Composite Quantum Systems
Like bits, qubits can be put together for more computational power.
Table of Contents
1. Tensor Product
First, some preliminaries about tensor product.
Tensor product of vectors:
\[ \begin{pmatrix}a\\b\end{pmatrix}\otimes\begin{pmatrix}c\\d\end{pmatrix}=\begin{pmatrix}ac\\ad\\bc\\bd\end{pmatrix} \]
Tensor product of matrices:
\[ A \otimes B=\begin{pmatrix}a_{00}B & a_{01}B\\a_{10}B & a_{11}B\end{pmatrix}=\begin{pmatrix}a_{00}b_{00}&a_{00}b_{01}&a_{01}b_{00}&a_{01}b_{01}\\a_{00}b_{10}&a_{00}b_{11}&a_{01}b_{10}&a_{01}b_{11}\\a_{10}b_{00}&a_{10}b_{01}&a_{11}b_{00}&a_{11}b_{01}\\a_{10}b_{10}&a_{10}b_{11}&a_{11}b_{10}&a_{11}b_{11}\end{pmatrix} \]
Tensor product operation satisfies the following properties. For any \(c\in\mathbb{C}\) and arbitrary \(\ket{\psi},\ket{\phi}\):
- \( c(\ket{\psi}\otimes\ket{\phi})=(c\ket{\psi})\otimes\ket{\phi}=\ket{\psi}\otimes(c\ket{\phi}) \).
- \( (\ket{\psi} + \ket{\psi'})\otimes\ket{\phi}=\ket{\psi}\otimes\ket{\phi}+\ket{\psi'}\otimes\ket{\phi} \)
- \( (\bra{\psi}\otimes\bra{\psi'})(\ket{\phi}\otimes\ket{\phi'})=\braket{\psi|\phi}\braket{\psi'|\phi'} \)
For matrices \(A,B\), we have \( (A \otimes B)(\ket{\psi}\otimes\ket{\phi})=A\ket{\psi}\otimes B\ket{\phi} \)
The bra-ket notation for two-qubit state is a 4D vector. Here \(\ket{xy}\) is short for \(\ket{x}\otimes\ket{y}\). \( \{\ket{00},\ket{01},\ket{10},\ket{11}\} \) forms the computational basis for 2 qubits.
\[ \ket{\psi} = a_{00}\ket{00} + a_{10}\ket{10} + a_{01}\ket{01} + a_{11}\ket{11} = \begin{pmatrix}a_{00}\\a_{01}\\a_{10}\\a_{11}\end{pmatrix} \]
A two-qubit state \(\ket{\psi}\) is called a product state, if there exists single-qubit state \(\ket{\phi_{1}}, \ket{\phi_{2}}\) such that
\[ \ket{\psi}=\ket{\phi_{1}}\otimes\ket{\phi_{2}} \]
2. Two-Qubit Gates
Tensor product gates \(U \otimes V\) with \(U,V\) being single-qubit gates forms a two-qubit gate. It does in parallel \(U\) on qubit 1 and \(V\) on qubit 2.
Sometimes we write \(X_{1}Y_{2}=X \otimes Y\) using the convention \(X_{2}=I \otimes X\) and \(X_{1} = X \otimes I\)
2.1. Controlled-NOT (CNOT) Gate
By definition, CNOT Gate is
\[ \mathrm{CNOT}\ket{00}=\ket{00}\\ \mathrm{CNOT}\ket{01}=\ket{01}\\ \mathrm{CNOT}\ket{10}=\ket{11}\\ \mathrm{CNOT}\ket{11}=\ket{10}\\ \]
Equivalently, we can write CNOT as
\[ \mathrm{CNOT}=\mathrm{CNOT}_{1,2}=\ket{0}\bra{0} \otimes I + \ket{1}\bra{1} \otimes X \]
CNOT Gate represents “do \(X\) on qubit 2 conditioning on qubit 1 begin \(\ket{1}\)”.
2.2. Controlled Gates
In general, a controlled gate “if control bit = 1, do \(U\) on target bit” can be represented as
\[ C(U) = \ket{0}\bra{0} \otimes I + \ket{1}\bra{1}\otimes U \]
Since we have identities \(HXH=Z, HZH=X\) and CNOT gate is \(C(X)\), we can implement other gates with only CNOT and single-qubit gates.
Any 2-qubit controlled-gate can be decomposed into a circuit that uses at most 2 CNOTs.
3. Entangled States
A quantum state \(\ket{\psi}\) is entangled, if it’s not product.
Bell Basis. The most important entangled states.
\[
\begin{aligned} \ket{B_{0}} &= \frac{\ket{00}+\ket{11}}{\sqrt{2}} \\ \ket{B_{1}} &= (X \otimes I) \ket{B_{0}} \\ \ket{B_{2}} &= (Y \otimes I) \ket{B_{0}} \\ \ket{B_{3}} &= (Z \otimes I) \ket{B_{0}} \end{aligned}\]
The Bell states \(\{B_{i}\}\) form an O.N.B1 for 2 qubits.
A gate may not be either product or entangled, e.g., SWAP gate.
4. Two-Qubit Measurement
Two-qubit measurement works in a very similar way as the single-qubit case. For an arbitrary basis \(\{\ket{\phi_{1}},\ket{\phi_{2}},\ket{\phi_{3}},\ket{\phi_{4}}\}\), the Born’s rule still holds.
4.1. Measuring Entangled States, Partial Measurement
Born’s Rule for Partial Measurement
The probability of getting outcome \(x\) when partially measuring any quantum state \(\ket{\psi}\) in basis \(\{\ket{\phi_{x}}\}_{x}\) is
\[ P(x) = \Braket{\psi | \left( \ket{\phi_{x}}\bra{\phi_{x}} \otimes I \right) | \psi } \]
Footnotes:
orthonormal basis