QuantumCore

Hands-on lab · 2 steps

Hit entanglement-entropy targets

Compute the entanglement entropy S(ρA)=−∑kλklog⁡2λkS(\rho_A)=-\sum_k\lambda_k\log_2\lambda_k of a Schmidt state from its Schmidt coefficients, then prepare a three-qubit state whose marginals ρ{0}\rho_{\{0\}}, ρ{0,1}\rho_{\{0,1\}} and ρ{2}\rho_{\{2\}} each carry exactly one bit of entropy. All entropies are in bits; qubit 0 is the leftmost bit.

0/2 steps passedBest score 0/100

Step 1: Entropy of a Schmidt state

Give the entanglement entropy, in bits, of

∣ψ⟩=32∣00⟩+12∣11⟩,\lvert\psi\rangle=\tfrac{\sqrt3}{2}\lvert00\rangle+\tfrac12\lvert11\rangle ,

that is S(ρ0)S(\rho_0) with ρ0=Tr1∣ψ⟩⟨ψ∣\rho_0=\mathrm{Tr}_1\lvert\psi\rangle\langle\psi\rvert. Answers are accepted within 10−310^{-3}; an expression such as 2 - 0.75*log2(3) is also accepted.

Enter exact expressions or decimals, e.g. 1/sqrt(2), pi/3. Values must be real.

enter a value
Fill every field with a valid expression.

Step 2: One bit on every cut

Using at most five gates from {H,CNOT,X,RY}\{H,\mathrm{CNOT},X,R_Y\}, prepare a three-qubit state from ∣000⟩\lvert000\rangle with

S(ρ{0})=1,S(ρ{0,1})=1,S(ρ{2})=1(bits),S(\rho_{\{0\}})=1,\qquad S(\rho_{\{0,1\}})=1,\qquad S(\rho_{\{2\}})=1\quad\text{(bits)},

where ρQ\rho_Q is the reduced state on the qubits in QQ (all other qubits traced out). Each entropy is graded with absolute tolerance 10−610^{-6}.

  • 3 qubits
  • allowed gates: H, CNOT, X, RY
  • at most 5 gates
  • no measurements

One gate per line or separated by ;, e.g. H(0); CNOT(0,1); RY(1, pi/3); measure(1 -> 0). Qubit 0 is the top wire (most significant bit); controls come first. # starts a comment.

Measure into a classical bit with measure(q -> c); apply a gate only when a bit is set with X(2) if c1==1 (several bits, first = most significant: if c[0,1]==3).

Circuit diagram of your answerEmpty circuitq0q1q2
Statevector amplitudes
BasisAmplitude α|α|Phase arg αProbability
|000⟩1
1.0000
0
100.00%
7 basis states with zero amplitude hidden