- #1
- 10,876
- 423
I would be interested in seeing a correct statement and proof of the Schmidt decomposition theorem that (I think) says that if x\otimes y\in H_1\otimes H_2, there exists a choice of bases for these Hilbert spaces, such that
x\otimes y=\sum_{n=1}^\infty \sqrt{p_n}\ a_n\otimes b_n
with \sum_n\sqrt{p_n}=1. I'm not even sure that this is what the theorem says. Maybe it applies to all members of the tensor product space, and not just members of the form x\otimes y. Wikipedia has a proof for the finite-dimensional case (here), but I'm more interested in the infinite-dimensional case. (I haven't made the effort to try to understand the finite-dimensional case yet). I found this, but I don't see how to relate it to what I wrote above. Maybe it's a completely different theorem, but there are some similarities with Wikipedia's approach (they both talk about eigenvalues of operators of the form (T*T)1/2) that suggest that this is in fact the right theorem, and that I just need to figure out how to use it. (The pages that can't be read at Google Books can be read at Amazon).
x\otimes y=\sum_{n=1}^\infty \sqrt{p_n}\ a_n\otimes b_n
with \sum_n\sqrt{p_n}=1. I'm not even sure that this is what the theorem says. Maybe it applies to all members of the tensor product space, and not just members of the form x\otimes y. Wikipedia has a proof for the finite-dimensional case (here), but I'm more interested in the infinite-dimensional case. (I haven't made the effort to try to understand the finite-dimensional case yet). I found this, but I don't see how to relate it to what I wrote above. Maybe it's a completely different theorem, but there are some similarities with Wikipedia's approach (they both talk about eigenvalues of operators of the form (T*T)1/2) that suggest that this is in fact the right theorem, and that I just need to figure out how to use it. (The pages that can't be read at Google Books can be read at Amazon).
Last edited:
