# Tensor product of qubits Tensor product of two vector fields $V,W$ is formal pair of $v\otimes w$ for every $v\in V,w\in W$ . Linearity is extend to this pair by $(v+v^\prime)\otimes w = v\otimes w+v^\prime\otimes w$ and $(\alpha v)\otimes w=v\otimes(\alpha w) = \alpha(v\otimes w)$ where $\alpha$ is a scalar. same linearity is true for $w$. Tensor of $|\psi\rangle$ and $|\phi\rangle|$ is $|\psi\rangle \otimes|\phi\rangle$ . This is also written as $|\psi,\phi\rangle$ or $|\psi\phi\rangle$. ## Examples 1) $\ket1\otimes\ket0=\ket 1\ket 0= \ket {10}=\ket 2=\begin{pmatrix}1 \\ 0 \end{pmatrix}\otimes \begin{pmatrix}0 \\ 1 \end{pmatrix}$ 2) $\ket+\otimes\ket0=\frac{1}{\sqrt{2}}(|0\rangle+|1\rangle)\otimes\ket0=\frac{1}{\sqrt{2}}(\ket{00}+\ket{10})$ 3) $ \begin{align} \ket+\otimes\ket+= &\bigg(\frac{1}{\sqrt{2}}(|0\rangle+|1\rangle)\bigg)\otimes\bigg( \frac{1}{\sqrt{2}}(|0\rangle+|1\rangle)\bigg) = \\ &=\bigg(\frac{1}{\sqrt{2}}(|0\rangle+|1\rangle)\bigg)\otimes \frac{1}{\sqrt{2}}|0\rangle + \bigg(\frac{1}{\sqrt{2}} (|0\rangle+|1\rangle)\bigg)\otimes |1\rangle\bigg) \\ &= \frac{1}{2}(|00\rangle + |10\rangle) + \frac{1}{2}(|01\rangle+|11\rangle)) \\ &= \frac{1}{2}(|00\rangle + |01\rangle+|10\rangle+|11\rangle) \end{align} $ 4. we show that $\frac{1}{\sqrt{2}}\big(|00\rangle+|11\rangle\big) =\frac{1}{\sqrt{2}}\big(|++\rangle+|--\rangle\big)$ . This identity is used in the [[CHSH game]]. $ \begin{align} &\frac{1}{\sqrt{2}}\big(|++\rangle+|--\rangle\big)=\\ &=\frac{1}{\sqrt{2}}\bigg(\frac{1}{\sqrt{2}}(\ket0+\ket1)\otimes\frac{1}{\sqrt{2}}(\ket0+\ket1)+\frac{1}{\sqrt{2}}(\ket0-\ket1)\otimes\frac{1}{\sqrt{2}}(\ket0-\ket1)\bigg) \\ &=\frac{1}{\sqrt{2}}\big(\ket{00}+\ket{11}\big) \end{align} $ ## Exercise 1 - compute: $\bigg(\frac{1}{\sqrt{2}}(|0\rangle+|1\rangle)\bigg)\otimes\bigg( \frac{1}{\sqrt{2}}(|0\rangle-|1\rangle)\bigg)$ qubits can be grouped to `registers` (similar to hardware registers that are a logic group of bits) e.g., the first qubit is register 1 and the 2nd register 2. This is a logic grouping that helps to communicate a computation. ## Inner product in Tensor space We define $\langle v_1\otimes w_1, v_2\otimes w_2 \rangle:=\langle v_1,v_2\rangle\cdot \langle w_1,w_2\rangle$ . We get $\|v\otimes w\|^2=\langle v,v\rangle\cdot\langle w,w\rangle=\|v\|^2\cdot\|w\|^2$ and so $\|v\otimes w\|=\|v\|\cdot\|w\|$. This means that qubits norm of 1 is kept under tensor product. ## Linear operators on tensors Let $A$ be a linear operator on vector space $V$ and $B$ operator on $W$. $(A\otimes B) (v\otimes w):= (Av)\otimes(Bw)$ for $v\in V,w\in W$. Properties: - $(A\otimes B)\circ(C\otimes D):= (A\circ C)\otimes(B\circ D)$ - $I_{V\otimes W} := I_V\otimes I_W$ - $(A\otimes I)\circ (I\otimes B) =(I\otimes B)\circ (A\otimes I) = (A\otimes B)$ - $(A\otimes B)^{-1}=A^{-1}\otimes B^{-1}$ - $(A\otimes B)^\dagger=A^\dagger\otimes B^\dagger$. This means that if $A$ and $B$ are hermitian than also $A\otimes B$. Also, that if $A$ and $B$ are correct physical operator then also their tensor product. - $(A)^{\otimes n} := A\otimes\cdots\otimes A$ ## Examples - $(I\otimes Z)|11\rangle=(I\otimes Z)(|1\rangle\otimes|1\rangle)=I|1\rangle\otimes Z|1\rangle = |1\rangle\otimes|-1\rangle=-|11\rangle$ - $(X\otimes I)(|0\rangle\otimes|1\rangle)=(X|0\rangle)\otimes(I|1\rangle)=|1\rangle\otimes|1\rangle=|11\rangle.$ - $(H\otimes X)|00\rangle=(H|0\rangle)\otimes(X|0\rangle)=\frac{1}{\sqrt{2}}(|0\rangle+|1\rangle)\otimes|1\rangle=\frac{|01\rangle+|11\rangle}{\sqrt{2}}$ - Constructing 2-qubits with uniform distribution amplitudes i.e., the probability to measure any of the 4 results is the same i.e., it is equal to 0.25. $\begin{align} H^{\otimes 2}|0\rangle^{\otimes 2}&= (H\otimes H)(|0\rangle\otimes|0\rangle) \\ &= H|0\rangle\otimes H|0\rangle\\ &= \frac{1}{\sqrt 2}(|0\rangle+|1\rangle)\otimes\frac{1}{\sqrt 2}(|0\rangle+|1\rangle) \\ &=\frac{1}{2}( |0\rangle\otimes|0\rangle+ |0\rangle\otimes|1\rangle+ |1\rangle\otimes|0\rangle+ |1\rangle\otimes|1\rangle) \\ &=\frac{1}{2}(|00\rangle+|01\rangle+|10\rangle+|11\rangle)\\ &=\frac{1}{\sqrt{2^2}}\sum_{x\in\{0,1\}^2}|x\rangle \end{align}$ ## Exercise 2: compute $H^{\otimes 3}|0\rangle^{\otimes 3}=\frac{1}{\sqrt{2^3}}\sum_{x\in\{0,1\}^3}|x\rangle$ . This is a generalization of the last example. Conclude for general $n$: $H^{\otimes n}|0\rangle^{\otimes n}=\frac{1}{\sqrt{2^n}}\sum_{x\in\{0,1\}^n}|x\rangle$ --- Examples with vector notation 1) The [[X-gate (NOT gate)]] $X=\begin{bmatrix} 0 & 1 \\ 1& 0\end{bmatrix}$ . $X\otimes I= \begin{bmatrix} 0\cdot I & 1\cdot I \\ 1\cdot I & 0\cdot I \end{bmatrix}= \begin{bmatrix} 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ \end{bmatrix} $ 2) [[Z-gate (Relative phase gate)]], $Z=\begin{bmatrix} 1 & 0 \\ 0 & -1\end{bmatrix}$ $ I\otimes Z=\begin{bmatrix} 1Z & 0Z \\ 0Z& 1Z \end{bmatrix}=\begin{bmatrix} 1 &0 &0& 0\\ 0&-1&0&0\\ 0&0&1&0 \\ 0&0&0&-1 \end{bmatrix} $ Tag: #tech #math ## Created 2026-01-14 11:50