A Schmidt decomposition - How do I find the matrix related to the state?

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The discussion centers on understanding the Schmidt decomposition and its application in determining the entanglement of states. The user seeks clarity on how to represent a quantum state as a matrix to apply Singular Value Decomposition (SVD) and extract Schmidt coefficients. They reference a specific exercise involving a mapping from C^2⊗C^3 to C^6, which includes 36 unspecified constants, raising concerns about the completeness of the problem statement. The user expresses confusion over the multiple possibilities for the constants and their implications for the exercise. Overall, the thread highlights the challenges in applying Schmidt decomposition to complex quantum states.
Arquimedes
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How to writte the matrix associated to the state, so I can apply the Singular Value Decomposition to that matrix and get the Schmidt coefficients.
Hello, I am currently studying the Schmidt decomposition and how to use it to determine if a state is entangled or not and I can't understand how to write the state as a matrix so I can apply the Singular Value Decomposition and find the Schmidt coefficients. The exercise I am trying to complete is this one:;

problem25.png
 
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A general mapping from ##\mathbb{C}^2\otimes\mathbb{C}^3## to ##\mathbb{C}^6## may be written,

##v_i = \sum_{k=1}^2 \sum_{m=1}^3 \alpha_{i k m} u_k w_m##

there are 36 unspecified constants, ##\alpha_{i k m}##. So there are many possibilities. I think the problem statement is incomplete?
 
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Thread 'Angular Momentum Vector and Its Magnitude'

For the quantum state ##|l,m\rangle= |2,0\rangle## the z-component of angular momentum is zero and ##|L^2|=6 \hbar^2##. According to uncertainty it is impossible to determine the values of ##L_x, L_y, L_z## simultaneously. However, we know that ##L_x## and ## L_y##, like ##L_z##, get the values ##(-2,-1,0,1,2) \hbar##. In other words, for the state ##|2,0\rangle## we have ##\vec{L}=(L_x, L_y,0)## with ##L_x## and ## L_y## one of the values ##(-2,-1,0,1,2) \hbar##. But none of these...
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