9
votes
Accepted
Is it true that if $U$ sends computational basis states to product states, then it sends product states to product states?
TL;DR: The claim is false, i.e. it is not true that if $U|x\rangle$ is a product state for all computational basis states $|x\rangle$, then $U$ sends product states to product states. Counterexamples ...
- 20.7k
4
votes
Is it true that if $U$ sends computational basis states to product states, then it sends product states to product states?
No - the CNOT gate is a simple counterexample. It maps computational basis states to computational basis states, but it can be used to create entangled states (e.g. when acting on $|+\rangle|0\rangle$)...
- 5,647
2
votes
Accepted
Matrix representation for biproduct mixed states
I think you pretty much got it right. Yes you use the "Kronecker product representation" to represent tensor products as matrices/vectors for generic operators/vectors, and thus in ...
glS♦
- 23.3k
2
votes
Calculate the product state/quantum register back into its tensor product
You know that:
$$\begin{pmatrix}a_1\\a_2\end{pmatrix}\otimes\begin{pmatrix}b_1\\b_2\end{pmatrix}=\begin{pmatrix}a_1b_1\\a_1b_2\\a_2b_1\\a_2b_2\end{pmatrix}=\frac{1}{\sqrt{2}}\begin{pmatrix}\frac12+\...
- 5,324
2
votes
Accepted
For bipartite mixed state, if one part is pure, then the global mixed state is a product state?
Yes. Any mixed state $\rho$ is a convex combination of pure states, that is
$$
\rho = \sum_i \lambda_i |\phi_i\rangle\langle\phi_i|
$$
where $\lambda_i >0$, $\sum_i\lambda_i=1$. The partial trace ...
- 6,711
1
vote
Is my interpretation about the outer product representation correct?
If it is said that an $n$-qubit system is initially prepared in the product state $$ \rho_0 = |0\rangle \langle0|^{\otimes n} $$
Does that mean $\rho $
Here, you have switched to using the symbol $\...
- 848
1
vote
Simple proof that entangled pure states are not separable
simple proof that pure, entangled states are not separable
You won't find a proof. It's a matter of definition. We define entanglement by the statement "a state is entangled if it is not ...
- 55.7k
1
vote
Is factoring of a product state unique?
It's unique up to global phase.
$|0\rangle \otimes |0\rangle = (i |0\rangle) \otimes (-i|0\rangle)$
I think the easiest way to prove it is to go in reverse: start by assuming the factors are same/...
- 33.2k
1
vote
How to show that the QFT satisfies $\frac1{\sqrt N}\sum_j\prod_le^{2\pi i j_l k/2^l}|j_1...j_n⟩=\bigotimes_l \frac1{\sqrt2}(|0⟩+e^{2\pi i k/2^l}|1⟩)$?
Where every j mapped like:
So:
notice that:
Keep also this in mind:
- 1,174
1
vote
How to show that the QFT satisfies $\frac1{\sqrt N}\sum_j\prod_le^{2\pi i j_l k/2^l}|j_1...j_n⟩=\bigotimes_l \frac1{\sqrt2}(|0⟩+e^{2\pi i k/2^l}|1⟩)$?
We can transform the second expression as follows
$$
\begin{align}
\bigotimes_{l=1}^{n} \frac{1}{\sqrt 2}\left(|0\rangle + e^{2\pi i k/2^l} |1\rangle\right)
&=\frac{1}{\sqrt{2^n}}\bigotimes_{l=1}^{...
- 20.7k
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