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18 votes
Accepted

What is the Helstrom measurement?

The Helstrom measurement is the measurement that has the minimum error probability when trying to distinguish between two states. For example, let's imagine you have two pure states $|\psi\rangle$ ...
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11 votes
Accepted

What is the relation between POVMs and observables (as Hermitian operators)?

One way of looking at the relationship between POVMs and observables arises from identifying their counterparts in the theory of probability of which quantum mechanics can be thought of as an ...
  • 14.8k
9 votes
Accepted

Are three POVM measurements on a single qubit physically realizable?

Three outcomes amounts to more than one bit if the outcomes are all deterministic, and give you information about the original qubit. But suppose I have a coin (that is either heads or tails). I ...
6 votes

What is an example of a measurement that is LOCC but not separable?

Such an example is given in Bennett et al., Quantum Nonlocality without Entanglement, Phys. Rev. A. 59, 1070 (1999).
6 votes
Accepted

What are examples of extremal non-projective POVMs?

It is indeed possible to have extremal, non-projective POVMs. Examples can be drawn from SIC-POVMs, as suggested in a comment. For example, as mentioned in the Wiki page, the only possible SIC-POVM in ...
  • 19.6k
6 votes
Accepted

Does a basis of maximally entangled states exist for two-qubit or two-qutrit system so that the density matrices of the basis states don't commute?

No such (orthonormal) basis can exist. An orthonormal basis $\{|\psi_i\rangle\}$ requires $\langle \psi_i | \psi_j \rangle = 0$ for $i\neq j$, and so clearly \begin{align} [\rho_i, \rho_j] &= |\...
  • 5,067
5 votes
Accepted

What does the POVM corresponding to single-qubit state tomography look like?

Quantum state tomography owes its power and flexibility to the fact that it supports a wide class of measurements. Any informationally complete POVM, i.e. one whose elements span the space $L_H(\...
  • 14.8k
5 votes

POVM three-qubit circuit for symmetric quantum states

This is not the unitary that you have to implement: you need a two-qubit unitary $$ \frac{1}{\sqrt{3}}\left(\begin{array}{cccc} 1 & 1 & 1 & 0 \\ 1 & \omega & \omega^2 & 0 \\ 1 &...
  • 48.2k
5 votes
Accepted

What are examples of non-trivial POVM measurements?

Here's a few examples POVMs with two components These are POVMs involving only two matrices $M_1,M_2\ge0$ with $M_1+M_2=I$. This implies that they are mutually diagonalisable, as discussed e.g. in ...
  • 19.6k
4 votes
Accepted

How do you embed a POVM matrix in a Unitary?

I'm not sure that I agree with what is presented as the solution (although the final answer seems OK). Let me explain what I would do. That task gives you 3 states $|A\rangle$, $|B\rangle$ and $|C\...
  • 48.2k
4 votes

Why are POVMs useful? Are they just an axiomatic way to define measurement?

If you want to prove a no-go theorem in quantum mechanics, such as the impossibility of distinguishing two non-orthogonal states given a single unknown state, you want to prove that even if you add ...
4 votes

Give an explicit example of a $d = 4$ SIC-POVM

As indicated by Danylo in his anwser, eq. (32) in arXiv: 1103.2030 presents the sixteen vectors ("ignoring overall phases and normalisation") \begin{equation} \left( \begin{array}{cccc} x & 1 &...
4 votes

How to express a probability distribution $P(x,y,z)= \sum_\lambda P(x|y,\lambda)P(y|\lambda,z)P(z)P(\lambda)$ in terms of a trace of a density matrix?

First, recall that $\mathrm{tr} A = \sum_i \langle i|A|i \rangle$. Each equation is then a sum where all terms are products of $P(z)$ and three other quantities. Further, the sum in the first equation ...
  • 14.8k
4 votes

Are SIC-POVMs optimal for quantum state reconstruction?

First of all, here's a short disclaimer: I'm not an in-depth expert in this field, I'm just currently getting in contact with tomography more and more often :) So take the following with a grain of ...
4 votes
Accepted

Why is $\| M|\psi\rangle \| \leq 1$ for POVM $M$?

Here's the simplest proof I could come up with. First note that by definition we have $M \leq I$ where $I$ is the identity operator. Now use $A \leq B \implies X^\dagger A X \leq X^\dagger B X$ with $...
  • 4,252
4 votes

Why is $\| M|\psi\rangle \| \leq 1$ for POVM $M$?

By the spectral theorem, there is an orthonormal basis $|\phi_k\rangle$ in which $M=\mathrm{diag}(\lambda_1,\dots,\lambda_n)$ with eigenvalues $\lambda_k\in[0,1]$ since $M$ is a POVM element. Let $a_k\...
  • 14.8k
3 votes
Accepted

How to find the POVM that optimally distinguishes between two given states?

The optimal probability of guessing correctly is $$ \frac12 + \frac12 \Big\|\frac23 \rho_0 - \frac13 \rho_1 \Big\|_1 $$ where $\| X \|_1 = \mathrm{Tr}[\sqrt{X^* X}]$ is the Schatten 1-norm. This ...
  • 4,252
3 votes
Accepted

Why are POVMs useful? Are they just an axiomatic way to define measurement?

For me, generalised measurements cover everything (obviously, that's why they're generalised), with projective measurements being a simple case that covers what we usually want to be doing. So, yes, ...
  • 48.2k
3 votes
Accepted

Why are entanglement breaking channels, defined as $\Phi(\rho)=\sum_a \operatorname{Tr}(\mu(a)\rho)\sigma_a$, entanglement breaking?

Minimalist formal proof (I'll use $\mu_a\equiv \mu(a)$): $\textrm{(A)}\Rightarrow\textrm{(B)}:$ Let $\Gamma\ge0$. Then, $$ (\Phi_A\otimes I_B)(\Gamma_{AB}) = \sum (\sigma_a)_A\otimes\mathrm{tr}_A[((\...
3 votes
Accepted

What are the matrices in the POVM for measuring the first $m$ qubits?

Well, since these are projective measurements on the subspace of the first $m$ qubits, we can just list all projectors on the computational basis of this first subspace and 'pad' them with $I$'s on ...
  • 4,898
3 votes
Accepted

How do I efficiently implement a POVM using a fixed universal gate set and the ability to measure in the standard basis?

What is the guarantee this implementation is efficient? Is there any rule regarding when implementing such POVMs is efficient? The implementation of such a gate will only depend on the parameter $k$ (...
  • 48.2k
3 votes

Give an explicit example of a $d = 4$ SIC-POVM

You can find it here Symmetric Informationally Complete Quantum Measurements or here SIC-POVMs: A new computer study, in the appendix B. Update Given a single fiducial vector $v = (a_1,a_2,a_3,a_4)^...
  • 6,013
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
Accepted

Does the dilation in Naimark's theorem produce a state?

First of all, $\|A|\psi\rangle\|^2 = (A|\psi\rangle)^\dagger(A|\psi\rangle) = \langle \psi | A^\dagger A|\psi\rangle = 1$. To see why your last expression equals to $1$ note that $$ \langle\psi\rvert ...
  • 6,013
3 votes
Accepted

Are projective measurements the only optimal measurements to discriminate between two states?

Suppose you are given either $\rho_1$ or $\rho_2$, and you also know that the probabilities you got one or the other are $p_1$ and $p_2$, respectively. If you have no prior knowledge of the ...
  • 19.6k
3 votes

How to distinguish between two very similar pure quantum states?

The claim does not specify what protocols for distinguishing quantum states are acceptable. In particular, it does not state whether we are allowed to err or reserve judgment. Below, we note success ...
  • 14.8k
3 votes
Accepted

Can any rank-$n$ POVM be realized as a rank-one POVM?

Is this an exercise? The answer is yes, you can do it, but I'm not going to tell you what the equivalent rank-1 POVM is. In general, any higher-rank POVM is equivalent to a rank-1 POVM if you relabel ...
2 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 $\sum_iP_i^\dagger P_i$, ...
  • 48.2k
2 votes

POVM three-qubit circuit for symmetric quantum states

Check this $$\frac{1}{\sqrt{3}}\left(\begin{array}{ccc} 1 & 1 & 1 \\ 1 & \omega & \omega^2 \\ 1 & \omega^2 & \omega \\ \end{array}\right) = \left(\begin{array}{cc} H & 0 \\ ...
  • 6,013
2 votes

Why are POVMs useful? Are they just an axiomatic way to define measurement?

POVMs model the most general way to measure something about a quantum state. I'd argue POVMs differ from "generalized measurements" in that they are tools built to answer different questions....
  • 19.6k

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