I Basic question on meaning of momentum operator

[Kashmir]

I don't know what your symbol $\hat{P}$ means. You have three momentum-component operators. In the product notation it's defined as
$$\hat{P}_x=\hat{p}_x \otimes \hat{1} \otimes \hat{1}, \quad \hat{P}_y=\hat{1} \otimes \hat{p}_y \otimes \hat{1}, \quad \hat{P}_z=\hat{1} \otimes \hat{1} \otimes \hat{p}_z.$$
The momentum operators in the position representation are $\hat{p}_j=-\mathrm{i} \hbar \partial/\partial x_j$. This you can derive from the commutation relations $[\hat{x}_j,\hat{p}_k]=\mathrm{i} \hbar \delta_{jk} \hat{1}$ with $j,k \in \{x,y,z \}$.

I meant this $\hat{p}=\left(\hat{p}_{x} \otimes 1 \otimes 1\right)+\left(1 \otimes \hat{p}_{y} \otimes 1\right)+\left(1 \otimes 1 \otimes \hat{p}_{z}\right)$

 

[PeroK]

Since $\langle x |$ needs to be in the tensor product space just like
$\langle r|=\langle x| \otimes\langle y| \otimes\langle z|$

What is $\langle x|$ here in the lhs of above quoted equation?

I think I was a bit sloppy in using $\langle x|$ to pick out the first component in the tensor product. I think your suggestion to use $\hat p_x$ instead is better and makes more sense.

 

[PeroK]

Shouldn't this have been $$\langle r|\hat p_x |\psi \rangle = -i \hbar \frac{\partial}{\partial x} \psi(x, y, z)$$$$\langle r |\hat p_y |\psi \rangle = -i \hbar \frac{\partial}{\partial y} \psi(x, y, z)$$$$\langle r |\hat p_z |\psi \rangle = -i \hbar \frac{\partial}{\partial z} \psi(x, y, z)$$ ?

Yes, I think that's better. The key point is that an inner product produces a single complex number (for any Hilbert space). The underlying equality must, therefore, be three separate equalities, that we want to write in more compact notation.

Any vectorization of these three equalities is, therefore, a symbolic/notational thing.

 

[Kashmir]

Thank you.

In the position representation, Hamiltonian of a free particle in 3D is usually given as :
$$-\frac{h^{2}}{2 m} \nabla^{2}$$

To prove this, is the below correct :

$$
\hat{H}=\frac{\hat{p}^{2}}{2 m}=\frac{1}{2 m} \hat{p} \cdot \hat{p}
$$

Then the position representation is

$$
\left\langle r\left|\frac{1}{2 m} \hat{p} \cdot \hat{p}\right| \psi\right\rangle
$$

Where we note that
$$
\langle r|=\langle x| \otimes\langle y| \otimes\langle z|
$$ and $$\hat{p}=\left(\hat{p}_{x} \otimes 1 \otimes 1\right)+\left(1 \otimes \hat{p}_{y} \otimes 1\right)+\left(1 \otimes 1 \otimes \hat{p}_{z}\right)$$ and solve this

 

[vanhees71]

The last equation is of course not correct. It doesn't make any sense, or what do you think adding the three components of the momentum wrt. some arbitrary cartesian basis should physically mean? What makes sense is the notation
$$\hat{\vec{p}}=\begin{pmatrix}\hat{p}_x \\ \hat{p}_y \\ \hat{p}_z \end{pmatrix}.$$
It is the quantum version of the same mathematical description of the momentum components in classical mechanics you also usually put in a column-vector notation.

Than it's also immediately clear that, analogous to classical mechanics (and in this case even without any "operator-ordering problem", because the components of the momentum operator all commute)
$$\hat{\vec{p}}^2=\hat{p}_x^2+\hat{p}_y^2 + \hat{p}_z^2.$$

 

[Kashmir]

How did you go from this

$$\hat{\vec{p}}=\begin{pmatrix}\hat{p}_x \\ \hat{p}_y \\ \hat{p}_z \end{pmatrix}.$$

To this?

$$\hat{\vec{p}}^2=\hat{p}_x^2+\hat{p}_y^2 + \hat{p}_z^2.$$

It's like you treated it like a euclidean vector

 

[andresB]

It's like you treated it like a euclidean vector

The momentum operator (and the position) are vector operators . Loosely speaking, they have components along each axis. Under a rotation, the components behave/transform pretty much like the components of a Euclidean vectors.

In this case, the $P^{2}$ operator is invariant under rotations, just like the magnitude of an euclidean vector is invariant under rotations. See Messiah ch XIII $10