3 n-Qubit Systems
Moving beyond 2 qubits unlocks the full richness of quantum entanglement. With 3 or more qubits, we encounter genuinely different classes of entanglement—states that cannot be converted to each other using local operations.
3.1 Tensor Products for n Qubits
3.1.1 The n-Qubit Hilbert Space
An \(n\)-qubit system lives in a Hilbert space of dimension \(2^n\).
Computational basis: \(\{|b_1b_2\cdots b_n\rangle : b_i \in \{0,1\}\}\)
For 3 qubits: \(|000\rangle, |001\rangle, |010\rangle, |011\rangle, |100\rangle, |101\rangle, |110\rangle, |111\rangle\) (8 states)
For \(n\) qubits: \(2^n\) basis states
3.1.2 General n-Qubit State
\[ |\psi\rangle = \sum_{b_1,\ldots,b_n \in \{0,1\}} c_{b_1\cdots b_n} |b_1\cdots b_n\rangle \]
with normalization: \(\sum_{b_1,\ldots,b_n} |c_{b_1\cdots b_n}|^2 = 1\)
Note: The number of complex amplitudes grows exponentially: \(2^n\) complex numbers!
- 10 qubits: 1,024 amplitudes
- 20 qubits: 1,048,576 amplitudes
- 50 qubits: \(\sim 10^{15}\) amplitudes (more than can be stored classically!)
This exponential growth is both the power and the challenge of quantum computing.
3.2 GHZ States: Maximal Multipartite Entanglement
3.2.1 Definition
The GHZ state (Greenberger-Horne-Zeilinger) for \(n\) qubits is:
\[ |\text{GHZ}_n\rangle = \frac{1}{\sqrt{2}}(|0\rangle^{\otimes n} + |1\rangle^{\otimes n}) = \frac{1}{\sqrt{2}}(|00\cdots0\rangle + |11\cdots1\rangle) \]
Examples: - \(n=2\): \(|\text{GHZ}_2\rangle = \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle) = |\Phi^+\rangle\) (Bell state) - \(n=3\): \(|\text{GHZ}_3\rangle = \frac{1}{\sqrt{2}}(|000\rangle + |111\rangle)\) - \(n=4\): \(|\text{GHZ}_4\rangle = \frac{1}{\sqrt{2}}(|0000\rangle + |1111\rangle)\)
3.2.2 Properties of GHZ States
Perfectly correlated: Measuring any qubit in the Z-basis gives 0 or 1, and all other qubits will have the same value.
Fragile: Measuring even one qubit completely destroys the multipartite entanglement, leaving a product state.
Maximal bipartite entanglement: For any bipartition (splitting qubits into two groups), the entanglement entropy is maximal.
3.2.3 GHZ Correlations
For \(|\text{GHZ}_3\rangle = \frac{1}{\sqrt{2}}(|000\rangle + |111\rangle)\):
Measuring all qubits in Z-basis: - \(P(000) = \frac{1}{2}\) - all measure 0 - \(P(111) = \frac{1}{2}\) - all measure 1 - \(P(\text{any other outcome}) = 0\) - impossible!
This perfect 3-way correlation has no classical analogue.
3.2.4 Entanglement Entropy of GHZ
For a bipartition \(A|B\) where \(A\) has \(k\) qubits and \(B\) has \(n-k\) qubits:
\[ S_{A|B} = 1 \text{ bit} \]
regardless of how we partition! (as long as both groups are non-empty)
3.3 W States: Robust Entanglement
3.3.1 Definition
The W state for \(n\) qubits is an equal superposition of all basis states with exactly one qubit in \(|1\rangle\):
\[ |W_n\rangle = \frac{1}{\sqrt{n}}(|10\cdots0\rangle + |01\cdots0\rangle + \cdots + |0\cdots01\rangle) \]
Examples: - \(n=3\): \(|W_3\rangle = \frac{1}{\sqrt{3}}(|100\rangle + |010\rangle + |001\rangle)\) - \(n=4\): \(|W_4\rangle = \frac{1}{2}(|1000\rangle + |0100\rangle + |0010\rangle + |0001\rangle)\)
3.3.2 Properties of W States
Robust: If you measure one qubit and get 0, the remaining qubits are still entangled in a W state with \(n-1\) qubits.
Partial bipartite entanglement: For bipartition \(1|23\cdots n\), the entanglement is less than maximal.
Symmetric: All qubits play equivalent roles (state is unchanged by permuting qubits).
3.3.3 W State Resilience
For \(|W_3\rangle = \frac{1}{\sqrt{3}}(|100\rangle + |010\rangle + |001\rangle)\):
Measuring qubit 1 in Z-basis:
- Outcome 1 (probability \(\frac{1}{3}\)):
- State collapses to \(|100\rangle\) (separable)
- Outcome 0 (probability \(\frac{2}{3}\)):
- State collapses to \(\frac{1}{\sqrt{2}}(|010\rangle + |001\rangle) = |0\rangle \otimes \frac{1}{\sqrt{2}}(|10\rangle + |01\rangle)\)
- Qubits 2 and 3 are still entangled in a Bell state!
Key difference from GHZ: W states preserve entanglement when losing qubits.
3.3.4 Entanglement Entropy of W
For \(|W_3\rangle\) with bipartition \(1|23\):
The state can be written:
\[ |W_3\rangle = \frac{1}{\sqrt{3}}|1\rangle|00\rangle + \frac{\sqrt{2}}{\sqrt{3}}|0\rangle \cdot \frac{1}{\sqrt{2}}(|10\rangle + |01\rangle) \]
Schmidt coefficients: \(\lambda_1 = \frac{1}{\sqrt{3}}, \lambda_2 = \sqrt{\frac{2}{3}}\)
Entropy:
\[ S = -\frac{1}{3}\log_2\frac{1}{3} - \frac{2}{3}\log_2\frac{2}{3} \approx 0.918 \text{ bits} \]
This is less than 1 bit - W states are not maximally entangled for bipartitions!
3.4 GHZ vs W: Different Entanglement Classes
GHZ and W states represent fundamentally different types of entanglement.
3.4.1 LOCC Equivalence
Two states are LOCC equivalent (Local Operations and Classical Communication) if one can be transformed into the other using: - Local quantum operations on individual qubits - Classical communication between parties
Physical interpretation: - GHZ: Fragile, maximal correlations, “all-or-nothing” entanglement - W: Robust, partial correlations, “distributed” entanglement
3.4.2 Summary of Differences
| Property | GHZ State | W State |
|---|---|---|
| Form | \(\frac{1}{\sqrt{2}}(|00\cdots0\rangle + |11\cdots1\rangle)\) | \(\frac{1}{\sqrt{n}}\sum_i |0\cdots010\cdots0\rangle_i\) |
| Bipartite entropy | \(S = 1\) (maximal) | \(S < 1\) (sub-maximal) |
| Robustness | Fragile to loss | Robust to qubit loss |
| Correlations | Perfect all-way | Partial pairwise |
| LOCC class | Distinct | Distinct |
3.5 Multipartite Entanglement: Beyond 2 Parties
3.5.1 Genuine Multipartite Entanglement
A state has genuine multipartite entanglement if it is entangled across all possible bipartitions.
Example: \(|\text{GHZ}_3\rangle\) is genuinely tripartite entangled: - Bipartition \(1|23\): entangled ✓ - Bipartition \(2|13\): entangled ✓ - Bipartition \(3|12\): entangled ✓
3.5.2 Biseparable States
A state is biseparable if it can be written as a mixture of states that are separable across some bipartition.
Example: \(|\psi\rangle = |0\rangle_1 \otimes |\Phi^+\rangle_{23}\) is biseparable (qubit 1 is separable from qubits 2-3).
This has entanglement but not genuine tripartite entanglement.
3.6 Measuring n-Qubit States
3.6.1 Partial Measurements
When measuring \(k\) qubits of an \(n\)-qubit state, the remaining \(n-k\) qubits collapse to a state determined by the measurement outcome.
General formula: For state \(|\psi\rangle = \sum_{i,j} c_{ij}|i\rangle_A|j\rangle_B\) where \(A\) represents the measured qubits and \(B\) the unmeasured:
Measuring \(A\) and getting outcome \(i\):
\[ |\psi\rangle \xrightarrow{\text{measure } A=i} |i\rangle_A \otimes \frac{1}{\sqrt{P(i)}}\sum_j c_{ij}|j\rangle_B \]
where \(P(i) = \sum_j |c_{ij}|^2\).
3.6.2 Example: Measuring GHZ\(_3\)
Start with: \(|\text{GHZ}_3\rangle = \frac{1}{\sqrt{2}}(|000\rangle + |111\rangle)\)
Measure qubit 1 in Z-basis:
Outcome 0 (probability \(\frac{1}{2}\)): \[ |\text{GHZ}_3\rangle \to |0\rangle_1 \otimes |00\rangle_{23} \] Qubits 2 and 3 are now in a product state \(|00\rangle\) (no entanglement!)
Outcome 1 (probability \(\frac{1}{2}\)): \[ |\text{GHZ}_3\rangle \to |1\rangle_1 \otimes |11\rangle_{23} \] Again, qubits 2 and 3 are in a product state \(|11\rangle\).
Key observation: GHZ entanglement is completely destroyed by measuring a single qubit!
3.6.3 Example: Measuring W\(_3\)
Start with: \(|W_3\rangle = \frac{1}{\sqrt{3}}(|100\rangle + |010\rangle + |001\rangle)\)
Measure qubit 1 in Z-basis:
Outcome 0 (probability \(\frac{2}{3}\)): \[ |W_3\rangle \to |0\rangle_1 \otimes \frac{1}{\sqrt{2}}(|10\rangle + |01\rangle)_{23} = |0\rangle_1 \otimes |\Psi^+\rangle_{23} \] Qubits 2 and 3 are in a Bell state (still entangled!)
Outcome 1 (probability \(\frac{1}{3}\)): \[ |W_3\rangle \to |1\rangle_1 \otimes |00\rangle_{23} \] Product state (no entanglement).
Key observation: W entanglement survives the loss of one qubit (with probability \(\frac{2}{3}\)).
3.7 Tensor Network Visualization
For large \(n\)-qubit states, we can use tensor network diagrams to visualize entanglement structure.
3.7.1 Product State
|0⟩ —— |0⟩ —— |0⟩ —— |0⟩No connections → no entanglement.
3.7.2 GHZ State
╱─────╲
╱ ╲
|ψ⟩ ●─────────●
╲ ╱
╲─────╱Central node connected to all qubits → maximal multipartite entanglement.
3.7.3 W State
|ψ⟩ ───┬───┬───┬───
│ │ │
q1 q2 q3 q4Distributed structure → robust, partial entanglement.
(Note: These are simplified diagrams. Full tensor networks are more complex.)
3.8 Computational Complexity
3.8.1 State Vector Simulation
Classical simulation of \(n\)-qubit states requires: - Memory: \(O(2^n)\) complex numbers - Operations: \(O(2^n)\) per gate
Limitations: - ~45 qubits: Maximum on world’s best supercomputers (\(\sim 35\) TB RAM) - 50+ qubits: Classically intractable to simulate in general
This is where quantum computers achieve advantage!
3.8.2 Tensor Network Methods
For certain states (e.g., matrix product states), we can simulate efficiently using tensor networks: - Memory: \(O(n \chi^2)\) where \(\chi\) is bond dimension - Works for: 1D systems, weakly entangled states, ground states of local Hamiltonians
Limitation: Highly entangled states (like GHZ, W) require exponential resources even for tensor networks.
3.9 Creating n-Qubit States in QPL
In the Quantum Process Language, you can create GHZ and W states directly:
from qpl import QPLProgram
program = QPLProgram("n-Qubit Entanglement")
# Create 4 qubits
q0, q1, q2, q3 = [program.create_system() for _ in range(4)]
# GHZ state (default)
ghz4 = program.entangle(q0, q1, q2, q3)
print(f"GHZ entropy: {ghz4.entanglement_entropy:.3f}") # 1.000
# W state
w4 = program.entangle(q0, q1, q2, q3, state_type="w")
print(f"W entropy: {w4.entanglement_entropy:.3f}") # < 1.000This relations-first approach treats entanglement as a primitive operation, not a derived concept!
3.10 Summary
In this chapter, you learned:
✅ n-qubit systems live in \(2^n\)-dimensional Hilbert spaces ✅ GHZ states: maximal, fragile, all-or-nothing entanglement ✅ W states: robust, partial, distributed entanglement ✅ GHZ and W are LOCC-inequivalent (different entanglement classes) ✅ Genuine multipartite entanglement involves all bipartitions ✅ Partial measurements affect entanglement differently for GHZ vs W ✅ Classical simulation becomes intractable beyond ~45 qubits
Part I Complete! You now have the quantum foundations needed to understand MBQC and QPL.
3.11 Exercises
3.11.1 Exercise 3.1: GHZ State Verification
For \(|\text{GHZ}_4\rangle = \frac{1}{\sqrt{2}}(|0000\rangle + |1111\rangle)\):
(a) Calculate the probability of measuring outcome \(|0101\rangle\) in the Z-basis.
(b) If you measure qubits 1 and 2 and get outcomes 0 and 1 respectively, what is the state of qubits 3 and 4?
3.11.2 Exercise 3.2: W State Properties
For \(|W_4\rangle = \frac{1}{2}(|1000\rangle + |0100\rangle + |0010\rangle + |0001\rangle)\):
(a) Calculate \(P(1000)\) when all qubits are measured in the Z-basis.
(b) After measuring qubit 1 and getting 0, prove that qubits 2, 3, 4 are in state \(|W_3\rangle\).
3.11.3 Exercise 3.3: Entanglement Entropy Calculation
Calculate the bipartite entanglement entropy for \(|W_3\rangle\) with partition \(12|3\) (qubits 1-2 vs qubit 3).
Hint: First find the Schmidt decomposition for this bipartition.
3.11.4 Exercise 3.4: LOCC Impossibility
Explain intuitively why you cannot convert \(|\text{GHZ}_3\rangle\) to \(|W_3\rangle\) using only local operations and classical communication.
3.11.5 Exercise 3.5: Multipartite Correlation
Consider the state:
\[ |\psi\rangle = \frac{1}{2}(|000\rangle + |011\rangle + |101\rangle + |110\rangle) \]
(a) Is this state genuinely tripartite entangled? Check all bipartitions.
(b) Calculate the entanglement entropy for partition \(1|23\).
3.11.6 Exercise 3.6: Fragility vs Robustness
Design an experiment (in words) that demonstrates the difference in robustness between GHZ and W states when qubits are lost.
3.11.7 Exercise 3.7: n-Qubit Generalization
Generalize the formula for \(|W_n\rangle\) and prove that measuring \(k\) qubits (with all outcomes 0) leaves the remaining qubits in state \(|W_{n-k}\rangle\) (properly normalized).
Next: Part II: Measurement-Based Quantum Computing
Part I complete! You’re now ready to explore MBQC and QPL.