Reading Week 4 - Multiqubit Gates, Quantum Entanglement

2.11: Tensors (for me)

The Tensor Product or the Kronecker product, or the direct product is defined as follows. A new set of basis vectors $n m$ is defined by the tensor product $i \otimes j$ of the basis vectors. The space is bilinear, so linear in both $V, W$ :

\begin{aligned} | V ⟩ \otimes | W ⟩ & = \sum_{i = 1}^{n} v_{i} | i ⟩ \otimes \sum_{j = 1}^{m} w_{j} | j ⟩ \\ = \sum_{i = 1}^{n} \sum_{j = 1}^{m} v_{i} w_{j} (| i ⟩ \otimes | j ⟩) \end{aligned}

View this as a space $| V ⟩ \otimes | W ⟩$ with basis vectors $| i ⟩ \otimes | j ⟩$ whose coefficients are $v_{i} w_{j}$ :

| V ⟩ \otimes | W ⟩ = (v_{1}, . . ., v_{n}) \otimes (w_{1}, . . ., w_{n}) = [\begin{matrix} v_{1} w_{1} \\ ⋮ \\ v_{1} w_{m} \\ v_{2} w_{1} \\ ⋮ \\ v_{2} w_{m} \\ ⋮ \\ v_{n} w_{m} \end{matrix}]

As an example, if $| V ⟩ = (1, 2)$ and $| W ⟩ = (3, 4, 5)$ (as column vectors) then:

| V ⟩ \otimes | W ⟩ = [\begin{matrix} 1 \cdot 3 \\ 1 \cdot 4 \\ 1 \cdot 5 \\ 2 \cdot 3 \\ 2 \cdot 4 \\ 2 \cdot 5 \end{matrix}] = [\begin{matrix} 3 \\ 4 \\ 5 \\ 6 \\ 8 \\ 10 \end{matrix}]

For a tensor product of 2x2 matrices, we just repeat entries into submatrices:

[\begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{matrix}] \otimes [\begin{matrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{matrix}] = [\begin{matrix} a_{11} [\begin{matrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{matrix}] & a_{12} [\begin{matrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{matrix}] \\ a_{21} [\begin{matrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{matrix}] & a_{22} [\begin{matrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{matrix}] \end{matrix}] = [\begin{matrix} a_{11} b_{11} & a_{11} b_{12} & a_{12} b_{11} & a_{12} b_{12} \\ a_{11} b_{21} & a_{11} b_{22} & a_{12} b_{21} & a_{12} b_{22} \\ a_{21} b_{11} & a_{21} b_{12} & a_{22} b_{11} & a_{22} b_{12} \\ a_{21} b_{21} & a_{21} b_{22} & a_{22} b_{21} & a_{22} b_{22} \end{matrix}]

Exercise

Notice:

\begin{aligned} ⟨ a | \otimes ⟨ b |) (| c ⟩ \otimes | d ⟩ & = ? \end{aligned}

if $| i ⟩ = (i_{1}, . . ., i_{n})^{*}$ for each $i = a, b, c, d$ then:

| c ⟩ \otimes | d ⟩ = | c d ⟩ = (c_{1} d_{1}, c_{1} d_{2}, . . ., c_{1} d_{n}, c_{2} d_{1}, . . ., c_{n} d_{n})^{*}

while:

⟨ a | \otimes ⟨ b | = ⟨ a b | = (a_{1} b_{1}, a_{1} b_{2}, . . ., a_{1} b_{n}, a_{2} b_{1}, . . ., a_{n} b_{n})

Thus:

Our equation = ⟨ a b | c d ⟩ = \sum_{i = 1}^{n} \sum_{j = 1}^{n} a_{i} b_{j} c_{i} d_{j} = \sum_{i = 1}^{n} a_{i} c_{i} \sum_{j = 1}^{n} b_{j} d_{j} = ⟨ a | c ⟩ ⟨ b | d ⟩

5.4 (cont.) Multiqubit States (no Gates)

To catch up see Reading Week 3 - Ending Quantum Superposition, Starting Qubits#5.4 Quantum Gates

5.4.4: Multiqubit Gates

We'll learn how to represent a system of two or more qubits and the gate operations which are performed on two or more qubits.

5.4.5: Representing a Multi Qubit State

A system of $n$ -qubits has $2^{n}$ orthonormal computational basis states, denoted $| x_{1} . . . x_{n} ⟩$ where $x_{i} \in 0, 1$ . The state is a superposition of all $2^{n}$ states with probability amplitudes $a_{x}$ :

| ψ ⟩ = \sum_{x \in {0, 1}^{n}} a_{x} | x ⟩

where we require:

\sum_{x \in {0, 1}^{n}} | a_{x} |^{2} = 1

Recollect the tensor math from Reading Week 4 - Multiqubit Gates, Quantum Entanglement#2.11 Tensors (for me). Assume $| ψ_{1} ⟩ = a_{1} | 0 ⟩ + b_{1} | 1 ⟩$ and $| ψ_{2} ⟩ = a_{2} | 0 ⟩ + b_{2} | 1 ⟩$ . Then:

\begin{aligned} | ψ ⟩ & = | ψ_{1} ⟩ \otimes | ψ_{2} ⟩ \\ = a_{1} a_{2} | 00 ⟩ + a_{1} b_{2} | 01 ⟩ + b_{1} a_{2} | 10 ⟩ + b_{1} b_{2} | 11 ⟩ \end{aligned}

We require that $⟨ ψ | ψ ⟩ = 1$ (orthonroamality) so then:

\sum_{i = 1}^{n = 2} \sum_{j = 1}^{n = 2} | a_{i} b_{j} |^{2} = 1

We use the $| i j ⟩$ notation to combine our tensor product vector $| i ⟩ \otimes | j ⟩ = | i j ⟩$ . As a result:

| 00 ⟩ = [\begin{matrix} 1 \\ 0 \\ 0 \\ 0 \end{matrix}], | 01 ⟩ = [\begin{matrix} 0 \\ 1 \\ 0 \\ 0 \end{matrix}], . . .

Sometimes, to save space, we contract the $\otimes$ and just put the ket's or the bra's together:

| 0 ⟩ \otimes | 0 ⟩ = | 0 ⟩ | 0 ⟩ = | 00 ⟩

As with a single qubit, the probability of measuring a single eigenvalue is via:

P_{i}^{z z} = | ⟨ λ_{i} | ψ ⟩ |^{2}

where here $z$ indicates the direction of measurement and $λ_{i}$ is one of the possible eigenvalues.

The probability of measuring a 0 in the first qubit is:

p_{1} (0) = P (00) + P (01) = | a_{1} a_{2} |^{2} + | a_{1} b_{2} |^{2}

A partial measurement of the system results in a new superposition state. We calculate the state by removing terms which are no longer applicable, and we must normalize the new superposition state since the sum of probabilities may not be 1:

| ψ ⟩ = \frac{a_{1} a_{2} | 00 ⟩ + a_{1} b_{2} | 01 ⟩}{\sqrt{| a_{1} a_{2} |^{2} + | a_{1} b_{2} |^{2}}}

similarly for measuring 1 in the first qubit:

p_{1} (1) = P (10) + P (11) = | b_{1} a_{2} |^{2} + | b_{1} a_{2} |^{2}

| ψ ⟩ = \frac{b_{1} a_{2} | 10 ⟩ + b_{1} b_{2} | 11 ⟩}{\sqrt{| b_{1} a_{2} |^{2} + | b_{1} b_{2} |^{2}}}

5.6: Quantum Entanglement

In the last section, we looked at a quantum system with $2^{n}$ dimensions, where $n$ is the number of qubits we have. There is no correlation between qubits. If we somehow find a correlation between qubits and can describe the system as a single system, then the qubits are entangled.

Consider the following QC:

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We have two qubits in the default $| 0 ⟩$ state (per usual). We apply an $H$ Gate (see Reading Week 3 - Ending Quantum Superposition, Starting Qubits#5.4.1.3 The Hadamard Gate, $H$ -Gate) to q[0] and perform a CNOT (see later section on multi-qubit systems) gate with target q[1] and q[0] being the control bit. We then measure q[0],q[1] at the and put on the classical wire.

Step 1: Apply $H$ Gate

H {| 0 ⟩}_{0} = \frac{1}{\sqrt{2}} [\begin{matrix} 1 \\ 1 \end{matrix}] = \frac{1}{\sqrt{2}} (| 0 ⟩ + | 1 ⟩)

apply this in a tensor product to bring back to a 2 qubit system:

\begin{array}{r} \to \frac{1}{\sqrt{2}} [\begin{array}{c} 1 \\ 1 \end{array}] \otimes [\begin{array}{c} 1 \\ 0 \end{array}] = \frac{1}{\sqrt{2}} ([\begin{array}{c} 1 \\ 0 \\ 0 \\ 0 \end{array}] + [\begin{array}{c} 0 \\ 0 \\ 1 \\ 0 \end{array}]) = \frac{1}{\sqrt{2}} (| 00 ⟩ + | 10 ⟩) \end{array}

Step 2: Apply CNOT

[\begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{matrix}] \frac{1}{\sqrt{2}} ([\begin{matrix} 1 \\ 0 \\ 0 \\ 0 \end{matrix}] + [\begin{matrix} 0 \\ 0 \\ 1 \\ 0 \end{matrix}]) = \frac{1}{\sqrt{2}} [\begin{matrix} 1 \\ 0 \\ 0 \\ 1 \end{matrix}] = \frac{1}{\sqrt{2}} (| 00 ⟩ + | 11 ⟩) = | ϕ^{+} ⟩

Note that we cannot "factor" out the $| 0 ⟩$ or anything from $| ϕ^{+} ⟩$ , namely we cannot express it in terms of terms q[0] and q[1]. This resultant state is an entangled state and the qubits are said to be maximally entangled. The entangled qubits are the EPR Pair.

Here $| Φ^{+} ⟩$ is one of the Bell States.

\begin{aligned} | 00 ⟩ \to \frac{1}{\sqrt{2}} (| 00 ⟩ + | 11 ⟩) & = | ϕ^{+} ⟩ \\ | 01 ⟩ \to \frac{1}{\sqrt{2}} (| 00 ⟩ - | 11 ⟩) & = | ϕ^{+} ⟩ \\ | 10 ⟩ \to \frac{1}{\sqrt{2}} (| 01 ⟩ + | 10 ⟩) & = | Ψ^{+} ⟩ \\ | 11 ⟩ \to \frac{1}{\sqrt{2}} (| 01 ⟩ + | 10 ⟩) & = | Ψ^{-} ⟩ \end{aligned}

Some example code to see entanglement:

q = QuantumRegister(2, 'q')
c = ClassicalRegister(2, 'c')
qc = QuantumCircuit(q, c)
qc.h(q[0])
qc.cx(q[0], q[1])
qc.measure(q[0], c[0])
qc.measure(q[1], c[1])

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