Linked Questions

5 votes
2 answers
7k views

Hamiltonian operator in polar coordinates with momentum operators

The Hamiltonian operator for a free non-relativistic particle looks like $$ \hat{H} = \frac{\hat{p}^2}{2m} = -\frac{\hbar^2}{2m} \nabla^2. $$ In polar coordinates, the Laplacian expands to $$ \...
Kasper's user avatar
  • 2,010
1 vote
1 answer
5k views

Derivation of the radial momentum operator

I have been studying Quantum Mechanics and when my book was going through the Hydrogen wave equation, it was talking about this equation: $$ \frac{p_r^2}{2\mu} +\frac{L^2}{2\mu r^2}+V(r)=E$$ I ...
Phantom101's user avatar
3 votes
3 answers
1k views

Imagining zero orbital angular momentum for s-orbitals

Orbital Angular momentum of a s-orbital is always zero. One can easily imagine why this is so: QM says $\hat{p}=-i\hbar \nabla_{r}$, and since the s-wave functions are radially symmetric, the momentum ...
k.kulkarni19952's user avatar
1 vote
1 answer
2k views

Expected momentum of ground state hydrogen $<p>$

I am trying to calculate the expected momentum of an electron in the ground state of hydrogen atom. This is the wave function. So far I have done this:$$\iiint_V \Psi^* (-i\hbar) \frac {d\Psi} {dr} ...
Warda Anis's user avatar
2 votes
1 answer
2k views

How to prove radial momentum operator is hermitian?

Technically, if we want to prove an operator $\hat{A}$ to be an hermitian,we should prove $ \left< \psi \hat{A}|\psi \right> =\left< \psi |\hat{A}\psi \right>$. It works well in Cartesian ...
Fadelis Hu's user avatar
3 votes
1 answer
2k views

Clarification in deriving the radial momentum operator $p_r$

In deriving an expression for $p_r$, a particle's radial momentum, I am unsure what is happening at a certain step. The derivation given in The Physics of Quantum Mechanics by Binney and Skinner is as ...
Nick Chapman's user avatar
5 votes
1 answer
611 views

Momentum of a hydrogen system is imaginary. Why?

The hydrogen ground eigenstate: $$ \psi(r) = \frac{1}{\sqrt{\pi r_0^3}}\exp\left(\frac{-r}{r_0}\right) $$ Notice how nice: $$ \frac{\partial \psi}{\partial r}(r) = \frac{1}{\sqrt{\pi r_0^3}}\frac{-1}{...
Physicist137's user avatar
  • 3,334
1 vote
1 answer
800 views

Canonical commutation relation for spherical coordinates?

What coordinates systems can the canonical commutation relation be generalised to? I also ask specifically for spherical coordinates. This is because I want to prove $\hat{\bf p}\cdot{\bf e}_r-{\bf e}...
user107224's user avatar
0 votes
0 answers
490 views

Is the radial component of $p^2$ a hermitian operator?

The kinetic energy of a system $p^2/2m$ in real space representation takes the form $-\hbar^2\nabla^2/2m$. I want to express this in cylindrical coordinates via the representation of the laplace ...
Suppenkasper's user avatar
0 votes
0 answers
582 views

Momentum operator in spherical coordinates

In the position representation the momentum operator takes the form of the gradient, $-i\hbar \nabla$. It is understood that its components denote $p_{x}, p_{y}, p_{z}$ respectively; but, when ...
IdentityOne's user avatar
1 vote
2 answers
222 views

Hermiticity of a radial momentum operator $\hat{p}_r$ and the spectral theorem

In Nolting's QM book (Theoretical Physics 7), in the chapter on central potentials, a radial momentum operator $\hat{p}_r$ is defined as \begin{equation} \hat{p_r} = -i \hbar \Big( \frac{\partial}{\...
EM_1's user avatar
  • 920
-1 votes
1 answer
91 views

How do we determine the conjugate of radial component of linear momentum?

How do we describe the radial part of linear momentum? Here I found the following description: Classically $$p_r=\hat{D}_r = \frac{\hat{r}}{r} \cdot\hat{p} = \frac{\hbar}{i}\frac{\partial}{\partial r}...
Ashutosh Aman 5-Yr IMD Physics's user avatar