A pure quantum state in a bipartite system, which is an operator $$\rho = \langle\psi \,,\, \cdot \,\rangle \, \psi \in \mathcal{L}(H_1 \otimes H_2)$$ for some $\psi \in H_1 \otimes H_2$, is factorizable (i.e. not entangled) iff the reduced density matrices $\rho_s$ are such that $\rho_s^2 = \rho_s$, where $\rho_s$ is the partial trace of $\rho$ over $H_s$ for $s= 1$ or $2$.

To prove this result for finite dimensional Hilbert spaces $H_1$ and $H_2$ , we use Schmidt decomposition in those spaces. And the decomposition exists because of the SVD theorem.

I want to know if it is also true for separable Hilbert spaces in general, where a lot of quantum mechanics happen. I believe it is, but haven't found a proof. So, is SVD theorem valid for "infinite matrices"? Or, if it is not, is there another way to prove there exists a Schmidt decomposition (a series) in this case?

  • $\begingroup$ What exactly do you mean by "Schmidt decomposition" and "infinite matrices"? Hilbert-Schmidt operators might be what you are looking for, but without further information it's hard to say. $\endgroup$
    – MaoWao
    Commented Oct 27, 2019 at 9:06
  • $\begingroup$ If I have two Hilbert spaces $H_{A}$ and $H_{B}$ and a basis in each one of them, any element $v$ in $H_{A} \otimes H_{B}$ can be written as $v = \sum \alpha_{ij} a_{i} \otimes b_{j}$. Then you can prove that there exists a basis for $H_{A}$ and another for $H_{B}$ such that this double sum turns into a simple one : $v = \sum \gamma_{i} m_{i} \otimes n_{i}$. Even if one space has more dimensions than the other. That is the Schmidt decomposition I'm talking about. This is a direct consequence of writing the matrix $\alpha$ of the coefficients as $udv$. $\endgroup$
    – Sahdo
    Commented Oct 27, 2019 at 11:03
  • $\begingroup$ And when I say that I want to know if this works for infinite dimensional spaces, I mean two separable Hilbert spaces for which I can define a countable ordered basis. That' why if SVD theorem works for infinite matrices, it solves my problem. $\endgroup$
    – Sahdo
    Commented Oct 27, 2019 at 11:09

1 Answer 1


First note that the Hilbert space tensor product $H_A\otimes H_B$ can be isometrically identified with the space of Hilbert-Schmidt operators from $H_A$ to $H_B$ via $$ x\otimes y\mapsto \langle \,\cdot\,,x\rangle y. $$ Every Hilbert-Schmidt operator $T\colon H_A\to H_B$ is compact and therefore has a singular value decomposition $$ T=\sum_{n} s_n\langle\,\cdot\,,e_n\rangle f_n $$ with ONBs $(e_n)$ of $H_A$, $(f_n)$ of $H_B$ and a positive sequence $(s_n)\in\ell^2$. Thus every $v\in H_A\otimes H_B$ has a representation of the form $$ v=\sum_{n} s_n e_n\otimes f_n $$ with $(e_n),(f_n),(s_n)$ as above.

The proofs can be found in virtually any introductory text on functional analysis.

  • 1
    $\begingroup$ I revisited this question and it wasn't making much sense, in fact it had some wrong affirmations, so i corrected it. Even so, you managed to understand what I wanted to know at the time haha, thank you :) $\endgroup$
    – Sahdo
    Commented Jun 20, 2020 at 20:50

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .