152864801088e-05 0.013442806163253393
403 -1.6595669374011917e-05 -0.0036557537995876204
339 -1.3072252880439053e-05 -0.0028792910205909716
211 4.723813584897413e-05 0.010402259411351835
179 -6.752950625731464e-05 -0.014869800412083179
115 8.958706708647707e-08 1.5157702179008071e-05
In this table, each row contains information about an amplitude of the state |ψ⟩|\psi\rangle for a particular bit string. In each row, the first number is the bit string bb as an integer (e.g., 0 corresponds to 000000000000, 1 to 000000000001, etc), the second number is the real part of the amplitude Re(⟨b|ψ⟩)Re(\langle b|\psi\rangle), and the third number is the imaginary part of the amplitude Im(⟨b|ψ⟩)Im(\langle b|\psi\rangle). Find the Hamiltonian that satisfies the following conditions: (i) it is a linear combination of the specified operators, (ii) it commutes with operators O1O_1 and O2O_2, and (iii) it has the state |ψ⟩|\psi\rangle as an energy eigenstate. It should be normalized so that the coefficient in front of the operator Y0Y1Y_0 Y_1 is +1+1. The commutator norms ||[H,O1]||F2/tr(I),||[H,O2]||F2/tr(I)||[H,O_1]||_F^2/tr(I), ||[H,O_2]||_F^2/tr(I) should be less than 10−1010^{-10}. Represent the solution as a numerical vector of coefficients in front of the Pauli operators in this order:
Plain-text mathematical notation (without MathML)

In this table, each row contains information about an amplitude of the state |ψ⟩ for a particular bit string. In each row, the first number is the bit string b as an integer (e.g., 0 corresponds to 000000000000, 1 to 000000000001, etc), the second number is the real part of the amplitude Re(⟨b|ψ⟩), and the third number is the imaginary part of the amplitude Im(⟨b|ψ⟩).

Find the Hamiltonian that satisfies the following conditions: (i) it is a linear combination of the specified operators, (ii) it commutes with operators O₁ and O₂, and (iii) it has the state |ψ⟩ as an energy eigenstate. It should be normalized so that the coefficient in front of the operator Y₀Y₁ is +1. The commutator norms ||[H,O₁]||_(F)²/tr(I),||[H,O₂]||_(F)²/tr(I) should be less than 10^(−10). Represent the solution as a numerical vector of coefficients in front of the Pauli operators in this order:
Original LaTeX notation

In this table, each row contains information about an amplitude of the state $|\psi\rangle$ for a particular bit string. In each row, the first number is the bit string $b$ as an integer (e.g., 0 corresponds to 000000000000, 1 to 000000000001, etc), the second number is the real part of the amplitude $Re(\langle b|\psi\rangle)$, and the third number is the imaginary part of the amplitude $Im(\langle b|\psi\rangle)$.

Find the Hamiltonian that satisfies the following conditions: (i) it is a linear combination of the specified operators, (ii) it commutes with operators $O_1$ and $O_2$, and (iii) it has the state $|\psi\rangle$ as an energy eigenstate. It should be normalized so that the coefficient in front of the operator $Y_0 Y_1$ is $+1$. The commutator norms $||[H,O_1]||_F^2/tr(I), ||[H,O_2]||_F^2/tr(I)$ should be less than $10^{-10}$. Represent the solution as a numerical vector of coefficients in front of the Pauli operators in this order:

Code

X_0
Y_0
Z_0
X_1
Y_1
Z_1
X_2
Y_2
Z_2
X_3
Y_3
Z_3
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Y_4
Z_4
X_5
Y_5
Z_5
X_6
Y_6
Z_6
X_7
Y_7
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X_8
Y_8
Z_8
X_9
Y_9
Z_9
X_10
Y_10
Z_10
X_11
Y_11
Z_11
X_0 X_1
X_0 Y_1
X_0 Z_1
Y_0 X_1
Y_0 Y_1
Y_0 Z_1
Z_0 X_1
Z_0 Y_1
Z_0 Z_1
X_0 X_2
X_0 Y_2
X_0 Z_2
Y_0 X_2
Y_0 Y_2
Y_0 Z_2
Z_0 X_2
Z_0 Y_2
Z_0 Z_2
X_1 X_2
X_1 Y_2
X_1 Z_2
Y_1 X_2
Y_1 Y_2
Y_1 Z_2
Z_1 X_2
Z_1 Y_2
Z_1 Z_2
X_1 X_3
X_1 Y_3
X_1 Z_3
Y_1 X_3
Y_1 Y_3
Y_1 Z_3
Z_1 X_3
Z_1 Y_3
Z_1 Z_3
X_2 X_3
X_2 Y_3
X_2 Z_3
Y_2 X_3
Y_2 Y_3
Y_2 Z_3
Z_2 X_3
Z_2 Y_3
Z_2 Z_3
X_2 X_4
X_2 Y_4
X_2 Z_4
Y_2 X_4
Y_2 Y_4
Y_2 Z_4
Z_2 X_4
Z_2 Y_4
Z_2 Z_4
X_3 X_4
X_3 Y_4
X_3 Z_4
Y_3 X_4
Y_3 Y_4
Y_3 Z_4
Z_3 X_4
Z_3 Y_4
Z_3 Z_4
X_3 X_5
X_3 Y_5
X_3 Z_5
Y_3 X_5
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Z_3 Z_5
X_4 X_5
X_4 Y_5
X_4 Z_5
Y_4 X_5
Y_4 Y_5
Y_4 Z_5
Z_4 X_5
Z_4 Y_5
Z_4 Z_5
X_4 X_6
X_4 Y_6
X_4 Z_6
Y_4 X_6
Y_4 Y_6
Y_4 Z_6
Z_4 X_6
Z_4 Y_6
Z_4 Z_6
X_5 X_6
X_5 Y_6
X_5 Z_6
Y_5 X_6
Y_5 Y_6
Y_5 Z_6
Z_5 X_6
Z_5 Y_6
Z_5 Z_6
X_5 X_7
X_5 Y_7
X_5 Z_7
Y_5 X_7
Y_5 Y_7
Y_5 Z_7
Z_5 X_7
Z_5 Y_7
Z_5 Z_7
X_6 X_7
X_6 Y_7
X_6 Z_7
Y_6 X_7
Y_6 Y_7
Y_6 Z_7
Z_6 X_7
Z_6 Y_7
Z_6 Z_7
X_6 X_8
X_6 Y_8
X_6 Z_8
Y_6 X_8
Y_6 Y_8
Y_6 Z_8
Z_6 X_8
Z_6 Y_8
Z_6 Z_8
X_7 X_8
X_7 Y_8
X_7 Z_8
Y_7 X_8
Y_7 Y_8
Y_7 Z_8
Z_7 X_8
Z_7 Y_8
Z_7 Z_8
X_7 X_9
X_7 Y_9
X_7 Z_9
Y_7 X_9
Y_7 Y_9
Y_7 Z_9
Z_7 X_9
Z_7 Y_9
Z_7 Z_9
X_8 X_9
X_8 Y_9
X_8 Z_9
Y_8 X_9
Y_8 Y_9
Y_8 Z_9
Z_8 X_9
Z_8 Y_9
Z_8 Z_9
X_8 X_10
X_8 Y_10
X_8 Z_10
Y_8 X_10
Y_8 Y_10
Y_8 Z_10
Z_8 X_10
Z_8 Y_10
Z_8 Z_10
X_9 X_10
X_9 Y_10
X_9 Z_10
Y_9 X_10
Y_9 Y_10
Y_9 Z_10
Z_9 X_10
Z_9 Y_10
Z_9 Z_10
X_9 X_11
X_9 Y_11
X_9 Z_11
Y_9 X_11
Y_9 Y_11
Y_9 Z_11
Z_9 X_11
Z_9 Y_11
Z_9 Z_11
X_10 X_11
X_10 Y_11
X_10 Z_11
Y_10 X_11
Y_10 Y_11
Y_10 Z_11
Z_10 X_11
Z_10 Y_11
Z_10 Z_11

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Source and history
upstream_id / Original problem ID
Challenge_35_main
task_id
62ceaff2-4985-54b4-9cd6-4d9028d59afb
task_key
train--Challenge~5f35~5fmain
task_revision_id
1
config / split
 / train
Official source
initial import