Quantum Mechanics – Understanding Outer Product as an Operator in Infinite Dimensional 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

For any $\psi\in H$, you can define an operator $P_\psi$ by $P_\psi \phi:=\langle \psi, \phi\rangle_H \psi$ for all $\phi\in H$. It is easily verified that $P_\psi$ is a linear bounded operator, which in quantum mechanics is often written as $P_\psi=|\psi\rangle\langle \psi|$.

You don't have to argue with tensor products and so on a priori.

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However, if you insist, then you can consider the space of all Hilbert-Schmidt operators on $H$, denoted by $\mathcal B_2(H)$. A linear bounded operator $A$ is in this Hilbert-Schmidt space if $\sum\limits_{n\in \mathbb N} \|Ae_n\|_H^2 <\infty $, where $(e_n)_{n\in \mathbb N}\subset H$ is an orthonormal basis of $H$.

It can also be shown that $\mathcal B_2(H)$ is a Hilbert space with inner product defined by $$\langle A_1,A_2\rangle_{\mathcal B_2(H)}:=\mathrm{Tr}_H\,A_1^*A_2 \quad . $$

Of course, you first have to show that all of these expressions are well-defined and so on.

Finally, it can be shown that $H\otimes H^\prime$ is naturally isomorphic to $\mathcal B_2(H)$, and with that, it indeed makes sense to view $P_\psi$ as $|\psi\rangle\otimes\langle \psi|$.