# Wave functionsWatch

#1
What is the difference between the most probable position of a particle and the expectation value??
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1 year ago
#2
Think of it as the mode vs the mean. The most probable value is the point at which the probability density is maximised. The expectation is a weighted average of all the possible positions.

They can be the same, most notably if the distribution is symmetric, but they needn't be.
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#3
(Original post by Rinsed)
Think of it as the mode vs the mean. The most probable value is the point at which the probability density is maximised. The expectation is a weighted average of all the possible positions.

They can be the same, most notably if the distribution is symmetric, but they needn't be.

What does it mean for the probabilty density to be maximised? Is it just the peak point of the distribution?
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#4
Also another question: how do I find the potential that causes a wavefunction?
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#5
bump
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1 year ago
#6
(Original post by IDontKnowReally)
What does it mean for the probabilty density to be maximised? Is it just the peak point of the distribution?
Yes, that's exactly right. The point where the pdf is at its highest is the point which is more likely than any other individual point. So you just find the maximum like with any other function.

Also another question: how do I find the potential that causes a wavefunction?
So I'm guessing you've been given the wavefunction, and have been asked to infer the potential? You should be able to plug it into the Schroedinger equation and solve for the potential.
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#7
(Original post by Rinsed)
Yes, that's exactly right. The point where the pdf is at its highest is the point which is more likely than any other individual point. So you just find the maximum like with any other function.

So I'm guessing you've been given the wavefunction, and have been asked to infer the potential? You should be able to plug it into the Schroedinger equation and solve for the potential.
Yes but I have been asked to draw it and Im not too sure how since rearranging the formula involves constants like m and E which I don't have values for
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1 year ago
#8
(Original post by IDontKnowReally)
Yes but I have been asked to draw it and Im not too sure how since rearranging the formula involves constants like m and E which I don't have values for
So I think it's perfectly acceptable to give the potential in terms of constants like E and m. Physicists generally dislike messing around with actual numbers unless they have to.

If you've already done that, I'd probably work out the important points of the potential function (i.e. maxima, minima, inflection points or generally places where stuff happens) in terms of those constants and mark them on the axes. Your drawing doesn't need to be perfect as long as the form looks broadly correct, but I'd imagine they want you to point out the key features.
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#9
(Original post by Rinsed)
So I think it's perfectly acceptable to give the potential in terms of constants like E and m. Physicists generally dislike messing around with actual numbers unless they have to.

If you've already done that, I'd probably work out the important points of the potential function (i.e. maxima, minima, inflection points or generally places where stuff happens) in terms of those constants and mark them on the axes. Your drawing doesn't need to be perfect as long as the form looks broadly correct, but I'd imagine they want you to point out the key features.
I see! Thats really helpful, thank you.
One more question: would proving that a wave function is not normalisable be sufficient enough to show that a wave function is not a solution to the TISE for a free particle?
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1 year ago
#10
(Original post by IDontKnowReally)
I see! Thats really helpful, thank you.
One more question: would proving that a wave function is not normalisable be sufficient enough to show that a wave function is not a solution to the TISE for a free particle?
The solution to the TISE for a free particle is a plane wave, and I think plane waves are technically non-normalisable. So my answer would probably be no...

Can you not just plug the wavefunction in and show it doesn't work? Or is this a more general question?
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#11
(Original post by Rinsed)
The solution to the TISE for a free particle is a plane wave, and I think plane waves are technically non-normalisable. So my answer would probably be no...

Can you not just plug the wavefunction in and show it doesn't work? Or is this a more general question?
It was a more general question since plugging it into the equation without the values for the other constants seems a bit pointless- how can I be sure it is the right wavefunction?
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1 year ago
#12
(Original post by IDontKnowReally)
It was a more general question since plugging it into the equation without the values for the other constants seems a bit pointless- how can I be sure it is the right wavefunction?
Check it's an eigenfunction of the Hamiltonian.
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#13
(Original post by alow)
Check it's an eigenfunction of the Hamiltonian.
Sorry if this a dumb question but how do I do that?
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1 year ago
#14
(Original post by IDontKnowReally)
Sorry if this a dumb question but how do I do that?
Apply the Hamiltonian operator to the wavefunction.
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