7 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 ...
  • 19.5k
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 ...
  • 3,114
6 votes
Accepted

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 ...
  • 4,898
6 votes
Accepted

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 ...
  • 361
5 votes
Accepted

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} | \...
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 ...
  • 19.5k
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-...
  • 5,978
5 votes
Accepted

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 ...
5 votes
Accepted

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 ...
5 votes
Accepted

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\...
5 votes
Accepted

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_-)...
5 votes
Accepted

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 ...
  • 14.7k
5 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=...
  • 19.5k
4 votes
Accepted

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 ...
  • 48.1k
4 votes
Accepted

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 ...
  • 5,978
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 ...
  • 48.1k
4 votes
Accepted

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 ...
4 votes
Accepted

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, ...
  • 19.5k
4 votes
Accepted

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+ \...
  • 14.7k
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}...
  • 3,553
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 \...
  • 19.5k
3 votes

What is the relation between POVMs and projective measurements?

POVMs are more general than projective measurements. Thus, every projective measurement is also a POVM, by choosing $E_i=P_i$.
3 votes

Projecting $\lvert ++ \rangle$ on Bell Basis

The four Bell states are $$ |\Phi_{\pm}\rangle=(|00\rangle\pm|11\rangle)/\sqrt{2}\qquad |\Psi_{\pm}\rangle=(|01\rangle\pm|10\rangle)/\sqrt{2}. $$ So, let's consider what happens then we try and ...
  • 48.1k
3 votes
Accepted

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}...
  • 5,978
3 votes
Accepted

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 ...
  • 48.1k
3 votes

Find the unitary implementing the transformation $|0\rangle\to\frac1{\sqrt2}(|0\rangle+|1\rangle),|1\rangle\to\frac1{\sqrt2}(|0\rangle-|1\rangle)$

Another way to solution: Hadamard gate changes $|0\rangle$ to $\frac{1}{\sqrt{2}}(|0\rangle + |1\rangle) = |+\rangle$ and $|1\rangle$ to $\frac{1}{\sqrt{2}}(|0\rangle - |1\rangle) = |-\rangle$. In ...
3 votes
Accepted

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). ...
3 votes

When discussing error correction, what are the objects in the expression $PE_i^\dagger E_j P=\alpha_{ij} P$?

Generally speaking, if a quantum channel $\Phi$ sends operators in $\mathcal X$ into operators in $\mathcal Y$, and its Kraus representation reads $\Phi(X)=\sum_a A_a X A_a^\dagger$, then the Kraus ...
  • 19.5k
3 votes
Accepted

Are projections determined by their action on a full-rank density matrix?

Yes. Clearly, ${\rm Ker}(P) \subset {\rm Ker}(P\rho P)$. On the other hand, since $\rho$ is a density operator and has full rank, we have $\rho > 0$, that is, $\langle x | \rho | x \rangle > 0$ ...
  • 5,978
2 votes

Quantum error correction using bit-flip code for the amplitude damping channel

Firstly, I'm presuming that when you write $E_0^3$ it corresponds to $E_0^{q_1} \otimes E_0^{q_2} \otimes E_0^{q_3}$. In question 2. you've written out how the Superoperator of the Kraus operators ...

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