I am trying to understand the Schmidt Decomposition, currently in my QC class. We had a tutorial where we were told if $|\psi\rangle$ is a pure state of a composite system A then there exists $|i_A\rangle$ and $|i_B\rangle$ which are orthonomral states for systems A and B respectively.

$$|\psi\rangle = \sum_i \lambda_i |i_A\rangle|i_B\rangle$$

I more or less understand that we write $|\psi\rangle$ as:

$$ |\psi\rangle = \sum_{j,k} c_{jk}|a_j\rangle|b_k\rangle$$

Where $|a_j\rangle_j$ and $|b_k\rangle_k$ are orthonormal bases for A and B and $C=(c_{jk})$ a complex matrix. Then proceed by using singular value decomposition on C. My question is how should I move on to apply the Schmidt decomposition on a numerical example such as the $|\beta_{00}\rangle$. Some pointers would be lovely as I struggle to understand, and would prefer it if there were no direct answers as I want to struggle through it myself as much as possible. Thanks!


1 Answer 1


First suggestion: don't try it on a Bell state! This is already in the Schmidt basis, and lots of things that you do to the state will keep in Schmidt decomposed as well!

Instead, why not try an example such as $$ |\psi\rangle=\frac{3}{5\sqrt{2}}|00\rangle+\frac{3}{5\sqrt{2}}|01\rangle+\frac{4}{5\sqrt{2}}|10\rangle-\frac{4}{5\sqrt{2}}|11\rangle $$ So, as you say, work out what the matrix $C$ is, and perform a singular value decomposition on it. From there, try to extract $|i_A\rangle, |i_B\rangle$ and $\lambda_i$ values. (is this all you were after? since you don't want too many hints, I'm not sure what else to say.)

  • $\begingroup$ I was working through the given examples and first one was the bell states. Now it makes sense, I will try doing the one example you gave me! Thanks a lot. $\endgroup$ Jun 16, 2020 at 13:39
  • $\begingroup$ Just to make sure: $\sum_{i=0,1}= \frac{1}{\sqrt(2)}|i\rangle|i\rangle$ would be the decomposition of $|\beta_{00}\rangle$ right? Ah thanks so much everything is proper now I was stuck at such a weird place. $\endgroup$ Jun 16, 2020 at 13:44
  • $\begingroup$ Yes, that's right. $\endgroup$
    – DaftWullie
    Jun 16, 2020 at 14:05

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