commit 33e2e1e8523d81df8ac6fbe798cd0aaa313210f4
parent 335c29161c326aa9012f8cc2c6e44254a1708aad
Author: miksa <milutin@popovic.xyz>
Date: Sun, 20 Jun 2021 15:15:32 +0200
done sehs5
Diffstat:
7 files changed, 921 insertions(+), 0 deletions(-)
diff --git a/sesh4/main.tex b/sesh4/src/main.tex
diff --git a/sesh4/uni.bib b/sesh4/src/uni.bib
diff --git a/sesh5/src/main.bbl b/sesh5/src/main.bbl
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diff --git a/sesh5/src/main.pdf b/sesh5/src/main.pdf
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+
+
+\markright{Popovic, Vogel\hfill Detection of Quantum Entanglement with MUBs \hfill}
+
+\title{Universität Wien\\ Fakultät für Physik\\
+\vspace{1.25cm}Labcours Theoretical Physik 2021S \\ Detection of Quantum
+Entanglement with MUBs
+}
+\author{Milutin Popovic \& Tim Vogel \vspace{1cm}\\ Betreuerin: Beatrix C. Hiesmayr}
+\date{20. Juni, 2021}
+
+\begin{document}
+\maketitle
+
+\noindent\rule[0.5ex]{\linewidth}{1pt}
+\begin{abstract}
+ In this lab course we go through the QM problem of detecting entangled and
+ separable states. Even thought given a density matrix we cannot always know if
+ the state is separable or entangled. Thus a new concept is introduced, a
+ witness function build upon by the so called mutually unbiased bases short
+ MUBs. With help of the witness we can experimentally test to prove
+ entangled Bell states.
+\end{abstract}
+\noindent\rule[0.5ex]{\linewidth}{1pt}
+
+\tableofcontents
+
+\section{Background}
+\subsection{Heisenberg's uncertainty relation-Robertson version}
+Given an observable $\mathcal{A}$ we can define a hermitian operator $\hat{A}$
+, given a state $\psi$, we can define the expectation value $\langle \hat{A} \rangle _\psi
+= \text{Tr}(\hat{A}\psi)$ and thus a standard derivation $(\Delta
+\hat{A})^2_\psi = \langle \hat{A}^2 \rangle_\psi - \langle \hat{A}
+\rangle_\psi^2$, where any such operator needs to satisfy.
+\begin{align}
+ \langle \hat{A}^\dagger \hat{A} \rangle_\psi = \langle
+ \psi|\hat{A}^\dagger\hat{A} \rangle = \langle \hat{A}\psi | \hat{A}\psi
+ \rangle \geq 0.
+\end{align}
+Furthermore two arbitrary hermitian operators $\hat{A}$ and $\hat{B}$ hold the
+following inequality
+\begin{align}
+ (\Delta \hat{A})_\psi \cdot (\Delta \hat{B})_\psi \geq \frac{1}{2} |\langle
+ \hat{A}, \hat{B}\rangle_\psi|
+\end{align}
+for any state $\psi$. This uncertainty is called the Heisenberg's uncertainty
+principle and forms a fundamental basis for quantum mechanics the
+unpredictability of quantum mechanics.
+
+\subsection{Entropic Uncertainty Relations-Quantum Information Theoretical
+Formulation}
+In quantum-information theory the entropic uncertainty is defined as the
+following
+\begin{align}
+ H(\hat{O}_n) + H(\hat{O}_m) \geq - \log_2\bigg(\max_{i,j}\{|\langle
+ \chi_n^i|\chi_m^j \rangle|^2\} \bigg) = log_2(|\frac{1}{\sqrt{2}}|^2)
+\end{align}
+where $H(\bar{0})_n$ is the binary entropy for a pure state $\psi$
+\begin{align}
+ H(\bar{O}_n) = -p(n)\log_2(p(n)) -(1-p(n))\log_2(1-p(n))
+\end{align}
+and $p(n) = |\langle \chi_n | \psi \rangle|^2$ is the probability for the
+outcome $n$ of $\hat{O}_n$ for $\psi$.
+
+The entropic uncertainty relation can be extended for an arbitrary number of
+outcomes, $d$, with the von-Neumann Entropy $S(\hat{O}_n)$
+\begin{align}
+ &S(\hat{O}_n) + S(\hat{O}_m) \geq - \log_d\bigg(\max_{i,j}\{|\langle
+ \chi_n^i|\chi_m^j \rangle|^2\} \bigg)\\
+ \nonumber\\
+ &S(\hat{O}_n) = -\sum^{d-1}_{i=0} p_n(i)\ln(p_n(i)) \;\;\;\ \text{or}\\
+ &S(\hat{O}_n) = -\text{Tr}(\hat{O} \ln(\hat{O})). \label{eq:vn}
+\end{align}
+
+\subsection{Mutually Unbiased Bases (MUBs)}
+A ONB of a $d$-dimensional Hilbert space is $B = \{|i \rangle\} = \{|0\rangle ,\dots,
+|(d-1)\rangle \}$. In quantum information theory a set of orthonormal bases
+$\{B_1, \dots, B_m\}$ (each an ONB of the $d$-dimensional Hilbertspace $H^d$) is called mutually
+unbiased if
+\begin{align}
+ |\langle i_k| j_{k'}\rangle|^2 = \delta_{k,k'}\delta_{i,j}
+ (1-\delta_{k,k'})\frac{1}{d}
+\end{align}
+
+Thus the maximum of the entropy uncertainty relation is
+
+\begin{align}
+ S(\hat{O}_n) + S(\hat{O}_m) \geq - \log_d(\frac{1}{d})
+\end{align}
+
+\subsection{Construction of MUBs \label{sec:mubs}}
+In this section we will show how to construct mutually unbiased bases (MUBs)
+using the Hadamard Matrix $\mathbb{H}$. In fact two orthonormal basies are
+connected by the Hadamard Matrix (unitary)
+\begin{align}
+ \mathbb{H}=\sum_{i,j} \frac{1}{\sqrt{d}} e^{i\phi_{ij}}|i\rangle\langle j|.
+\end{align}
+where $\phi_{i,j}$ is a phase chosen such that the $\mathbb{H}$ is unitary. A
+simple choice $e^{i\phi_{i,j}} = \omega^{-ij} = e^{\frac{2\pi i}{d}}$ always
+works. In this case the matrix is called the Fourier matrix
+\begin{align}
+ \mathbb{H}=\sum_{i,j} \frac{1}{\sqrt{d}} \omega^{-ij}|i\rangle\langle j|.
+\end{align}
+Furthermore the Hadamard matrix is directly related to the generalized
+Pauli-matrices.
+\begin{align}
+ &\sigma_{\mathbb{Z}} = \sum_i \omega^i |i\rangle\langle i|\\
+ &\sigma_{\mathbb{X}} = \mathbb{H}\sigma_{\mathbb{Z}} \mathbb{H} = \sum_i
+ |i+1\rangle\langle i|\\
+ &\sigma_{\mathbb{X}}\sigma_{\mathbb{Z}} = i \sigma_{\mathbb{Y}}.
+\end{align}
+
+All this means, the problem of finding MUBs, essentially narrows down, to finding
+these Hadamad matrices.
+\newline
+
+A second way of constructing MUBs is the so called Heisenberg-Weyl
+construction. If $d$ is prime, the eigenvectors of the operators, form a MUB,
+which looks like:
+\begin{align}
+ (\sigma_\mathbb{Z},\sigma_\mathbb{X},\sigma_\mathbb{X}.
+ \sigma_\mathbb{Z},\sigma_\mathbb{X}.\sigma^2_\mathbb{Z},...,\sigma_\mathbb{X}.
+ \sigma^{d-1}_\mathbb{Z})
+\end{align}
+
+\textbf{Examples:}
+\newline
+
+MUBs for qubits (d=2)
+\begin{align}
+ &B_1 = \{|0_1\rangle, |1_1\rangle\} = \{|0\rangle, |1\rangle\}\\
+ &B_2 = \{|0_2\rangle, |1_2\rangle\} = \frac{1}{\sqrt{2}} \{|0\rangle
+ +|1\rangle , |0\rangle - |1\rangle\}\\
+ &B_2 = \{|0_3\rangle, |1_3\rangle\} = \frac{1}{\sqrt{2}} \{|0\rangle
+ +i|1\rangle , |0\rangle - i|1\rangle\}\\
+\end{align}
+
+MUBs for qutrits (d=3)
+\begin{align}
+ &B_1 = \{|0_1\rangle, |1_1\rangle, |2_1\rangle\} =
+ \bigg\{
+ \begin{pmatrix}1 \\ 0\\0\end{pmatrix},
+ \begin{pmatrix}0 \\ 1\\0\end{pmatrix},
+ \begin{pmatrix}0 \\ 0\\1\end{pmatrix}
+ \bigg\}
+ \\
+ &B_2 = \{|0_2\rangle, |1_2\rangle, |2_2\rangle\} =
+ \frac{1}{\sqrt{3}}
+ \bigg\{
+ \begin{pmatrix}1 \\ 1\\1\end{pmatrix},
+ \begin{pmatrix}1 \\ \omega\\\omega^2\end{pmatrix},
+ \begin{pmatrix}1 \\ \omega^2\\\omega\end{pmatrix}
+ \bigg\}
+ \\
+ &B_3 = \{|0_3\rangle, |1_3\rangle, |2_1\rangle\} =
+ \frac{1}{\sqrt{3}}
+ \bigg\{
+ \begin{pmatrix}1 \\\omega\\\omega\end{pmatrix},
+ \begin{pmatrix}1 \\ \omega^2\\1\end{pmatrix},
+ \begin{pmatrix}1 \\ 1\\\omega^2\end{pmatrix}
+ \bigg\} \\
+ &B_4 = \{|0_4\rangle, |1_1\rangle, |2_1\rangle\} =
+ \frac{1}{\sqrt{3}}
+ \bigg\{
+ \begin{pmatrix}1 \\\omega^2\\\omega^2\end{pmatrix},
+ \begin{pmatrix}1 \\ \omega\\1\end{pmatrix},
+ \begin{pmatrix}1 \\ 1\\\omega\end{pmatrix}
+ \bigg\}
+\end{align}
+
+With these bases we can define an bell state seed $\Omega_{0,0}$ with $P_{0,0}
+= |\Omega_{0,0}\rangle\langle \Omega_{0,0}|$,
+\begin{align}\label{eq:arb}
+ |\Omega_{0,0}\rangle = \frac{1}{\sqrt{d}} \sum_{s=0}^{d-1} |ss\rangle
+\end{align}
+extending this with the Wely operators $W_{kl}$ we can arrive at an arbitrary
+bell state $P_{i,j}$
+\begin{align}
+ &|\Omega_{k,l}\rangle = W_{kl} \otimes \mathbbm{1}|\Omega_{0,0}\rangle\\
+ \nonumber\text{where:}\\
+ &W_{kl} = \sum_{j=0}^{d-1} \omega^{j\cdot k} |j\rangle \langle j+l|
+\end{align}
+where $\omega = e^{\frac{2\pi i}{d}}$ and $\sum_{j=0}^{d-1} \omega^j = 0$.
+
+\subsection{Detecting Entanglement via MUBs}
+One of the most important aspects of quantum theory, is the prediction of
+entanglement, and furthermore finding ways to construct experiments, that, with
+minimal effort allow the creation of so called entanglement witnesses for
+entanglement detection. Because, the bigger a system gets, the more
+measurements are needed, which for huge systems is often straight up impossible
+to realize. So, essentially, quantum theory tries to witness entanglement with
+as few measurements as possible, and without resorting to full state
+tomography.
+\begin{table}[h!]
+ \centering
+\begin{tabular}{||c|c|c || c|c||}
+\hline
+ & \multicolumn{2}{|c||}{Lower Bounds} &\multicolumn{2}{|c||}{Upper Bounds}\\
+\hline
+ m & $L_{m,2}^{MUB}$ &$L_{m,3}^{MUB}$&$U_{m,2}^{MUB}$ &$U_{m,3}^{MUB}$ \\
+\hline
+ 2 & 1/2 &0.211 &3/2 & 4/3\\
+\hline
+ 3 & 1 &1/2 &2 & 5/3\\
+\hline
+ 4 & &1 & & 2\\
+\hline
+\end{tabular}
+ \caption{Lower $L$ and upper $U$ bounds for the MUB witness for $d = 2, 3$
+ and $m=1, \dots, d+1$ \label{tab:1}}
+\end{table}
+
+\newpage
+\section{Exercises}
+
+\begin{MyExercise}
+ \textbf{Compute the Heisenberg uncertainty relation for $\hat{A} =
+ \hat{\sigma}_{1}$ and $\hat{B} = \hat{\sigma}_{2}$ (Pauli matrices) for an
+ arbitrary pure state $|\psi \rangle = \cos\frac{\theta}{2} |\Uparrow\rangle
+ + \sin\frac{\theta}{2} e^{i\phi} |\Downarrow\rangle$. Furthermore compute
+ the quantum-information theoretical version of the inequality for
+ $\hat{O}_{n,m} = \hat{\sigma}_{1, 2}$}.
+ \newline
+
+ To start of, the Pauli matrices are
+ \begin{align}
+ \sigma_1 =
+ \begin{pmatrix}
+ 0 & 1\\ 1& 0
+ \end{pmatrix} \;\;\;\;\;
+ \sigma_2 =
+ \begin{pmatrix}
+ 0 & -i \\ i & 0
+ \end{pmatrix} \;\;\;\;\;
+ \sigma_3 =
+ \begin{pmatrix}
+ 1&0\\ 0& -1
+ \end{pmatrix} \;\;\;\;\;
+ \end{align}
+ Now we have a straight forward calculation
+ \begin{align}
+ &\langle \sigma_1\rangle^2_\psi = \sin^2 \theta \cos^2 \phi\\
+ &\langle (\sigma_1)^2\rangle_\psi = 1\\
+ \nonumber \\
+ &\langle \sigma_2\rangle_\psi^2 = \sin^2\theta \sin^2\phi\\
+ &\langle (\sigma_2)^2\rangle_\psi = 1\\
+ \nonumber \\
+ &\frac{1}{2} |\langle[\sigma_1, \sigma_2]\rangle = cos\theta
+ \end{align}
+ after some basic algebra with trigonometric functions we arrive at the
+ following inequality
+ \begin{align}
+ \sin^4\theta \sin^2(2\phi) \geq 0
+ \end{align}
+ which holds true for all $\theta, \phi$.
+
+ For the quantum-theoretical version of the inequality we use Equation
+ \ref{eq:vn} to calculate the von Neumann entropy. The maximum of the right
+ hand side is $\frac{1}{2}$
+ \begin{align}
+ S(\sigma_1) = -\text{Tr}(\sigma_1\ln(\sigma_1)) = 0\\
+ S(\sigma_2) = -\text{Tr}(\sigma_2\ln(\sigma_2)) = \pi
+ \end{align}
+ thus the inequality is
+ \begin{align}
+ \pi \geq 1
+ \end{align}
+
+ Since the Heisenberg's uncertainty principle is mathematically correct,
+ because it holds true for all hermitian operators, a violation of
+ the principle would put the basis of functional analysis and/or
+ the axioms of quantum mechanics at question.
+
+ The quantum information theoretical approach to the uncertainty principal
+ is convenient since the right hand side does not depend on any particular
+ state.
+\end{MyExercise}
+
+\begin{MyExercise}\label{ex:2}
+ \textbf{Compute
+ \begin{align}
+ &I_m^{MUB} = \sum_{k=1}^m\sum_{i=0}^{d-1}
+ \text{Tr}((|i_k\rangle\langle i_k| \otimes |i_k\rangle \langle i_k|)
+ \varrho) \;\;\;\;\;\; \text{and}\\
+ &I_m^{MUB} = \sum_{k=1}^m\sum_{i=0}^{d-1}
+ \text{Tr}((|i_k\rangle\langle i_k| \otimes (|i_k\rangle \langle
+ i_k|)^*)
+ \varrho)
+ \end{align}
+ for two qubits ($d=2$), for $m=1, 2, 3$ and $|\psi\rangle =
+ cos\alpha|00\rangle + sin\alpha |11>$. Here $|i_k\rangle$ is the eigenvector
+ of the Pauli matrix $\sigma_k$.}
+
+ The strategy to calculate the witness is to use the computer to loop over
+ $d$ and $m$ for $m = 1, \dots, d+1$ then we compare the results with table
+ \ref{tab:1}. Note that $|i_k\rangle\langle i_k|$ is a $d$-dimensional
+ matrix, the density matrix is a $d^2$-dimensional matrix and
+ thus the matrix inside the trace is $d^2$.
+
+ We start of with $I_m^{MUB}$ without conjugation
+ \begin{align}
+ &I^{MUB}_{m=1} = \cos^2\alpha\\
+ &I^{MUB}_{m=2} = \frac{1}{4}(-\sin(2\alpha) + 2\cos(2\alpha) + 3)\\
+ &I^{MUB}_{m=3} = \cos^2(\alpha) + \frac{1}{2}
+ \end{align}
+ For $m=2$ entangled states for lower bound $\alpha = \frac{\pi}{4}$. For
+ $m = 3$ entangled states for lower bound $\alpha = \frac{3\pi}{4}$.
+
+ with conjugation we get
+ \begin{align}
+ &I^{MUB}_{m=1} = \cos^2\alpha\\
+ &I^{MUB}_{m=2} = \frac{1}{4}(\sin(2\alpha) + 2\cos(2\alpha) + 3)\\
+ &I^{MUB}_{m=3} = \frac{1}{\sqrt{2}} \sin(2\alpha + \frac{\pi}{4}) +1
+ \end{align}
+ For $m=2$ entangled states for lower bound $\alpha = \frac{\pi}{4}$. For
+ $m = 3$ entangled states for lower bound $\alpha = -\frac{\pi}{8}$.
+\end{MyExercise}
+
+\begin{MyExercise}
+ \textbf{Compute the same as in exercise \ref{ex:2} for
+ the isotropic state
+ \begin{align}
+ \varrho^{iso}_d (p) = (1-p)\cdot\frac{1}{d^2}\mathbbm{1}_{d^2} + p
+ P_{i,j}
+ \end{align}
+ for a freely chosen bell state $P_{i,j}$, and for both $d=2$ qubits and for
+ $d=3$ qutrits. For $p\in [-\frac{1}{d^2-1}, 1]$
+ we have the positivity condition and for $p\in [-\frac{1}{d^2 -1},
+ \frac{1}{d+1}]$ we have a separable state else entangled.
+ }
+
+ We choose $P_{i,j} = P_{0,0} = |\Omega_{0,0}\rangle \langle \Omega_{0,0}|$.
+ To calculate $\Omega$ we use the equation \ref{eq:arb} and use the MUBs given
+ in section \ref{sec:mubs}.
+
+ For $d=2$ we have the following for the standard $I^{MUB}$
+ \begin{align}
+ &I^{MUB}_{m=1} = \frac{1}{4}(3p+1) \\
+ &I^{MUB}_{m=2} = \frac{1}{2}(3p+1) \\
+ &I^{MUB}_{m=3} = \frac{1}{4}(5p+3) \\
+ \end{align}
+ For $m=2$ we have entanglement on the upper bound for $p = \frac{2}{3}$.
+ For $m=3$ we have entanglement on the upper bound for $p = 1$.
+
+
+ with conjugation we get
+ \begin{align}
+ &I^{MUB}_{m=1} = \frac{1}{4}(3p+1) \\
+ &I^{MUB}_{m=2} = \frac{1}{2}(3p+1) \\
+ &I^{MUB}_{m=3} = \frac{1}{4}(9p+3) \\
+ \end{align}
+ For $m=2$ we have entanglement on the upper bound for $p = \frac{2}{3}$.
+ For $m=3$ we have entanglement on the upper bound for $p = \frac{4}{9}$.
+
+ For $d=3$ we have the following
+ \begin{align}
+ &I^{MUB}_{m=1} = \frac{1}{9}(16p+2) \\
+ &I^{MUB}_{m=2} = \frac{1}{9}(23p+4) \\
+ &I^{MUB}_{m=3} = 3p + \frac{1}{3}\\
+ &I^{MUB}_{m=4} = \frac{1}{9}(31p + 8)
+ \end{align}
+ For $m=2$ we have entanglement on the upper bound for $p = \frac{8}{23}$.
+ For $m=3$ we have entanglement on the upper bound for $p = \frac{4}{9}$.
+
+ with conjugation we get
+ \begin{align}
+ &I^{MUB}_{m=1} = \frac{1}{9}(16p+2) \\
+ &I^{MUB}_{m=2} = \frac{1}{9}(32p+4) \\
+ &I^{MUB}_{m=3} = \frac{16}{3}p + \frac{2}{3}\\
+ &I^{MUB}_{m=4} = \frac{1}{9}(64p + 8)
+ \end{align}
+ For $m=3$ we have entanglement on the upper bound for $p = \frac{5}{3}$.
+\end{MyExercise}
+\newpage
+\begin{MyExercise}
+ \textbf{
+ Compute $I_m^{MUB}$ with conjugation and without for the Werner states
+ for $d=2, 3$ and $m=1,\dots, d+1$
+ \begin{align}
+ \varrho_W(q) = q \frac{P_{sym}}{d(d+1)} + (1-q) \frac{P_{asym}}{d(d-1)}
+ \end{align}
+ where $P_{sym} = (\mathbbm{1} + \mathbb{P})$ and $P_{asym} =
+ (\mathbbm{1} - \mathbb{P})$ for $\mathbb{P} = \sum_{ij}
+ |ji\rangle\langle ij|$. The state is separable for $q\in [0,\frac{1}{2}]$ and
+ entangled for $q\in [\frac{1}{2}, 1]$
+ }
+
+ First we calculate for $d=2$ we choose the basis $B_1$ to calculate the projection
+ operator. And note that $|ij\rangle = |i\rangle \otimes |j\rangle$ we need
+ the tensor product here.
+
+ Straightforward computation gives
+ \begin{align}
+ &I^{MUB}_{m=1} = \frac{q}{3}\\
+ &I^{MUB}_{m=2} = \frac{2q}{3}\\
+ &I^{MUB}_{m=3} = q
+ \end{align}
+ For $m=3$ we have entanglement on the lower bound for $p = 1/2$.
+
+ with conjugation
+ \begin{align}
+ &I^{MUB}_{m=1} = \frac{q}{3}\\
+ &I^{MUB}_{m=2} = \frac{2q}{3}\\
+ &I^{MUB}_{m=3} = \frac{q}{3} + \frac{1}{2}
+ \end{align}
+ For $m=3$ we have entanglement on the lower bound for $p = 0$.
+
+ For $d=3$ we choose the basis $B_1$ to calculate the projection operator
+ and straightforward computation gives
+ \begin{align}
+ &I^{MUB}_{m=1} = \frac{q}{3}\\
+ &I^{MUB}_{m=2} = \frac{2q}{3}\\
+ &I^{MUB}_{m=3} = q \\
+ &I^{MUB}_{m=3} = \frac{4q}{3}
+ \end{align}
+ For $m=2$ we have entanglement on the lower bound for $p = 0.3165$.
+ For $m=3$ we have entanglement on the lower bound for $p = \frac{1}{2}$.
+
+ with conjugation
+ \begin{align}
+ &I^{MUB}_{m=1} = \frac{q}{3}\\
+ &I^{MUB}_{m=2} = \frac{1}{12}(5q + 2)\\
+ &I^{MUB}_{m=3} = \frac{1}{36}(15q + 14)\\
+ &I^{MUB}_{m=4} = \frac{1}{36}(15q + 22)\\
+ \end{align}
+ For $m=2$ we have entanglement on the lower bound for $p = 0.1064$.
+ For $m=3$ we have entanglement on the lower bound for $p = \frac{4}{15}$.
+
+ A simple comparison with exercise 3, we arrive at the conclusion that for
+ the Werner states we detect entanglement only on the lower bound and for the
+ isotropic states we detect entanglement only on the upper bound.
+\end{MyExercise}
+
+
+
+\nocite{cite1}
+\nocite{cite2}
+\nocite{cite3}
+\nocite{cite4}
+\nocite{cite5}
+\printbibliography
+
+
+
+\end{document}
diff --git a/sesh5/src/uni.bib b/sesh5/src/uni.bib
@@ -0,0 +1,71 @@
+@article{cite1,
+ title = {Uncertainty in Quantum Measurements},
+ author = {Deutsch, David},
+ journal = {Phys. Rev. Lett.},
+ volume = {50},
+ issue = {9},
+ pages = {631--633},
+ numpages = {0},
+ year = {1983},
+ month = {Feb},
+ publisher = {American Physical Society},
+ doi = {10.1103/PhysRevLett.50.631},
+ url = {https://link.aps.org/doi/10.1103/PhysRevLett.50.631}
+}
+
+
+@article{cite2,
+ title = {Generalized entropic uncertainty relations},
+ author = {Maassen, Hans and Uffink, J. B. M.},
+ journal = {Phys. Rev. Lett.},
+ volume = {60},
+ issue = {12},
+ pages = {1103--1106},
+ numpages = {0},
+ year = {1988},
+ month = {Mar},
+ publisher = {American Physical Society},
+ doi = {10.1103/PhysRevLett.60.1103},
+ url = {https://link.aps.org/doi/10.1103/PhysRevLett.60.1103}
+}
+
+@article{cite3,
+ title={Complementarity reveals bound entanglement of two twisted photons},
+ volume={15},
+ ISSN={1367-2630},
+ url={http://dx.doi.org/10.1088/1367-2630/15/8/083036},
+ DOI={10.1088/1367-2630/15/8/083036},
+ number={8},
+ journal={New Journal of Physics},
+ publisher={IOP Publishing},
+ author={Hiesmayr, Beatrix C and Löffler, Wolfgang},
+ year={2013},
+ month={Aug},
+ pages={083036}
+}
+
+@article{cite4,
+author = {Hiesmayr, Beatrix and Löffler, Wolfgang},
+year = {2013},
+month = {09},
+pages = {},
+title = {Mutually Unbiased Bases and Bound Entanglement},
+volume = {2014},
+journal = {Physica Scripta},
+doi = {10.1088/0031-8949/2014/T160/014017}
+}
+
+
+@article{cite5,
+ title={Entanglement detection via mutually unbiased bases},
+ volume={86},
+ ISSN={1094-1622},
+ url={http://dx.doi.org/10.1103/PhysRevA.86.022311},
+ DOI={10.1103/physreva.86.022311},
+ number={2},
+ journal={Physical Review A},
+ publisher={American Physical Society (APS)},
+ author={Spengler, Christoph and Huber, Marcus and Brierley, Stephen and Adaktylos, Theodor and Hiesmayr, Beatrix C.},
+ year={2012},
+ month={Aug}
+}
