When performing the Schmidt decomposition of a bipartite state $\left|\psi\right>_{AB}=\sum_{ij}c_{ij}\left|v_i\right>_{A}\left|w_j\right>_{B}$ is computing the eigenvectors of the reduced density matrices $\rho_{A}=CC^{\dagger}$ and $\rho_{B}=C^{\dagger}C$ to the eigenvalues $\sigma(CC^{\dagger})=\sigma(C^{\dagger}C)=\{\lambda_{i}\}_{i}$ (singular values of $C$), denoted by $\{\left|a_i\right>_{A}\}$, $\{\left|b_i\right>_{B}\}$, in order to decompose the state like $\left|\psi\right>_{AB}=\sum_{i}\lambda_{i}\left|a_i\right>_{A}\left|b_i\right>_{B}$ a general procedure which always works? Do I have to try in a different manner when having multiple singular values? The state I'm tryng to decompose is $\left|\psi\right>_{AB}=\frac{1}{2}(\left|00\right>+\left|01\right>+\left|10\right>-\left|11\right>)$, for which this method doesn't seem to work.

$\left|\psi\right>_{AB}=\frac{1}{2}(\left|00\right>+\left|01\right>+\left|10\right>-\left|11\right>)$ leads to the coefficients matrix

$ C= \left[ {\begin{array}{cc} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & -\frac{1}{2} \\ \end{array} } \right] $,

which in turn results in the reduced density matrices

$ \rho_{A}=CC^{\dagger}= \left[ {\begin{array}{cc} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \\ \end{array} } \right]=C^{\dagger}C=\rho_{B} $.

Thus, the reduced density matrices share the eigenvectors $\left|0\right>$ and $\left|1\right>$ for the repetead eigenvalue $\frac{1}{2}$. I thought then the bipartite state could have been decomposed like $\left|\psi\right>_{AB}=\sum_{i}\lambda_{i}\left|a_i\right>_{A}\left|b_i\right>_{B}$, using the singular values and the eigenvectors of $\rho_{A}$ and $\rho_{B}$. However, $\frac{1}{\sqrt{2}}\left|0\right>\otimes\left|0\right>+\frac{1}{\sqrt{2}}\left|1\right>\otimes\left|1\right>$ does not yield the state I wanted to decompose.

Thanks in advance

  • 2
    $\begingroup$ Why does it not work? You have to add details, your question is unclear as it is now. $\endgroup$ – Norbert Schuch Dec 30 '19 at 22:08
  • $\begingroup$ it's not clear what you are trying to do. To find the Schmidt decomposition you need only find the SVD of $C$ (which in this case is also symmetric, making the task easier). You cannot reconstruct the state from the reduced density matrices, like you seem to be trying to do $\endgroup$ – glS Jan 9 at 1:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.