Uncovering the Mystery of Wave Function Symmetry

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Discussion Overview

The discussion revolves around the properties of wave functions in a potential well, particularly focusing on the symmetry of the wave function and the implications of changing the coordinate system. Participants explore how these changes affect the form of the wave functions over time, specifically addressing the use of sine and cosine functions in relation to boundary conditions.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant presents the initial wave function for a potential well and discusses the need to find its form at a general time, emphasizing the importance of symmetry.
  • Another participant notes that for an infinite potential well, the wave function must vanish at the boundaries, leading to the conclusion that sine functions are appropriate for certain conditions.
  • There is a question about why, after changing the coordinate system to achieve symmetry, sine functions are still used instead of cosine functions, given the new boundary conditions.
  • Some participants argue that changing the coordinates does not fundamentally alter the problem, suggesting that the wave functions remain sine waves when transformed back to the original coordinates.
  • One participant expresses confusion about the implications of using sine versus cosine functions and the nature of superposition under the given assumptions.
  • Another participant highlights that the distinction between sine and cosine functions relates to a phase shift and depends on the choice of origin in the coordinate system.
  • There is a recognition that understanding the relationship between sine and cosine functions is crucial for resolving the confusion about the wave functions.

Areas of Agreement / Disagreement

Participants express differing views on the implications of changing the coordinate system and the resulting wave functions. There is no consensus on the necessity or correctness of using sine versus cosine functions in the context of the problem, indicating ongoing debate and exploration of the topic.

Contextual Notes

Participants discuss the boundary conditions and the nature of wave functions without reaching a definitive conclusion on the appropriateness of sine or cosine functions in the transformed coordinate system. The discussion reflects uncertainty regarding the implications of phase shifts and superposition in wave functions.

Who May Find This Useful

This discussion may be of interest to students and professionals in physics, particularly those studying quantum mechanics, wave functions, and boundary conditions in potential wells.

noamriemer
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Hello again!

Say I have a potential well, between 0 and a. I also know how the wave function looks like for (t=0):
\psi(x,0)= \frac {2bx} {a} for 0<x<\frac {a} {2}
and
\psi(x,0)= 2b(1- \frac {x} {a} ) for \frac {a} {2} <x<a

Now, I wish to find the wave function of a general time, t.
I know I need to find symmetry. Therefore I move the x axis:
x\rightarrow x', x'\equiv x-\frac {a} {2}

So now I have the same well, only symmetrical.

That means:

\varphi_n =\sqrt {\frac {2} {a} } sin(\frac {\pi lx&#039;} {a}) \rightarrow l=2n<br /> \varphi_n =\sqrt {\frac {2} {a} } cos(\frac {\pi lx&#039;} {a}) \rightarrow l=2n+1

For the question before the axis transition, I know I have to look for a sin- solution.
But now, when I have this symmetry, I should look for a solution which vanishes at x=-a/2 instead of x=0

Meaning, the cosine must stay. But in this problem's solution, I see that the sine stays...
why is that so?
Thank you!
 
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noamriemer said:
Hello again!

Say I have a potential well, between 0 and a ...
Meaning, the cosine must stay. But in this problem's solution, I see that the sine stays...
why is that so?
Thank you!

I assume you mean outside of [0, a], the potential is infinity (an infinite potential well).

If that's the case, the wave function must vanishes at x=0 and x=a. \varphi_n = \sin \frac{\pi n x}{a} do this, \varphi_m = \cos \frac{\pi m x}{a} do not.
 
mathfeel said:
I assume you mean outside of [0, a], the potential is infinity (an infinite potential well).

If that's the case, the wave function must vanishes at x=0 and x=a. \varphi_n = \sin \frac{\pi n x}{a} do this, \varphi_m = \cos \frac{\pi m x}{a} do not.

Yes. of course. What I don't understand is why- if I change the axis so the problem becomes symmetric, do I still use sine instead of cosine.

For this axis, the wave function has to be zero for a/2 and -a/2, instead of 0 and a...
 
noamriemer said:
Yes. of course. What I don't understand is why- if I change the axis so the problem becomes symmetric, do I still use sine instead of cosine.

For this axis, the wave function has to be zero for a/2 and -a/2, instead of 0 and a...

You just did a mathematical trick to re-define your coordinates .. nothing about the problem has actually changed. If you transform back to your original coordinates, all of your wavefunctions will be sine waves. As to why you still have some sine solutions in the new coordinates, look closely at the conditions where the wavefunctions are sine solutions. Does that suggest anything to you?
 
Thank you... but no... it doesn't suggest anything. I thought that there can't be a superposition of the two functions under such assumptions...
 
noamriemer said:
Thank you... but no... it doesn't suggest anything. I thought that there can't be a superposition of the two functions under such assumptions...

You only have sine functions for *even* values of n. Where are the nodes for such a sine wave? How does that match up with your boundary conditions? What if you tried to use cos functions for the even values of n?
 
Why should it matter?

once I know how the wave function looks like for t=0 I know it had collapsed into this form. Therefore- If I had sine function- it will not change...

Or did I misunderstood you?
Thank you so much for your enlightening answers...
 
noamriemer said:
Why should it matter?

once I know how the wave function looks like for t=0 I know it had collapsed into this form. Therefore- If I had sine function- it will not change...

Or did I misunderstood you?
Thank you so much for your enlightening answers...

You are missing the point of my answers ... for a given n (i.e. a given frequency), the only difference between a sine and a cosine is a phase shift of \frac{\pi}{2}. So, whether or not a function is a sine or a cosine depends on your choice of origin, which you changed when you changed your coordinates. Now, the question you have to answer is what happens to the phase of the sine waves describing the eigenstates in your initial coordinate system when you shift to the new coordinate system. I suggest that you work it out on paper if it is not clear to you ...
 
Oh... I did miss the point. Of course that depends on a phase only. I know that. I just can't get a clear answer about general things I don't understand.

But writing the questions (and, of course, reading the answers) make everything more clear...
Thank you so much!
 

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