Tensor product in quantum mechanics?

$\lvert A\rangle \langle B \rvert$ is the tensor of a ket and a bra (well, duh). This means it is an element of the tensor product of a Hilbert space $H_1$ (that's where the kets live) and of a dual of a Hilbert space $H_2^\ast$, which is where the bras live. Although for Hilbert spaces their duals are isomorphic to the original space, this distinction should be kept in mind. So we can "feed" a ket $\lvert \psi\rangle$ from $H_2$ to the bra in $\lvert \phi\rangle\otimes \langle\chi\rvert \in H_1\otimes H_2^\ast$, and are left with a state in $H_1$ given by $\langle \chi \vert \psi\rangle \lvert \phi\rangle$. The usual use case for such a tensor product is when $H_1=H_2$ to construct a map from $H_1$ to itself, e.g. the projector onto a state $\lvert \psi \rangle$ is given by $\lvert\psi\rangle \langle \psi \rvert$.

In general, a tensor in $H_2 \otimes H_1^\ast$ corresponds to a linear operator $H_1\to H_2$. In the finite-dimensional case, these are all linear operators, in the infinte-dimensional case, this is no longer true, e.g. $H^\ast \otimes H$ are precisely the Hilbert-Schmidt operators on $H$.

In constract, a tensor $\lvert A\rangle\otimes \lvert B\rangle$ (also just written $\lvert A \rangle \lvert B\rangle$) in $H_1\otimes H_2$, although it corresponds to a bilinear map $H_1\times H_2\to\mathbb{C}$ by definition, is usually not meant to denote an operator, but a state. Given two quantum systems $H_1$ and $H_2$, $H_1\otimes H_2$ is the space of the states of the combined system (as for why, see this question).

The notion of tensor product is independent from the Hilbert space structure, it is defined for vector spaces on the field $\mathbb K$ (usually $\mathbb R$ or $\mathbb C$). A formal definition is given below (there are many equivalent approaches).

First, if $V$ is a vector space, $V^*$ denotes its algebraic dual space, namely the vector space of the linear maps $f: V \to \mathbb K$ with vector structure defined by: $$(af+bg)(u) := af(u) + bg(u)\quad \forall u \in V \tag 0$$ if $f,g \in V^*$. It turns out that $\textrm{dim} \,V = \textrm{dim} \,V^*$ if $\textrm{dim}\, V$ is finite, the proof being elementary. However, the given definition of $V^*$ does not require finiteness of the dimension of $V$.

To go on, notice that $V$ identifies with a subspace of $(V^*)^*$ by means of the injective linear map $$\imath: V \ni v \mapsto \imath(v) \quad \mbox{where $\imath(v)(f) := f(v)$ if $f \in V^*$}\tag 1$$ The linear embedding $\imath : V\to (V^*)^*$ is a natural vector space isomorphism provided, again, $\textrm{dim}\, V$ is finite, the proof being evident as the embedding is a linear and injective map between spaces with equal finite dimension.

The embedding (1) permits us to define a vector space called the tensor product $$V_1 \otimes \ldots \otimes V_n$$ of vector spaces $V_1,\ldots, V_n$, with the common field of scalars $\mathbb K$.

The tensor product is a subspace of the vector space ${\cal L}(V^*_1,\ldots, V^*_n)$ of multi-linear maps $F$ with $$F: V^*_1\times \cdots \times V^*_n \ni (f_1,\ldots, f_n) \mapsto F(f_1,\ldots, f_n)\:.$$ The vector space structure on ${\cal L}(V^*_1,\ldots, V^*_n)$ is defined along an evident generalisation of (0).

In fact, if we pick out $v_i \in V_i$ for $i=1,\ldots, n$ we can construct the multilinear map over $V^*_1\times \cdots \times V^*_n$ called the tensor product of vectors $v_i \in V_i$ as $$v_1\otimes \cdots \otimes v_n : (f_1,\ldots, f_n) \mapsto f_1(v_1)\cdots f_n(v_n)\:.$$

Definition. $V_1 \otimes \ldots \otimes V_n$ is the subspace of ${\cal L}(V^*_1,\ldots, V^*_n)$ spanned by all finite linear combinations of tensor products $v_1\otimes \cdots \otimes v_n$ for $v_i \in V_i$ for $i=1,\ldots, n$.

It turns out that, if $dim V_i$ is finite for every $i$, then $\dim (V_1 \otimes \ldots \otimes V_n) = \prod_{i=1}^n dim V_i$ and $$V_1 \otimes \ldots \otimes V_n = {\cal L}(V^*_1,\ldots, V^*_n)\:.$$ (There is a property of the pair $(V_1 \otimes \ldots \otimes V_n, \otimes)$ called universality property which characterize the notion of tensor product at the level of the theory of categories, but I do not think is necessary to describe it here.)

Let us come to the Hilbertian tensor product of Hilbert spaces. Consider a finite number of (complex) Hilbert spaces $H_1, \ldots, H_n$ with respective Hermitian scalar products $\langle \cdot| \cdot \rangle_1, \ldots, \langle \cdot| \cdot \rangle_n $. Relying upon the above definition, we can first define their algebraic tensor product $$H_1 \otimes \cdots \otimes H_n\:.$$ This is not a Hilbert space yet. However it is possible (not so easy) to prove that $H_1 \otimes \cdots \otimes H_n$ admits an Hermitian scalar product induced by the ones of each $H_i$. This scalar product $\langle \cdot| \cdot \rangle$ it is the unique right-linear and left-antilinear extension of $$\langle \psi_1 \otimes \cdots \otimes\psi_n| \phi_1 \otimes \cdots \otimes\phi_n\rangle = \prod_{i=1}^n \langle \psi_i|\phi_i\rangle_i \:.\tag 2$$ The said (anti)linear extension is necessary because $\psi_1 \otimes \cdots \otimes\psi_n$ is not the generic element of $H_1 \otimes \cdots \otimes H_n$, the generic element is a finite linear combination of these elements!

It turns out that the unique (anti)liner extension of (2) defines an Hermitian scalar product on $H_1 \otimes \cdots \otimes H_n$, in particular the extension is positively defined.

Definition. The Hilbertian tensor product of (complex) Hilbert spaces $H_1, \ldots, H_n$ is the (complex) Hilbert space $H_1 \otimes_H \cdots \otimes_H H_n$ given as the completion of the algebraic tensot product $H_1 \otimes \cdots \otimes H_n$ with respect to the Hermitian scalar product $\langle\cdot |\cdot \rangle$ which uniquely (anti)linearly extends (2).

The completion $\overline{V}$ of a vector space $V$ equipped with an Hermitian scalar product $\langle \cdot |\cdot \rangle$ is the complete (Hilbert) space of the equivalence classes of the Cauchy sequences in $V$ equipped with the unique continuous extension of $\langle \cdot |\cdot \rangle$. So It is uniquely defined (up to Hilbert space isomorphisms) and $V$ is dense in $\overline{V}$.

A fundamental (also in QM) result is that,

Proposition. If $\{\psi_{i,j_i}\}_{j_i\in I_j} \subset H_i$ is a Hilbert basis (even uncountable) of the Hilbert space $H_i$ then $$\{\psi_{1,j_1} \otimes \cdots \otimes \psi_{n,j_n}\}_{j_1\in I_1, \ldots,j_n\in I_n } \subset H_1 \otimes_H \cdots \otimes_H H_n $$ is a Hilbert basis of $H_1 \otimes_H \cdots \otimes_H H_n$. Therefore $H_1 \otimes_H \cdots \otimes_H H_n $ is separable if each $H_i$ is separable.

A second important result, very used in QM, in case $H_i = L^2(X_i, \mu_i)$ where $\mu_i$ is $\sigma$-finite (as for the standard Lebesgue measure over $\mathbb R^n$) reads as follows.

Proposition. Assume $H_i = L^2(X_i, \mu_i)$, where $\mu_i$ is $\sigma$-finite. Then the map $$L^2(X_1, \mu_1)\otimes_H \cdots \otimes_H L^2(X_n, \mu_n) \ni \psi_1\otimes \cdots \otimes \psi_n \mapsto \psi_1 \cdots \psi_n \in L^2(X_1\times \cdots \times X_n, \mu_1 \otimes \cdots \otimes \mu_n)$$ uniquely continuously and linearly extends to a Hilbert space isomorphism.

Above $\psi_1 \cdots \psi_n$ is the standard pointwise product $$\psi_1 \cdots \psi_n(x_1,\ldots,x_n) := \psi_1(x_1) \cdots \psi_n(x_n)\:.$$

N.B: I henceforth denote by $H_1 \otimes \cdots \otimes H_n$ the Hilbertian tensor product omitting the index $_H$, thus adopting the standard notation in Quantum Mechanics textbooks.

I am now in a position I can rigorously answer the question. First notice that the topological dual space $H'$ of a Hilbert space $H$, that is the subspace $H' \subset H^*$ made of continuous linear maps $f : H \to \mathbb C$ is a Hilbert space in its own right.

Indeed, the celebrated Riesz' theorem states that

Theorem. If $H$ is a Hilbert space with scalar product $\langle \cdot | \cdot \rangle $, the map $$H \ni \psi \mapsto \langle \psi | \cdot \rangle \in H'$$ is anti-linear and bijective. Thus, every element $f$ of the topological dual space $H'$ is represented by $\langle \psi_f | \cdot \rangle$ with $\psi_f \in H$ uniquely determined by $f$.

Obviously $H'$ is the space of "bra" vectors shortening $\langle \psi | \cdot \rangle$ to $\langle \psi|$.

It is clear that, in view of the stated result, $H'$ turns out to be a Hilbert space as soon as we define the scalar product $$\langle f | g \rangle' := \overline{\langle \psi_f | \psi_g\rangle}\:.$$

This feature of $H'$ permits us to define the Hilbertian tensor product $$H_1 \otimes H_2'$$ the elements are linear combinations (also infinite provided they converge in the natural topology of the space) of elementary tensor products $$\psi \otimes f = |\psi \rangle \otimes \langle \phi| = |\psi \rangle \langle \phi|$$ where I employed also some common notations used in physics textbooks.

The difference between $|\psi \rangle \otimes |\phi \rangle$ (an element of $H_1 \otimes H_2$) and $|\psi \rangle \otimes \langle \phi|$ (an element of $H_1 \otimes H_2'$) should be now evident.

It is also clear that $|\psi \rangle \otimes \langle \phi|$ defines a continuous operator $H_2 \to H_1$. Also infinite linear combinations of these operators, assumed to be convergent with respect to the natural scalar product on $H_1 \otimes H_2'$, define continuous linear operators $H_2 \to H_2$. These operators are compact (they transform bounded sets into compact sets) and satisfy a further property regarding the notion of trace which characterizes them as Hilbert-Schmidt operators from $H_2$ to $H_1$.

As a final remark, notice that $I = \sum_k |\psi_k \rangle \langle \psi_k|$ referred to a Hilbert basis $\{\psi_k\}_{k\in K}$ does not belong to $H_1 \otimes H_2'$ in spite of the notation if $K$ is not finite! This is because that vector has no finite norm in $H_1 \otimes H_2'$ and the convergence of the series has to be interpreted exploiting another topology, the so-called strong operator topology.