If the bound states to the infinite square well are stationary waves, why does the ground state consist of half a wavelength?
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For a stationary wave, you need two counter-propagating travelling waves that 'reflect' off the boundaries of the well at either end then superimpose in the middle to produce a stationary wave.
If this is the case, then, for the infinite square well why is the ground state a travelling wave?
I call it a travelling wave as it doesn't have a long enough wavelength to 'reflect' of the walls and produce it's own standing wave. The image below may put my confusion into context:
The text in green at the end says it all, and it confuses me even more as it just raises yet another question: Do all of the bound states (1st excited, 2nd excited etc.) need 2 counter-propagating (travelling) waves to produce their standing waves?
This question has been bothering me for some time now, and I have searched the internet but can't get a straight answer. I even looked at Born-Von Karman boundary condition in solid state; the density of states argument along with the Karman boundary condition tells us the ground state cannot be as it is shown in the image above.
But, that was solid state physics, not quantum mechanics.
hope it will help u if u like my ans. plzzzz mark me as brainileast
thnx dear
For a stationary wave, you need two counter-propagating travelling waves that 'reflect' off the boundaries of the well at either end then superimpose in the middle to produce a stationary wave.
If this is the case, then, for the infinite square well why is the ground state a travelling wave?
I call it a travelling wave as it doesn't have a long enough wavelength to 'reflect' of the walls and produce it's own standing wave. The image below may put my confusion into context:
The text in green at the end says it all, and it confuses me even more as it just raises yet another question: Do all of the bound states (1st excited, 2nd excited etc.) need 2 counter-propagating (travelling) waves to produce their standing waves?
This question has been bothering me for some time now, and I have searched the internet but can't get a straight answer. I even looked at Born-Von Karman boundary condition in solid state; the density of states argument along with the Karman boundary condition tells us the ground state cannot be as it is shown in the image above.
But, that was solid state physics, not quantum mechanics.
hope it will help u if u like my ans. plzzzz mark me as brainileast
thnx dear
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