I Vector operator acting on a ket gives a ket out of the state space

[Kashmir]

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?

 

[PeterDonis]

We also have vector operators

Vector operators are not single operators. They are sets of three operators. Each of the three operators acting on a ket gives another ket in the state space.

 

[Kashmir]

Vector operators are not single operators. They are sets of three operators. Each of the three operators acting on a ket gives another ket in the state space.

So one cannot call the set of operator itself as an operator?

 

[PeterDonis]

So one cannot call the set of operator itself as an operator?

The term "vector operator" means "a set of 3 operators".

 

[vanhees71]

It's a set of three operators $\hat{V}_i$ that has the commutation relation
$$[\hat{V}_i,\hat{J}_j]=\mathrm{i} \epsilon_{ijk} \hbar \hat{V}_k$$
with the total-angular-momentum operators $\hat{J}_j$. The Einstein summation convention applies here.

It's easy to prove that for finite rotations, represented by the unitary operator
$$\hat{U}(\vec{n},\varphi) = \exp(-\mathrm{i} \varphi \vec{n} \cdot \hat{\vec{J}}/\hbar),$$
you get
$$\hat{U}(\vec{n},\varphi) \hat{V}_j \hat{U}^{\dagger}(\vec{n},\varphi)=R_{jk}(\vec{n},\varphi) \hat{V}_k.$$
Here $R_{jk}(\vec{n},\varphi)$ is the rotation matrix for a rotation by an angle, $\varphi$, around the axis given by the unit vector $\vec{n}$.

Important examples are, of course, the position and momentum operators $\hat{x}_i$ and $\hat{p}_i$ and the angular momentum operators $\hat{J}_i$ themselves.

 

[Kashmir]

It's a set of three operators $\hat{V}_i$ that has the commutation relation
$$[\hat{V}_i,\hat{J}_j]=\mathrm{i} \epsilon_{ijk} \hbar \hat{V}_k$$
with the total-angular-momentum operators $\hat{J}_j$. The Einstein summation convention applies here.

It's easy to prove that for finite rotations, represented by the unitary operator
$$\hat{U}(\vec{n},\varphi) = \exp(-\mathrm{i} \varphi \vec{n} \cdot \hat{\vec{J}}/\hbar),$$
you get
$$\hat{U}(\vec{n},\varphi) \hat{V}_j \hat{U}^{\dagger}(\vec{n},\varphi)=R_{jk}(\vec{n},\varphi) \hat{V}_k.$$
Here $R_{jk}(\vec{n},\varphi)$ is the rotation matrix for a rotation by an angle, $\varphi$, around the axis given by the unit vector $\vec{n}$.

Important examples are, of course, the position and momentum operators $\hat{x}_i$ and $\hat{p}_i$ and the angular momentum operators $\hat{J}_i$ themselves.

so when we apply such a " vector operator " we actually mean three separate operations?

 

[DrClaude]

so when we apply such a " vector operator " we actually mean three separate operations?

You would get a Cartesian vector of kets,
$$
\hat{\mathbf{J}} = \mathbf{i} \hat{J}_x + \mathbf{j} \hat{J}_y + \mathbf{k} \hat{J}_z
$$
so
$$
\hat{\mathbf{J}} \ket{\phi} = \mathbf{i} \hat{J}_x \ket{\phi} + \mathbf{j} \hat{J}_y \ket{\phi}+ \mathbf{k} \hat{J}_z \ket{\phi}
$$

It is very rare to have to calculate the application of such vector operators directly. They usually appear in dot products, for instance
$$
\hat{J}^2 = \hat{\mathbf{J}} \cdot \hat{\mathbf{J}} = \hat{J}_x^2 + \hat{J}_y^2 + \hat{J}_z^2
$$

 

[Kashmir]

You would get a Cartesian vector of kets,
$$
\hat{\mathbf{J}} = \mathbf{i} \hat{J}_x + \mathbf{j} \hat{J}_y + \mathbf{k} \hat{J}_z
$$
so
$$
\hat{\mathbf{J}} \ket{\phi} = \mathbf{i} \hat{J}_x \ket{\phi} + \mathbf{j} \hat{J}_y \ket{\phi}+ \mathbf{k} \hat{J}_z \ket{\phi}
$$

It is very rare to have to calculate the application of such vector operators directly. They usually appear in dot products, for instance
$$
\hat{J}^2 = \hat{\mathbf{J}} \cdot \hat{\mathbf{J}} = \hat{J}_x^2 + \hat{J}_y^2 + \hat{J}_z^2
$$

You wrote "$\hat{\mathbf{J}} \ket{\phi} = \mathbf{i} \hat{J}_x \ket{\phi} + \mathbf{j} \hat{J}_y \ket{\phi}+ \mathbf{k} \hat{J}_z \ket{\phi}$ " on the rhs you've multiplied each ket by a euclidean vector which isn't an element of the field over which our ket space lies.

Can you please elaborate on that.

 

[DrClaude]

You wore "$\hat{\mathbf{J}} \ket{\phi} = \mathbf{i} \hat{J}_x \ket{\phi} + \mathbf{j} \hat{J}_y \ket{\phi}+ \mathbf{k} \hat{J}_z \ket{\phi}$ " on the rhs you've multiplied each ket by a euclidean vector which isn't an element of the field over which our ket space lies.

Can you please elaborate on that.

It is what happens when you mix our 3D physical world with the Hilbert space of quantum states. Yes, $\hat{\mathbf{J}} \ket{\psi}$ doesn't correspond to a state ket, but it is not something that appears by itself in QM.

 

[vanhees71]

It's just a short-hand notation for three kets you get by operating with each of the $\hat{J}_k$ operators on that ket.

Further the kets are NOT the wave functions but abstract operators in an abstract Hilbert space. The wave functions are the "components" wrt. to the generalized position-operator eigenvectors, $|\vec{x} \rangle$, which are the common eigenvectors of the three components of the position operator $\hat{x}_j$. The wave function related to the state ket $|\psi(t) \rangle$ is
$$\psi(t,\vec{x})=\langle \vec{x}|\psi(t) \rangle,$$
where I assumed that we use the Schrödinger picture of time evolution.

 

[Kashmir]

It's just a short-hand notation for three kets you get by operating with each of the $\hat{J}_k$ operators on that ket.

Further the kets are NOT the wave functions but abstract operators in an abstract Hilbert space. The wave functions are the "components" wrt. to the generalized position-operator eigenvectors, $|\vec{x} \rangle$, which are the common eigenvectors of the three components of the position operator $\hat{x}_j$. The wave function related to the state ket $|\psi(t) \rangle$ is
$$\psi(t,\vec{x})=\langle \vec{x}|\psi(t) \rangle,$$
where I assumed that we use the Schrödinger picture of time evolution.

"Further the kets are NOT the wave functions but abstract operators in an abstract Hilbert space".

Aren't Kets vectors of Hilbert space rather than operators?

 

[vanhees71]

Of course, it must read "abstract vectors". Sorry for the confusion.