4
votes
What does $\omega$ mean in SHM?
$\omega$ is the "angular frequency." If you plot the particle's velocity and position as a function of time, the particle moves around a circle (see the comments for a discussion of the ...
- 4,347
3
votes
Accepted
Quantum mechanical harmonic oscillator - where does the number operator come from?
You've got the logic of the derivation wrong, I think. I recommend re-reading it, but here's an outline of the process, and you'll see that it's not circular:
Show that $\hat{H}$ can be factored as $...
- 6,105
3
votes
Accepted
Integral arising from the proof that the quantum harmonic oscillator satisfies Heisenberg's uncertainty principle
You are missing the forest for the trees, and tobogganing into error. Set $\sqrt{m\omega/\hbar}=1, \implies x=\xi$.
Recall the basic recursions,
$$ \bbox[yellow]{
\xi H_n(\xi)= \tfrac{1}{2} H_{n+1}(\...
- 55.4k
2
votes
Accepted
Difference between Displacement from Equilibrum and Amplitude of SHM
The maximum displacement of a particle from its mean position in SHM is defined as the amplitude of the SHM, whereas displacement is just the shift or movement of the particle from its equilibrium ...
- 152
2
votes
Accepted
Ground eigenstate of the quantum harmonic oscillator with the interacting vacuum $| \Omega \rangle$
There's a little bit of confusion here between notions from quantum mechanics and notions from quantum field theory, but this is a very interesting question that leads in interesting directions!
The ...
- 71
2
votes
Integral arising from the proof that the quantum harmonic oscillator satisfies Heisenberg's uncertainty principle
Calculating integrals with polynomials is probably not the easiest way to find $\langle x^2\rangle$ and $\langle p^2\rangle$. In my opinion, it is better to use ladder operators. Or an even simper ...
- 3,670
2
votes
Applying the position operator several times to a harmonic oscillator state $\hat x^m |n\rangle =$ ______?
A foolproof approach is doing this in position representation (where it is just multiplication by $x^m$) using your favorite methods of dealing with Hermit polynomials (e.g., Schiff has a clear ...
- 54.8k
1
vote
Accepted
Does the formula $ V =\omega r$ holds in angular frequency
Angular frequency is the magnitude of angular velocity, so yes. Although as far as I'm aware, we typically only speak of angular frequency when the angular velocity (or speed) is constant.
- 10.6k
1
vote
What does $\omega$ mean in SHM?
Simple harmonic motion and circular motion are mathematically closely related.
On a unit circle (circle with radius 1), every point has the coordinates $(\cos\theta,\sin\theta)$ for some angle $\theta$...
- 2,468
1
vote
Difference between Displacement from Equilibrum and Amplitude of SHM
Amplitude of a particle performing SHM is the maximum displacement from the equilibrium position it can achieve during its motion while displacement from mean/equilibrium position suggests its set off ...
- 11
1
vote
Accepted
Equilibrium position for double spring hanging vertically
The only force pulling $m_2$ up is the tension in spring $k_2$ (as you should be able to see by drawing a free body diagram). $m_1$ on the other hand is pulled up by tension in spring 1, but pulled ...
- 10.6k
1
vote
Will a positively charged rod undergo simple harmonic motion when displaced by a small amount perpendicular to the axis of a positively charged ring?
If you examine in detail the field lines of the ring, you will find that it is not so simple: the field has a radial component directed inwards in the central region and outwards after . So it is not ...
- 4,291
1
vote
Will a positively charged rod undergo simple harmonic motion when displaced by a small amount perpendicular to the axis of a positively charged ring?
Either this situation will be unstable or it will be stable. If it is stable, it will undergo harmonic motion. If it is unstable, a small perturbation from equilibrium will exponentially grow over ...
- 4,347
1
vote
Quantum mechanical harmonic oscillator - where does the number operator come from?
First of all, I stress that the ladder operators are not defined as
$$
\hat{a}|n\rangle = \sqrt{n}|n-1\rangle \\
\hat{a}^\dagger|n\rangle = \sqrt{n+1}|n+1\rangle.
$$
Rhey are instead defined in terms ...
- 65.2k
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