Talk:Particle in a box/Archive 1

Time-independent equation
Thanks! A nice, detailed article.

Maybe we should mention that we are using the time-independent Schrödinger equation here, which is different from the one on Schrodinger wave equation. Or maybe that page should be changed.

Also, is it a coincidence that all &psi;n are real-valued functions? Shouldn't complex numbers show up? --AxelBoldt


 * The eigenfunctions of the time independent Schrodinger equation can always be made real. Phys 16:01, 17 Aug 2004 (UTC)

Thanks for the compliment on the article. I just added a statement that we are using the time-independent Schroödinger equation. I think it would be useful to add this equation to the Schrodinger wave equation page and explain when its use is appropriate (probably a task for someone more knowledgeable on this subject than myself).

As for the absence of complex numbers, my understanding is that complex numbers, indicate the "phase" of the system and arise in the time-dependent equation. Since the particle in a box is a time-independent problem and corresponds to a standing wave there is no time-dependent portion and hence no complex component to the solution. If anyone more knowledgeable than myself on this subject can confirm or deny this explanation, I would greatly appreciate it.--Matt Stoker

-

About the doubt above: the wavefunctions for time-independent quantum mechanics problems ARE complex in general, and not necessarily real. The example in this article is simple and compelling because they are real, but in general they can be complex. A simple example of a complex standing wave are the electron states in a hidrogen atom. - Ernesto.


 * The hydrogen atom electron states can also be made real, by appropriate linear combination. Pfalstad 12:12, 8 January 2006 (UTC)

In this case, the wavefunctions are real because the problem is one-dimensional. Generically, one-dimensional solutions to the S. wave equation have no phase variation, so we might as well take them to be real. -- CYD


 * For bound states, yes, but the free particle in one dimension has phase variation, and it's easier to solve that using complex numbers. Same with the periodic potential.  There are also real solutions, though.  Pfalstad 12:12, 8 January 2006 (UTC)


 * in the case of a particle in a box, the 'eigenfunctions' of the Hamiltonian can be chosen to be real. In general, the wave function describing the state of a particle is a superposition of these eigenfunctions which can be complex. for example $$\Psi(x)=\frac{1}{\sqrt{2}}(\psi_1(x)+i\psi_2(x))$$ where $$\psi_1$$ and $$\psi_2$$ are the eigenfunctions corresponding to energies E1 and E2 respectively. --V. 05:14, 16 February 2007 (UTC)

Math markup
The math markup on this article is ghastly. The hbar looks horrible, the fractions are done using (num)/(den) which also looks horrible when it is done everywhere, and the integral limits look horrible as well. I would like to Tex-ify this article when I have time, or someone else can do it, unless there is large protest against it. --dave

As for: I would like to Tex-ify this article
 * The (fortunately!) few instances of TeX-urication in Wikipedia seem to make the fonts of their articles ghastly inconsistent, both in size and in type.


 * I've seen this complaint before, but I don't really buy it. The fonts for the formulas are bigger than the text, but the formulas are all the same size.  So I don't really see any "inconsistency".  The font size of the formulas are just bigger than the text, which makes them stand out.  It's no big deal if the formula stands out.  For the tex-math stuff inline, this can screw up the spacing between lines, but but only for certain math things.  Just simple greek symbols don't cause any problem.

As for: The hbar looks horrible
 * How does one set one's Wikipedia:Preferences for how HTML operators (such as "&lt;strike&gt;") are implemented and rendered? And: why might this article require any particular "struck-out unit" anyways?
 * Best regards, Frank  W ~@) R 20:34 Apr 2, 2003 (UTC)


 * I'm not sure if I understand you, but the "struck-out" unit is necessary in this case to adhear to convention. It is h-bar which is a standard unit in quantum mechanics.  It prevents having to write out 2*Pi everywhere.  --dave

Ok, about the above stuff....I think tex for the indented formulas is great. Much better than the html. But tex is definitely bad for inline stuff. But I think using html in the inline stuff is ok, and yes that makes the font size inconsistent of variables with the main formula, but it's okay, it's still the same letters and it can still be italic, so I don't see this as a huge factor. At least, I think having the main formula in TeX looks so much nicer that it outweighs any concern for the font sizes being inconsistent. --dave

Radical is slanted
Does anyone know why formula in 18 and 19 the square root sign is slanted in one, but not the other? Thanks. --dave
 * pure speculaion from my part: it's because 'y' is slighty longer than 'x'; if you replace:
 * $$Y_{n_y}=\sqrt{\frac{2}{L_y} \sin \left( \frac{n_y \pi y}{L_y} \right) } \quad (19)$$
 * by
 * $$Y_{n_y}=\sqrt{\frac{2}{L_y} \sin \left( \frac{n_y \pi y}{L_x} \right) } \quad (19)$$
 * then both are slanted! -- looxix 20:30 Apr 3, 2003 (UTC)
 * then both are slanted! -- looxix 20:30 Apr 3, 2003 (UTC)
 * then both are slanted! -- looxix 20:30 Apr 3, 2003 (UTC)


 * Thanks looxix, I tried fixing it using \vphantom{y} but it doesn't support that command for some reason.... --dave
 * So it's solved now (was my fault sorry). -- looxix 00:11 Apr 4, 2003 (UTC)
 * Oh, you mean because the root sign shouldn't have been over the sine? I thought I was the one that did that... oh well  --dave

Analysis and shapes
"Particle in a box" analysis is said to be invalid for certain shapes.

Chemists, led by graduate student Heng Yu and William E. Buhro, Ph. D., professor of chemistry in Arts & Science, at Washington University in St. Louis have published an article in Nature Materials, August 2003,which is described in a press release as "the first comprehensive comparison relating shape to the phenomenon known as "quantum confinement"

http://www.sciencedaily.com/releases/2003/10/031014071540.htm

The release states: "After compiling all of the data graphically, the group found that the famous "particle-in-a-box" calculation -- a quantum "model" both simple enough for college freshmen and rigorous enough to explain observed quantum phenomena – fell short of predicting absolute values for individual shapes, due to its simplification of the nature of diameter dependence.

"However, they were encouraged to discover that particle-in-a-box approximated the ratio of the slopes of wires to dots, a fact verified by their experimental and theoretical data. This is a particularly advantageous finding because it gives other researchers a method of assessing 2-D confinement in their own quantum wires without the need of a supercomputer."

Ken A.

Scienceworld Links
For one, I think that link to wolfram should be removed. Anyone look at that page?


 * I think I fixed it. Wikipedia should have its own finite well page. PAR 17:48, 1 Apr 2005 (UTC)

Technical language cleanup request
I don't expect the main body of the article to be comprehensible to the average reader. However, the introduction could be made more accessible by using less jargon, explaining things in more detail, and explaining the significance of things more. -- Beland 17:58, 3 September 2005 (UTC)

I added an informal introduction to attempt to alieviate this problem. This was not simple, as this is a technical mathematical excersize. Suggestions as to how it can be further improved are welcome.

kyle, PitOfBabel 20:03, 1 November 2005 (UTC) -

PAR and Pfalstad fixed it up till it works. Nice job! PitOfBabel 20:55, 1 November 2005 (UTC)

Looks great! -- Beland 21:41, 13 November 2005 (UTC)

which one is first?
--HydrogenSu 19:41, 17 February 2006 (UTC)

L
Since this specifies linear momentum, Ρ x = -iħδ/dδ in terms of x the component, and since the wave function is also specified to the quantum number n x  shouldn't L actually be L x ?--64.12.116.73 14:28, 18 February 2006 (UTC)

come to think of it
Why isn't linear momentum specified explicitly? Based on the derivation shown, the starting point was obviously Ρ x , yet it just shows the Ĥn x  = -(ħ2/2m)*(δ2/dδ2), without even mentioning that this is derived from the linear momentum operator, Ρ x = -iħδ/dδ, why is that?--64.12.116.73 14:34, 18 February 2006 (UTC)


 * Which part of the article are you talking about? Pfalstad 15:51, 18 February 2006 (UTC)
 * Particle in a 1-D box, it uses the Ĥn x  = -(ħ2/2m)*(δ2/dδ2) notation, without even mentioning what it is dervied from, ie.. the Ĥ*Ψ(x)=Ĥ*E expression, with Ĥ = (1/2)P x 2*m + V(x), where the linear momentum operator P x = -iħδ/dδ, all of which seems to be omitted, I was just wondering if there was a reason for it's omission, or if was simply an oversight--64.12.116.73 18:10, 18 February 2006 (UTC)


 * Brevity? That section is about a specific solution of the Schrodinger equation, so it gets right to the point of solving it, not for deriving that particular form of the schrodinger equation.  One of my textbooks, Bransden and Joachain, does similarly.  It introduces the time-independent Schrodinger eq in that form, in fact, and only later talks about the Hx=Ex form, or p operator form.  Pfalstad 20:28, 18 February 2006 (UTC)

Wrong statement !!!
Attention to all:

The statement that " The absolute value of the momentum can also be determined from an energy eigenfunction: $$|p_n| = \hbar k = \frac{nh}{2L}$$ is absolutely wrong !!! A localized particle can't have a well determined momentum - this is one of the cornerstones of QM ! The abovementioned relation holds only for a plane wave (story with Bohr orbitals is a subject of separate discussion). Another argument: The Hamiltonian for a particle in a box does not commute with momentum operator. Recall the university QM course (if you had any) and try to figure out how to calculate momentum correctly. If you'll encounter any difficulties with this task, I can help you next weekend :-)

Elektrik 05:31, 21 February 2006 (UTC)


 * I agree; I was puzzled by that too. Pfalstad 20:15, 8 March 2006 (UTC)


 * The particle does not have a well-defined momentum, it has a well-defined absolute value of momentum. Its momentum wave function consists of two delta functions, one at nh/2L and one at -nh/2L. It follows that the uncertainty in momentum is nh/2L. PAR 00:50, 9 March 2006 (UTC)


 * I disagree that the momentum wave function consists of two delta functions. If you combine the two momentum eigenstates as you describe, you get a wave function that extends throughout space.  It matches the nth energy eigenstate inside the box, but outside the box it doesn't vanish as it should; it's periodic.  Pfalstad 04:03, 9 March 2006 (UTC)


 * Oops - Yes, you are right. That absolute value stuff needs to be changed. I will do it tomorrow unless you do it first. PAR 05:14, 9 March 2006 (UTC)


 * I just removed all the momentum stuff for now. We could derive the wave function in momentum space and show how it has two peaks.  Not sure if that would be necessary or useful in a basic article like this, but feel free to add it.  I'd rather graphs of the position wave functions, a description of how the quantum picture differs from the classical, etc.  Pfalstad 03:39, 10 March 2006 (UTC)

particle in one dimensional box
my question is " using particle in one dimensional box model explain why an electron could not be held in the nucleus of an atom   assume mass of electron=10 to raise -30Kg    and   size of nucleus=10 to raise  -14m ". I shall wait at my e mail address for answer please please hurry to send answer I shall be very thankfull to you for this snd me at   cavandish2005@yahoo.com

A request to experts
After the mathematical derivation of the solution to the one dimensional infinite potential well, the physical meaning of the solution is not explained. The wave-function and energy eigenstate is calculated, but it is not explained in what way these cause observational deviation from classical predictions.


 * That should be fixed. We should also have some graphs of the solutions.  Pfalstad 20:34, 9 March 2006 (UTC)

Also, what is this time-independent Schroedinger equation? In the books I've read, Schrodinger's equation is formulated as relating the time derivative of the wave-function to the canonically quantized Hamiltonian of the wave-function. As far as I know, the time-dependence is a crucial character of the Schrodinger equation that makes it non-relativistic (due to non-invariance of time frame in special relativity), something that was corrected by Dirac in his relativistic equation describing wave-function evolution of an electron.

Thank you. Loom91 09:19, 9 March 2006 (UTC)


 * See here. Pfalstad 20:34, 9 March 2006 (UTC)

In equations 10 and 21 the $$\frac{\hbar^2\pi^2}{2m} $$ has been canceled out leaving just $$\frac{h^2}{8m} $$, apologies if I am wrong but is this not possible also in equation 23? Would it be better to have them all the same, either one way or the other? sijarvis 15:01, 31 May 2006 (UTC)

Good Article nomination has failed
The Good article nomination for has failed
 * lead-in paragraph is info of its own; just one reference is not enough; text is only for university-level. Wim van Dorst (Talk)'' 21:25, 16 June 2006 (UTC).

I'm not sure what you mean by "lead-in paragraph is info of its own." That doesn't make sense to me. As for text seeming to be university level, that may not be something that can and should be fixed. I originally started the opening paragraphs to frame the problem. I'm not sure what can be done about the body, though I"ll look.PitOfBabel 19:02, 23 June 2006 (UTC)

Errr ... The wavefunctions in the figure are wrong!!!

It shows the groundstate labled (correctly) as n=1, but what is labled n=2 is n=4 and what is labled n=3 is n=9 and n=16 is labled as n=4. Someone seems to get confused with the Energylevels, these go indeed with n squared. It's easy to see that the wavefunctions are wrong: According to (Sturms) Node rule each exitation has one more node the next lower laying energy level. So where is the wavefunction with just 1 node in the middle, whcih is the real first exiatation and therefore n=2?

That's a bit embarrassing for the scientific reputation of wikipedia, isn't it?

Low Dimensional Systems
I clarified some of the earlier derivation and added a section on low-dimensional systems. I feel this is a very important topic as it considers a 1D quantum well constructed in the real world rather than in an artificial 1D universe. Please have a look and see if everything's correct. Thanks --djsik 11:54, 22 July 2006 (UTC)


 * I removed the section on Low dimensional systems because:
 * It has a number of errors - The Schroedinger needs $$\hbar$$ squared, not $$\hbar$$, and the $$k_y^2+k_z^2$$ term needs to be included in the muliplication, etc. All these errors are propagated through the derivation (I think).
 * This whole section could be accomplished by taking the limit of the finite-dimensional box as $$L_y$$ and $$n_y$$ go to infinity while $$n_y/L_y$$ remains finite, using the already-established formula $$k_y=n_y\pi/L_y$$ (same for the z coordinate). We don't need to go through the whole derivation all over again.
 * This is not to say there shouldnt be something on low dimensional systems, it just seems that this was erroneous, and maybe too repetitive.PAR 14:25, 22 July 2006 (UTC)


 * Sorry about the errors in the derivation - I wrote it from the top of my head and obviously didn't check it thoroughly. They can all be corrected very easily.  However, I still feel that the section would be useful, if only for introducing the concept of energy sub-bands replacing discrete energy levels.  If you agree that it would be useful, I'll put the section back in with an abbreviated (and corrected!) derivation if required. --djsik 14:53, 22 July 2006 (UTC)


 * I think it would be useful, but I think it should be a short section, with only the new ideas given any real space. Look, if we just take equations 22 and 23 and substitute $$k_i=n_i\pi/L_i$$ for i=y,z we get:


 * $$\psi_{n_x,n_y,n_z} = \sqrt{\frac{8}{L_x L_y L_z}} \sin \left( \frac{n_x \pi x}{L_x} \right) \sin \left( k_y y \right) \sin \left( k_z z \right)$$


 * $$E_{n_x,n_y,n_z} = \frac{\hbar^2\pi^2}{2m} \left[ \left( \frac{n_x}{L_x} \right)^2 + \left( \frac{k_y}{\pi} \right)^2 + \left( \frac{k_z}{\pi} \right)^2 \right]$$


 * and these are almost the final answers we are looking for. The $$\sin \left( k_y y \right)$$ term is the same as $$e^{ik_y y}$$ except for a meaningless phase difference, and we can multiply the plane wave by $$\sqrt{L_yL_z}$$ and still have a plane wave, but with a different normalization. Now we just let the n 's and L 's go to infinity while n/L remains finite, and thats it. PAR 17:00, 22 July 2006 (UTC)

Image:Particle1D.jpg
I like the idea behind Image:Particle1D.jpg in this article. However, I was thinking that it might be better to clarify the distinction between energy and wavefunction in the image. For example, perhaps a black horizontal line could be drawn at each of the energy levels (1, 4, 9, 16). (As it is currently shown, it is possible that someone unfamiliar with the subject might interpret the wavefunction curves to be energy.)--GregRM 13:00, 25 August 2006 (UTC)
 * After looking at it again, it seems like the wavefunction curves are incorrect. The curve labeled n=2 is actually the wavefunction for n=4, the curve labeled n=3 is actually the wavefunction for n=9, and I am guessing that the curve labeled n=4 is actually the wavefunction for n=16. I am going to temporarily remove the image until these issues can be sorted out.--GregRM 13:08, 25 August 2006 (UTC)
 * Agreed, good catch. Pfalstad 13:45, 25 August 2006 (UTC)

Degeneracy in the 3D Particle in a Box
Under the section "3 dimensional box and higher", the article states that "degeneracy results from symmetry in the system." However, in three-dimensional QM, degeneracy is sometimes the result of symmetry (as the article suggests) and sometimes it is simply inherent. In the case of a cubical box, $$L_{x}=L_{y}=L_{z}$$, for instance, the energy $$E_{116}=E_{235}$$ is not a result from any symmetry in the system. I believe these are termed "accidental degeneracies". In sum, I believe that from the perspective of completeness and accuracy in the article a mention of the fact that degeneracy in 3D QM is sometimes the result of symmetry in the system, and sometimes not, is warranted.Leoj.Cire 01:06, 23 December 2006 (UTC)

Good article
This does not merit being described as a good article. It fails to describe correctly what this is about for the general reader. you can show-off maths and logic skills later in the article. In the meantime you should improve the communication of this concept - to the layman.

Perhaps put a cat in the box. —Preceding unsigned comment added by 150.101.126.88 (talk) 06:13, 29 September 2007 (UTC)

Separation of variables
Derivation of the two- and three-dimensional cases start with the assumption that the wave function is separable, i.e. can be written as eq.(12). What I miss here is an argument, why this actually covers all solutions. Without such an argument, we can only argue that the solutions we gain from this assumption are quantized, which is a fairly weak statement. There could be an infinite number of other solutions, in particular, continuous ones. To make concise that quantization is really a physically existing phenomenon, we need to argue that we have derived a complete set of solutions. What would be a suitable argument to use here? --62.80.17.84 (talk) 09:43, 23 April 2008 (UTC)

Assessment comment
Substituted at 15:31, 1 May 2016 (UTC)