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1 vote

Representing a von Neumann measurement as $[\mathcal{I} \otimes P_i] U(\rho_s \otimes \rho_a)U^{-1} [\mathcal{I} \otimes P_i]$, how do we choose $U$?

No such representation exists, unless the underlying spaces were trivial to begin with. To see this let us first simplify notation by defining $\rho_{sa}:=U(\rho_s\otimes\rho_a)U^{-1}$ (which is ...
Frederik vom Ende's user avatar
2 votes

How to take partial trace of a $n - 1$ qubit subsystem from a $n$ qubit system

While the comment of glS contains the mathematical tools needed here, this particular problem features a bit more structure. For simplicity I will treat the case $n=2$ but all that follows does ...
Frederik vom Ende's user avatar
0 votes

Exercise 11.7 in Nielsen & Chuang and basic properties of Shannon entropy

If my calculations are not completely off the mark, the strategy of employing a flat distribution as our $\tilde p(y)$ sometimes works (I'm adding a tilde to make clear that it has nothing to do with $...
atlantropa's user avatar
2 votes

The matrix norm $\|A\|=\max_{\langle u|u\rangle=1}|\langle u|A|u\rangle|$ in the proof of Lieb's theorem

The short answer is that there are two problems with your argument: The partial derivative of $u^TAu$ you state is wrong More gravely, the whole approach is flawed because you're treating $u$ as a ...
Frederik vom Ende's user avatar
1 vote

Show that while calculating partial traces the probability is independent of the basis of one of the measurements

Independence of the slightly more general expression $\sum_n\langle x\otimes \beta_n|\rho_{AB}|y\otimes \beta_n\rangle$ from the orthonormal basis $\{\beta_n\}_n$ boils down to the fact that the same ...
Frederik vom Ende's user avatar
1 vote

How do I show that a reduced density matrix of $1$ is $\rho_{12}^{1} = \text{Tr}_{2}[\rho_{12}] = \sum_{i}\langle i_{2} | \rho | i_{2} \rangle$?

Complementing FDGod's comment, this question would make sense if you had first defined the partial trace via the duality ${\rm tr}({\rm tr}_1(\rho)\omega)={\rm tr}(\rho({\bf1}\otimes\omega))$ as is ...
Frederik vom Ende's user avatar
0 votes

Exercise 11.7 in Nielsen & Chuang and basic properties of Shannon entropy

It's just a thought experiment inspired by John Watrous's deleted comment. Suppose we have a probability distribution $p(x,y)$ and we want to approximate it with $q(x,y)$. Suppose that we know $p(x)$, ...
MonteNero's user avatar
  • 2,491
3 votes
Accepted

Bound on success Probability for Regev's factoring algorithm

This is just Markov's inequality that states that for a positive random variable $X$ and $a > 0$ then $$ P( X \geq a ) \leq \frac{E(X)}{a}. $$ see the wikipedia: https://en.wikipedia.org/wiki/...
Frederik Ravn Klausen's user avatar
4 votes
Accepted

Why can't the eigenvalues of a unitary matrix have the form $e^{i\theta}$?

in case of $e^{2\pi\cdot i\cdot \theta}$ the values of $\theta\in [0,1]$, in the case of $e^{i\cdot \theta}$ the values of $\theta \in [0,2\pi]$. it's just a different convention but they are ...
Sezzart's user avatar
  • 160
0 votes

Is $|A\rangle = \frac{1}{\sqrt2} |00\rangle + \frac{1}{\sqrt2} |01\rangle$ a valid quantum state?

Given that we are talking about the state of 2 qubits, the canonical basis of the state space is $$(|00\rangle,|01\rangle,|10\rangle,|11\rangle)$$ A physically valid quantum state is just a vector in ...
Pierre-Paul T.'s user avatar
2 votes

An Introduction to Quantum Computing - Exercise 6.4.1

The proof of this would be quite empirical in nature. You can rewrite your state in terms of product as follows: $$ \frac{1}{\sqrt{2^n}} \prod_{i=1}^n (|0\rangle + (-1)^{x_i}|1\rangle) $$ We can write ...
Zee's user avatar
  • 361
1 vote

An Introduction to Quantum Computing - Exercise 6.4.1

A way to see how this identity comes about is to simply work out the first few terms from your Eq.(2) upon expanding the terms inside brackets. You will notice that qubit $k$ will contribute with a ...
Ezequiel Rodriguez Chiacchio's user avatar
1 vote

Help finding mistake when modifying $T$ injection protocols

After the second $H\otimes H$ operator, once the mistake with the CNOT is corrected, the qubits should be in this state: $$\frac{1}{\sqrt{2}}\left(\alpha\left|00\right\rangle + \beta\omega\left|01\...
AG47's user avatar
  • 476
2 votes
Accepted

Help finding mistake when modifying $T$ injection protocols

When applying the CNOT(1 $\rightarrow$ 2) you got confused in the last row. The $(1-\omega)|1\rangle$ flips the second qubit so $(\alpha+\beta)|0\rangle$ turns into $(\alpha+\beta)|1\rangle$ (and the ...
David Dentelski's user avatar
0 votes

Is $|A\rangle = \frac{1}{\sqrt2} |00\rangle + \frac{1}{\sqrt2} |01\rangle$ a valid quantum state?

To add to the answer from Yet another Random Guy, you can see for yourself by recreating the state programmatically (here using the Amazon Braket SDK): ...
Cody Wang's user avatar
  • 1,203
6 votes

Is $|A\rangle = \frac{1}{\sqrt2} |00\rangle + \frac{1}{\sqrt2} |01\rangle$ a valid quantum state?

Yes, the state $ |A\rangle = \frac{1}{\sqrt{2}} |00\rangle + \frac{1}{\sqrt{2}} |01\rangle $ is indeed a valid quantum state. In quantum mechanics, a valid quantum state can be any normalized linear ...
Yet another Random Guy's user avatar

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