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$\newcommand{\ket}[1]{\vert#1\rangle}$ First, write $\ket\psi$ and $\ket{\tilde\psi}$ in their Schmidt decomposition: $$ \begin{aligned} \ket\psi &= \sum \lambda_i \ket{a_i}\ket{b_i}\ , \\ \ket{\tilde\psi} & = \sum \tilde\lambda_i \ket{\tilde a_i}\ket{\tilde b_i}\ . \end{aligned} $$ Let us assume for simplicty that the $\lambda_i$ are non-degenerate. ...


4

Operator applied from $t_{0}$ to $t_{1}$ is $CNOT \otimes I$, i.e. \begin{equation} U_1= \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 &...


3

Suppose you have 2-qubit state $|\psi\rangle_{AB}$. Entangled or not, you can always write it as $$|\psi\rangle_{AB}=a_0|0\rangle_A|\psi_0\rangle_{B}+a_1|1\rangle_A|\psi_1\rangle_{B}$$ If you measure qubit $A$ in state $|0\rangle$, then the qubit $B$ is in state $|\psi_0\rangle$, and you can compute the probability of qubit $B$ being in state $|0\rangle$ as $...


3

Your comments talk about an exam setting. Probably the most reliable thing to do is calculate the partial trace of each qubit. If any of them is mixed, there is entanglement in the system. (I’m not sure that a yes/no question about the existence of entanglement is a good question in a system of more than 2 qubits, but...) In your particular examples, you can ...


2

The definition of an entangled state is one that cannot be decomposed into an arbitrary Kronecker product of individual state vectors corresponding to individual quantum systems. Mathematically speaking, an un-entangled state $|\psi\rangle$ is one that can be written in the following way: $$|\psi\rangle \ = \ \displaystyle\bigotimes_n |\psi_n\rangle$$ ...


2

Another way to see this is remembering that the density matrix of a bipartite pure state is essentially its covariance matrix, in the sense that if a state $|\psi\rangle$ has components $\psi_{ij}$ and $\rho=|\psi\rangle\!\langle\psi|$, then $\rho=\psi\psi^\dagger$ (where we are thinking of $\psi$ as a matrix, so that $(\psi\psi^\dagger)_{ij}=\sum_\ell \psi_{...


2

It is due to Schmidt decomposition. For some $|\psi \rangle_{AB} \in H_A \otimes H_B$, there exists a decomposition in terms of the orthonormal basis (Schmidt bases) of system A and B. $\lambda_i$ are the Schmidt coefficients calculated from $Tr_B(|\psi\rangle \langle\psi|_{AB})$ whose eigenvalues are $\lambda^2_i$. Given below is the Schmidt decomposition, {...


2

Let's say that you're working in the energy eigenbasis of a quantum harmonic oscillator $\dagger$ and label the basis states as $\{|0⟩_H,|1⟩_H,|2⟩_H,…\}$, and you have the basis $\{|0⟩_Q,|1⟩_Q\}$ for your qubit. $d$ would be $\text{min}(𝑁,2)=2$ in this case. Then say you choose the basis $$\{|0_Q0_H⟩,|0_Q2_H⟩,|1_Q0_H⟩,|1_Q2_H⟩\}$$ for $\Bbb C^2\otimes \Bbb ...


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There is a more direct characteristic that makes the state of an entangled pair of qubits distinct from a non-entangled pair (which is also known as a separable state). When two qubits are not entangled, the state of the one qubit can be described without any knowledge of the state of the other qubit. Mathematically we can write: \begin{equation} |\psi\...


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In this paper, the authors used Knot theory to define what they call 'Path Model Representation'. In a later section they convert this representation to qubits by switching to binary.


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The probability of finding an entangled state while randomly sampling (from the uniform distribution) over pure states is one: almost all pure states are entangled. This is shown in this answer on physics.SE. This paper might also be of interest. The gist is that separable states correspond to rank-1 matrices, which have zero measure in the set of all ...


1

Even for two qubits, there are infinite number of entangled states: $$ \left| \psi \right\rangle_2 = \alpha \left| 00 \right\rangle + \beta \left| 11 \right\rangle $$ where $\left| \alpha \right|^2$ and $\left| \beta \right|^2$ can be any probabilities that satisfy $\left| \alpha \right|^2 + \left| \beta \right|^2 = 1 $. And even this $\left| \psi \right\...


1

If you are interested only in the measurements of the system that comprises only 2nd and 3rd qubits, then you could compute the reduced density matrix $\rho_{23}$ on those qubits from the total density matrix $\rho_{1234}=|\psi\rangle \langle\psi|$, i.e. $$ \rho_{23} = \text{Tr}_{14}(\rho_{1234}) $$ Density matrix carry all information that is need to ...


1

Judging from your computations, the Bell states are $$|B_{00}\rangle = |\Phi^+\rangle = \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle)$$ $$|B_{10}\rangle = |\Phi^-\rangle = \frac{1}{\sqrt{2}}(|00\rangle - |11\rangle)$$ $$|B_{01}\rangle = |\Psi^+\rangle = \frac{1}{\sqrt{2}}(|01\rangle + |10\rangle)$$ $$|B_{11}\rangle = |\Psi^-\rangle = \frac{1}{\sqrt{2}}(|01\...


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Intraphoton entanglement uses the degrees of freedom from one photon only to create entanglement. So, here either polarization and linear momentum or polarization and angular momentum can be used to create entanglement. Interphoton entanglement is the entanglement created between 2 spatially separated photons. So, naturally latter is less stable than former. ...


1

Considering your question is not specific, you can take a look at Chapters 10 (classical entropy) and 11 (quantum entropy) of this book. Quantum entanglement entropy as we know it comes from classical Shannon theory. Claude Shannon introduced entropy in order to quantify information. You can read his paper here.


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