0

Suppose I have an infinite, separable Hilbert space $\mathcal{H}$ over $\mathbb{C}$ and bounded linear operators $D$ and $T$ on $\mathcal{H}$ such that $D$ is a diagonal operator given by

$$\begin{bmatrix} d_{1} \\ & d_{2} &\dots \\ &\vdots \end{bmatrix}=\mathrm{diag}[d_{k}]_{k=1}^{\infty},$$

where $d_{k}\in\mathbb{C}$ for $k=1,2,...$. Consider the tensor product $\otimes$ such that

$$D\otimes T = \mathrm{diag}[d_{k}T]_{k=1}^{\infty}.$$

It is my understanding that $D\otimes T\in\mathcal{B}(\mathcal{H}\otimes \mathcal{H})$. But my question is, is $\mathcal{H}\otimes\mathcal{H}$ isomorphic to $\mathcal{H}$? And if so, how does the operator $D\otimes T$ act on $\mathcal{H}$?

Clearly if $\mathcal{H}$ is finite dimensional then $\mathcal{H}\otimes\mathcal{H}$ cannot be isomorphic to $\mathcal{H}$. But what about the case when $\mathcal{H}$ is infinite dimensional?

----- UPDATE -----

As Qiaochu Yuan rightly states in the comments, if $\mathcal{H}$ is finite dimensional with dimension equal to zero or one, then $\mathcal{H}\otimes\mathcal{H}$ is indeed isomorphic to $\mathcal{H}$.

MathematicalNovice
  • 1
  • Yes, it's another infinite-dimensional separable Hilbert space. Your claim about the finite-dimensional case is (slightly) false; $H \otimes H \cong H$ if $H$ is either $0$ or $1$-dimensional. – Qiaochu Yuan Mar 12 '25 at 18:36
  • Thank you Qiaochu, I forgot about the finite case of dimensions 0 and 1 of course! Is this new infinite-dimensional separable Hilbert space just H again though? And does the operator D\tensor T act on H in the usual manner? That is, for e_{m} the mth basis vector, does D\tensor T e_{m} = d_{m}Te_{m}? – MathematicalNovice Mar 12 '25 at 19:10
  • There's only one infinite-dimensional separable Hilbert space (up to isomorphism); any two of them are isomorphic. But how $D \otimes T$ acts depends on the choice of isomorphism. It is probably not helpful to use this isomorphism here. – Qiaochu Yuan Mar 12 '25 at 20:11

1 Answers1

1

Yes, if a Hilbert space $H$ is infinite dimensional, then $H\otimes H$ is isomorphic to $H$.

Recall that Hilbert spaces are completely determined by the cardinality of their orthonormal basis up to isomorphism, i.e. if a Hilbert space $K$ has an orthonormal basis $\{e_i\}_{i\in I}$, then $K\cong \ell^2(I)$, where $\ell^2(I)$ is the space of those $f\colon I\to\mathbb{K}$ with $\sum_{i\in I}|f(i)|^2<\infty$.

Moreover, if $I,J$ are two sets and $\phi:I\to J$ is a bijection, then we get an induced unitary operator $U:\ell^2(I)\to\ell^2(J)$ satisfying $U\delta_i:=\delta_{\phi(i)}$ for all $i\in I$, where $\{\delta_i\}_{i\in I}$ is the canonical orthonormal basis of $\ell^2(I)$, namely $\delta_i(i')=1$ for $i'=i$ and $\delta_i(i')=0$ for $i'\ne i$.

Now if $H$ is infinite dimensional and $\{e_i\}_{i\in I}$ is an orthonormal basis for $H$, then $\{e_i\otimes e_j\}_{(i,j)\in I\times I}$ is an orthonormal basis for $H\otimes H$. As $I\times I$ and $I$ have the same cardinality (for which we assume the axiom of choice -- see here), we see that $H\otimes H\cong H$.

As for how $D\otimes T$ acts on $H$, this depends on the choice of unitary operator between $H\otimes H$ and $H$ that you choose; these are in a correspondence with the bijections between $I\times I$ and $I$, which are of course many.

Just dropped in
  • 16,664