$H$ (a Hadamard gate) is a single qubit unitary
Also two qubit unitary gates (CNOT here)
Measurements
Model of universal quantum computation
Example of discrete time, many body dynamics
(Mostly) concerned with unitary circuits made from unitary gates
Gate is $n$-qubit unitary $U_{s_1\ldots s_n,s’_1,\ldots, s’_n}$
$$ \sum_{s_1'\ldots s_N'}U_{s_1\ldots s_n,s'_1,\ldots, s'_n} U^\dagger_{s'_1\ldots s'_n,s''_1,\ldots, s''_n}=\delta_{s_1,s_1''}\ldots \delta_{s_N,s_N''} $$
$$ \ket{\Psi} = \sum_{s_{1:N}\in \{0,1\}^N} \Psi_{s_1\ldots s_N}\ket{s_1}_1\ket{s_2}_2\cdots \ket{s_N}_N $$
Write $\ket{s_1}_1\ket{s_2}_2\cdots \ket{s_N}_N =\ket{s_1\cdots s_N}=\ket{s_{1:N}}$
for brevity
Operator on $N$ qubits has matrix elements
$$ \mathcal{O}_{s_{1:N},s'_{1:N}} = \bra{s_{1:N}}\mathcal{O}\ket{s'_{1:N}} $$
Multiplication by a Pauli matrix: $X$, $Y$, or $Z$.
General case $U = a_0\mathbb{1} + \vectorbold{a}\cdot(X,Y,Z)$ with $|a_0|^2+|\vectorbold{a}|^2=1$
Other special cases used in quantum information e.g. Hadamard gate
$$ H = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} $$
Usually write in basis $\ket{00}$, $\ket{01}$, $\ket{10}$, $\ket{11}$
Simplest example: SWAP gate
$$ \operatorname{SWAP}=\left[\begin{array}{llll} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right] $$
Takes product state to product state in computational basis
$$ \operatorname{SWAP}\ket{10} = \ket{01} $$
$$ \sqrt{\operatorname{SWAP}}=\left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & \frac{1}{2}(1+i) & \frac{1}{2}(1-i) & 0 \\ 0 & \frac{1}{2}(1-i) & \frac{1}{2}(1+i) & 0 \\ 0 & 0 & 0 & 1 \end{array}\right] $$
$$ \sqrt{\operatorname{SWAP}}\ket{10} = \frac{1}{2}\left[(1+i)\ket{10}+(1-i)\ket{01}\right] $$
Conserves number of 1s and 0s (in fact fully rotationally invariant)
$\sqrt{\operatorname{SWAP}}$ and single qubit unitaries are universal gate set
\begin{equation} U = e^{i \phi} (u_+ \otimes u_-) V[J_x, J_y, J_z] (v_- \otimes v_+), \end{equation}
\begin{align} V[J_x, J_y, J_z] &= \exp \left[-i\left(J_x \sigma^x \otimes \sigma^x + J_y \sigma^y \otimes \sigma^y+ J_z \sigma^z \otimes \sigma^z\right)\right]\\ &= \begin{bmatrix} e^{-i J_z} \cos(J_-) & 0 & 0 & -i e^{-i J_z \sin(J_-)} \\ 0 & e^{iJ_z} \cos(J_+) & -ie^{i J_z} \sin(J_+) & 0 \\ 0 & -ie^{i J_z} \sin(J_+) & e^{iJ_z} \cos(J_+) & 0 \\ -i e^{-i J_z \sin(J_-)} & 0 & 0 & e^{-i J_z} \cos(J_-) \\ \end{bmatrix} \end{align}
Time evolution operator $U=\exp(-iHt)$
If $H=\sum_j h_j$ a sum of single qubit terms
$$ U = \exp(-iHt) = \prod_j \exp(-ih_j) = \prod_j U_j $$ $$ U_j=\mathbb{1}\otimes \ldots \otimes\mathbb{1} \otimes \overbrace{u_j}^{j\text{th factor}} \ldots \otimes\mathbb{1} $$
$$ \begin{align} h_{12} &= J\left[X\otimes X+Y\otimes Y+Z\otimes Z\right] =J\left[X_1X_2+Y_1Y_2 + Z_1Z_2\right]\\ &=2\operatorname{SWAP} - 1 \end{align} $$ $$ U(J) = \exp(-ih_{12}) = e^{iJ}\left[\cos (2J) \mathbb{1} - i\sin (2J) \operatorname{SWAP}\right] $$
$$ U(\pi/4)=\operatorname{SWAP} $$ $$ U(\pi/8)=\sqrt{\operatorname{SWAP}} $$
$H=\sum_{i,j} h_{i,j}$ a sum of two qubit terms with $[h_{i,j},h_{j,k}]\neq 0$
$U\neq \prod_{i,j} \exp(-ih_{i,j})$. More complicated!
Suzuki–Trotter expansion: decompose $H=H_A + H_B$
$$ U = \exp(-iH) = \left[\exp\left(-\frac{iH}{n}\right)\right]^n \sim \left[e^{-iH_A/n} e^{-iH_B/n}\right]^n $$
$$ H = \sum_j h_{j,j+1} $$ $$ H_A = \sum_j h_{2j, 2j+1}\qquad H_B = \sum_j h_{2j-1, 2j} $$ $$ e^{-iH_A/n}=\prod_j U_{2j,2j+1}\qquad e^{-iH_B/n} = \prod_j U_{2j-1,2j} $$
$$ \begin{aligned} H_{\text{KIM}}(t) = H_\text{I}[\mathbf{h}] + \sum_{m}\delta(t-n)H_\text{K}\\ H_\text{I}[\mathbf{h}]=\sum_{j=1}^L\left[J Z_j Z_{j+1} + h_j Z_j\right],\qquad H_\text{K} &= b\sum_{j=1}^L X_j, \end{aligned} $$
$$ \begin{aligned} U(n_+) &= \left[U(1_+)\right]^n,\qquad U(1_-) = K I_\mathbf{h}\\ I_\mathbf{h} &= e^{-iH_\text{I}[\mathbf{h}]}, \qquad K = e^{-iH_\text{K}} \end{aligned} $$
$$ \begin{aligned} \mathcal{K} &= \exp\left[-i b X\right]\\ \mathcal{I} &= \exp\left[-iJ Z_1 Z_2 -i \left(h_1 Z_1 + h_2 Z_2\right)/2\right]. \end{aligned} $$
Normally matrix-vector multiplication is $O(\operatorname{dim}^2)=2^{2N}$
Gates are sparse so $O(\operatorname{dim})=2^{N}$, but still exponentially hard
For low depth $T<N$ move horizontally instead
Evaluate $\bra{\Psi}\mathcal{O}\ket{\Psi}=\bra{\Psi_0}U^\dagger\mathcal{O}U\ket{\Psi_0}$ for local $\mathcal{O}$
If $\Psi_0$ is product state top and bottom indices match
After folding lines correspond to two indices / 4 dimensions
Semicircle denotes $\delta_{ab}$
$$ \rho_A = \operatorname{tr}_B\left[\ket{\Psi}\bra{\Psi}\right]=\operatorname{tr}_B\left[U\ket{\Psi_0}\bra{\Psi_0}U^\dagger\right] $$
$\mathcal{H}=\mathcal{H}_A\otimes\mathcal{H}_B$
any state $\Psi_{AB}$
can be written$$ \ket{\Psi_{AB}} = \sum_{\alpha=1}^{\min(\operatorname{dim} \mathcal{H}_A, \operatorname{dim} \mathcal{H}_B)} \lambda_\alpha \ket{u_\alpha}_A\otimes\ket{v_\alpha}_B $$
$\ket{u_\alpha}$ and $\ket{v_\alpha}$ orthonormal; $\lambda_\alpha\geq 0$
$\lambda_\alpha$ quantify entanglement between A and B
$$ \begin{align} \rho_A &= \operatorname{tr}_B\left[\ket{\Psi}\bra{\Psi}\right] \\ &= \sum_\alpha \lambda_\alpha^2 \ket{u_\alpha}\bra{u_\alpha} \end{align} $$
$\operatorname{rank}=\min(\operatorname{dim} \mathcal{H}_A, \operatorname{dim} \mathcal{H}_B)=2^{\min(2t-2, N_A)}$
Here $t=4$, $N_A=4$
$$ \begin{align} S_A &= -\operatorname{tr}\left[\rho_A\log \rho_A\right]\\ &=-\sum_\alpha p_\alpha \log p_\alpha \end{align} $$
$$ S_A \leq \min(2t-2, N_A)\log 2 $$
$$ \lim_{L\to\infty} S_A =\min(2t-2,N_A)\log 2, $$
$$ S^{(n)}_A = \frac{1}{1-n}\log \text{tr}\left[\rho^n\right]=\frac{1}{1-n}\sum_\alpha p_\alpha^n $$
$$ p_\alpha = \left(\frac{1}{2}\right)^{\min(2t-2,N_A)} $$
After $N_A/2 + 1$ steps, reduced density matrix is $\propto \mathbb{1}$
All expectations (with $A$) take on infinite temperature value
$$ \begin{aligned} \mathcal{K} &= \exp\left[-i b X\right]\\ \mathcal{I} &= \exp\left[-iJ Z_1 Z_2 -i \left(h_1 Z_1 + h_2 Z_2\right)/2\right]. \end{aligned} $$
$$ \rho_0=\overbrace{\frac{\mathbb{1}}{2}\otimes \frac{\mathbb{1}}{2} \cdots }^{t-1} \otimes\overbrace{|Z_1\rangle\langle Z_1|\otimes |Z_2\rangle\langle Z_2| \cdots }^{N_A-2t+2 } \otimes \overbrace{\frac{\mathbb{1}}{2}\otimes \frac{\mathbb{1}}{2} \cdots }^{t-1} $$