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9 votes
Accepted

How to check if a $n$-qubit unitary is the tensor product of single-qubit unitaries

TL;DR: A two-qubit unitary operator $U$ is a product operator, i.e. $U=U_1\otimes U_2$ for some single-qubit unitaries $U_1$ and $U_2$ if and only if $U$ has Schmidt rank one $$ U = U_1\otimes U_2 \...
Adam Zalcman's user avatar
7 votes
Accepted

Schmidt decomposition for tripartite system $ABC$ with vanishing mutual information between $A$ and $C$

TL;DR: The key observation is that Schmidt basis on a subsystem consists of eigenvectors of the reduced state of that subsystem. Consequently, if the reduced state is a product state then its Schmidt ...
Adam Zalcman's user avatar
5 votes
Accepted

Open neighborhood of an entangled state with non-decreasing Schmidt rank

TL;DR: Yes. The set of pure bipartite quantum states with Schmidt rank at least $r$ is open$^1$. Let $S_{r}$ denote the set of all pure bipartite quantum states in $H_A\otimes H_B$ with Schmidt rank ...
Adam Zalcman's user avatar
4 votes

Is it possible to derive a Schmidt decomposition for a mixed state?

Usually, the Schmidt rank of a mixed state $\rho$ is defined as the minimum of the maximal Schmidt rank of an element in any decomposition of $\rho$ into a mixture of pure states (there are many ...
Danylo Y's user avatar
  • 7,442
4 votes
Accepted

How to calculate the Schmidt decomposition of a state without SVD

Why do you need to calculate the Schmidt decomposition without SVD? Doing it in other ways will just lead to the same result. That aside, the problem with the calculation here is that knowing the ...
glS's user avatar
  • 25.5k
4 votes

What's the Schmidt decomposition of $|\psi\rangle = 1/ \sqrt{3}( |0\rangle| 0\rangle + |0\rangle |1\rangle + |1\rangle |1\rangle)$?

One way of computing the decomposition is through density matrices, but then you will have to diagonalize those density matrices. This requires the eigenvalue decomposition of each density matrix. ...
MonteNero's user avatar
  • 2,684
4 votes
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Does proximity of two bipartite states in a norm force high overlap between the elements of the Schmidt bases?

No. Here is an example without small Schmidt coefficients. To this end, consider $$ \lvert\phi\rangle = a\lvert0\rangle\lvert0\rangle + b \lvert1\rangle\lvert1\rangle\ , $$ and $$ \lvert\psi\rangle = ...
Norbert Schuch's user avatar
4 votes
Accepted

Prove that there are infinitely many two-qubit entanglement classes under LU

TL;DR: This is an application of Schmidt decomposition followed by local basis change on each qubit. By Schmidt decomposition any two-qubit state $|\psi\rangle$ may be written in the form $$ |\psi\...
Adam Zalcman's user avatar
3 votes

SWAPing Schmidt vectors

TL;DR: In general, such "inner product" is not well-defined. This can be remedied by appropriate choice of $A$ and $B$. Still, the inner product depends on the global phase and thus cannot ...
Adam Zalcman's user avatar
3 votes
Accepted

Truncating the bond dimension of an MPS -- how good is the approximation?

The answer is found in F. Verstraete and J.I. Cirac, Matrix product states represent ground states faithfully, Lemma 1 on pg. 5 in the arxiv version. It states that the total error in approximating a ...
Norbert Schuch's user avatar
2 votes

For tetrapartite state, and another way of decomposition, is the Schmidt basis separable?

This is immediately disproved by the structure of the counterexamples to your previous question Does proximity of two bipartite states in a norm force high overlap between the elements of the Schmidt ...
Norbert Schuch's user avatar
2 votes

Does proximity of two bipartite states in a norm force high overlap between the elements of the Schmidt bases?

TL;DR: No such relation exists, because the upper bound on the norm fails to impose any constraints whatsoever on the basis elements corresponding to very small Schmidt coefficients $\sqrt{p_i}$ and $\...
Adam Zalcman's user avatar
2 votes

A separable pure bipartite quantum state must be a product state

Definitions By definition, a bipartite quantum state $\rho$ is a product state if $\rho=\sigma\otimes\tau$. By another definition, $\rho$ is separable if it can be written as a convex combination of ...
Adam Zalcman's user avatar
2 votes

Why does it matter that Schmidt number is invariant under unitary transformations?

Regarding your first question. Consider having the following bipartite state: $$ |\psi\rangle_{AB} = \frac{1}{2}|00\rangle_A|00\rangle_B+ \frac{\sqrt{3}}{6}|01\rangle_A|01\rangle_B+ \frac{\sqrt{6}}{6}...
diemilio's user avatar
  • 472
2 votes

Why does it matter that Schmidt number is invariant under unitary transformations?

Crucially, "the Schmidt number is preserved under unitary transformations on system A or system B alone". So Schmidt numbers are invariant under local operations $U_A\otimes U_B$, but not ...
Frederik vom Ende's user avatar
1 vote

How to calculate the Schmidt decomposition of a state without SVD

When a question says "without using the SVD" it may be implying that there are easier ways, at least in this specific case. It's always worth just having a stare at the state for a minute (...
DaftWullie's user avatar
  • 59.3k
1 vote
Accepted

How to write Schmidt decomposition for pure tripartite state?

Assuming the Schmidt decomposition exists, it would be written as: $$\sum_{i=1}^rp_i\left|i_A\right\rangle\left|i_C\right\rangle\left|i_B\right\rangle$$ With $\left\{\left|i_A\right\rangle\right\}_i$, ...
Tristan Nemoz's user avatar
  • 6,692
1 vote
Accepted

Why does Schmidt decomposition (2 qubits) requires density matrix of each system?

The reason why they go from the pure state to the individual density matrices of the subsystems is that this is going to help you find the Schmidt coefficients and the orthonormal bases for both sites....
DaftWullie's user avatar
  • 59.3k

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