22

TL;DR: This is probably going to be disappointing. If a cat enters a superposition and we lose track of the relative phase $\phi$ then there is only one deterministic operation that returns to the $|\text{alive}\rangle$ state: the state preparation channel. In other words, we have to get a new cat. Let us represent the states of the cat on the Bloch sphere ...


14

The minimum overlap is zero and the maximum overlap is $\frac{1}{d}$. The overlap is a linear function of $\rho$ and the set $S$ of separable states is convex, so the overlap is both minimized and maximized at extreme points. Extreme points of $S$ are the states of the form$^1$ $\rho = \overline\sigma\otimes\tau$. The reason we choose to define $\sigma$ as ...


7

Yes. Intuitively, the set of pure product states has lower dimension than the set of all pure states. Therefore, almost all pure two-qubit states are entangled. Let $\mathcal{F}$ denote the set of all pure states of two qubits and $\mathcal{S}$ denote the set of all pure product states of two qubits. Note that $\mathcal{S}$ can be thought of as the Cartesian ...


5

There are several, these are the ones I have seen: $|+\rangle_y,|-\rangle_y$ a bit lazy but easy to remember $|+i\rangle,|-i\rangle$ the same as before but you replace the sub index with an imaginary unit $i$ $|\circlearrowleft\rangle,|\circlearrowright\rangle$ this notation is borrowed from light polarization, as you can use photons for light too, circular ...


5

As the other answer mentioned, they are often denoted as $$|+i\rangle= \dfrac{|0\rangle + i|1\rangle}{\sqrt{2}} = \dfrac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ i\end{pmatrix} \ \ \ \textrm{and} \ \ \ |-i\rangle = \dfrac{|0\rangle - i|1\rangle}{\sqrt{2}} = \dfrac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ -i\end{pmatrix} $$ but sometime you might also see them denoted as ...


5

The answer is no. Define X=[[0,1,0],[0,0,1],[1,0,0]] Z=[[1,0,0],[0,w,0],[0,0,w^2]], w^3=1 Then the Pauli group is generated by X and Z and is of order 27. With H being your matrix, you can check that H'XH and H'ZH are not in the group. Calculations like this are easy to do in gap The dim=3 counterpart of the Hadamard gate is the 3 dimensional Fourier ...


5

OK, honestly I did not follow the later part of your post (where you asked the questions) -- it was too confusing. But I suspect that your confusion arises because you were trying to go between abstract bra-ket notation and matrix notation (which entails choosing some basis to express the operators in). Maybe this will help. Let $$ \hat{\rho} = \sum_i p_i |\...


4

If you measure a state $|\psi\rangle = a|0\rangle + b|1\rangle$, the post-measurement state will be either $|0\rangle$ or $|1\rangle$ with $|a|^2$ and $|b|^2$ probability, respectively. The important part here is that the values for $a$ and $b$ are lost in the post-measurement state and you end up with a basis state $\{|0\rangle, |1\rangle\}$ with an ...


4

Your reasoning is correct, this is indeed the resulting state they would get. More generally, you can think about it as applying a $X\otimes I$ gate on the whole system, where $I$ is the identity gate. However, do note that if Alice and Bob do not communicate, there is nothing Alice can do to influence the measurement she will get, since its qubit is in the ...


4

Notice that it is somewhat a coincidence of that particular Bell state and choice of basis. The states $|0\rangle$ and $|1\rangle$ are in the $z$ axis of the Bloch sphere and $|+\rangle$,$|-\rangle$ are in the $x$-axis. The state you chose is a sum of products of single states that are the same, and it turns out that the same is true when you convert it to ...


4

Qubits are more flexible than bits in a way that's difficult to summarize. But one key difference is that qubits support "phase kickback", and bits have no concept of phase kickback. Here is a puzzle. Fill in the blanks to make these two circuits equal: With bits, this is impossible. There is no single-input single-output process that can reverse ...


4

We know that $$ \begin{gather} |0\rangle = \frac{|+\rangle+|-\rangle}{\sqrt{2}} \\ |1\rangle = \frac{|+\rangle-|-\rangle}{\sqrt{2}} \end{gather} $$ Thus, we can rewrite the $GHZ$ state as $$ \begin{align} |GHZ\rangle &= \frac{1}{\sqrt{2}}\left(|0\rangle|00\rangle+|1\rangle|11\rangle\right) \\ &=\frac{1}{2}\left(|+\rangle|00\rangle+|-\rangle|00\rangle+...


4

It depends on what you want to take as your definition of maximally entangled. But, here's a good one: Given a Bell state, I can convert it into any other two-qubit entangled state using only local operations and classical communication Given that local operations and classical communication cannot increase entanglement, the possibility of performing $|\...


3

Source of the problem The purported contradiction arises due to the use of incorrect assumptions for Klein equality $$ S(\rho||\sigma) \ge 0. $$ The inequality does not require any particular relationship$^1$ between the support of $\rho$ and the support of $\sigma$. However, it does require that $\rho$ and $\sigma$ be states, i.e. unit trace positive ...


3

The answer is: no, it is not true that any $n$ exchangeable state is a linear combination of density matrices of states in the symmetric subspace (that is supported on the symmetric subspace). Actually, there are even pure state counterexamples when $n=2$. Consider the state $$ \rho = |\phi\rangle\langle \phi|, $$ where $$ |\phi\rangle = \frac{1}{\sqrt{2}}(|...


3

The definition of information is context dependent and has different interpretations. See How to better define information forum (Physics Insights) As you did not provide a paper, I will assume a very standard context. In general when people talk about information loss in quantum mechanics due to collapse, it is related to irreversibility of collapse. If you ...


2

As you are asking specifically for the evaluation of the energy only, I will be brief. I will assume that you have a init_state (a quantum circuit) that produces the the Hartree-Fock wavefunction or any other wavefunction you like to test. I could not find a Qiskit function that provides the energy expectation value of a given wavefunction, given some ...


2

They are commonly denoted as $$ \left\{ |+i\rangle = \frac{|0\rangle + i|1\rangle}{\sqrt{2}}, |-i\rangle = \frac{|0\rangle - i|1\rangle}{\sqrt{2}}\right\} $$ So, you could use $|\pm i\rangle = \frac{|0\rangle \pm i|1\rangle}{\sqrt{2}}$, which would be very similar to the case of $X$, i.e., $Y|\pm i\rangle = \pm|\pm i\rangle$.


2

Error correcting codes work the same on entangled qubits as any other qubit. All Alice and Bob have to do is separately encode their qubit into a Shor code. They each run an encoding circuit, apply noise, then run a decoding circuit and apply the corrections it inferred.


2

As you say, the difference is in the global phase. Let me explain using the first of your examples, $$ \left[\begin{array}{cc} \frac{1-i}{2} & \frac{1+i}{2} \end{array}\right]. $$ Mathematically, this is the same as $$ \left[\begin{array}{cc} \frac{e^{-i\pi/4}}{\sqrt{2}} & \frac{e^{i\pi/4}}{\sqrt{2}} \end{array}\right]=\left[\begin{array}{cc} e^{-i\...


2

One strong element of the intuition is related to the fact that it is maximally entangled. One definition of a pure state $|\psi\rangle$ being maximally entangled is that the individual systems have density matrices $$ \rho_A=\text{Tr}_B(|\psi\rangle\langle\psi|)=\frac{I}{d} $$ where $d$ is the dimension of $A$'s Hilbert space. Now, one thing that the ...


2

Yet another derivation Applying a local unitary $U^A$ on the first subsystem of a bipartite maximally entangled state $|\psi^{AB}\rangle$ is equivalent to applying a possibly different unitary $V^B$ on the second subsystem $$ (U^A\otimes I)|\psi^{AB}\rangle = (I\otimes V^B)|\psi^{AB}\rangle\tag1. $$ In the specific case of the Bell state $(|00\rangle+|11\...


2

Overview The short answer is no you cannot prepare this state efficiently; but, infact your question is a special case of a more general case I prove below which I will outline first as it applicable to a more general audience. After I will show that your question is indeed a special case of theorem. Finally, I will address the comments under your original ...


2

Calculating $\rho_1$ Let $N=2^n$ denote the dimension of the Hilbert space where $|\psi\rangle$ lives. For $i=0,\dots,N-1$, let $V_i$ be any unitary that maps $|i\rangle$ to $|0\rangle$. The action of $V_i$ on other computational basis states $|j\rangle$ with $j\ne i$ is irrelevant. Exploiting the invariance of the Haar measure to absorb $V_i$ into the ...


2

$U_f$ is defined as $U_f: |x\rangle|y\rangle \rightarrow |x\rangle|y\oplus f(x)\rangle$. Now, let's write the product state of the complete system of two qubits before applying $U_f$. We can do this with tensor products as follows: $$ \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle) \otimes |0\rangle = \frac{1}{\sqrt{2}}(|0\rangle\otimes|0\rangle + |1\rangle\otimes|...


2

Entanglement is analogous to correlation. Correlation over a property simply exists where simultaneous determination is not equivalent to individual determination of a property. What does it mean? Say you have 2 coins - one black and one white. Let's say if you have a black coin you get a $+1$ colour value whereas with white you get a $-1$ colour value. If ...


2

TL;DR: Active and passive transformations The dichotomy between the two types of unitary transformations is real and is an example of a division of transformations into active and passive types. This duality is inherent to any use of coordinates and arises from the fact that there is a degree of arbitrariness in the way coordinates are assigned to objects ...


1

In the specific $|\psi \rangle$ there is symmetry on the order of bits, so we must only show that one bit is entangled and it then follows that they all are. Notice we may write: $$|\psi \rangle = \frac{1}{2}\Big(|0 \rangle \big(|001 \rangle +|010 \rangle +|100 \rangle \big) + |1 \rangle \big(|000 \rangle \big)\Big)=\frac{\sqrt3 |0\rangle |A \rangle + |1\...


1

An intuitive way to think about it is that $E[M]=E[X_1 \otimes Z_2]=E[X_1 \otimes \mathbb{1}]E[\mathbb{1} \otimes Z_2]$ If we only think about $E[\mathbb{1} \otimes Z_2]$, it is just the expectation value of $Z_2$ on the second qubit. Consider that our second Qubit in the entangled state $\frac{| 00\rangle + | 11\rangle}{\sqrt{2}}$ is measured to be $\frac{+\...


1

Taking the last two terms of last expression you gave, we can do the following $$ \begin{align} M \left(\frac{|00\rangle+|11\rangle}{\sqrt{2}}\right) &= X_1\otimes Z_2\left(\frac{|00\rangle+|11\rangle}{\sqrt{2}}\right) \\ &= \left(\frac{X_1|0\rangle \otimes Z_2|0\rangle+X_1|1\rangle \otimes Z_2|1\rangle}{\sqrt{2}}\right) \\ &= \left(\frac{|1\...


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