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2 votes

How can I calculate the measuring probabilities of a two qubit state along a certain axis?

First, you need to construct the measurement projection operators. For measuring spin along $\vec{n}$, the two projection operators will be given by $$ P_{\pm} = \frac{I \pm \vec{n}\cdot\vec{\sigma}}{...
FDGod's user avatar
  • 1,833
2 votes

How to calculate pseudo-threshold of the [[5,1,3]] code analytically to match simulation?

In accordance with what @Peter-Jan said, what I was failing to understand is that if an error is "within weight-1" of a stabilizer element then it is correctable. For example suppose the ...
Eric Kubischta's user avatar
3 votes
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Can post-measurement states have entropy larger than the original state?

The stronger inequality holds if $\rho$ is a pure state. You can see it from the fact that $\sqrt M \rho \sqrt M$ has unit rank if $\rho$ does. Thus in all such cases you have $S(\rho)=S(\rho_i)=0$ (...
glS's user avatar
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2 votes
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How to calculate pseudo-threshold of the [[5,1,3]] code analytically to match simulation?

To find an exact expression for the blue line you could one by one check for each physical error if it leads to a logical error. For a five qubit code with a code capacity noise model this is doable ...
Peter-Jan's user avatar
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1 vote
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What is the explicit action of the following circuit?

I am assuming the two control symbols connecting |𝜓⟩ and |+⟩ are entangling the states somehow? Yes, this represents a CZ gate. Then I don't understand the top right measurement. It is not in a ...
Peter-Jan's user avatar
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1 vote

What does it mean that the elements of a generalized measurement are operators?

Forgetting for a second the way you formalise generalised measurements (that is, POVM), ask yourself how a "measurement" should be formalised in quantum mechanics. A measurement is ...
glS's user avatar
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1 vote
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What does it mean that the elements of a generalized measurement are operators?

$\newcommand{\bra}[1]{\left<#1\right|}\newcommand{\ket}[1]{\left|#1\right>}\newcommand{\bk}[2]{\left<#1\middle|#2\right>}\newcommand{\bke}[3]{\left<#1\middle|#2\middle|#3\right>}$ I ...
FDGod's user avatar
  • 1,833
1 vote

Performing a projective measurement, is the resulting expectation value $\langle \Psi|M|\Psi\rangle$ bounded between $+1$ and $-1$?

The expectation value of an observable is not constrained to be between +1 and -1. You can define $M$ to be any observable - i.e. a Hermitian operator. See https://physics.stackexchange.com/questions/...
Balint Pato's user avatar
2 votes

Performing a projective measurement, is the resulting expectation value $\langle \Psi|M|\Psi\rangle$ bounded between $+1$ and $-1$?

The measurement operators $\{M_i\}$ obey two conditions, firstly they are positive operators, $M_i\geq 0$, which is $\forall |\psi \rangle,\, \langle \psi | M_i |\psi \rangle \geq 0$. Secondly, they ...
Ethan's user avatar
  • 61
2 votes
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Measuring one register of the state $\frac{1}{2^{m}} \sum_{x}\sum_{k} (-1) ^ {x\cdot k} |k\rangle |f(x)\rangle$

Your state before measurement is $$ |\psi\rangle=\frac{1}{2^m}|0\rangle\sum_x|f(x)\rangle+\frac{1}{2^m}\sum_x\sum_{k\neq 0}(-1)^{x\cdot k}|k\rangle|f(x)\rangle. $$ How do we calculate the probability ...
DaftWullie's user avatar
2 votes

Why does measuring a GHZ state in the computational basis destroy the entanglement?

The two-qubit GHZ state (Bell state) case is special: any measurement in a basis $$ \cos\theta X+\sin\theta Z $$ will return perfectly correlated results between the two parties. In the case of a ...
DaftWullie's user avatar

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