We know that if $T_1$ is a linear bounded operator on a Hilbert space $H_1$ and $T_2$ is a linear bounded operator on a Hilbert space $H_2$ there exists a unique linear bounded operator $T$ on $H_1 \otimes H_2$ such that $$ T (x_1 \otimes x_2) = T_1x_1 \otimes T_2x_2$$ for all $x_1$ in $H_1$ and $x_2$ in $H_2$. This operator is called a tensor product of operators $T_1$ and $T_2$ and denoted by $T_1 \otimes T_2$.

We can extend this to the arbitrary finite tensor product of Hilbert spaces and even infinite one which works with respect to some stabilising sequence.

How about the converse? Is it true that if $T$ is a linear bounded operator on a Hilbert space $H_1 \otimes H_2$ (here of course $H_1$, $H_2$ also Hilbert spaces) then we can find operators (unique?) $T_1$ and $T_2$ on $H_1$, $H_2$ respectively, such that $Tu= (T_1 \otimes T_2)u$ for every $u \in H_1 \otimes H_2$. If not I would be grateful for any counterexamples. If it is true, how about the arbitrary tensor product and what about inifinite one?

Thank you for the help. Any suggestions for good books are welcome.


2 Answers 2


Given a bounded linear operator $T$ on a tensor product $H_1 \otimes H_2$, the operator $T$ is not necessarily of the form $T_1 \otimes T_2$, where $T_i:H_i \to H_i$. Even in finite dimensions, $T$ is not necessarily of this form.

Here is a simple counterexample. Let $H_1 = H_2 = \mathbb C^n$. Let $T$ be the "swap" map, the linear extension of $$ T(x \otimes y)= y \otimes x. $$ After a little bit of thought, it should become intuitively clear that $T \neq T_1 \otimes T_2$ for any maps $T_1, T_2$; the maps $T_i$ depend only on the space $H_i$, but the map $T$ needs to "work with" both inputs $x$ and $y$ simultaneously.

You can easily obtain a proof by fixing orthonormal bases $(e_i)$ for the Hilbert spaces $H_i$. If $T_1$ corresponds to the matrix $(t_{ij}^1)$ and $T_2$ to the matrix $(t_{kl}^2)$, then $T_1 \otimes T_2$ corresponds to the matrix $(t_{ij}^1t_{kl}^2)$. We will then have, $$(T_1 \otimes T_2)(e_s \otimes e_t) =\sum_{is}t_{ij}^1t_{kt}^2 \ e_i \otimes e_k. $$ If this is to equal $e_t \otimes e_s$, we must have $t_{is}^1 = \delta_{it}$ and $t_{kt}^2=\delta_{ik}$. But these constraints need to hold for every choice of $s$ and $t$ simultaneously and are thus contradictory. (For example, the conditions imposed on $t_{ij}^1$ by insisting that $e_1 \otimes e_2 \mapsto e_2 \otimes e_1$ and also that $e_1 \otimes e_3 \mapsto e_3 \otimes e_1$ are contradictory.)

Another way of thinking about this is as follows. Let $L(H)$ denote the bounded linear operators on a Hilbert space $H$. We have an inclusion $L(H_1) \otimes L(H_2)$ into $L(H_1 \otimes H_2)$ (as described in the first paragraph of your question). If the spaces $H_i$ are finite-dimensional, we even have $$ L(H_1) \otimes L(H_2) = L(H_1 \otimes H_2). $$ But not every element in $L(H_1) \otimes L(H_2)$ is an elementary tensor, i.e., of the form $T_1 \otimes T_2$. Even when $H_i$ is finite-dimensional, the most we can ask for in general is that $T$ is of the form $$ T=\sum_i T_{1,i} \otimes T_{2,i}. $$

  • $\begingroup$ Should not the last expression be of the similar form as for vectors, i.e. $T=\sum_i \sum_j c_{ij} T_{1,i} \otimes T_{2,j}$? $\endgroup$
    – Wildcat
    Oct 2, 2014 at 15:50
  • $\begingroup$ Just a remark, this relates to the notion of separable density operators in quantum mechanics. When an operator can be written as a convex combination of tensor products then it is called separable. $\endgroup$
    – user135520
    Jan 6 at 16:21

When dealing with infinite dimension we have to be careful with our notation since although the sets do not coincide, the closures (with respecto to any of the topologies below) do coincide.

Denoting with $\otimes _a$ the algebraic tensor product and without the subindex the closure w.r.t. any of the weak, $\sigma$-weak, strong, $\sigma$-strong, strong*, $\sigma$-strong* topologies, it is immediate that $L(H) \otimes_a L(H) \subset L(H \otimes H) \Rightarrow L(H) \otimes L(H) \subset L(H \otimes H)$.

Now, take a $X\in L(H \otimes H)$. We will show that there exists a sequence in $L(H) \otimes_a L(H)$ weakly convergent to$X$. Consider $\{e_i \otimes e_j \}_{ij}$ an orthonormal basis of $H \otimes H$. For any $n$ define the trunked operator \begin{align*}X_n&=\sum_{i,j,i^\prime,j^\prime}^n |e_{i^\prime} \otimes e_{j^\prime} \rangle \langle e_{i^\prime} \otimes e_{j^\prime}| X |e_{i} \otimes e_{j}\rangle \langle e_{i} \otimes e_{j}|\\ &=\sum_{i,j,i^\prime,j^\prime}^n \langle e_{i^\prime} \otimes e_{j^\prime}, Xe_{i} \otimes e_{j} \rangle |e_{i^\prime} \rangle \langle e_i | \otimes |e_{j^\prime} \rangle \langle e_j |\in L(H) \otimes_a L(H). \end{align*}

For arbitrary simple tensors $u\otimes v,z\otimes w\in H \otimes H$, \begin{align*}\langle u \otimes v ,X_n z \otimes w \rangle& =\sum_{i,j,i^\prime,j^\prime}^n \big\langle e_{i^\prime} \otimes e_{j^\prime},X e_i \otimes e_j \big\rangle \big\langle u\otimes v, |e_{i^\prime}\rangle\langle e_i |z \otimes | e_{j^\prime} \rangle \langle e_j | w \big\rangle \\ &= \bigg \langle\sum_{i^\prime,j^\prime}^n \langle e_{i^\prime} \otimes e_{j^\prime},u\otimes v \rangle e_{i^\prime} \otimes e_{j^\prime} , X \sum_{i,j}^n \langle e_{i} \otimes e_{j}, z\otimes w \rangle e_{i} \otimes e_{j} \bigg \rangle, \end{align*} from where it follows that $\langle u \otimes v ,X_n z \otimes w \rangle \xrightarrow[n]{\ \ \ \ \ } \langle u \otimes v ,X z \otimes w \rangle $ since $X$ is bounded.

While for general tensors $s,t\in H \otimes H$, $s=\sum_{m^\prime l^\prime} s_{m^\prime ,l^\prime}\ e_{m^\prime} \otimes e_{l^\prime}$, $t=\sum_{m,l} t_{m l}\ e_m\otimes e_l$, where $s_{m^\prime,l^\prime}$ y $t_{ml}$ are complex numbers such that \begin{align*}\sum_{m^\prime, l^\prime} |s_{m^\prime, l^\prime}|^2<\infty,\ \ \ \ \sum_{m, l} |t_{m l}|^2<\infty,\end{align*} \begin{align*}\langle s, X_n t \rangle &=\lim_{h^\prime,h\to\infty}\sum_{m^\prime, l^\prime, m,l}^{h^\prime,h} \overline{s}_{m^\prime l^\prime} t_{m l} \big\langle e_{m^\prime} \otimes e_{l^\prime}, X_n e_{m} \otimes e_{l} \big\rangle \\ &=\lim_{h^\prime,h\to\infty}\sum_{m^\prime, l^\prime, m,l}^{h^\prime,h} \overline{s}_{m^\prime l^\prime} t_{m l}\bigg \langle \sum_{i^\prime,j^\prime}^n \langle e_{i^\prime} \otimes e_{j^\prime}, e_{m^\prime} \otimes e_{l^\prime} \rangle e_{i^\prime} \otimes e_{j^\prime}, X \sum_{i,j}^n \langle e_i \otimes e_j, e_m \otimes e_l \rangle e_i \otimes e_j \bigg \rangle \\ &=\sum_{i,j,i^\prime,j^\prime}^n \bigg \langle \langle e_{i^\prime} \otimes e_{j^\prime},s \rangle e_{i^\prime} \otimes e_{j^\prime}, X \langle e_i\otimes e_j,t \rangle e_i\otimes e_j \bigg \rangle \xrightarrow[n]{\ \ \ \ \ } \langle s, X t \rangle, \end{align*} since $X$ is continuos. Hence $L(H\otimes H) \subset \overline{ L(H) \otimes_a L(H)}=L(H) \otimes L(H)$.

So not every operator $T$ can be written like $T_1 \otimes T_2$ , but as the limit of linear combination of simple tensors.

  • $\begingroup$ Do you have a reference? Thanks. $\endgroup$
    – Evian
    Apr 14, 2020 at 20:31

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