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1 vote

Proability of measuing a qubit in a two qubit system without having a perpendicular basis

Lets presume your example is correct then the give homework answer follows from the example. Born's rule in quantum mechanics is a fundamental principle used to calculate the probability of measuring ...
Bram's user avatar
  • 564
1 vote

Prove that there is no polynomial size quantum algorithm for a Simon's problem with no promise on the input

Consider the set of values taken by $f$; that is, consider the range of $f$. For any arbitrary function $f$, this range can be partitioned as follows: There could be a first set of images $A$ that ...
Mark Spinelli's user avatar
4 votes
Accepted

To understand the notation of $| a, a \oplus b \rangle$

As you correctly wrote $|a,a\oplus b\rangle:=|a\rangle\otimes|a\oplus b\rangle$ where $a$ as well as $a\oplus b$ are either $0$ or $1$. The reason for the latter is that the logic symbol $\oplus$ ...
Frederik vom Ende's user avatar
-1 votes

Prove that there is no polynomial size quantum algorithm for a Simon's problem with no promise on the input

Assume You have a function $f(x) := \begin{cases} 1 & \text{if } x = \widehat{x} \\ 0 & \text{else}\end{cases} $ for an arbitrary value $\widehat{x}$....
Sezzart's user avatar
  • 160
0 votes

Creating a uniform superposition of a subset of basis states

Please refer to Algorithm 1 in https://arxiv.org/abs/2306.11747. This is the most efficient deterministic method of preparing the uniform superposition states. Instead of using the probabilistic ...
Quantum_Brains's user avatar
1 vote

Prove the fidelity can be written in terms of Pauli expectation values as ${\rm tr}(\rho\sigma)=\sum_k \chi_\rho(k)\chi_\sigma(\rho)$

A more general perspective on this is offered thinking in terms of operator frames. Suppose we're working in a $\mathbb{C}^d$. Let $(\mathcal O_i)_{i=1}^n$ be a set of Hermitian operators that spans ...
glS's user avatar
  • 24.8k
5 votes

Are measurements in $X$ and $Z$ bases being equal enough to prove equality of quantum states?

No, it is not possible, as you can clearly see from @CraigGidney's example. However, I would like to also provide some algebra as to why that is the case. The most general single qubit state has three ...
FDGod's user avatar
  • 2,391
5 votes
Accepted

Are measurements in $X$ and $Z$ bases being equal enough to prove equality of quantum states?

Consider $\frac{1}{\sqrt{2}}\big(|0\rangle + i|1\rangle\big)$ vs $\frac{1}{\sqrt{2}}\big(|0\rangle - i|1\rangle\big)$.
Craig Gidney's user avatar
  • 36.7k
0 votes

What is the IQ plane?

I know it a bit late for this, but in case I can help future people searching for this, this paper explains it quite well at the start: Efficient Z-gates for quantum computing Link: http://dx.doi.org/...
Guillermo Abad Lopéz's user avatar
1 vote

Show that quantum channels act as affine transformations in the Bloch sphere

There is already mistake in the very first line of your computation: When you write $\mathcal{E}(\rho)=[\ldots]=\frac12(I+\sum_l E_l(\vec{r}.\vec{\sigma})E_l^\dagger)$ you forgot to apply the channel ...
Frederik vom Ende's user avatar
6 votes

How can quantum error correction correct small rotations/continuous errors?

The trick is to rewrite your continuous rotations as a perturbative sum. For example, consider applying $R_Z(\theta)$ to all data qubits of an $n$-qubit code. You can rewrite: $$R_Z(\theta) = I \cos \...
Craig Gidney's user avatar
  • 36.7k
3 votes
Accepted

How can quantum error correction correct small rotations/continuous errors?

When you encode in an error correcting code, you select a subspace that your encoded qubit sits in. You can identify this with some projector $P$. For instance, if you know your logical 0 and logical ...
DaftWullie's user avatar
  • 57.9k
1 vote

Show that all extensions of $\rho$ can be obtained as a channel applied to its purification

Any purification of a state $\rho$ can be written as $\operatorname{vec}(\sqrt\rho V^\dagger)$ for some partial isometry $V$. More specifically, $V$ is an isometry on the support of $\rho$, so an ...
glS's user avatar
  • 24.8k
1 vote

How to convert one state of a quantum system to a desired state?

Here is a quantum circuit that first prepares your initial state and then transforms it to the second state. The circuit is partitioned into 6 parts by barriers (vertical lines) for providing a ...
Mathias's user avatar
  • 133
2 votes

How to generate the Bell state $\frac{1}{\sqrt{2}}(|01\rangle+|10\rangle)$ from the state $|00\rangle$ using Qiskit?

The Bell state $\frac{1}{\sqrt{2}}(|01\rangle + |10\rangle)$ can be generated from $|00\rangle$ in Qiskit as follows: ...
Nick Mertes's user avatar
2 votes

How to generate the Bell state $\frac{1}{\sqrt{2}}(|01\rangle+|10\rangle)$ from the state $|00\rangle$ using Qiskit?

Qiskit uses little-endian ordering for both classical bits and qubits. So, the least significant bit (LSB) is the rightmost bit in the binary representation of numbers. That means, you should use <...
Egretta.Thula's user avatar
8 votes

Why is the coefficient-squared the probability, and not just the coefficient itself?

Scott Aaronson describes quantum mechanics as "statistics but with the L2-norm". States are L2-norm unit vectors (sum of squared amplitudes is 1) instead of L1-norm unit vectors (sum of ...
Craig Gidney's user avatar
  • 36.7k
0 votes

How to create a qudit of any dimension $d$ in qiskit?

Qiskit has some support for qudits via the dims parameter in Statevector as described here https://docs.quantum.ibm.com/api/...
Nick Mertes's user avatar
3 votes

How can I get the state of my quantum circuit like save_statevector() in qiskit 1.0.2?

Aer is not gone, the namespace integration into qiskit where you could import it ...
Steve Wood's user avatar
  • 1,463
0 votes

How can I get the state of my quantum circuit like save_statevector() in qiskit 1.0.2?

You can use from qiskit.providers.fake_provider import GenericBackendV2, for example (I have slightly changed your code) ...
Roy Elkabetz's user avatar
1 vote

How to create a qudit of any dimension $d$ in qiskit?

In order to do that on a real hardware you will need to create your own discriminator for classifying higher energy states. An old example can be found in here for the case of qutrits (d=3). However, ...
Roy Elkabetz's user avatar
1 vote

Getting rid of extra qubits without changing phase

Just apply $\text{CNOT}$ from the first qubit to every other qubit. Let $|\psi\rangle$ be the initial state. $$|\psi\rangle = \frac{1}{\sqrt{2}}|0000\rangle + \frac{e^{i\phi}}{\sqrt{2}}|1111\rangle\,.$...
FDGod's user avatar
  • 2,391
2 votes

Can Interaction with the environment cause $|0\rangle$ to evolve into a superposition like $|+\rangle$?

It entirely depends on what "environment" and "interaction" you're talking about. As an extreme example, your "environment" could be a qubit in the state $|+\rangle$, and ...
glS's user avatar
  • 24.8k
2 votes
Accepted

Can Interaction with the environment cause $|0\rangle$ to evolve into a superposition like $|+\rangle$?

If the qubit is in connection with an environment, most likely it will develop quantum correlations (entanglement) and will become a mixed state. If the state is pure, the most general state is a ...
Zarathustra's user avatar
2 votes

Show that all extensions of $\rho$ can be obtained as a channel applied to its purification

Changing your notation for clarity, for a fixed state $\rho_A$, we are given $\sigma_{AE'}\in \text{D}(\mathcal{H}_{AE'})$ and $|\psi_{AE}\rangle \in \mathcal{H}_{AE}$ such that \begin{equation} \text{...
forky40's user avatar
  • 6,718
1 vote

Equivalence of quantum circuits

No, these two circuits won't be equivalent in general. Reasoning with states might not be the easiest way to understand it, I found expressing $A$ and $B$ in terms of matrices more clear. Consider $A =...
AG47's user avatar
  • 439
1 vote

Equivalence of quantum circuits

The circuits are not equivalent in general. As an example, take $A = B = X$. The unitary matrices representing the circuits are different: ...
Nick Mertes's user avatar
3 votes
Accepted

Equivalence of quantum circuits

Your problem is that your notation is leading you astray. In your first way of writing the circuit, you have the gate $B$ achieving the change in amplitudes $b'c\rightarrow b'c''$ while in the second ...
DaftWullie's user avatar
  • 57.9k
4 votes
Accepted

Existence of a two-outcome measurement $M$ such that the induced distributions differs between different density matrices

Yes, you can use $\rho - \sigma$ to construct a measurement where the measurement statistics differ for $\rho$ compared to $\sigma$. The issue is that $\rho - \sigma$ is not necessarily positive, ...
forky40's user avatar
  • 6,718
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

What is the difference between symmetric states and geometrically uniform states?

This is only a partial answer. Finding an actual counterexample would be cool! A set of symmetric quantum states is geometrically uniform, but the converse is not necessarily true. Note that $\left\{\...
Tristan Nemoz's user avatar
  • 6,162
1 vote

Efficient computation of probabilities for an N-particle state

For $\phi_p=0$, the normalization equals the permanent, which is hard to compute. This suggests that even the probability of one string $x_1,\dots,x_N$ is hard to compute. I'm not entirely sure what ...
Norbert Schuch'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
2 votes

Is it always possible to write the state corresponding to a set of stabilizer generators?

You can do this easily using Qiskit as follows: ...
Egretta.Thula's user avatar
6 votes
Accepted

Is it always possible to write the state corresponding to a set of stabilizer generators?

Given an $n$-qubit system and $n$ generators $g_i$ (which commute and square to identity), then the state that you are after satisfies $$ g_i|\psi\rangle=|\psi\rangle. $$ (This defines it uniquely, up ...
DaftWullie's user avatar
  • 57.9k
3 votes

Deriving the choi matrix definition of the quantum depolarizing channel

For $\mathcal E(X)=p\mathrm{tr}(X)\,\frac{\mathbb I}{2}+(1-p)X$, then $$\begin{align} \sigma &= (\mathcal E \otimes \mathbb I)(|\Omega\rangle\langle \Omega|)\\ &= \sum_{ij} \mathcal E(|i\...
HerrWarum's user avatar
  • 131
2 votes
Accepted

How can I shift elements around in my state vector according to a specific pattern?

In block notation, we can write $|\psi_1\rangle$ and $|\psi_2\rangle$ as $[A \quad B \quad C \quad D]^T$ and $[D \quad A \quad B \quad C ]^T$ respectively. So, it's a cyclic permutation of four ...
Egretta.Thula's user avatar

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