How Does SHM Relate to Probability Distributions in Quantum Mechanics?

[Claire84]

Hey there.

As part of our quantum assignment we;ve to also look at a simple classical oscillator (it's part b to a question about the one dimensional harmonic oscillator). Problem is, I can hardly remember a thing that isn't to do wih quantum! So any help here would be appreciated. We've to find an expession for the velocioty of the particle as a function of positon, and I think I'm going okay with it, it's just the limits that I don't know about (maximum displacements are a and -a).

I've got it down to v(dv/dx)= (-w^2)x which gives me vdv=(-w^2)xdx, but I don't know which way to integrate between the limits. Is the amplitude at the bottom or the top? The example of it I looked at online as x as the upper limit and the amplitude as the bottom one, but I don't get that...

Secondly, we've to (as a result fo the first bit), show that the probability of locating the particle in an interval dx between the maximum displacements a and -a is given by the ewt P(x)dx=dx/(pi(a^2 - x^2)^(1/2)). I'd like to take a stab at it but I don't know how to work out the probability for it - is it related to the probability that we use in quantum mechanics? Sorry, I sound so dumb here but I've no idea how to work out the probability. I've checked the net but to no avail. Please help (it would be much appreciated!)!

Claire

 

[quantumdude]

Hey there.

Hi, and welcome to PF!

I've got it down to v(dv/dx)= (-w^2)x which gives me vdv=(-w^2)xdx, but I don't know which way to integrate between the limits. Is the amplitude at the bottom or the top? The example of it I looked at online as x as the upper limit and the amplitude as the bottom one, but I don't get that...

If your initial displacement is the amplitude, then yes the lower limit of integration should be at the bottom. The reason you integrate to an upper limit of x is because you're trying to find a function of x. So, you let the upper limit take on a variable value. If you integrated to another constant, your answer would be a constant.

Secondly, we've to (as a result fo the first bit), show that the probability of locating the particle in an interval dx between the maximum displacements a and -a is given by the ewt P(x)dx=dx/(pi(a^2 - x^2)^(1/2)).

Sorry, what's an "ewt"?

I'd like to take a stab at it but I don't know how to work out the probability for it - is it related to the probability that we use in quantum mechanics?

Short answer: No. You don't calculate it using QM at all.

Longer answer: Sorta. The probability density for the classical SHO is the large-n limit for the probability density of the QM SHO. This must be the case, as per the correspondence principle.

 

[Claire84]

LOL, eqt, not ewt. You've just got to love my typing. ;)

Think I've sorted it out ok-ish now. Discoverd that P(x)=(B^2)/v(x) where B^2 is a constant.

Thanks for your help btw! :)

 

[quantumdude]

LOL, eqt, not ewt. You've just got to love my typing. ;)

Ha Ha, I thought you were trying to say exp(ωt), and I was wondering how that got in there!

Think I've sorted it out ok-ish now. Discoverd that P(x)=(B^2)/v(x) where B^2 is a constant.

That's right. P(x) is inversely proportional to speed, which makes sense because the faster you are going in an interval from x to x+dx, the less time you spend there, and the less likely it is that you'll be found there.

All you have to do is find B by normalizing P(x) to 1.