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Consider a system in the state $|\Psi\rangle=\frac{1}{2}\left(|00\rangle+|01\rangle+|10\rangle+|11\rangle\right)$. This state is easily seen to not be an entangled state, since

$$|\Psi\rangle=\frac{1}{\sqrt2}(|0\rangle+|1\rangle)\otimes\frac{1}{\sqrt2}(|0\rangle+|1\rangle).$$

But if I want to calculate the Schmidt decomposition of the state $|\Psi\rangle$ I do not obtain this result.

The state $|\Psi\rangle$ can be written as $|\Psi\rangle=\mathrm{diag}(\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{1}{4})$.

Now, you have to calculate the singular value decomposition of this matrix. But I don't understand how I know which basis I have to choose for $V$ in order to obtain the Schmidt representation (decomposition) of this vector. How do I know which basis I have to choose?

My second question is how I can apply the Schmidt decomposition to Operators or Matrices. Apparently this is possible, but I do not know how this is doable.

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  • $\begingroup$ For the Schmidt decomposition you first partition your space in two pieces, say A and B. The Schmidt decomposition is then the (unique) expression $|\psi \rangle = \sum_\alpha \lambda_\alpha |\psi_{A,\alpha}\rangle \otimes |\psi_{B,\alpha}\rangle$ where $\{ |\psi_{A,\alpha} \rangle \}_\alpha $ forms an orthonormal basis on subspace A (and similarly for B). So by virtue of you rewriting $|\psi \rangle$ as product state between site one and two, you have in fact calculated that the Schmidt decomposition is given by $\lambda_0 = 1$, $\lambda_1 = 0$, and [continued] $\endgroup$ – Ruben Verresen Apr 23 '16 at 12:01
  • $\begingroup$ [continuing] $|\psi_{A,0}\rangle = \frac{1}{\sqrt{2}} \left( |0\rangle + |1 \rangle \right)$, $|\psi_{A,1}\rangle = \frac{1}{\sqrt{2}} \left( |0\rangle - |1 \rangle \right)$, and similarly for the second site ('B'). $\endgroup$ – Ruben Verresen Apr 23 '16 at 12:02
  • $\begingroup$ Thanks for your answer. But I don't really see how I can apply this in a more General case. In this example, it was pretty easy to guess the orthonormal basis of A and B. But how can you find this basis in a more general and more complex example? $\endgroup$ – anonymous Apr 23 '16 at 12:26
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Okay, let me elaborate on my comment to show how you would calculate the Schmidt decomposition in general. This might also answer your second question.

As I said in my comment, the Schmidt decomposition requires you to subdivide your system in two parts, A and B. It is then the (unique) decomposition $|\psi \rangle = \sum_\alpha \lambda_\alpha |\psi_{A,\alpha} \rangle \otimes |\psi_{B,\alpha} \rangle $ (where the components define orthonormal bases in A and B). This can be considered the decomposition that minimally entangles the two subsystems (the entanglement being given by the Schmidt values $\lambda_\alpha$). To calculate this decomposition, one rewrites the state as a matrix and then applies the SVD decomposition.

It is in fact simple to rewrite the state as a matrix: one pretends that the wavefunction indices concerning subsystem A are the row indices, and the indices for subsystem B are the column indices. For example, for your state $|\psi\rangle = \frac{1}{2}\left( |00 \rangle +|01\rangle + |10 \rangle + |11\rangle \right)$, we can write this as $|\psi\rangle = \sum_{ij} A_{ij} |ij\rangle$ with $A_{00} = A_{01} = A_{10} = A_{11} = \frac{1}{2}$. We can consider this to define a matrix $$A = \frac{1}{2}\left( \begin{array}{cc} 1 & 1 \\ 1 & 1 \end{array} \right).$$ Now if you look up the definition of the SVD decomposition, it is not hard to see that this then exactly gives us what we want for the Schmidt decomposition. In this case SVD gives us $$A = \frac{1}{\sqrt{2}}\left( \begin{array}{cc} 1 & 1 \\ 1 & -1 \end{array} \right) \; \left( \begin{array}{cc} 1 & 0 \\ 0& 0 \end{array} \right) \; \frac{1}{\sqrt{2}}\left( \begin{array}{cc} 1 & 1 \\ 1 & -1 \end{array} \right)$$ This then exactly gives us the $\lambda_\alpha$ and $|\psi_{A,\alpha}\rangle$ that I wrote down in my comment to your original post.

More generally, if our state is $|\psi\rangle = \sum_{ij} A_{ij} |i_A\rangle \otimes |j_B\rangle$, then the Schmidt decompositions is given by the SVD decomposition as $$A = \left( \begin{array}{cccc} |\psi_{A,0}\rangle & \cdots & |\psi_{A,\alpha} \rangle & \cdots\end{array} \right) \; \left( \begin{array}{cccc} \lambda_0 & 0 & 0 & 0 \\ 0& \ddots & 0 & 0 \\ 0 & 0 & \lambda_\alpha & 0 \\ 0 & 0 & 0 & \ddots \end{array} \right) \; \left( \begin{array}{c} |\psi_{B,0}\rangle \\ \vdots \\ |\psi_{B,\alpha} \rangle \\ \vdots\end{array} \right) $$

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  • $\begingroup$ Hey, thanks for your answer. I think I understood now how to find a Schmidt decomposition, but I still didn't get how I could apply this to Operators. I can't find any sources or something on how to apply the Schmidt decomposition for Operators. $\endgroup$ – anonymous Apr 24 '16 at 18:53
  • $\begingroup$ Where did you read this is possible? I was thinking that maybe you confused it with the fact that to calculate the Schmidt decomposition for a state, you calculate the SVD decomposition for a matrix, which is an operator (this is what I showed above). If you mean something else, then where did you read this? $\endgroup$ – Ruben Verresen Apr 24 '16 at 20:55
  • $\begingroup$ I read it here njohnston.ca/2014/06/… and here arxiv.org/ftp/arxiv/papers/1006/1006.3412.pdf $\endgroup$ – anonymous Apr 25 '16 at 5:07

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