# Vector operator acting on Kets takes them out of state space

Definition of linear operator in quantum mechanics

A linear operator $$A$$ associates with every ket $$|\psi\rangle \in \mathcal{E}$$ another ket $$\left|\psi^{'}\right\rangle \in \mathcal{E}$$, the correspondence being linear

We also have vector operators $$\hat{A}$$ (such as a position operator $$\hat{r}$$ )

$$\hat{A}=\left(\begin{array}{l}\hat{A}_{1} \\ \hat{A}_{2} \\ \hat{A}_{3}\end{array}\right)$$

their action on ket is : $$\hat{A}|u\rangle=\left(\begin{array}{c}\hat{A}_{1}|u\rangle \\ \hat{A}_{2}|u\rangle \\ \hat{A}_{3}|u\rangle\end{array}\right)=\left(\begin{array}{c}\left|u_{1}\right\rangle \\ \left|u_{2}\right\rangle \\ \left|u_{3}\right\rangle\end{array}\right)$$

But this operator upon acting on a ket didn't give another ket belonging to the same space.

What am I missing?

• You have said that $A_1 |u\rangle = |u_1\rangle$ etc. In what way does $|u_1\rangle$ not belong to the same space? Aug 23, 2022 at 13:09
• What do you mean with your notation exactly?- What is $(|u_1\rangle, |u_2\rangle, |u_3\rangle)^T$? Aug 23, 2022 at 13:10
• BTW in Quantum Mechanics the concept of "vector operator" is well-defined and has nothing to do with grouping 3 operators in a column (as a 3x1 matrix). Aug 23, 2022 at 13:13
• @DanielC yes... either that or the OP is thinking of a vector of operators... Aug 23, 2022 at 13:14
• @JasonFunderberker, Isn't that what you get when you apply a vector operator on a ket? How else can on write it? Aug 23, 2022 at 13:14

The position operator should not be thought of as a column of operators. Rather, you should think of three distinct operators $$X_1,X_2,X_3$$. The sense in which they constitute a vector operator is that the action of a unitary rotation operator $$U_R$$ which corresponds to a rotation matrix $$R$$ on the position operators is given by $$X_i \mapsto X'_i = U_R X_i U_R^{\dagger} = \sum_j R_{ij} X_j$$ in analogy with the way that the components of vectors transform in classical physics. Equivalently, the commutators of the position operators with the angular momentum operators (which are the infinitesimal generators of rotations) are $$[L_i,X_j] = i\hbar \sum_k\epsilon_{ijk} X_k$$

The operators $$X_i$$ act on kets individually, and there is no sense in which we would act with the entire collection $$(X_1,X_2,X_3)^T$$ at once. It may help to interpret "vector operator" as "vector (of) operator(s)."

• Thank you so much. Aug 23, 2022 at 16:13
• So $\mathbf{\hat{x}} |\mathbf{x}' \rangle = \mathbf{\hat{x}} |\mathbf{x} \rangle$ means nothing but three separate eigenvalue equations? The vector operator is just an abbreviation to condense the three equations? Aug 23, 2022 at 16:28
• In above comment i meant $\mathbf{\hat{x}} |\mathbf{x}' \rangle = \mathbf{{x}} |\mathbf{x} \rangle$ Aug 24, 2022 at 6:25

A "vector operator" is not an operator, it is a collection of operators. Writing $$\vec A = \begin{pmatrix} A_1 \\ A_2 \\ A_3 \end{pmatrix}$$ doesn't imply that this "$$\vec A$$" can be applied meaningfully to a state vector - as you've already discovered, it does not make sense to apply this "$$\vec A$$" to a state vector (for formalizations of what $$\vec A$$ is, see this question and its answers, but that's really beside the point - it doesn't really matter). What you're "missing" is that you insist on interpreting terminology in a strict sense when that's just not how it works - a "vector operator" is not an "operator", a "manifold with boundary" is not a "manifold", that's just the way it is.

What we mean when we speak about vector operators is that the components $$A_i$$ of $$\vec A$$ transform e.g. under rotations like a vector would, and that is why it they are a useful notion; for instance $$\vec A \cdot \vec A = A_1^2 + A_2^2 + A_3^2$$ is an operator in the ordinary sense and is invariant under rotations by construction.

• Thank you so much. I request you for a little more clarification here please: sakurai defines an translation operator as $\mathscr{J}\left(d \mathbf{x}^{\prime}\right)\left|\mathbf{x}^{\prime}\right\rangle=\left|\mathbf{x}^{\prime}+d \mathbf{x}^{\prime}\right\rangle$. It's commutator with position operator equals $\mathbf {dx'}$ where sakurai says "$\mathbf {dx'}$" is a number multiplied by a identity operator. So what does $\mathbf {dx'}$. $\mathbf 1$ $|\mathbf u\rangle$ =$\mathbf {dx'}$. $|\mathbf u\rangle$ =? Aug 23, 2022 at 16:38
• @Kashmiri Please do not ask unrelated questions in comments; ask a new question if you have a new question. Aug 23, 2022 at 16:41
• Oh sorry. I thought it might be related. Aug 23, 2022 at 16:41

A vector operator is defined by the commutation relations of its components with the angular momentum operators: $$[L_i,A_j]=i\epsilon_{ijk} A_k \tag{1}$$ The components $$A_k$$ act on kets in the regular way: $$A_k\vert u\rangle=\sum_{\alpha}\vert \alpha \rangle\langle \alpha \vert A_k\vert u\rangle$$ where $$\{\vert\alpha\rangle\}$$ spans the HIlbert space.

What makes this a vector (or more generally a tensor operator) is that the action of components on kets is related by (1), i.e.
\begin{align} i \langle v\vert A_z\vert u\rangle = \langle v\vert [L_x,A_y]\vert u\rangle = \langle v\vert L_xA_y\vert u\rangle - \langle v\vert A_yL_x\vert u\rangle\ ,\\ =\sum_{\alpha}\langle v\vert L_x\vert \alpha\rangle\langle \alpha \vert A_y\vert u\rangle+ \sum_{\alpha}\langle v\vert A_y\vert \alpha\rangle\langle \alpha \vert L_x\vert u\rangle \tag{2} \end{align} clearly showing that the matrix elements of different components are not independent but linked through matrix elements of the angular momenta.

Normally, the kets $$\vert u\rangle, \vert \alpha\rangle$$ and $$\vert v\rangle$$ will have some known angular momentum properties, v.g. $$\vert u\rangle\mapsto \vert \gamma_u,\ell_u,m_u\rangle$$, $$\vert \alpha\rangle\mapsto \vert \gamma_\alpha \ell_\alpha m_\alpha\rangle$$ and $$\vert v\rangle\mapsto \vert \gamma_v \ell_v m_v\rangle$$, where $$\gamma_a$$ is whatever other quantum number is required to completely specify the state, so that the action of the $$L_x$$ (or any other component of angular momentum) can be evaluated explicitly, making the linear relation between $$\langle v\vert A_z\vert u\rangle$$ and $$\langle\alpha \vert A_y\vert u\rangle$$ and $$\langle v\vert A_y\vert \alpha\rangle$$ even more explicit.

• Thank you for your help :) Aug 23, 2022 at 16:16