Outer product as an operator in an infinite Hilbert space

Question

The outer product between a bra-ket $|a\rangle\langle a|$ where if $|a\rangle\in\mathcal{H}$ and $\langle a|\in\mathcal{H}_{dual}$ is a vector in the tensor vector space formed by the Hilbert space and its dual. Then it can be related to a linear operator due to the isomorphism with the space of all endomorphism of the Hilbert space:$$\mathcal{H}\otimes\mathcal{H}_{dual}\simeq End(\mathcal{H})$$ This means that for any vector in the tensor vector space I will be able to find a one-to-one corrspondence with a linear transformation $\rho:V\to V$ where $\rho\in End(\mathcal{H})$. This one-to-one correspondence arises because in finite dimensions, the space $\mathcal{H}\otimes\mathcal{H}_{dual}$ captures all bilinear forms on $\mathcal{H}$ and the set of linear transformations $End(\mathcal{H})$ also represents all linear maps on $\mathcal{H}$. However, in infinite-dimensional spaces, this correspondence should break down because the potential presence of additional elements in $\mathcal{H}\otimes\mathcal{H}_{dual}$ that do not correspond to linear transformations on $\mathcal{H}$. These additional elements arise due to the complexities of infinite-dimensional spaces, such as the lack of a finite basis, the non-separability of the space, and the divergence of certain series. Then why in quantum mechanics we still use $|a\rangle\langle a|$ as an operator even if the Hilbert space has infinite dimensions? Is my reasoning that the infinite dimensions break the isomorphism wrong?

Answer 1

The isomorphism breaks for infinite-dimensional vector spaces, but in the opposite way to what you are thinking: the tensor product $\mathcal{H} \otimes \mathcal{H}^*$ is too small in this case, not too big.

More formally, we can still define map from $\mathcal{H} \otimes \mathcal{H}^*$ to the space $\mathrm{B}(\mathcal{H})$ of bounded linear operators on $\mathcal{H}$. To do this, we assign to every rank $1$ tensor $\left|a\rangle\langle b\right|$ the linear map $\left|x\right> \mapsto \left|a\rangle\langle b|x\right>$, and then extend using linearity and continuity. This map is injective, but for infinite-dimensional vector spaces there exist linear operators which cannot be obtained from tensors, for example, the identity operator.

The linear operators in $\mathrm{B}(\mathcal{H})$ which correspond to tensors in the algebraic tensor product are exactly operators of finite rank, and the linear operators which correspond to tensors in the topological tensor product are called Hilbert-Schmidt operators.