# How do you find a Schmidt basis, and how can the Schmidt decomposition be used for operators?

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.

• 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] – Ruben Verresen Apr 23 '16 at 12:01
• [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'). – Ruben Verresen Apr 23 '16 at 12:02
• 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? – anonymous Apr 23 '16 at 12:26

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 $$A = \left( \begin{array}{cccc} |\psi_{A,0}\rangle & \cdots & |\psi_{A,\alpha} \rangle & \cdots\end{array} \right) \; \left( \begin{array}{cccc} \lambda_0 & 0 & 0 & 0 \\ 0& \ddots & 0 & 0 \\ 0 & 0 & \lambda_\alpha & 0 \\ 0 & 0 & 0 & \ddots \end{array} \right) \; \left( \begin{array}{c} |\psi_{B,0}\rangle \\ \vdots \\ |\psi_{B,\alpha} \rangle \\ \vdots\end{array} \right)$$