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Suppose that $x,y\in\mathbb H$, where $\mathbb H$ is a separable Hilbert space. Define the tensor $x\otimes y:\mathbb H\to\mathbb H$ by setting $(x\otimes y)(z)=\langle z,y\rangle x$ for $z\in\mathbb H$. We can also define $A=x\otimes y\tilde\otimes x\otimes y: S_2(\mathbb H)\to S_2(\mathbb H)$ in an analogous way, where $S_2(\mathbb H)$ is the space of the Hilbert-Schmidt operators, which is also a Hilbert space. I am interested in the trace of $A$. On one hand, $$ \operatorname{Tr}A=\operatorname{Tr}[x\otimes y\tilde\otimes x\otimes y]=\operatorname{Tr}[x\otimes y]\operatorname{Tr}[x\otimes y]=|\langle x,y\rangle|^2. $$ On the other hand, \begin{align*} \operatorname{Tr}A &=\sum_{k,l\ge1}\langle{A(e_k\otimes e_l),e_k\otimes e_l}\rangle_{\mathrm{HS}}\\ &=\sum_{k,l\ge1}\langle\langle e_k\otimes e_l,x\otimes y\rangle_{\mathrm{HS}}(x\otimes y),e_k\otimes e_l\rangle_{\mathrm{HS}}\\ &=\sum_{k,l\ge1}\langle x\otimes y,e_k\otimes e_l\rangle_{\mathrm{HS}}\langle e_k\otimes e_l,x\otimes y\rangle_{\mathrm{HS}}\\ &=\|x\otimes y\|_{\mathrm{HS}}^2\\ &=\|x\|^2\|y\|^2. \end{align*} Where am I making a mistake? Which is the correct answer?

Any help is much appreciated!

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  • $\begingroup$ For Hilbert-Schmidt operators $S,T$ one defines the scalar product by $<S,T>_{HS}=Tr (ST^{*})$ and so if You view $x\otimes y$ as a Hilbert-Schmidt operator ($S_2(\mathbb{H})\cong \mathbb{H}\otimes\mathbb{H}$) then $||x\otimes y||_{HS}^2=Tr ((x\otimes y)(x\otimes y)^{*})$. I can't see how this is equal to $||x||^2||y||^2$ $\endgroup$ Sep 24, 2018 at 11:52
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    $\begingroup$ @PeterMelech $\|x\otimes y\|_{\mathrm{HS}}^2=\sum\|(x\otimes y)(e_k)\|^2=\|x\|^2\sum|\langle e_k,y\rangle|^2=\|x\|^2\|y\|^2$ (see here). $\endgroup$
    – Cm7F7Bb
    Sep 24, 2018 at 11:56
  • $\begingroup$ Sure! I can't find a mistake either. Good question! $\endgroup$ Sep 24, 2018 at 11:59

1 Answer 1

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Alright to first clear up how maps of the form $(x\otimes y)\tilde\otimes(z\otimes w)$ for arbitrary $x,y,w,z\in\mathbb H$ act on general $\varphi\in S_2(\mathbb H)$, it goes like $$ [(x\otimes y)\tilde\otimes(z\otimes w)](\varphi):=\langle \varphi,z\otimes w\rangle_\text{HS} (x\otimes y) $$ as defined by OP in another question. One readily verifies $\langle \varphi,z\otimes w\rangle_\text{HS} =\langle \varphi(w),z\rangle_{\mathbb H}$. The above map admits the trace $$ \operatorname{tr}[(x\otimes y)\tilde\otimes(z\otimes w)]=\sum_{k,l\geq 1}\langle [(x\otimes y)\tilde\otimes(z\otimes w)](e_k\otimes e_l),e_k\otimes e_l\rangle_\text{HS}=\sum_{k,l\geq 1} \langle (e_k\otimes e_l)(w),z\rangle_{\mathbb H}\langle x,(e_k\otimes e_l)(y)\rangle_{\mathbb H}=\sum_{k,l\geq 1} \langle w,e_l\rangle_{\mathbb H}\langle e_k,z\rangle_{\mathbb H}\langle x,e_k\rangle_{\mathbb H}\langle e_l,y\rangle_{\mathbb H}=\langle w,y\rangle_{\mathbb H}\langle x,z\rangle_{\mathbb H} $$ by basis expansion. On the other hand, $\operatorname{tr}(a\otimes b)=\langle a,b\rangle_{\mathbb H}$ for all $a,b\in\mathbb H$ (can also be easily seen by basis expansion). Piecing things together, one gets the somewhat peculiar identity $$ \boxed{\operatorname{tr}[(x\otimes y)\tilde\otimes(z\otimes w)]=\operatorname{tr}(x\otimes z)\operatorname{tr}(w\otimes y)} $$ so the trace identity in your answer is not correct. This is solely due to the above definition of the tensor product $\tilde\otimes$ to go from $S_2(\mathbb H)$ to operators on $S_2(\mathbb H)$ which does not coincide with the usual tensor product (as in that case one would in fact have $\operatorname{tr}(A\tilde\otimes B)=\operatorname{tr}(A)\operatorname{tr}(B)$ given $A,B$ are trace-class). To see how $(x\otimes y)\tilde\otimes (x\otimes y)$ should behave on $\mathbb H\otimes\mathbb H$ to mimic the usual tensor product of linear Hilbert space operators, check my old answer below.


OLD ANSWER:

To identify the Hilbert-Schmidt-Operators with $\mathbb H\otimes\mathbb H$ (as Peter Melech pointed out in his comment) and to not raise confusion regarding the actual Hilbert space tensor product $\otimes_{\mathbb H}=:\otimes$, I will "rename" your map $x\otimes y$ to $\langle \cdot,y\rangle x:\mathbb H\to\mathbb H$ (but the action is the same!). Consider the map $$ A=(\langle \cdot,y\rangle x)\otimes(\langle \cdot,y\rangle x):\mathbb H\otimes \mathbb H\to\mathbb H\otimes \mathbb H $$ defined via $$ a\otimes b\mapsto (\langle a,y\rangle x)\otimes(\langle b,y\rangle x)=\langle a,y\rangle\langle b,y\rangle\, x\otimes x=\langle a\otimes b,y\otimes y\rangle\, x\otimes x $$ and its linear extension onto all of $\mathbb H\otimes\mathbb H$, i.e., $$\boxed{A(z)=\langle z,y\otimes y\rangle x\otimes x}$$ for all $z\in\mathbb H\otimes\mathbb H$. [Edit: Thus in your notation and given your definition, the identity $$(x\otimes y)\tilde\otimes(x\otimes y)=(x\,\tilde\otimes \,x)\otimes(y\,\tilde\otimes \,y)$$ has to hold to mimic the usual tensor product of linear operators.] Now on the one hand, $$ \operatorname{tr}(A)=\operatorname{tr}(\langle \cdot,y\otimes y\rangle\, x\otimes x)=\langle x\otimes x,y\otimes y\rangle=\langle x,y\rangle^2\,. $$ On the other hand, let $\lbrace e_k\rbrace_{k\geq 1}$ be an orthonormal basis of $\mathbb H$, then $\lbrace e_k\otimes e_l\rbrace_{k,l\geq 1}$ is an orthonormal basis of $\mathbb H\otimes\mathbb H$. By definition of the trace $$ \operatorname{tr}(A)=\sum_{k,l\geq 1}\langle A(e_k\otimes e_l),e_k\otimes e_l\rangle= \sum_{k,l\geq 1}\langle e_k\otimes e_l,y\otimes y\rangle\langle x\otimes x,e_k\otimes e_l\rangle\\ =\Big\langle x\otimes x, \sum_{k,l\geq 1}\langle y\otimes y,e_k\otimes e_l\rangle e_k\otimes e_l\Big\rangle=\langle x\otimes x,y\otimes y\rangle=\langle x,y\rangle^2 $$ where in the last step we just used the basis expansion in $\mathbb H\otimes\mathbb H$.

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  • $\begingroup$ Did'nt You change from $A=x\otimes y \tilde{\otimes}x\otimes y$ to $x\otimes x\tilde{\otimes}y\otimes y$? $\endgroup$ Sep 25, 2018 at 13:34
  • $\begingroup$ Curiously for that map You could as well compute $Tr(x\otimes x\tilde{\otimes}y\otimes y)=Tr(x\otimes x)Tr(y\otimes y)=<x,x><y,y>=||x||^2||y||^2$! $\endgroup$ Sep 25, 2018 at 13:48
  • $\begingroup$ Yes I guess in OP's notation the key result I used is $A=(x\otimes y)\tilde\otimes(x\otimes y)=(x\tilde\otimes x)\otimes(y\tilde\otimes y)$ ($\neq (x\otimes x)\tilde\otimes(y\otimes y)$ (!)). But this is why I personally am not too fond of this notation - it at least for me is quite easy to confuse $x\otimes y$ (the $\langle\cdot,y\rangle x$ pairing) and actual tensor products $x\tilde\otimes y$ in $\mathbb H\tilde\otimes\mathbb H$... $\endgroup$ Sep 25, 2018 at 16:07
  • $\begingroup$ However the OP's map acts like $A(e_k\otimes e_l)=<e_k\otimes e_l,x\otimes y>x\otimes y$ whereas Your map acts like $A(e_k\otimes e_l)=<e_k\otimes e_l,y\otimes y>x\otimes x$ and the point is that for both(!) maps one gets two different results for the trace using the two different methods. I don't think it is a matter of misleading notation, there must be some more subtle mistake. $\endgroup$ Sep 26, 2018 at 8:39
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    $\begingroup$ Your peculiar identity looks quite strange (but is true I think) and it is really counterintuitive that the "mutiplicity" of the trace with the "tensor product" does not hold, as well it is unclear what a usual tensor product could be. However I agree that the mistake lies exactly here:+1 $\endgroup$ Sep 27, 2018 at 10:35

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