Kronecker Product

Not to be confused with Kronecker Delta

The Kronecker product is a way to compute the Tensor Product of two Matrix. It outputs a matrix as well.

Let there be a spin-1/2 system and two particles living HAH_A and HBH_B respectively which spans ∣0⟩A,∣1⟩A\left\lvert 0 \right\rangle_A, \left\lvert 1 \right\rangle_A and ∣0⟩B,∣1⟩B\left\lvert 0 \right\rangle_B, \left\lvert 1 \right\rangle_B respectively. The tensor product basis for HA⊗HBH_A\otimes H_B is

{∣0⟩A⊗∣0⟩B,∣0⟩A⊗∣1⟩B,∣1⟩A⊗∣0⟩B,∣1⟩A⊗∣1⟩B}\{\left\lvert 0 \right\rangle_A\otimes\left\lvert 0 \right\rangle_B, \left\lvert 0 \right\rangle_A\otimes\left\lvert 1 \right\rangle_B, \left\lvert 1 \right\rangle_A\otimes\left\lvert 0 \right\rangle_B, \left\lvert 1 \right\rangle_A\otimes\left\lvert 1 \right\rangle_B\}

This can be rewritten as

∣00⟩≜∣0⟩A⊗∣0⟩B\left\lvert 00 \right\rangle\triangleq\left\lvert 0 \right\rangle_A\otimes\left\lvert 0 \right\rangle_B ∣01⟩≜∣0⟩A⊗∣1⟩B\left\lvert 01 \right\rangle\triangleq\left\lvert 0 \right\rangle_A\otimes\left\lvert 1 \right\rangle_B ∣10⟩≜∣1⟩A⊗∣0⟩B\left\lvert 10 \right\rangle\triangleq\left\lvert 1 \right\rangle_A\otimes\left\lvert 0 \right\rangle_B ∣11⟩≜∣1⟩A⊗∣1⟩B\left\lvert 11 \right\rangle\triangleq\left\lvert 1 \right\rangle_A\otimes\left\lvert 1 \right\rangle_B

Let ∣0⟩,∣1⟩\left\lvert 0 \right\rangle, \left\lvert 1 \right\rangle be a rank-1 tensor, i.e., a vector. Recall from Ket (State) and quantum systems that it is defined as

∣0⟩=(10)∣1⟩=(01)\left\lvert 0 \right\rangle=\begin{pmatrix} 1 \\ 0 \end{pmatrix} \quad \left\lvert 1 \right\rangle=\begin{pmatrix} 0 \\ 1 \end{pmatrix}

Note that

∣0⟩A=(10)∣0⟩B=(10)\left\lvert 0 \right\rangle_A=\begin{pmatrix} 1 \\ 0 \end{pmatrix} \quad \left\lvert 0 \right\rangle_B=\begin{pmatrix} 1 \\ 0 \end{pmatrix}

But when we compute the tensor product (for matrices this is known as the Kronecker product), we get

∣0⟩A⊗∣0⟩B=(10)⊗(10)=(1⋅(10)0⋅(10))=(1000)|0\rangle_A \otimes |0\rangle_B = \begin{pmatrix} 1 \\ 0 \end{pmatrix} \otimes \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \cdot \begin{pmatrix} 1 \\ 0 \end{pmatrix} \\ 0 \cdot \begin{pmatrix} 1 \\ 0 \end{pmatrix} \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix}

We can see the full basis of the new Hilbert space HA⊗HBH_A\otimes H_B as

∣00⟩=(1000),∣01⟩=(0100),∣10⟩=(0010),∣11⟩=(0001)|00\rangle = \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix},\quad |01\rangle = \begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \end{pmatrix},\quad |10\rangle = \begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \end{pmatrix},\quad |11\rangle = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix}