6
votes
Accepted
Is 216 qumodes photonic quantum processor equivalent to 216 qubits superconducting quantum processor, in terms of computational power?
My name is Isaac and I work at Xanadu.
That’s a really good question! Generally speaking, you can argue that 216 qumodes are more powerful than 216 superconducting (SC) qubits since qumodes encode ...
5
votes
Is there a quantum gate that can turn any superposition $|\varphi \rangle $ into a unit column vector $|00\cdots 1\cdots 0\rangle $
It depends on what you mean.
For a given state $|\varphi\rangle$, it is always possible to find a gate $U_\varphi$ such that $U_\varphi|\varphi\rangle=|i\rangle_{10}$ for some given $i$.
However, it ...
4
votes
Accepted
Visualize representation of a qubit
One certainly can give examples of the form you're asking for. Here's one:
$$
\frac{1}{\sqrt{2}}(e^{i\pi/3}|0\rangle+e^{-i\pi/3}|1\rangle).
$$
However, note that you will essentially never see an ...
3
votes
How to find initial quantum states from the density matrix?
In general, for a given $\rho$ there are many ensembles, i.e. sets of pure states $|\psi_i\rangle$ and probabilities $p_i$ such that $\rho=\sum_i p_i|\psi_i\rangle\langle\psi_i|$. For example, if $\...
3
votes
Accepted
In what contexts are different notations used for indicating measurement outcomes?
It's an interesting question as to how to the bijection between the measurement eigenvalues $\{+1,-1\}$ and the bits $\{0,1\}$ or the qubits $\{|0\rangle,|1\rangle\}$ are reflected or intuited in the ...
3
votes
Accepted
Difficulty of creating arbitrary amplitude-encoded quantum states
Here's a crude argument (which one can undoubtedly pick holes in, but I think helps with intuition)...
Let's say you want to create an arbitrary state of $n$ qubits, so $N=2^n$. This means that you ...
3
votes
What are the best-known lower bounds on the number of measurements required for quantum state tomography?
This preprint is just submitted a few days ago:
An Improved Sample Complexity Lower Bound for Quantum State Tomography by Henry Yuen.
It shows that $\Omega(rd/\epsilon)$ copies of an unknown rank-$r$, ...
3
votes
What is the actual Hilbert space of a $N$-qubit system?
First off, the Bloch sphere is the complex projective line $\mathbb{C}P^1$, which is homeomorphic to $S^2$, while $SU(2)$ is homeomorphic to $S^3$. $SU(2^N)$ is the group of operators on pure states, ...
3
votes
Accepted
What is the actual Hilbert space of a $N$-qubit system?
Just a small remark for part of the question: Letting two Hilbert spaces $\mathcal{H}_1$ and $\mathcal{H}_2$ (this can be generalized to any linear space) the tensor product $\mathcal{H}_1 \otimes \...
2
votes
How does a Hadamard gate impact the initial/previous values of a Qubit?
I think you might just be misunderstanding the Hadamard gate slightly. In the computational basis, $H= \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1 \\ 1 & -1\end{pmatrix}$ maps $|0\rangle \...
2
votes
When starting from Qubit State |0>, which operations when applied (from right to left) will result in the qubit state |->
You can start by writing everything down in matrix-vector form. The $|0\rangle$ state is represented in the $Z$-basis by the vector $\begin{bmatrix}
1\\
0
\end{bmatrix}$. Applying the Hadamard-gate to ...
2
votes
Can the phase in a qubit be expressed as a phase on another qubit in a larger system?
We can write the total state as a tensor product of two qubits, but there is only one wavefunction of the total system. In fact $\vert1\rangle \otimes (e^{i\theta}\vert\psi\rangle)$ is exactly the ...
2
votes
Accepted
Can someone explain to me what is T gate and tdg like im five?
tdg is the method used to apply $T^\dagger$ (read T dagger). Thus, there are no differences between these two.
For a quantum gate $U$, $U^\dagger$ is the inverse of ...
2
votes
Accepted
what does it mean to "measure a qubit" if measurement is defined on operators
Measurement is defined with respect to a measurement basis. Given a state $\rho$, and a measurement basis $\{|u_k\rangle\}_k$, the measurement results in the outcome probabilities $p_k = \langle u_k|\...
glS♦
- 18.7k
2
votes
what does it mean to "measure a qubit" if measurement is defined on operators
I'm not sure if your question is about "what is a measurement", or what does "measuring a qubit" mean if we don't specify the operator or basis.
For the first answer, a generic ...
2
votes
Accepted
What is the difference between "maximally entangled" and "entangled" states?
A maximally entangled state is a state that maximises some entanglement measure.
In the case of bipartite states, this generally means a state that maximises the entanglement entropy, that is, the von ...
glS♦
- 18.7k
1
vote
How to tell if a state is entangled?
$\alpha\gamma|00\rangle + \alpha\lambda|01\rangle +\beta\gamma|10\rangle+\beta\lambda|11\rangle = (\alpha|0\rangle+\beta|1\rangle)\otimes(\gamma|0\rangle+\lambda|1\rangle)$ is a separable state, so it'...
1
vote
How to write a code to separate two qubits?
Maybe it should be useful tu use qutip to better understand what you desire (if you are familiar with python) by visualizing exactly the 2 qubits and its state vectors (which I believe is what you ...
1
vote
Is there a gate sending $|0\rangle^{\otimes n}$ to a state where some amplitudes are zero?
you can check out qiskit textbook,
https://qiskit.org/textbook/ch-gates/phase-kickback.html
https://qiskit.org/textbook/ch-quantum-hardware/density-matrix.html#5.-Mixed-States-in-the-Bloch-Sphere--
...
1
vote
When starting from Qubit State |0>, which operations when applied (from right to left) will result in the qubit state |->
To answer your question, the correct answer is option $(i)ZH$. This is because mathematically, the gates are applied in the right-to-left order and hence you apply the $H$ gate first taking us from ...
1
vote
Accepted
What is the difference between $|+\rangle$ and $|-\rangle$?
States with different complex amplitudes are simply different states with different observable consequences. They are not the same. We need them because they describe the outcomes of measurements.
To ...
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