Schmidt Decomposition: Is It Enough to Find States?

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Schmidt decomposition is essential for understanding bipartite systems, as it indicates whether a state is entangled based on the number of Schmidt coefficients. Every bipartite state can be decomposed, but only entangled states have more than one coefficient. The lack of responses may be due to the complexity of the topic or timing of the posts. Engaging discussions on technical subjects can sometimes take longer to elicit replies. Overall, Schmidt decomposition remains a critical concept in quantum mechanics for identifying entanglement.
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Is entangle enough to find that given states have Schmidt decomposition?
 
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Why didn't anyone reply me?
 
Hello,
Schmidt decomposition is a usefull tool for bipartite systems: if it admits a decomposition with more than one Schmidt coefficient, then the overall state is an entangled one. So every bipartite state has a Schmidt decomposition, but only if it's entangles there's more than one coefficinet in the expansion.
 
ber70 said:
Why didn't anyone reply me?

are you cute? you waited 6hours only?
 
I'm sorry you waited so long, but I just signed in yesterday and life is so involving that someone sometimes has to wait a little...:-)
 

Thread 'Two equivalent statements of time reversal symmetric Hamiltonian'

Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?
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