New answers tagged quantum-state
0
I wrote a small package for these kinds of operations, QM, that is available on GitHub here.
Here is a couple of examples (the code above will import directly the main file from GitHub without installing it, this is enough here but some other things will require that you install the full package):
Get["https://raw.githubusercontent.com/lucainnocenti/QM/...
2
Let's define the kets,
ket0 = {{1},{0}};ket1 = {{0},{1}};
This function produces input from a string,
f[x_?StringQ] := ToExpression[StringJoin["ket",x]];
This function produces the diagonal matrix corresponding to a string "000",
matrixFunc[x_]:= KroneckerProduct@@ f/@ StringPartition[x,1] .
ConjugateTranspose[KroneckerProduct@@ f/@ StringPartition[x,...
2
I am not following all of the calculations in your post (for one thing because your first displayed equation does not mention $C_1$), but I know why the goal of the exercise is true. In fact it is true for any code at all, not just a CSS code. If $\mathcal{C} \subseteq \mathcal{H}_\text{qubit}^{\otimes n}$ is any code, then you get an equivalent code $\...
3
The question presupposes a misconception that the vector form of a state $|\psi\rangle$ exists independently of its density operator form $|\psi\rangle\langle\psi|$, which is often described as secondary. In reality, the density operator of a state is all that truly exists --- and even then, it only exists as statistical information. In fact, you can ...
3
There isn't. A density matrix encodes all the knowledge available about a state, therefore if two states are described by the same density matrix, they are indistinguishable.
Ket vectors differing by only a global phase have always the same density matrix, and represent the same physical state.
3
We can begin this question by comparing two small cases, a qubit with a "rebit", the latter being a qubit confined to real amplitudes. The set of pure state of a rebit makes a great circle in the Bloch sphere, say a vertical circle. The corresponding measure on the probability interval is given by projecting uniform measure on the circle onto the $z$ ...
1
(I worked on this answer before the edit to the question, where the OP considers the real case with qutrits with the rather awesome plots. I'm not sure how the answer fits in with the edit, and this may be very wrong. If it is I'll kill it.)
To role up the comments, thinking only about the real case and only with a single qubit, I think we can say ...
4
There is a geometric interpretation that you certainly can take seriously, but the geometry that you get is not as clean as you might have hoped.
Trace distance between operator states is an example of a Banach norm on a vector space $V$. The rules for such a norm are that $||v|| > 0$ when $0 \ne v \in V$, $||\lambda v|| = |\lambda|\cdot||v||$ for $\...
3
There is one main key point in the description of your question: Is $s$ meant to be a classical secret or a quantum secret?
If $s$ is meant to be a classical secret, then the answer is yes, but there is not really much quantum in the positive answer. If $s_A$, $s_B$, and $s_C$ are all $d$-state digits, then there is a simple construction that works in ...
2
Given the constraints of the problem, we have for $U$ unitary,
$$
\newcommand{\ketbra}[1]{\lvert#1\rangle\!\langle #1\rvert}
\newcommand{\ket}[1]{\lvert#1\rangle}
\newcommand{\bra}[1]{\langle#1\rvert}
\newcommand{\sqoverlap}[2]{|\langle #1|#2\rangle|^2}
U=\ketbra{\lambda}+\sum_k e^{i\varphi_k}\ketbra{\lambda_k},$$
for some orthonormal basis $\{\ket\lambda\}\...
2
Any "person" (x or y) can obtain the state of z using the quantum circuit consisting of CNOT and Hadamard gates and $Z$ gate; the circuit is actually a simplified version of quantum teleportation; let x be in the state $|0\rangle$, that is he wants to go to the floor 2 (state $|1\rangle$ means floor 3). Applying CNOT to z (control qubit) and x entangles z ...
1
Here is another way to think about it.
You can, in principle, store an infinite amount of information into a qubit, in the sense that you might need arbitrarily many bits to exactly pinpoint its state.
However, this is not as weird or surprising as one could think. You can make the same argument about a (classical) probability distributions.
Given any ...
8
I'm not sure what passage in Nielsen and Chuang you have in mind and I see all of this differently. I don't see any need to believe that it is "theoretically" possible store an infinite amount of information in a qubit. My answer to the paradox is that amplitudes aren't stored information. A qubit does not know its amplitudes any more than a randomized ...
4
Highly Relevant: (Physics SE) Informational capacity of qubits and photons
Here's how I rationalized it: You take all the information you want to store, put it in binary form, and make it the real component of $\alpha $ or $\beta$ (the coefficients of the computational basis states).
Yes, and then if you could then prepare a qubit precisely in the $\...
3
For the general case, $m>2$, do what DaftWullie said, except apply gates corresponding to the $m$-point DFT matrix instead of the Hadamard gate (which is the 2-point DFT matrix).
Edit Per Request
For the $m=3$, $n=3$ use a gate ($M$) corresponding to the 3-point DFT matrix instead of the Hadamard gate to effect the transformation $\vert j \rangle =\...
5
To get you started, the $m = 2$ (qubit) case:
Start with n qubits all in the 0 state. Apply Hadamard gates to the first n-1 qubits. Apply n-1 controlled nots, each one controlled by a different one of the hadamarded qubits, and each targeting the nth qubit (the one we didn’t hadamard).
Here's an example circuit for $n=3$:
To understand how this works, ...
2
You certainly need to know all the other eigenvectors and eigenvalues. It's also important to know $|\mu\rangle$ and the nature of the operator $U$. Say, if $U$ is unitary, then the eigenvectors form an orthonormal set and their eigenvalues have modulus $1$. It's easy to resolve $|\mu\rangle$ along these orthonormal vectors.
To be concrete, suppose the ...
2
In the linked post, $|\psi_{mn}\rangle$ is defined as
$$
|\psi_{mn}\rangle = a_{00}|0_m 0_n\rangle + a_{10}|1_m 0_n\rangle + a_{01}|0_m 1_n\rangle + a_{11}|1_m 1_n\rangle,
$$
and the action of a CNOT operating between $m$-th and $n$-th qubit is written as
$$
\mathrm{CNOT}^{(m,n)}|\psi_{mn}\rangle = a_{00}|0_m 0_n\rangle + a_{10}|1_m 0_n\rangle + a_{11}|0_m ...
4
Apparently $\vert W \rangle$ was first reported (and the naming convention first adopted) by Dür, Vidal and Cirac in this preprint on May 26, 2000 (version 1 of 2).
This is supported by the footnote on page 4 of this preprint on June 25, 2000 (version 3 of 3, this footnote did not appear in the earlier versions), which states (in part)
Very recently Dürr ...
4
$\mathrm{X}$ is not equivalent to a $\mathrm{CNOT}$ gate. The former is a 1-qubit gate whereas the 2nd is a 2-qubit gate (in essence, a controlled-$\mathrm{X}$). The $\mathrm{X}$ basically flips the state of qubit B i.e., $|0\rangle_B\to|1\rangle_B$ and $|1\rangle\to|0\rangle_B$, and does not depend on the state of qubit A.
5
The key to understanding many quantum protocols and circuits is in the following circuit:
This is especially true in the case where $U^2=I$, such that $U$ has eigenvalues $\pm1$. You can readily calculate that if the input, $|\psi\rangle$, of the second qubit has an amplitude $\alpha_+$ for being supported on the $+1$ eigenspace, then at the end of the ...
4
What that cSWAP test does (and doesn't) do
The important thing about the controlled-SWAP test is that what it does isn't just to SWAP, or to not SWAP, the two inputs. The controlled-SWAP test involves a control qubit which is in a superposition of $\def\ket#1{\lvert#1\rangle}\def\bra#1{\langle#1\rvert}\ket{0}$ and $\ket{1}$: that is, we measure the first ...
Top 50 recent answers are included
