Talk:Quantum harmonic oscillator

message: The momentum operator p, I think, should be (h_bar / i)*(d/dx), not (-h_bar * i)*(d/dx). Source: Modern Physics, Tipler, 5th Ed. page 244. Cheers, Justin.  —Preceding unsigned comment added by 67.160.211.137 (talk) 10:05, 15 December 2008 (UTC)
replies: text: They're the same thing.  1/i = -i. Reedbeta (talk) 20:42, 26 February 2011 (UTC)
thread_title: Untitled

message: I've deleted and transed it to copy at Wiki's reference desk.--HydrogenSu 12:10, 5 February 2006 (UTC)
thread_title:  2 questions 

message: On 13 June 2006, a user named User:Fanifol apparently reverted the article to the state of 10 March 2006. This was Fanifol's only contribution to the wikipedia, and his revision comment was "THE CERTAINTY PRINCIPLE WAR ". However, as in the last nine days no other wikipedia editor has bothered to revert this edit, I now hesitate to do so. Could somebody please provide some background on "THE CERTAINTY PRINCIPLE WAR"? — Tobias Bergemann 14:22, 22 June 2006 (UTC)
thread_title:  Fanifol revert 

message: I think one of the diagrams (the one captioned as "Wavefunction representations for the first six bound eigenstates...") is incorrect. I was plotting Ψ for n=0..5 in maple for some work i'm doing and i noticed that my plots for n=2 and n=3 are reflected in the x axis relative to those shown in the article. This website also agrees with my plots.
Have I missed something here, or are these actually wrong? If they are can some one fix them? Poobarb 17:15, 19 October 2006 (UTC)
replies: text: The only difference is a minus sign in the wave function. Such things are irrelevant in QM. David  09:06, 21 March 2007 (UTC)
thread_title:  error in diagram? 

message: The cubic potential is rather awkward...it will result in an unbounded system! I removed it for now. I think it's better to rewrite that section with a quartic potential instead, and probably in a different article. --HappyCamper 16:03, 14 April 2007 (UTC)
thread_title:  Anharmonic oscillator 

message: This is why I edited the example of the 3-dimensional harmonic oscillator:
:* A few small spelling mistakes. 
:* More importantly: a 3-dimensional harmonic oscillator is not necessarily isotropic. One can have different vibrational frequencies in different directions: \omega_x \ne \omega_y\ne \omega_z . 
:* I changed inline TeX letters to html italic, because for some reason, unknown to me, the inline TeX came out roman.
--P.wormer 08:53, 25 May 2007 (UTC)
thread_title:  3-dimensional harmonic oscillator 

message: The article states "the lowest achievable energy is not zero...It is not obvious that this is significant, because normally the zero of energy is not a physically meaningful quantity, only differences in energies."  If I'm not mistaken, the physically meaningful fact here is that the lowest energy is not equal to the minimum of the potential well.  This is why there is zero-point "motion" in a QHO as opposed to a classical oscillator.  We have set up the problem with the zero of energy at the bottom of the well, so when the ground state energy comes out as \hbar \omega / 2 all it means is that the ground state energy is higher than the bottom of the well - there is really no issue with the zero of energy not being physically meaningful.  Unless anyone objects, I'll edit the article to make this clear. Reedbeta (talk) 20:33, 26 February 2011 (UTC)
replies: text: I've made the change. Reedbeta (talk) 19:35, 27 February 2011 (UTC)
thread_title:  "the zero of energy is not a physically meaningful quantity" 

message: Note: Apologies for the first edit that accidentally resulted in removing entire sections from the article. I realized it just as it was corrected by Cuzkatzimhut.
Beside the ladder operator method, another useful way of solving the one-dimensional quantum harmonic oscillator is to directly solve the Schrödinger equation for the system.  We start by writing the equation:
:-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2}+\frac{1}{2}m\omega^2x^2\psi=E\psi.
Now we introduce dimensionless variables \xi=\sqrt{\frac{m\omega}{\hbar}}x and \epsilon=\frac{E}{\hbar\omega}, and rewrite the Schrödinger equation as:
:-\frac{\hbar^2}{2m}(\frac{d\xi}{dx}\frac{d}{d\xi})\psi+\frac{1}{2}m\omega^2(\sqrt{\frac{\hbar}{m\omega}}\xi)^2\psi=\epsilon\hbar\omega\psi,
which further simplifies to:
:-\frac{1}{2}\frac{d^2}{d\xi^2}\psi+\frac{1}{2}\xi^2\psi=\epsilon\psi,
and after moving terms becomes:
:\frac{d^2\psi}{d\xi^2}=(\xi^2-2\epsilon)\psi.
For very large values of \xi (i.e., at very large x), \xi^2 becomes much bigger than 2\epsilon, so we can approximated this equation as:
:\frac{d^2}{d\xi^2}\psi\approx\xi^2\psi,
Which has the general solution:
:\psi(\xi) = h(\xi)e^{-\frac{\xi^2}{2}}.
Now our goal is to find h(\xi). Let us start by finding the first and second derivatives:
:\psi'(\xi) = [h'(\xi) - \xi h(\xi)]e^{-\xi^2/2}
and
:\psi(\xi) = [h(\xi)-2\xi h'(\xi) + (\xi^2-1)h(\xi)]e^{-\xi^2/2}.
Note that according to the Schrödinger equation, \frac{d^2\psi}{d\xi^2}=(\xi^2-2\epsilon)\psi, so we substitute to achieve:
:h(\xi) - 2\xi h'(\xi) + (2\epsilon-1)h(\xi) = 0.
Let us try to find a solution in the form of power series of h(\xi):
:h(\xi) = \sum_{j=0}^{\infty}a_j \xi^j
By substituting this in the equation and collecting similar powers we get:
:\sum_{j=0}^{\infty}\xi^j[(j+1)(j+2)a_{j+2} - 2ja_j + (2\epsilon-1)a_j] = 0.
We can conclude that the coefficient of each power of \xi must be the same on both sides, thus we have:
:a_{j+2} = \frac{2j+1-2\epsilon}{(j+1)(j+2)}a_j, or:
:\frac{a_{j+2}}{a_j} = \frac{2j+1-2\epsilon}{(j+1)(j+2)}.
This recursion formula must terminate for some j=n, so that a_{n+2}=0. This will truncate either the series h_{even} or h_{odd} and the other must be zero from the beginning. This gives:
:\epsilon=n+\frac{1}{2}=\frac{E}{\hbar\omega},
Which is favorable because it gives the allowed energy levels for the system:
:E=\hbar\omega(n+\frac{1}{2})
The properly normalized eigenfunctions are
:\psi_n(x)=\frac{1}{\sqrt{2^n n!}}\left (\frac{m\omega}{\pi\hbar} \right )^{1/4}e^{-{m\omega x^2/2\hbar}}H_n\left (\sqrt{\frac{m\omega}{\hbar}}x\right ).
The first four hermite polynomials are:
:h_0(x)=1\,
:h_1(x)=2\xi\,
:h_2(x)=4\xi^2-2\,
:h_3(x)=8\xi^3-12\xi\,
And the rest can be calculated using the generating function:
:G(\xi,s)=e^{-s^2+2s\xi}=\sum_{n}h_n(\xi)\frac{s^n}{n!},
Or alternatively through the Rodrigues formula:
:h_n(\xi)=(-1)^ne^{\frac{+\xi^2}{2}}\frac{d^n}{d\xi^n}e^{-\xi^2}.
Reference
Unsigned talk comment by Mahrud, 15:40, 15 December 2014.
::I personally don't see the point of this mini-tutorial, as the point and the results are already summarized in black and white in section 1.1!  This looks like an attempt to introduce the novice to solving the Schr eqn, Hermite polynomials, the Rodrigues formula (not wikilinked??? WHY??) etc... at an introductory level. It is really a protracted explanatory footnote to section 1.1 to somebody shaky in QM.  Such sections are often put in hide boxes, and this, suitably condensed and wikilinked, could find its way in a hidebox at the end of section 1.1. However, It really detracts from the continuity and logical cohesion of the treatment... If sufficiently many editors assist in paring off repetitive statements in it and endorsed its insertion, in a hidebox, at the end of section 1.1, I could help stick it in. Cuzkatzimhut (talk) 16:33, 15 December 2014 (UTC)
thread_title:  Proposition: adding "Analytical method" 

message: The section "Example: 3D isotropic harmonic oscillator" correlates with the notion that the ground state has energy of 3/2.  (Permit me to omit the factor ħω.)  But, the following wave function [a] satisfies the wave equation, [b] normalizes, and [c] correlates with an energy of 1/2.
Ψ(r, θ, φ) ∝ r−1 exp(−(r2)/(s2)) Y0,0(θ, φ)
* Here, s has dimensions of length.
I am reluctant to try to edit the page itself, because …
* I think other people should set the scope for the page.
* I may be the discoverer of the above and other solutions (for isotropic quantum harmonic oscillators) [a] that people might say are "non-traditional" and [b] for which I find possibly physics-relevant applications.  If my findings are noteworthy, perhaps someone[s] else should report them on Wikipedia. 
* I have little experience directly editing Wikipedia pages.  Previously, I edited parts of only 1 page ["IDIQ"].
Some notes.
* Regarding  normalization, a relevant integral is ∫ Ψ* Ψ r2 dr.  The r2 factor allows for normalization.
* For each odd positive integer N with N≥3, there exist "non-traditional" solutions having energies of 1/2, 3/2, …, (N−2)/2, as well as the "traditional" solutions with energies N/2, (N+2)/2, … . 
* In some attempted research (in math, particle physics, astrophysics, and cosmology), I find use for non-traditional solutions such as I note above.
* Should someone want [a] more information about the math and/or [b] references to cite, … please feel free to consult (on-line [zero-cost download] or in print) the book "Theory of Particles plus the Cosmos."  The math I reprise above is in section 1.2.  The specific solution I show above paraphrases equation (1.2.68).  Here are links.  On-line - https://www.academia.edu/11329222/Theory_of_Particles_plus_the_Cosmos_Small_Things_and_Vast_Effects_Extract_-_Front_pages_through_Part_3_  Print - http://www.amazon.com/Theory-Particles-plus-Cosmos-Effects/dp/1508610339/
* Permit me to note that my attempted research also shows possible uses of similar findings correlating with linear representations (as well as, per above, with radial/spherical representations).  For example, for N=3, I find useful the non-traditional solution for which nx = −1, ny = 0, and nz = 0.  (book table 1.2.3 and section 2.1)  This solution correlates with an energy of 1/2.
* People might state concerns about using quantum numbers that people might say are negative numbers.  But, the mathematics "works."  And, the overall energies are all non-negative.  (And, the math may have applications.)
Thomasjbuckholtz (talk) 21:29, 15 March 2015 (UTC)
replies: text: You should be congratulated for your reluctance to edit, especially given the controlling WP:NOR rule. This is a threshold you must not cross.    Normalizability  is not the be–all and end–all, but regularity at the origin, which you are forfeiting, is. Martin has analyzed the issue of the neglected δ-function involved. The irregular solution, which Dirac knows about in his book and dismisses on physical grounds, does not solve the Schroedinger eqn for a regular potential like the oscillator, but matches a  δ-function add-on at the origin---and violates the virial theorem. While technically informative, and definitely not new with you!, the strictly formal phenomenon does not belong here and has little bearing on oscillators or physics.  Cuzkatzimhut (talk) 23:52, 15 March 2015 (UTC)
thread_title:  Possibly overlooked solutions - regarding multidimensional harmonic oscillators 

message: "The following commutators can be easily obtained by substituting the canonical commutation relation,"
These follow immediately (mathematically) from the preceding lines. Why give the impression that they depend on something presented as a physics concept in another article. Keith McClary (talk) 20:40, 31 January 2016 (UTC)
replies: text: Can you make clear what, exactly, you are proposing, and why? The reader who still needs reminding of where the following commutators came from is steered to use the canonical commutation relations linked to redo the elementary linear algebra to produce them. Cuzkatzimhut (talk) 21:07, 31 January 2016 (UTC)
thread_title:  Commutators 

message: Every formulation that I have seen for the quantum harmonic oscillator wave eqn has a term like 
(mω/πћ)(1/4).  Yet the units of this term before the root extraction are kg*sec-1/(m2*kg*sec-1), which cancels out to m-2.  After the root is taken, the units of the term are m-1/2.  Since the units cancel out in all the other terms of the wave eqn, the results of ψn(x), assuming x in meters or femtometers, have units 1/√m or 1/√fm.  How is that possible?  — Preceding unsigned comment added by CElliott316 (talk • contribs) 19:47, 15 March 2016 (UTC)
replies: text: ? What exactly is the problem? the probability density has then inverse units of length, so integrated over dx, it  yields a dimensionless probability. You did  notice, didn't you, the cube in the 3d formula in the suitable section? Taking a physics course should help. Cuzkatzimhut (talk) 20:08, 15 March 2016 (UTC)
thread_title:  Question: Units in Wave Eqn for Quantum Harmonic Oscillator 

message: I don't understand why ν is defined the way it is.  The factor of two does not make the equation any simpler.  It also makes conversion to natural units more annoying.
I would also like to propose rewriting the normalization constant in terms of the Γ function, because it is simpler and matches what is used by the orthogonality relations in the Laguerre polynomials article.
: Original:
:::\psi_{klm}(r,\theta,\phi) = N_{kl} r^{l}e^{-\nu r^2}L_k^{(l+{1\over 2})}(2\nu r^2) Y_{lm}(\theta,\phi)
::where
:::N_{kl}=\sqrt{\sqrt{\frac{2\nu^3}{\pi }}\frac{2^{k+2l+3}\;k!\;\nu^l}{
(2k+2l+1)!!}}~~ is a normalization constant; \nu \equiv {\mu \omega \over 2 \hbar}~;
: Proposed:
:::\psi_{k l m}(r, \theta, \phi) = N_{k l} r^l e^{-\gamma r^2 / 2} L_k^{(l + 1/2)}(\gamma r^2) Y_{l m}(\theta, \phi)
::where
:::N_{k l} = \sqrt{\frac{k! \, 2 \gamma^{l + 3/2}}{\Gamma(k + l + 3/2)}} = \sqrt{\sqrt{\frac{\gamma^3}{\pi}} \frac{2^{k + l + 2} \, k! \, \gamma^l}{(2 k + 2 l + 1)!!}}  is a normalization constant;
:::\gamma \equiv \frac{\mu \omega}{\hbar} ;
--Fylwind (talk) 05:05, 8 December 2017 (UTC)
thread_title:  Proposal: Replace ''ν'' with ''ν'' / 2 in [[Quantum harmonic oscillator#Example: 3D isotropic harmonic oscillator|#Example: 3D isotropic harmonic oscillator]] 

message: The animated plots showing the evolution of the wavefunction are really neat. It would be nice if they also showed the evolution of the probability density for the position, which is relevant when we measure the position of the particle over time.  — Preceding unsigned comment added by Pmpgl (talk • contribs) 13:14, 19 July 2020 (UTC)
thread_title:  Animated plots: improvement 

message: I propose that you remove the protection from the QHO page, it is really cringe. And this Page is not even worth protecting, it is pure stupidity and cringe.  So yeah, that is. I love hot girls. QHOorNAGATORO (talk) 15:04, 12 March 2023 (UTC)
thread_title:  Proposal: Remove Protection for za damn page.