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Questions tagged [schmidt-decomposition]

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1 vote
2 answers
53 views

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

I am reading Nielsen & Chuang and they say this: "The bases $|i_A\rangle$ and $|i_B\rangle$ are called the Schmidt bases for A and B, respectively, and the number of non-zero values $\...
researcher101's user avatar
2 votes
2 answers
397 views

How to calculate the Schmidt decomposition of a state without SVD

I have this state of two qubits here: $$ |\psi_{AB}\rangle = \frac{1}{2}(|0\rangle_A |0\rangle_B + |1\rangle_A |1\rangle_B + |1\rangle_A |0\rangle_B - |0\rangle_A |1\rangle_B) $$ Which means that the ...
Alessandro Romancino's user avatar
2 votes
1 answer
73 views

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

$|\psi\rangle = 1/ \sqrt{3}( |0\rangle| 0\rangle + |0\rangle |1\rangle + |1\rangle |1\rangle) $ I absolutely cannot figure out the Schmidt decomposition of this state. I have looked at a ton of ...
qityhd's user avatar
  • 21
0 votes
1 answer
43 views

How to write Schmidt decomposition for pure tripartite state?

Supposing a pure tripartite state, i wrote schmidt decomposition as follows $\sum_{}^{r}p_{ijk}|i\rangle|j\rangle|k\rangle$ Did i write it correctly? Like pure bipartite states If r>1 then the ...
reza's user avatar
  • 761
3 votes
1 answer
92 views

A separable pure bipartite quantum state must be a product state

I'm looking for the simple argument to prove that a separable pure bipartite quantum state is in fact a product state. This question comes from a statement in Wikipedia on separable states: In the ...
JMark's user avatar
  • 173
6 votes
1 answer
365 views

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

It is relatively simple to derive the Schmidt decomposition of a pure state $|{\psi}\rangle \in H_A \otimes H_B$ with the SVD decomposition theorem. There are plenty of examples (lecture notes, books, ...
JMark's user avatar
  • 173
0 votes
0 answers
33 views

is the main purpose of using Schmidt decompositon to visualize entanglement?

Is it correct to say that an entanglement state is entangled in any basis, but this fact may not be evident (like in the state $ \frac{1}{ 2} | 00 \rangle + \frac{1}{ 2}| 01 \rangle + \frac{1}{ 2}| 10 ...
Hadamard's user avatar
1 vote
0 answers
40 views

Can we use a Werner state for quantum teleportation? [duplicate]

Some background: The quantum teleportation protocol requires first that Alice and Bob share an entangled state, say a Bell state $|\psi^{+}\rangle_{AB}$. There is another state $|\psi\rangle_{A'}$ to ...
Physkid's user avatar
  • 518
1 vote
0 answers
35 views

schmidt coefficients are the square root of the eigenvalue of the two partial trace of a density matrix

Let $\psi\rangle_{AB} = \sum_{i=1}^{d}\lambda_{i}|i_{A}\rangle |i_{B}\rangle$ be a state vector of a pure bipartite syste. Now, $\rho_{AB} = |\psi\rangle\langle\psi| = \sum_{i=1}^{d} \lambda_{i}^{2}|...
Physkid's user avatar
  • 518
1 vote
0 answers
50 views

General Bell state expression: What condition for mixture of Bell states to be entangled?

Convention: $|qubit_{A}, qubit_{B}\rangle$ The general Bell state equation: $|\beta(a,b)\rangle = \frac{1}{\sqrt{2}}\sum_{k=0}^{1}(-1)^{ka}|k, k\oplus b\rangle = \frac{1}{\sqrt{2}}[|0,0 \oplus b\...
Physkid's user avatar
  • 518
1 vote
1 answer
96 views

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

This is in reference to Nielson & Chang (page 109). Schmidt decomposition: suppose $|\psi\rangle$ is a pure state of a composite syste, AB. Then there exists orthonormal state $|i_{A}\rangle$ for ...
Physkid's user avatar
  • 518
4 votes
1 answer
269 views

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

Let $\psi\in H_A \otimes H_B$ be an entangled state, which means that it has Schmidt rank $r \geq 2$. Does there exist some $\epsilon>0$ for which all states $\varphi$ with $\|\psi - \varphi\|< \...
QuantumHumanLearner's user avatar
0 votes
1 answer
63 views

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

Consider two tetrapartite quantum states $|\phi\rangle^{AA^\prime BB^\prime}$ and $|\psi_1\rangle^{AA^\prime}|\psi_2\rangle^{BB^\prime}$ in a finite dimentional Hilbert space $\mathcal{H}^A\otimes\...
Takimoto.R's user avatar
2 votes
2 answers
219 views

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

I want to know that there is a relation between the distance of two vectors and the corresponding elements of the Schmidt bases. We assume that two bipartite vectors $|\phi\rangle^{AB}$ and $|\psi\...
Takimoto.R's user avatar
2 votes
1 answer
155 views

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

Dur, 2000 states that (...)But even in the simplest systems, $|\psi\rangle$ and $|\phi\rangle$ are typically not related by LU, and continuous parameters are needed to label all equivalence classes. ...
Steve J.'s user avatar
  • 173

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