Is there a mathematically rigorous formalization of "operator-valued vectors" from quantum mechanics?

Question

I've seen in various quantum mechanics courses people define various "operator-valued vectors" for the case of three-dimensional systems. For example, people define momentum as $\hat{\vec{p}} = \hat{p}_{x}\vec{e}_{x} + \hat{p}_{y}\vec{e}_{y} + \hat{p}_{z}\vec{e}_{z}$. This is also invoked when we talk about angular momentum.

Shankar does this in his Principle of Quantum Mechanics book, for a specific example.

Is these a sense in which this can be mathematically formalized? What is the rigorous way to treat these kinds of objects?

Answer 1

There's really not much to define - a vector of $n$ operators is the same as a vector of $n$ numbers: just an element of the vector space $O^n$, where $O$ is your vector space of operators (usually some linear operators on a Hilbert space). Then you can say that $\vec p = (p_x,p_y,p_z)\in O^3$.

Answer 2

The way I see it is this : consider operators valued in the tensor product of the Hilbert space of states and some vector space, which transform in the right way under rotations (or more generally, under the symmetry transformations of your system). I don't know if this done properly in some textbook.

Let $\mathcal H$ be the Hilbert space of quantum states. We assume that it represents some system of particles in euclidean space $E_3$ (For simplicity, we consider only integer spin particles). For every rotation $R:E_3 \to E_3$, there is a corresponding unitary operator $U(R) : \mathcal H \to \mathcal H$, such that if $R,R'$ are two rotations, then : $$U(RR') = U(R)U(R')$$

Then, the position operator $\hat{X}$ is a map $\mathcal H\to E_3\otimes \mathcal H$, which we can define by : $$\forall x \in E_3, \hat X |x\rangle = x\otimes |x\rangle$$

For this collection of operators to truly be a vector, we need it to correctly transform under rotations. For any $x \in E_3$ and a rotation $R$ we have $U(R)|x\rangle=|Rx\rangle $ and therefore : \begin{align} \left(\mathbb I_{E_3}\otimes U(R)^\dagger\right)\hat XU(R)|x\rangle &= \left(\mathbb I_{E_3}\otimes U(R)^\dagger\right)\hat X|Rx\rangle \\ &= \left(\mathbb I_{E_3}\otimes U(R)^\dagger\right)(Rx\otimes |Rx\rangle) \\ &= Rx \otimes |x\rangle \\ &= \left(R\otimes \mathbb I_{\mathcal H}\right)\hat X|x\rangle \end{align} and therefore : $$\left(\mathbb I_{E_3}\otimes U(R)^\dagger\right)\hat X U(R) = \left(R\otimes \mathbb I_{\mathcal H}\right)\hat X$$

In the same way, we define a vector value momentum operator $\hat P : \mathcal H\to E_3 \otimes \mathcal H$. We can then give a basis invariant formulation of : $$[X^i, P^j] = i\hbar \delta^{ij}$$ Likewise, the angular momentum $\hat L = \frac{1}{m}\hat X\times \hat P$ can be defined in a basis invariant way from $\hat X$ and $\hat P$

Edit How to make sense of the fact that $\hat X$ is self adjoint ? Since the components of $\hat X$ in any basis are self-adjoint, we see that for any linear map $\ell :E_3 \to \mathbb R$, the operator $(\ell \otimes \mathbb I_{\mathcal H})\hat X : \mathcal H\to\mathcal H$ is self-adjoint.

Actually, since $E_3$ has a scalar product, we can directly define the adjoint of $\hat X$ as a map $\hat X^\dagger : E_3 \otimes \mathcal H \to \mathcal H$.

Then, if $\ell \in E_3^*$, we have : \begin{align} (\ell \otimes \mathbb I_{\mathcal H})\hat X &= \left((\ell \otimes \mathbb I_{\mathcal H})\hat X \right)^\dagger \\ &= \hat X^\dagger (\ell \otimes \mathbb I_{\mathcal H})^\dagger \\ &= \hat X^\dagger (\ell^\dagger \otimes \mathbb I_{\mathcal H}) \\ \end{align} where $\ell^\dagger$ is the unique vector in $E_3$ such that : $$\forall x \in E_3, (\ell^\dagger,x) = \ell(x)$$

Answer 3

In quantum field theory, it turns out that the correct formulation is through vector bundles. Their sections are vector fields. To be completely correct we also would have to use the language of distributions. Generally for a vector bundle $E$, we write $\Gamma E$ for its sheaf of sections.

Given a manifold $M$, there is a canonical vector bundle, the tangent bundle $TM$ and its vector fields are tangent fields. This also induces another canonical bundle, the cotangent bundle $T^*M$ and its vector fields are called the cotangent fields.

From this we can construct the exterior cotangent bundle $\wedge T^*M$. It's fields are the exterior cotangent fields - aka differential forms. The sheaf of differential forms is usually written as $\Omega^*(M)$. Thus we have:

$\Omega^*(M) = \Gamma (\wedge T^*M)$

Now, an $E$-valued differential form over the manifold $M$ is a field of $\wedge TM \otimes E$. We write the sheaf of these as:

$\Omega^*(M,E) := \Gamma( \wedge T^*M \otimes E)$

And an operator valued differential form is a $End(E)$-valued differential form. That is a section of $\wedge TM \otimes End(E)$. We write the sheaf of these as:

$\Omega^*(M,EndE) = \Gamma( \wedge T^*M \otimes EndE)$