8 votes

Confusion regarding projection operator

Does $|0\rangle\langle0|$ represent a tensor product or is it just matrix multiplication? You can think of $|0\rangle\langle0|$ as tensor product of $|0\rangle$ and $\langle0|$, or equivalently as ...
glS's user avatar
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6 votes

What does "measuring a state" mean?

Look like a lot of misunderstanding. If you measure a state $\alpha|0\rangle+\beta|1\rangle$ in computational basis, the state collapses either to $|0\rangle$ or $|1\rangle$. In Quantum Information ...
kludg's user avatar
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6 votes
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What's the observable when measuring multiple qubits in the computational basis?

Note that your current definitions of the projection matrices $\{P_{1},P_{2},...,P_{n}\}$ are actually not projection matrices, since $P_{i}^{2} = I \not= P_{i} \,\, \forall i$. What works 'better' is ...
JSdJ's user avatar
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6 votes
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How do I prove that $P_\pm=\frac12(1\pm U)$ if $U^2=I$?

First, we can start with $U = P_+ - P_-$, since the Hermitian is the sum of the projection operators of the eigenspaces scaled by their eigenvalues. If $U^2 = I$, that means $I = (P_+ - P_-)(P_+ - P_-)...
Joseph Geipel's user avatar
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Implement a projection operator as a quantum circuit

Consider the states $|\psi\rangle = a|0\rangle|\psi_0\rangle + b|1\rangle|\psi_1\rangle$, and $|\psi'\rangle = a|0\rangle|\psi_0'\rangle + b|1\rangle|\psi_1\rangle$, where $a$ and $b$ are non-zero and ...
GotCarter's user avatar
  • 401
6 votes

Proof that the projector onto the symmetric subspace of the Swap $F$, with $n=2$, equals $\frac{1}{2}(I+F)$

I'm assuming $F$ is the Swap operator here, acting on some finite-dimensional space $H\otimes H$. In bra-ket notation, this reads $F\equiv \sum_{ij} |ij\rangle\!\langle ji|$. Observe that if $\dim H=...
glS's user avatar
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6 votes

How does a quantum system identify hermitian and unitary matrices?

This is a bit like asking how your car identifies whether the vector you're giving it is its new position or its new velocity. There are various things you can do to quantum systems. All can be ...
Craig Gidney's user avatar
5 votes
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What can I conclude about $\langle \phi|\pi_1\pi_2|\phi\rangle$ if $\langle \phi|\pi_i|\phi\rangle\ge e$?

Based on those relations there's nothing more that you can conclude. Consider the two extremes. At one extreme $\pi_1$ and $\pi_2$ project onto the same subspace, in which case: $$\langle {\phi} | \...
Jonathan Trousdale's user avatar
5 votes

What does "measuring a state" mean?

Now If I chose the standard basis $|0\rangle,|1\rangle$ what will be the result I will get? If I measure with respect of the projection operator $\langle 0|$ I get $\alpha$ and If I measure with ...
glS's user avatar
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5 votes

Quantum channel representation of projective measurement

If $P$ is (ortho)projector, that is $P^2=P=P^\dagger$, then we can define unitary $U = I - 2P$. You can verify $$ UU^\dagger = U^2 = (I-2P)(I-2P) = I-4P+4P = I $$ Now we can express $P=\frac{1}{2}(I-...
Danylo Y's user avatar
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Weeding out qubit states with leftmost qubit as 1

I'm afraid it is not possible to remove states from a superposition that you do not want. At least, not easily. If it were, a lot of computationally hard problems would become much easier, since you ...
James Wootton's user avatar
5 votes
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Error syndromes and recovery procedure in bit flip code

You quoted the first part of the exercise; your question is exactly the second part of the exercise :-) Exercise 10.4 (2) asks you to show that the described recovery procedure (the one which ...
Mariia Mykhailova's user avatar
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Projecting $\lvert ++ \rangle$ on Bell Basis

The Hadamard gate is: $$\frac{1}{\sqrt 2} \left(|0\rangle \langle 0 | + |0\rangle\langle 1| + |1\rangle \langle 0| - |1\rangle \langle 1|\right)$$ And since $|+\rangle$ is $\frac{1}{\sqrt 2}\left(|0\...
Mahathi Vempati's user avatar
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Is a projective measurements over a superposition of eigenstates possible?

Yes, such a measurement is possible and the outcomes can be mapped to the outcomes of a computational basis measurement. This is accomplished by the use a unitary to transform from the basis we wish ...
Adam Zalcman's user avatar
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4 votes
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Quantum error correction using bit-flip code for the amplitude damping channel

Before starting, I should probably emphasise that, although useful for the practice of working through the maths of quantum error correction on a relatively simple case, amplitude damping combined ...
DaftWullie's user avatar
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4 votes
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Projection operators and positive operators

This is really a question about eigenvalues: A projector has eigenvalues 1 and 0. So, for a qubit, that could be eigenvalues $\{1,1\}$ or eigenvalues $\{1,0\}$. A positive operator is one for which ...
DaftWullie's user avatar
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Bob applies a projector - what happens to eigenvalues of Alice's reduced state?

Just notice that $$ \text{Tr}(\rho'_{AB}) = \text{Tr}(\Pi_B\rho_{B}). $$ One way to see this is to consider any decomposition $$ \rho_{AB} = \sum_i A_i \otimes B_i, $$ where $A_i, B_i$ just some ...
Danylo Y's user avatar
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How to code a projector operator in qiskit?

So, there actually is a method called to_operator() for the class Statevector which takes a statevector and converts it into a projector operator. Here is the code to write your specific projector ...
Winona's user avatar
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4 votes

What can I conclude about $\langle \phi|\pi_1\pi_2|\phi\rangle$ if $\langle \phi|\pi_i|\phi\rangle\ge e$?

Let's assume $$ \pi_i|\phi\rangle=e_i|\phi\rangle+\sqrt{1-e_i^2}|\phi_i^\perp\rangle, $$ where $\langle\phi|\phi_i^\perp\rangle=0$ and, for simplicity, let's assume the $e_i$ are real. We can ...
DaftWullie's user avatar
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4 votes
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Find the unitary implementing the transformation $|0\rangle\to\frac1{\sqrt2}(|0\rangle+|1\rangle),|1\rangle\to\frac1{\sqrt2}(|0\rangle-|1\rangle)$

Firstly, how rude of them to use the word 'obviously' in their solution! If it was obvious then we wouldn't be trying their exercises in the first place! But anyway, Adam Zalcman described the ...
sheesymcdeezy's user avatar
4 votes
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Relation between symmetric subspaces and $n$-exchangeable density matrices

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). ...
Zoltan Zimboras's user avatar
4 votes
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What is the state after a projective measurement?

The way I'd put it is that the state after the measurement is just $|v_m\rangle$, regardless of what's the phase of $\langle v_m|\psi\rangle$. This is because states are defined up to a global phase, ...
glS's user avatar
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4 votes
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How to write down product operators acting on non-adjacent subsystems?

You can exploit the distributive law. For example, the operator projecting onto the even subspace of qubits $1$ and $3$ can be written as $$ \langle 0|_1\otimes I_2\otimes\langle 0|_3\otimes I_4+ \...
Adam Zalcman's user avatar
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How to implement projective measurement from multiple measurements?

The point of the second register is to initialize it in some state with an equal probability of being found in any basis state $|i\rangle$. Then, conditioned on the state of the second register, one ...
Quantum Mechanic's user avatar
4 votes
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confusion on the LCU method regarding the normalization

You're correct, LCU will prepare a state proportional to $A|\psi\rangle$, with whatever renormalization is required for that to be true. Its possible that the normalization factor was omitted because ...
forky40's user avatar
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3 votes
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Prove that $P_{M_1}P_{M_2}= P_{M_2}P_{M_1}$ implies $\text{Pr}(\text{span}[M_1, M_2]) = \text{Pr}(M_1)+\text{Pr}(M_2)−\text{Pr}(M_1\cap M_2)$

It follows that $M_1$ is invariant subspace of operator $\text{Proj}_{M_2}$. Indeed, if $v \in M_1$ then $$\text{Proj}_{M_1}\text{Proj}_{M_2}v=\text{Proj}_{M_2}\text{Proj}_{M_1}v = \text{Proj}_{M_2}...
Danylo Y's user avatar
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3 votes

What can I conclude about $\langle \phi|\pi_1\pi_2|\phi\rangle$ if $\langle \phi|\pi_i|\phi\rangle\ge e$?

$$ \pi_1 = \begin{pmatrix} 1&0\\ 0&0\\ \end{pmatrix}\\ \pi_2 = \begin{pmatrix} 0&0\\ 0&1\\ \end{pmatrix}\\ \rho = \frac{1}{\sqrt{2}} \begin{pmatrix} 1\\ 1\\ \end{pmatrix}\\ e = \frac{1}...
AHusain's user avatar
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3 votes

Quantum channel representation of projective measurement

I'll cover a slightly more general case. Let $P_k$, $k=1,...,N$ a complete set of orthogonal projectors: $\sum_k P_k=I$ and $P_j P_k=\delta_{jk}P_j$. Consider the map $\mathcal E(\rho)=\sum_k P_k \...
glS's user avatar
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3 votes

Confusion regarding projection operator

A projection operator $P$ has two key properties: $$ P^\dagger=P\qquad P^2=P $$ A particularly simple instance of a projection operator is a rank 1 projector, $P=|\phi\rangle\langle\phi|$, which you ...
DaftWullie's user avatar
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3 votes

What is the relation between POVMs and projective measurements?

In the case of projective measurements, we have a set of projectors $\{P_i\}$ satisfying the completeness relation $$ \sum_iP_i=I. $$ Note that this also means they satisfy $\sum_iP_i^\dagger P_i$, ...
DaftWullie's user avatar
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