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?

  • $\begingroup$ My question is that the projector operator can be defined as the outer product of two vectors, as you said, because there exists the ismorphism that I mentioned. But if in infinte dimensions the ismorphism breaks, then why can we still define it in this way? $\endgroup$
    – Oscarcillo
    Feb 16 at 12:50
  • $\begingroup$ I have written an answer and removed my previous comments for that reason. $\endgroup$ Feb 16 at 17:05

1 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.

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 $|\psi\rangle\otimes\langle \psi|$.

  • $\begingroup$ I understand the connection between the tensor vector space and the Hilbert-Schmidt space, and that isomorphism is precisely the one that I was talking about. But as I suspected and explained in the link that you provided me (thank you!) this isomorphism is no longer true for the case where the dimensions of the Hilbert space are infinite. Then my question is: why in an infinite Hilbert space, $|\psi\rangle\otimes\langle\psi|$ can still be viewed as an operator? $\endgroup$
    – Oscarcillo
    Feb 16 at 17:05
  • $\begingroup$ @Fulltimelearner No, we are not talking about the same isomorphisms. The isomorphism I am talking about does indeed hold in infinite-dimensions. In infinite-dimensions, not all operators are HS. In finite-dimensions, all operators are HS however. I don't understand the last question in your comment, i.e. I don't understand where the difference to your question is at all: The answer to the question in your comment is exactly my answer. $\endgroup$ Feb 16 at 17:07
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    $\begingroup$ ...or let me put the first two or three sentences of my above comment like this: In infinite-dimensions, you have an isomorphism between $H\otimes H^\prime$ and $\mathcal B_2(H)$, but the latter is not the space of all operators (and not even the space of all bounded operators, for example). If $H$ is finite-dimensional, the isomorphism still holds and $\mathcal B(H)=\mathcal B_2(H)=\mathcal L(H)$, i.e. the HS space coincides with the space of bounded operators which itself coincides with the space of linear operators (on $H$). $\endgroup$ Feb 16 at 17:16
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    $\begingroup$ @ProphetX As I said, I haven't read the article myself, it was just to provide some starting point of a source. I sketched the proof of the canonical isomorphism in some comment above. Regarding your sought example: Take e.g. the identity operator. $\endgroup$ Mar 11 at 23:24
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    $\begingroup$ @ProphetX Well, the completeness relation makes sense when you interpret the sum in the strong operator topology, i.e. $I=\sum\limits_n |\psi_n\rangle\langle\psi_n|$ is defined to mean $I\psi:=\sum\limits_n \langle \psi_n,\psi\rangle \psi_n$ (which is just the basis expansion in the used orthonormal bases and convergence on the RHS is in the Hilbert space norm topology). The point is that the identity operator is not in $\mathcal B_2(H)$-- it has no finite norm. If you want to know more, I suggest to ask a separate and new question (after making sure it has not been asked before). $\endgroup$ Mar 11 at 23:28

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