The Schmidt decomposition theorem says:



> 
> If $H\_1,H\_2$ are Hilbert-spaces (for simplicity: of same dimension) and
>  $x\in H\_1\otimes H\_2$, then there exist orthonormal bases
>  $\alpha\_i,\beta\_i$ of $H\_1,H\_2$, and reals $\lambda\_i\geq0$ such that
>  $x=\sum\_i \lambda\_i\alpha\_i\otimes\beta\_i$.
> 
> 
> 


I know that this holds for **separable** Hilbert spaces (e.g., Breuer, Petruccione, The theory of open quantum systems).


Does this also hold for **nonseparable** Hilbert spaces? (If possible, please provide a reference to a paper or textbook.)



Schmidt decomposition on infinite-dimensional Hilbert spaces
Yes. Christian Remling's comment in essence already gave an affirmative answer, and this question is a bit old, but I think it's worthwhile to write a proof that does not require reducing to the separable situation.


I encountered the same question myself and have not found any reference---but with the polar decomposition and spectral theorems, the result follows with little fuss; this leads me to be believe it is almost certainly known, if not written down somewhere on some ancient scroll of functional analysis. (But the fact that I haven't seen it written down in general form compels me to put it out into the internet ether.)




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I take my Hermitian inner products to be antilinear in the first entry and linear in the second. If $\mathcal{H}$ is a Hilbert space, the continuous dual of a Hilbert space $\mathcal{H}$ will be denoted $\mathcal{H}^{\vee}$, the complex conjugate space will be denoted $\overline{\mathcal{H}}$. Every linear map $f: \mathcal{H} \rightarrow \mathcal{K}$ is functorially mapped to a linear map $\overline{f}: \overline{\mathcal{H}} \rightarrow \overline{\mathcal{K}}$. If you don't like conjugate spaces, just decorate the correct terms with ``anti".


First note that there is an isometric isomorphism of Hilbert spaces $I: \mathcal{H}\_{1} \otimes \mathcal{H}\_{2} \rightarrow \mathcal{H}\_{1} \otimes \overline{\mathcal{H}\_{2}}^{\vee}$ given by linearization and continuous extension of the map $\alpha \otimes \beta \mapsto \alpha \otimes \langle -, \beta \rangle\_{2}$. The Hilbert space $\mathcal{H}\_{1} \otimes \overline{\mathcal{H}\_{2}}^{\vee}$ is canonically identifiable with the space of Hilbert-Schmidt maps from $\overline{\mathcal{H}}\_{2}$ to $\mathcal{H}\_{1}$; in particular, it is canonically identifiable with a subspace of the space of bounded operators from $\overline{\mathcal{H}}\_{2}$ to $\mathcal{H}\_{1}$.


Define $X := I(x)$. By the polar decomposition theorem


\begin{align\*}
 X = U \sqrt{X^{\*} X}
\end{align\*}


for a partial isometry $U: \overline{\mathcal{H}}\_{2} \rightarrow \mathcal{H}\_{1}$. The operator $X^{\*} X: \overline{\mathcal{H}}\_{2} \rightarrow \overline{\mathcal{H}}\_{2}$ is positive semidefinite and trace class (with trace $\|x\|^2$), so, it is in particular self-adjoint and compact. By self-adjoint compact operator spectral theory we have that the non-vanishing spectrum $\sigma\_{\neq 0}$ of $X^{\*} X$ is countable, consists only of eigenvalues, and moreover the eigenspaces $E\_{\lambda}$ associated to each $\lambda \in \sigma\_{\neq 0}$ are finite dimensional and orthogonal. Positive semidefiniteness ensures that each eigenvalue is positive. By the spectral theorem


\begin{align\*}
 X^{\*} X = \sum\_{\lambda \in \sigma\_{\neq 0}} \lambda \mathrm{Proj}\_{\lambda},
\end{align\*}


where $\mathrm{Proj}\_{\lambda}$ is the orthogonal projection onto $E\_{\lambda}$. Choosing an orthonormal basis $(\beta^{\lambda}\_{i})\_{i = 1}^{\dim E\_{\lambda}}$ for each finite dimensional $E\_{\lambda}$


\begin{align\*}
 X^{\*} X &= \sum\_{\lambda \in \sigma\_{\neq 0}} \lambda \left[\sum\_{i\_{\lambda} = 1}^{\dim E\_{\lambda}} \beta^{\lambda}\_{i\_{\lambda}} \otimes (\beta^{\lambda}\_{i\_{\lambda}})^{\vee} \right],
\end{align\*}


(where $\omega^{\vee}$ denotes the linear form $\langle \omega, - \rangle$). Of course I can just reindex things here to get a sum


\begin{align\*}
 X^{\*} X=\sum\_{j = 1}^{r} \lambda\_{j} \beta\_{j} \otimes \beta\_{j}^{\vee},
\end{align\*}


where $r = \mathrm{rank}(X^{\*} X) = \sum\_{\lambda \in \sigma\_{\neq 0}} \dim E\_{\lambda} \leq \aleph\_{0}$, the $\lambda\_{j}$ are eigenvalues, and the $\beta\_{j} \in \overline{\mathcal{H}}\_{2}$ are associated eigenvectors with $\langle \beta\_{i}, \beta\_{j} \rangle = \delta\_{ij}$. Using our polar decomposition:


\begin{align\*}
 X = \sum\_{j = 1}^{r} \sqrt{\lambda\_{j}} (U \beta\_{j}) \otimes \beta\_{j}^{\vee}.
\end{align\*}


Because $U$ is a partial isometry, and one can verify $\ker(U) \leq \ker(X^{\*} X)^{\perp}$, the $U \beta\_{j} \in \mathcal{H}\_{1}$ remain orthonormal. Applying $I^{-1}$, then we simply have
\begin{align\*}
 x = \sum\_{j = 1}^{r} \sqrt{\lambda\_{j}} (U \beta\_{j}) \otimes \tilde{\beta}\_{j}.
\end{align\*}
where $\tilde{\beta}\_{j} \in \mathcal{H}\_{2}$ is the image of $\beta^{\vee}\_{j}$ under the canonical isomorphism $\overline{\mathcal{H}}^{\vee}\_{2} \overset{\sim}{\rightarrow} \mathcal{H}\_{2}$. It is an eigenvector of $\overline{X^{\*}X}: \mathcal{H}\_{2} \rightarrow \mathcal{H}\_{2}$ with eigenvalue $\lambda\_{j}$.




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A few side remarks:


* $U \beta\_{i}$ is an eigenvector of $X X^{\*}: \mathcal{H}\_{1} \rightarrow \mathcal{H}\_{1}$ associated to eigenvalue $\lambda\_{i}$. This follows by looking at the explicit construction of $U$ as in, e.g. Douglas' Lemma (a generalization of the polar decomposition theorem). I mention this because $\overline{X^{\*} X}$ and $X X^{\*}$ are the reduced density states associated to the pure state $x \otimes x^{\vee}$. The proof above is just a generalization of the usual finite-dimensional proofs that the Schmidt decomposition is given by tensor products of eigenstates of reduced density states.

 * Because Douglas' lemma constructs a unique $U$, the only choices involved in the procedure above are the choices of bases on each finite dimensional eigenspace $E\_{\lambda}$. Any two choices are then related by the action of $\prod\_{\lambda \in \sigma\_{\neq 0}} U(E\_{\lambda})$ (the infinite product of underlying groups). One can check that this group can be realized as a subgroup of $U(\bigoplus\_{\lambda \in \sigma\_{\neq 0}} E\_{\lambda})$, and hence $U(\mathcal{H}\_{2})$, or even $U(\mathcal{H}\_{1} \otimes \mathcal{H}\_{2})$ in a natural way. 

 * If one wants to be overly pedantic about the question you asked: you mentioned something about bases. Indeed, every Hilbert space has an orthornormal basis, so there exists a (possibly uncountable) basis $(e\_{k})\_{k \in K}$ for $\mathrm{span}(\beta\_{i})^{\perp} \leq \mathcal{H}\_{2}$ (I'm taking all spans to be closed) and a basis $(f\_{l})\_{l \in L}$ for $\mathrm{span}(U\beta\_{i})^{\perp} \leq \mathcal{H}\_{1}$; we can combine these with the bases we constructed on the spans to get orthonormal bases $(\tilde{e}\_{k})\_{k \in K'}$ and $(\tilde{f}\_{l})\_{l \in L'}$. Then, in a rather stupid way, we have

\begin{align\*}
 x = \sum\_{m \in M} \omega\_{m} \tilde{e}\_{m} \otimes \tilde{f}\_{m}
\end{align\*}
where $M$ is some (possibly uncountable) indexing set with $|M| = \mathrm{min}(|K'|,|L'|)$. Only countably many of the $\omega\_{m}$ are non-zero.


