# Does Schmidt decomposition exist on infinite dimensional Hilbert spaces?

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?

• 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. Commented Oct 27, 2019 at 9:06
• 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 $udv$. Commented Oct 27, 2019 at 11:03
• 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. Commented Oct 27, 2019 at 11:09

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.