--- jupytext: formats: ipynb,md:myst text_representation: extension: .md format_name: myst format_version: 0.13 jupytext_version: 1.14.0 kernelspec: display_name: Python 3 (phys-555) language: python name: phys-555 --- ```{code-cell} ipython3 :tags: [hide-cell] import mmf_setup;mmf_setup.nbinit() import logging;logging.getLogger('matplotlib').setLevel(logging.CRITICAL) %matplotlib inline import numpy as np, matplotlib.pyplot as plt ``` # Quantum Gates As discussed in [section 4.5.2], single qubit operations and ᴄɴᴏᴛ are universal for quantum computing. Single qubit operations are easily implemented, for example, using external magnetic fields for qubits made of spin-½ particles. The challenge is generally implementing a ᴄɴᴏᴛ gate: \begin{gather*} \mathsf{\small CNOT} = \begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0 \end{pmatrix} \end{gather*} For example, trying to find a Hamiltonian $\mat{H}$ such that \begin{gather*} \mathsf{\small CNOT} = e^{\mat{H}t/\I\hbar}, \qquad \frac{\mat{H} t}{\hbar} = \I\ln(\mathsf{\small CNOT}) = \frac{\pi}{2} \begin{pmatrix} 0\\ & 0\\ && -1 & 1\\ && 1 & -1 \end{pmatrix}. \end{gather*} :::{warning} The $\ln()$ function is multi-valued. For example: \begin{gather*} \I\ln(\mathsf{\small CNOT}) \equiv \frac{\pi}{2} \begin{pmatrix} 4\\ & 4\\ && -1 & 1\\ && 1 & -1 \end{pmatrix}. \end{gather*} also exponentiates to give ᴄɴᴏᴛ. ::: :::{admonition} Do it! Find all 2×2 $\mat{A}$ such that $\exp(\mat{A}) = \mat{1}$. :class: dropdown We first note that $e^{2\pi \I n} = 1$, thus: \begin{gather*} \mat{A} = 2\pi \I \mat{S}^{-1}\begin{pmatrix} m\\ & n\\ \end{pmatrix} \mat{S} \end{gather*} where $m$, and $n$ are integers, and $\mat{S}$ is any invertable matrix. For our purposes, we generally restrict $\mat{S} = \mat{U}$ to be unitary so that $\I\mat{A}$ is hermitian. Noting that any unitary matrix can be expressed as (see e.g. Eq. (B.10) in Appendix B of {cite:p}`Mermin:2007` or Eq. (4.8) of {cite:p}`Nielsen:2010`) \begin{gather*} \mat{U} = u_0\mat{1} + \I\vec{u}\cdot\vec{\mat{\sigma}}, \qquad u_0^2 + \abs{\vec{u}}^2 = 1, \end{gather*} we can write: \begin{gather*} \mat{A} = 2\pi \I \Bigl( m \mat{1} + n (u_0\mat{1} - \I\vec{u}\cdot\vec{\mat{\sigma}}) \mat{\sigma}_z (u_0\mat{1} + \I\vec{u}\cdot\vec{\mat{\sigma}}) \Bigr). \end{gather*} ::: This is often a bit tricky to realize, but if we can implement a single-qubit Hadamard gate, then we can construct ᴄɴᴏᴛ from a controlled-$Z$ gate (exercise 4.20 in {cite:p}`Nielsen:2010`): \begin{gather*} \mat{C}^Z = \begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & -1 \end{pmatrix}, \qquad \I\ln(\mat{C}^Z) = \begin{pmatrix} 0\\ & 0\\ && 0\\ && & \pi \end{pmatrix}. \end{gather*} This lays the background for trying to find an implementation of a complete set of gates. Now let's look at a specific physical example. ## Physical Implementation with Spins :::{margin} Working with spins like this is very feasible for small numbers of qubits, but may not be very scalable. ::: Consider two qubits implemented as spin-½ particles. Assume these can be separated physically and subject to independent magnetic fields $\vect{B}_{A}$ and $\vect{B}_{B}$. This allows us to implement arbitrary single-qubit unitary operations. Now, suppose that these two particles can be brought together so that they interact via a symmetric dipole interaction \begin{gather*} \mat{V} = \mu_{D} \vect{\op{S}}_{A} \cdot \vect{\op{S}}_{B} = \mu_{D}\begin{pmatrix} 1\\ & -1 & 2\\ & 2 & -1\\ &&& 1 \end{pmatrix} \end{gather*} Can we use this interaction to build a ᴄɴᴏᴛ gate? A quick check shows that this matrix itself is not enough :::{admonition} Do it! Compute $\mat{U} = \exp(\mat{V}t/\I\hbar)$. :class: dropdown The block structure makes this easy: we really only need to worry about the central block which we can express \begin{gather*} \mat{A} = \begin{pmatrix} -1 & 2\\ 2 & -1 \end{pmatrix} = -\mat{1} + 2\mat{\sigma}_x,\\ e^{\theta\mat{A}/2\I} = e^{\I\theta/2} e^{-\theta\I\mat{\sigma}_x} = e^{\I\theta/2} \left( \mat{1}\cos\theta - \I\mat{\sigma}_x\sin\theta \right). \end{gather*} The key here is to note that $\mat{\sigma}_x$ commutes with the identity, and $(-\I\mat{\sigma}_x)^2 = -\mat{1}$ acts like a matrix version of $\I$, allowing us to use the generalization of [Euler's formula] for [quaternion]s. Putting the pieces together: \begin{gather*} e^{\mat{V} t/\I\hbar} = e^{-\I\theta/2}\begin{pmatrix} 1\\ & e^{\I\theta}\cos\theta & -\I e^{\I\theta}\sin\theta\\ & -\I e^{\I\theta}\sin\theta & e^{\I\theta}\cos\theta\\ &&& 1 \end{pmatrix}, \qquad \theta = \frac{2t\mu_D}{\hbar}. \end{gather*} This does not have the appropriate structure to implement ᴄɴᴏᴛ. ::: ```{code-cell} ipython3 :tags: [hide-cell] from scipy.linalg import expm rng = np.random.default_rng(seed=2) mu_D, t, hbar = rng.random(size=3) V = mu_D * np.array([ [1, 0, 0, 0], [0,-1, 2, 0], [0, 2,-1, 0], [0, 0, 0, 1]]) theta = 2*t*mu_D/hbar c, s = np.exp(1j*theta)*np.array([np.cos(theta), 1j*np.sin(theta)]) U = np.exp(theta/2j) * np.array([ [1, 0, 0, 0], [0, c,-s, 0], [0,-s, c, 0], [0, 0, 0, 1]]) assert np.allclose(U, expm(V*t/1j/hbar)) ``` However, it does provide the right ingredients. In particular, a key role of ᴄɴᴏᴛ is to generate entanglement by producing Bell states: ```{code-cell} ipython3 :tags: [hide-input] from qiskit.quantum_info import Statevector, Operator from qiskit import QuantumCircuit qc = QuantumCircuit(2) qc.reset([0, 1]) qc.h(1) qc.cnot(0,1) display(qc.draw('mpl')) psi = Statevector.from_label('00') display(psi.evolve(qc).draw('latex')) #display(array_to_latex(Operator(qc).data)) ``` The effect of the ᴄɴᴏᴛ operation is to produce maximal entanglement from the appropriate input state. Note also that all the other Bell states can be obtained by single qubit operations. Thus, we can try to use $\mat{V}$ to generate maximum entanglement. At first it might seem that we need to explore a rather large parameter space, but noting that $\vec{\mat{S}}_{A}\cdot\vec{\mat{S}}_{B}$ is rotationally invariant, we can choose a frame in which the control qubit is $\ket{0}_{A}$ points along the $z$ axis. We may use the remaining rotational freedom about the $z$ axis to choose a frame in which the other qubit points somewhere in the $x-z$ plane. Physically, the result can only depend on the angle $\theta$ between the two vectors on the Bloch spheres representing these input vectors. Next, we must define a measure for the entanglement, and the von Neumann entropy will function well: \begin{gather*} S(\mat{\rho}) = -\Tr \mat{\rho}_A\ln\mat{\rho_A}, \qquad \mat{\rho}_A = \Tr_{B}\mat{\rho}, \end{gather*} where $\mat{\rho}_A$ is the reduced density matrix. \begin{gather*} \dot{\mat{\rho}} = \frac{\left[\mat{V}, \mat{\rho}\right]}{\I\hbar},\\ \ket{\theta} = \mat{R}^{y}_{\theta}\ket{0} = \begin{pmatrix} \cos\tfrac{\theta}{2}\\ \sin\tfrac{\theta}{2} \end{pmatrix},\\ \mat{\rho} = \ket{0}\bra{0}\otimes\ket{\theta}\bra{\theta}. \end{gather*} ```{code-cell} ipython3 _EPS = np.finfo(float).eps V = np.array([ [1, 0, 0, 0], [0,-1, 2, 0], [0, 2,-2, 0], [0, 0, 0, 1]]) def S(rho): assert np.allclose(rho, rho.T.conj()) lams = np.linalg.eigvalsh(rho) + _EPS assert np.all(lams >= 0) return -np.sum(lams * np.log(lams)) def tr_A(rho): return np.einsum('abAb', rho.reshape((2,)*4)) def get_dS(theta, V=V): """Return the change in von Neumann entropy due to V.""" ket_0 = np.array([1,0]) ket_theta = np.array([np.cos(theta/2), np.sin(theta/2)]) rho = np.einsum( 'a,A,b,B->abAB', ket_0, ket_0.conj(), ket_theta, ket_theta.conj()).reshape((4, 4)) assert np.allclose(rho, rho.T.conj()) drho = (V @ rho - rho @ V)/1j h = 1e-6 rho_As = [tr_A(rho + h*drho), tr_A(rho - h*drho)) dS = (S(rho_As[0]) - S(rho_As[1]))/h return dS get_dS(0.2) ``` A bit more generally, we can subject each spin to a different magnetic field $\vec{B}_A$ and $\vec{B}_B$, and add an overall energy shift, in addition to $\mat{V}$, so a physical Hamiltonian might be: \begin{align*} \mat{H} &= E\mat{1}\otimes\mat{1} - \frac{\hbar\mu_B}{2}\left( \vec{B}_A\cdot \vec{\mat{\sigma}}\otimes \mat{1} + \vec{B}_B\cdot \mat{1}\otimes \vec{\mat{\sigma}} \right) + \mat{V}\\ &= E\mat{1} + \vec{a}\cdot \vec{\mat{\sigma}}\otimes \mat{1} + \vec{b}\cdot \mat{1}\otimes\vec{\mat{\sigma}} + \frac{c}{2}(\vec{\mat{\sigma}}\otimes \mat{1}) \cdot (\mat{1}\otimes\vec{\mat{\sigma}})\\ &=\begin{pmatrix} d_0 & w^* & w^* &\\ w & d_1 & c & w^*\\ w & c & d_2 & w^*\\ & w & w & d_3 \end{pmatrix}, \qquad c = \frac{d_0 - d_1 - d_2 + d_3}{2}. \end{align*} where $d_i$ are real, and $w$ is complex. :::{admonition} Do it! Show that $\mat{H}$ has this form. :class: dropdown \begin{gather*} \mat{H} = E\mat{1} + \begin{pmatrix} a_z & & a_x-\I a_y &\\ & a_z & & a_x-\I a_y\\ a_x+\I a_y & & -a_z &\\ & a_x+\I a_y & & -a_z \end{pmatrix} + \\ + \begin{pmatrix} b_z & a_x-\I a_y & &\\ a_x+\I a_y & -b_z & &\\ & & b_z & a_x-\I a_y\\ & & a_x+\I a_y & -b_z \end{pmatrix} + \\ + \frac{c}{2}\begin{pmatrix} 1\\ & -1 & 2\\ & 2 & -1\\ &&& 1 \end{pmatrix},\\ = \begin{pmatrix} E+a_z + b_z + \frac{c}{2} & a_x-\I a_y & a_x-\I a_y &\\ a_x+\I a_y & E+a_z - b_z - \frac{c}{2} & c & a_x-\I a_y\\ a_x+\I a_y & c & E-a_z + b_z - \frac{c}{2} & a_x-\I a_y\\ & a_x+\I a_y & a_x+\I a_y & E-a_z - b_z + \frac{c}{2} \end{pmatrix}. \end{gather*} Grouping some terms, we can reduce this to the form \begin{gather*} \mat{H} = \begin{pmatrix} d_0 & w^* & w^* &\\ w & d_1 & c & w^*\\ w & c & d_2 & w^*\\ & w & w & d_3 \end{pmatrix}\\ d_0 = E + a_z + b_z + \frac{c}{2}, \qquad d_1 = E + a_z - b_z - \frac{c}{2}, \\ d_2 = E - a_z + b_z - \frac{c}{2}, \qquad d_3 = E - a_z - b_z + \frac{c}{2}, \\ 2c = d_0 - d_1 - d_2 + d_3\\ w = a_x+\I a_y, \end{gather*} where $a_i$ and $b$ are real, and $w$ is complex. ::: This can be used to implement the controlled-$Z$ gate. :::{admonition} Do it! :class: dropdown Note: you will have to use one of the more general forms of $\I\ln\mat{C}^Z$ and match the structure. ::: :::{margin} After trying to find a solution with brute-force optimization, I think it cannot be done directly, but as described in {cite:p}`Mermin:2007`, one can implement a ᴄᴢ gate. ::: How can these ingredients be used to implement a ᴄɴᴏᴛ gate? ## Spin-Spin Interaction ```{code-cell} ipython3 rng = np.random.default_rng(seed=1) # Pauli matrices I = np.eye(2) sigma_x, sigma_y, sigma_z = sigmas = np.array([ [[0, 1], [1, 0]], [[0, -1j], [1j, 0]], [[1, 0], [0, -1]], ]) A, B = rng.normal(size=(2, 2, 4)).view(dtype=complex) assert np.allclose( np.kron(A, B), np.einsum('ab,cd->acbd', A, B).reshape((4, 4))) S_A = [np.kron(_s, I) for _s in sigmas] S_B = [np.kron(I, _s) for _s in sigmas] V = np.einsum('abc,acd->bd', S_A, S_B) display(V) display(np.kron(sigma_x, I) + 2*np.kron(sigma_y, I)+ 3*np.kron(sigma_z, I)) display(np.kron(I, sigma_x) + 2*np.kron(I, sigma_y)+ 3*np.kron(I, sigma_z)) ``` ```{code-cell} ipython3 from scipy.linalg import logm CNOT = np.array([ [1, 0, 0, 0], [0, 1, 0, 0], [0, 0, 0, 1], [0, 0, 1, 0] ]) CZ = np.array([ [1, 0, 0, 0], [0, 1, 0, 0], [0, 0, 1, 0], [0, 0, 0, -1] ]) display((logm(CNOT)/1j/(np.pi/2)).round(5)) display((logm(CZ)/1j/(np.pi)).round(5)) ``` ## Symmetric Interactions Consider a more generic interaction $\op{V}_{AB}$ with the only restriction that $\op{V}_{AB} = \op{V}_{BA}$. If we group the indices appropriately, this implies: \begin{gather*} V_{ab;a'b'} = V_{ba;b'a'},\\ V = \begin{pmatrix} V_{00;00} & V_{00;01} & V_{00;10} & V_{00;11}\\ V_{01;00} & V_{01;01} & V_{01;10} & V_{01;11}\\ V_{10;00} & V_{10;01} & V_{10;10} & V_{10;11}\\ V_{11;00} & V_{11;01} & V_{11;10} & V_{11;11} \end{pmatrix}\\ V = \begin{pmatrix} d_0 & b & b & c\\ d & d_1 & e & f\\ d & e & d_1 & f\\ h & g & g & d_2 \end{pmatrix} \end{gather*} [section 4.5.2]: [Euler's formula]: [Quaternion]: