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Hiesmayr} 80 \date{20. Juni, 2021} 81 82 \begin{document} 83 \maketitle 84 85 \noindent\rule[0.5ex]{\linewidth}{1pt} 86 \begin{abstract} 87 In this lab course we go through the QM problem of detecting entangled and 88 separable states. Even thought given a density matrix we cannot always know if 89 the state is separable or entangled. Thus a new concept is introduced, a 90 witness function build upon by the so called mutually unbiased bases short 91 MUBs. With help of the witness we can experimentally test to prove 92 entangled Bell states. 93 \end{abstract} 94 \noindent\rule[0.5ex]{\linewidth}{1pt} 95 96 \tableofcontents 97 98 \section{Background} 99 \subsection{Heisenberg's uncertainty relation-Robertson version} 100 Given an observable $\mathcal{A}$ we can define a hermitian operator $\hat{A}$ 101 , given a state $\psi$, we can define the expectation value $\langle \hat{A} \rangle _\psi 102 = \text{Tr}(\hat{A}\psi)$ and thus a standard derivation $(\Delta 103 \hat{A})^2_\psi = \langle \hat{A}^2 \rangle_\psi - \langle \hat{A} 104 \rangle_\psi^2$, where any such operator needs to satisfy. 105 \begin{align} 106 \langle \hat{A}^\dagger \hat{A} \rangle_\psi = \langle 107 \psi|\hat{A}^\dagger\hat{A} \rangle = \langle \hat{A}\psi | \hat{A}\psi 108 \rangle \geq 0. 109 \end{align} 110 Furthermore two arbitrary hermitian operators $\hat{A}$ and $\hat{B}$ hold the 111 following inequality 112 \begin{align} 113 (\Delta \hat{A})_\psi \cdot (\Delta \hat{B})_\psi \geq \frac{1}{2} |\langle 114 \hat{A}, \hat{B}\rangle_\psi| 115 \end{align} 116 for any state $\psi$. This uncertainty is called the Heisenberg's uncertainty 117 principle and forms a fundamental basis for quantum mechanics the 118 unpredictability of quantum mechanics. 119 120 \subsection{Entropic Uncertainty Relations-Quantum Information Theoretical 121 Formulation} 122 In quantum-information theory the entropic uncertainty is defined as the 123 following 124 \begin{align} 125 H(\hat{O}_n) + H(\hat{O}_m) \geq - \log_2\bigg(\max_{i,j}\{|\langle 126 \chi_n^i|\chi_m^j \rangle|^2\} \bigg) = log_2(|\frac{1}{\sqrt{2}}|^2) 127 \end{align} 128 where $H(\bar{0})_n$ is the binary entropy for a pure state $\psi$ 129 \begin{align} 130 H(\bar{O}_n) = -p(n)\log_2(p(n)) -(1-p(n))\log_2(1-p(n)) 131 \end{align} 132 and $p(n) = |\langle \chi_n | \psi \rangle|^2$ is the probability for the 133 outcome $n$ of $\hat{O}_n$ for $\psi$. 134 135 The entropic uncertainty relation can be extended for an arbitrary number of 136 outcomes, $d$, with the von-Neumann Entropy $S(\hat{O}_n)$ 137 \begin{align} 138 &S(\hat{O}_n) + S(\hat{O}_m) \geq - \log_d\bigg(\max_{i,j}\{|\langle 139 \chi_n^i|\chi_m^j \rangle|^2\} \bigg)\\ 140 \nonumber\\ 141 &S(\hat{O}_n) = -\sum^{d-1}_{i=0} p_n(i)\ln(p_n(i)) \;\;\;\ \text{or}\\ 142 &S(\hat{O}_n) = -\text{Tr}(\hat{O} \ln(\hat{O})). \label{eq:vn} 143 \end{align} 144 145 \subsection{Mutually Unbiased Bases (MUBs)} 146 A ONB of a $d$-dimensional Hilbert space is $B = \{|i \rangle\} = \{|0\rangle ,\dots, 147 |(d-1)\rangle \}$. In quantum information theory a set of orthonormal bases 148 $\{B_1, \dots, B_m\}$ (each an ONB of the $d$-dimensional Hilbertspace $H^d$) is called mutually 149 unbiased if 150 \begin{align} 151 |\langle i_k| j_{k'}\rangle|^2 = \delta_{k,k'}\delta_{i,j} 152 (1-\delta_{k,k'})\frac{1}{d} 153 \end{align} 154 155 Thus the maximum of the entropy uncertainty relation is 156 157 \begin{align} 158 S(\hat{O}_n) + S(\hat{O}_m) \geq - \log_d(\frac{1}{d}) 159 \end{align} 160 161 \subsection{Construction of MUBs \label{sec:mubs}} 162 In this section we will show how to construct mutually unbiased bases (MUBs) 163 using the Hadamard Matrix $\mathbb{H}$. In fact two orthonormal basies are 164 connected by the Hadamard Matrix (unitary) 165 \begin{align} 166 \mathbb{H}=\sum_{i,j} \frac{1}{\sqrt{d}} e^{i\phi_{ij}}|i\rangle\langle j|. 167 \end{align} 168 where $\phi_{i,j}$ is a phase chosen such that the $\mathbb{H}$ is unitary. A 169 simple choice $e^{i\phi_{i,j}} = \omega^{-ij} = e^{\frac{2\pi i}{d}}$ always 170 works. In this case the matrix is called the Fourier matrix 171 \begin{align} 172 \mathbb{H}=\sum_{i,j} \frac{1}{\sqrt{d}} \omega^{-ij}|i\rangle\langle j|. 173 \end{align} 174 Furthermore the Hadamard matrix is directly related to the generalized 175 Pauli-matrices. 176 \begin{align} 177 &\sigma_{\mathbb{Z}} = \sum_i \omega^i |i\rangle\langle i|\\ 178 &\sigma_{\mathbb{X}} = \mathbb{H}\sigma_{\mathbb{Z}} \mathbb{H} = \sum_i 179 |i+1\rangle\langle i|\\ 180 &\sigma_{\mathbb{X}}\sigma_{\mathbb{Z}} = i \sigma_{\mathbb{Y}}. 181 \end{align} 182 183 All this means, the problem of finding MUBs, essentially narrows down, to finding 184 these Hadamad matrices. 185 \newline 186 187 A second way of constructing MUBs is the so called Heisenberg-Weyl 188 construction. If $d$ is prime, the eigenvectors of the operators, form a MUB, 189 which looks like: 190 \begin{align} 191 (\sigma_\mathbb{Z},\sigma_\mathbb{X},\sigma_\mathbb{X}. 192 \sigma_\mathbb{Z},\sigma_\mathbb{X}.\sigma^2_\mathbb{Z},...,\sigma_\mathbb{X}. 193 \sigma^{d-1}_\mathbb{Z}) 194 \end{align} 195 196 \textbf{Examples:} 197 \newline 198 199 MUBs for qubits (d=2) 200 \begin{align} 201 &B_1 = \{|0_1\rangle, |1_1\rangle\} = \{|0\rangle, |1\rangle\}\\ 202 &B_2 = \{|0_2\rangle, |1_2\rangle\} = \frac{1}{\sqrt{2}} \{|0\rangle 203 +|1\rangle , |0\rangle - |1\rangle\}\\ 204 &B_2 = \{|0_3\rangle, |1_3\rangle\} = \frac{1}{\sqrt{2}} \{|0\rangle 205 +i|1\rangle , |0\rangle - i|1\rangle\}\\ 206 \end{align} 207 208 MUBs for qutrits (d=3) 209 \begin{align} 210 &B_1 = \{|0_1\rangle, |1_1\rangle, |2_1\rangle\} = 211 \bigg\{ 212 \begin{pmatrix}1 \\ 0\\0\end{pmatrix}, 213 \begin{pmatrix}0 \\ 1\\0\end{pmatrix}, 214 \begin{pmatrix}0 \\ 0\\1\end{pmatrix} 215 \bigg\} 216 \\ 217 &B_2 = \{|0_2\rangle, |1_2\rangle, |2_2\rangle\} = 218 \frac{1}{\sqrt{3}} 219 \bigg\{ 220 \begin{pmatrix}1 \\ 1\\1\end{pmatrix}, 221 \begin{pmatrix}1 \\ \omega\\\omega^2\end{pmatrix}, 222 \begin{pmatrix}1 \\ \omega^2\\\omega\end{pmatrix} 223 \bigg\} 224 \\ 225 &B_3 = \{|0_3\rangle, |1_3\rangle, |2_1\rangle\} = 226 \frac{1}{\sqrt{3}} 227 \bigg\{ 228 \begin{pmatrix}1 \\\omega\\\omega\end{pmatrix}, 229 \begin{pmatrix}1 \\ \omega^2\\1\end{pmatrix}, 230 \begin{pmatrix}1 \\ 1\\\omega^2\end{pmatrix} 231 \bigg\} \\ 232 &B_4 = \{|0_4\rangle, |1_1\rangle, |2_1\rangle\} = 233 \frac{1}{\sqrt{3}} 234 \bigg\{ 235 \begin{pmatrix}1 \\\omega^2\\\omega^2\end{pmatrix}, 236 \begin{pmatrix}1 \\ \omega\\1\end{pmatrix}, 237 \begin{pmatrix}1 \\ 1\\\omega\end{pmatrix} 238 \bigg\} 239 \end{align} 240 241 With these bases we can define an bell state seed $\Omega_{0,0}$ with $P_{0,0} 242 = |\Omega_{0,0}\rangle\langle \Omega_{0,0}|$, 243 \begin{align}\label{eq:arb} 244 |\Omega_{0,0}\rangle = \frac{1}{\sqrt{d}} \sum_{s=0}^{d-1} |ss\rangle 245 \end{align} 246 extending this with the Wely operators $W_{kl}$ we can arrive at an arbitrary 247 bell state $P_{i,j}$ 248 \begin{align} 249 &|\Omega_{k,l}\rangle = W_{kl} \otimes \mathbbm{1}|\Omega_{0,0}\rangle\\ 250 \nonumber\text{where:}\\ 251 &W_{kl} = \sum_{j=0}^{d-1} \omega^{j\cdot k} |j\rangle \langle j+l| 252 \end{align} 253 where $\omega = e^{\frac{2\pi i}{d}}$ and $\sum_{j=0}^{d-1} \omega^j = 0$. 254 255 \subsection{Detecting Entanglement via MUBs} 256 One of the most important aspects of quantum theory, is the prediction of 257 entanglement, and furthermore finding ways to construct experiments, that, with 258 minimal effort allow the creation of so called entanglement witnesses for 259 entanglement detection. Because, the bigger a system gets, the more 260 measurements are needed, which for huge systems is often straight up impossible 261 to realize. So, essentially, quantum theory tries to witness entanglement with 262 as few measurements as possible, and without resorting to full state 263 tomography. 264 \begin{table}[h!] 265 \centering 266 \begin{tabular}{||c|c|c || c|c||} 267 \hline 268 & \multicolumn{2}{|c||}{Lower Bounds} &\multicolumn{2}{|c||}{Upper Bounds}\\ 269 \hline 270 m & $L_{m,2}^{MUB}$ &$L_{m,3}^{MUB}$&$U_{m,2}^{MUB}$ &$U_{m,3}^{MUB}$ \\ 271 \hline 272 2 & 1/2 &0.211 &3/2 & 4/3\\ 273 \hline 274 3 & 1 &1/2 &2 & 5/3\\ 275 \hline 276 4 & &1 & & 2\\ 277 \hline 278 \end{tabular} 279 \caption{Lower $L$ and upper $U$ bounds for the MUB witness for $d = 2, 3$ 280 and $m=1, \dots, d+1$ \label{tab:1}} 281 \end{table} 282 283 \newpage 284 \section{Exercises} 285 286 \begin{MyExercise} 287 \textbf{Compute the Heisenberg uncertainty relation for $\hat{A} = 288 \hat{\sigma}_{1}$ and $\hat{B} = \hat{\sigma}_{2}$ (Pauli matrices) for an 289 arbitrary pure state $|\psi \rangle = \cos\frac{\theta}{2} |\Uparrow\rangle 290 + \sin\frac{\theta}{2} e^{i\phi} |\Downarrow\rangle$. Furthermore compute 291 the quantum-information theoretical version of the inequality for 292 $\hat{O}_{n,m} = \hat{\sigma}_{1, 2}$}. 293 \newline 294 295 To start of, the Pauli matrices are 296 \begin{align} 297 \sigma_1 = 298 \begin{pmatrix} 299 0 & 1\\ 1& 0 300 \end{pmatrix} \;\;\;\;\; 301 \sigma_2 = 302 \begin{pmatrix} 303 0 & -i \\ i & 0 304 \end{pmatrix} \;\;\;\;\; 305 \sigma_3 = 306 \begin{pmatrix} 307 1&0\\ 0& -1 308 \end{pmatrix} \;\;\;\;\; 309 \end{align} 310 Now we have a straight forward calculation 311 \begin{align} 312 &\langle \sigma_1\rangle^2_\psi = \sin^2 \theta \cos^2 \phi\\ 313 &\langle (\sigma_1)^2\rangle_\psi = 1\\ 314 \nonumber \\ 315 &\langle \sigma_2\rangle_\psi^2 = \sin^2\theta \sin^2\phi\\ 316 &\langle (\sigma_2)^2\rangle_\psi = 1\\ 317 \nonumber \\ 318 &\frac{1}{2} |\langle[\sigma_1, \sigma_2]\rangle = cos\theta 319 \end{align} 320 after some basic algebra with trigonometric functions we arrive at the 321 following inequality 322 \begin{align} 323 \sin^4\theta \sin^2(2\phi) \geq 0 324 \end{align} 325 which holds true for all $\theta, \phi$. 326 327 For the quantum-theoretical version of the inequality we use Equation 328 \ref{eq:vn} to calculate the von Neumann entropy. The maximum of the right 329 hand side is $\frac{1}{2}$ 330 \begin{align} 331 S(\sigma_1) = -\text{Tr}(\sigma_1\ln(\sigma_1)) = 0\\ 332 S(\sigma_2) = -\text{Tr}(\sigma_2\ln(\sigma_2)) = \pi 333 \end{align} 334 thus the inequality is 335 \begin{align} 336 \pi \geq 1 337 \end{align} 338 339 Since the Heisenberg's uncertainty principle is mathematically correct, 340 because it holds true for all hermitian operators, a violation of 341 the principle would put the basis of functional analysis and/or 342 the axioms of quantum mechanics at question. 343 344 The quantum information theoretical approach to the uncertainty principal 345 is convenient since the right hand side does not depend on any particular 346 state. 347 \end{MyExercise} 348 349 \begin{MyExercise}\label{ex:2} 350 \textbf{Compute 351 \begin{align} 352 &I_m^{MUB} = \sum_{k=1}^m\sum_{i=0}^{d-1} 353 \text{Tr}((|i_k\rangle\langle i_k| \otimes |i_k\rangle \langle i_k|) 354 \varrho) \;\;\;\;\;\; \text{and}\\ 355 &I_m^{MUB} = \sum_{k=1}^m\sum_{i=0}^{d-1} 356 \text{Tr}((|i_k\rangle\langle i_k| \otimes (|i_k\rangle \langle 357 i_k|)^*) 358 \varrho) 359 \end{align} 360 for two qubits ($d=2$), for $m=1, 2, 3$ and $|\psi\rangle = 361 cos\alpha|00\rangle + sin\alpha |11>$. Here $|i_k\rangle$ is the eigenvector 362 of the Pauli matrix $\sigma_k$.} 363 364 The strategy to calculate the witness is to use the computer to loop over 365 $d$ and $m$ for $m = 1, \dots, d+1$ then we compare the results with table 366 \ref{tab:1}. Note that $|i_k\rangle\langle i_k|$ is a $d$-dimensional 367 matrix, the density matrix is a $d^2$-dimensional matrix and 368 thus the matrix inside the trace is $d^2$. 369 370 We start of with $I_m^{MUB}$ without conjugation 371 \begin{align} 372 &I^{MUB}_{m=1} = \cos^2\alpha\\ 373 &I^{MUB}_{m=2} = \frac{1}{4}(-\sin(2\alpha) + 2\cos(2\alpha) + 3)\\ 374 &I^{MUB}_{m=3} = \cos^2(\alpha) + \frac{1}{2} 375 \end{align} 376 For $m=2$ entangled states for lower bound $\alpha = \frac{\pi}{4}$. For 377 $m = 3$ entangled states for lower bound $\alpha = \frac{3\pi}{4}$. 378 379 with conjugation we get 380 \begin{align} 381 &I^{MUB}_{m=1} = \cos^2\alpha\\ 382 &I^{MUB}_{m=2} = \frac{1}{4}(\sin(2\alpha) + 2\cos(2\alpha) + 3)\\ 383 &I^{MUB}_{m=3} = \frac{1}{\sqrt{2}} \sin(2\alpha + \frac{\pi}{4}) +1 384 \end{align} 385 For $m=2$ entangled states for lower bound $\alpha = \frac{\pi}{4}$. For 386 $m = 3$ entangled states for lower bound $\alpha = -\frac{\pi}{8}$. 387 \end{MyExercise} 388 389 \begin{MyExercise} 390 \textbf{Compute the same as in exercise \ref{ex:2} for 391 the isotropic state 392 \begin{align} 393 \varrho^{iso}_d (p) = (1-p)\cdot\frac{1}{d^2}\mathbbm{1}_{d^2} + p 394 P_{i,j} 395 \end{align} 396 for a freely chosen bell state $P_{i,j}$, and for both $d=2$ qubits and for 397 $d=3$ qutrits. For $p\in [-\frac{1}{d^2-1}, 1]$ 398 we have the positivity condition and for $p\in [-\frac{1}{d^2 -1}, 399 \frac{1}{d+1}]$ we have a separable state else entangled. 400 } 401 402 We choose $P_{i,j} = P_{0,0} = |\Omega_{0,0}\rangle \langle \Omega_{0,0}|$. 403 To calculate $\Omega$ we use the equation \ref{eq:arb} and use the MUBs given 404 in section \ref{sec:mubs}. 405 406 For $d=2$ we have the following for the standard $I^{MUB}$ 407 \begin{align} 408 &I^{MUB}_{m=1} = \frac{1}{4}(3p+1) \\ 409 &I^{MUB}_{m=2} = \frac{1}{2}(3p+1) \\ 410 &I^{MUB}_{m=3} = \frac{1}{4}(5p+3) \\ 411 \end{align} 412 For $m=2$ we have entanglement on the upper bound for $p = \frac{2}{3}$. 413 For $m=3$ we have entanglement on the upper bound for $p = 1$. 414 415 416 with conjugation we get 417 \begin{align} 418 &I^{MUB}_{m=1} = \frac{1}{4}(3p+1) \\ 419 &I^{MUB}_{m=2} = \frac{1}{2}(3p+1) \\ 420 &I^{MUB}_{m=3} = \frac{1}{4}(9p+3) \\ 421 \end{align} 422 For $m=2$ we have entanglement on the upper bound for $p = \frac{2}{3}$. 423 For $m=3$ we have entanglement on the upper bound for $p = \frac{4}{9}$. 424 425 For $d=3$ we have the following 426 \begin{align} 427 &I^{MUB}_{m=1} = \frac{1}{9}(16p+2) \\ 428 &I^{MUB}_{m=2} = \frac{1}{9}(23p+4) \\ 429 &I^{MUB}_{m=3} = 3p + \frac{1}{3}\\ 430 &I^{MUB}_{m=4} = \frac{1}{9}(31p + 8) 431 \end{align} 432 For $m=2$ we have entanglement on the upper bound for $p = \frac{8}{23}$. 433 For $m=3$ we have entanglement on the upper bound for $p = \frac{4}{9}$. 434 435 with conjugation we get 436 \begin{align} 437 &I^{MUB}_{m=1} = \frac{1}{9}(16p+2) \\ 438 &I^{MUB}_{m=2} = \frac{1}{9}(32p+4) \\ 439 &I^{MUB}_{m=3} = \frac{16}{3}p + \frac{2}{3}\\ 440 &I^{MUB}_{m=4} = \frac{1}{9}(64p + 8) 441 \end{align} 442 For $m=3$ we have entanglement on the upper bound for $p = \frac{5}{3}$. 443 \end{MyExercise} 444 \newpage 445 \begin{MyExercise} 446 \textbf{ 447 Compute $I_m^{MUB}$ with conjugation and without for the Werner states 448 for $d=2, 3$ and $m=1,\dots, d+1$ 449 \begin{align} 450 \varrho_W(q) = q \frac{P_{sym}}{d(d+1)} + (1-q) \frac{P_{asym}}{d(d-1)} 451 \end{align} 452 where $P_{sym} = (\mathbbm{1} + \mathbb{P})$ and $P_{asym} = 453 (\mathbbm{1} - \mathbb{P})$ for $\mathbb{P} = \sum_{ij} 454 |ji\rangle\langle ij|$. The state is separable for $q\in [0,\frac{1}{2}]$ and 455 entangled for $q\in [\frac{1}{2}, 1]$ 456 } 457 458 First we calculate for $d=2$ we choose the basis $B_1$ to calculate the projection 459 operator. And note that $|ij\rangle = |i\rangle \otimes |j\rangle$ we need 460 the tensor product here. 461 462 Straightforward computation gives 463 \begin{align} 464 &I^{MUB}_{m=1} = \frac{q}{3}\\ 465 &I^{MUB}_{m=2} = \frac{2q}{3}\\ 466 &I^{MUB}_{m=3} = q 467 \end{align} 468 For $m=3$ we have entanglement on the lower bound for $p = 1/2$. 469 470 with conjugation 471 \begin{align} 472 &I^{MUB}_{m=1} = \frac{q}{3}\\ 473 &I^{MUB}_{m=2} = \frac{2q}{3}\\ 474 &I^{MUB}_{m=3} = \frac{q}{3} + \frac{1}{2} 475 \end{align} 476 For $m=3$ we have entanglement on the lower bound for $p = 0$. 477 478 For $d=3$ we choose the basis $B_1$ to calculate the projection operator 479 and straightforward computation gives 480 \begin{align} 481 &I^{MUB}_{m=1} = \frac{q}{3}\\ 482 &I^{MUB}_{m=2} = \frac{2q}{3}\\ 483 &I^{MUB}_{m=3} = q \\ 484 &I^{MUB}_{m=3} = \frac{4q}{3} 485 \end{align} 486 For $m=2$ we have entanglement on the lower bound for $p = 0.3165$. 487 For $m=3$ we have entanglement on the lower bound for $p = \frac{1}{2}$. 488 489 with conjugation 490 \begin{align} 491 &I^{MUB}_{m=1} = \frac{q}{3}\\ 492 &I^{MUB}_{m=2} = \frac{1}{12}(5q + 2)\\ 493 &I^{MUB}_{m=3} = \frac{1}{36}(15q + 14)\\ 494 &I^{MUB}_{m=4} = \frac{1}{36}(15q + 22)\\ 495 \end{align} 496 For $m=2$ we have entanglement on the lower bound for $p = 0.1064$. 497 For $m=3$ we have entanglement on the lower bound for $p = \frac{4}{15}$. 498 499 A simple comparison with exercise 3, we arrive at the conclusion that for 500 the Werner states we detect entanglement only on the lower bound and for the 501 isotropic states we detect entanglement only on the upper bound. 502 \end{MyExercise} 503 504 505 506 \nocite{cite1} 507 \nocite{cite2} 508 \nocite{cite3} 509 \nocite{cite4} 510 \nocite{cite5} 511 \printbibliography 512 513 514 515 \end{document}