@JohnRennie hi sir

@satan29 hi :-)

sir have a look at the previous question please

Give me a few minutes ...

@satan29 In my Griffiths 3rd edition 1.10 is a question about the digits of pi ...

sir in that excercise there is a question with parts a) b) c)..and we have to assume a potential 1/2 kx^2

there a graph given with two points a and b marked

oh oops

it was figure 1.10

problem 1.11

yes

The integration is going to be between the two turning points i.e. the points at which the KE is zero.

hmmm

i.e -b to b

It's just a classical SHO so the period is T = 2π√k/m

and the velocity you just get from the KE as usual.

sir wait

Yes ... ?

$$1/T * \int_{-b}^{b} \dfrac{dx}{\sqrt(k/m)*\sqrt(b^2-x^2)}$$

is the integral that gives the probability

which comes out to be $$ 1/T * \pi \sqrt{m/k} $$

No, it's just $$ \rho(x) = \frac{1}{v(x)T} $$

@JohnRennie yes, and $$v(x) = \sqrt{k/m}*\sqrt{b^2-x^2}$$

ah, and in that equation $T$ is the half period i.e. the time it takes get from one extreme to the other.

where $$b^2= 2E/k$$

@JohnRennie ohhhh yes

sir one more doubt, we cant we integrate between -inf and inf?

technically the probb is still 1.

Outside $\pm b$ the square root will become imaginary.

okay...

I think its because the velocity function is like a piecewise function:

v=0 (x in (-inf to -b), (b to inf) )

v= sqrt()(...) (x in -b to b)

anyways, my doubt is clear now, thanks :-)

:-)

ssir in 1.12

rho(P) will be simply m/pT ?

no wait

rho(p)= 1/ m a(x) T

@JohnRennie sir there?

I'm working I'm afraid ...

ok sir no problem> sir will you ping me when free>

@satan29 I guess we'd start as before with $\rho(p)dp = dt/T$

yes ssir.

Sir I solbved that question, i had another doubt

Then write $dt = dt/dp~dp$, and as you say dt/dp = 1/F, where F is the force.

yup.

And $F = kx$

yes

its just arithmetic now, to make everything a function of p...

So you end up with $\rho(p) = 1/kxT$

yes

@satan29 yes

the standard deviation of p comes out to be sqrt(mE).

the standard deviation in x was sqrt (E/k).

it then asks us to calculate s_p* s_x. this is ofcourse E/w

in classical M, E can be arbitrarty, but in a QM harmonic oscillator for eg, E is always <=\hbar w/2

so the product seems to be <= \hbar/2 always, verifying the uncertainty principle..

Which is remarkably cool really :-)

but sir my question was, sir assume this was a classical harmonic oscillator

as mentioned E can take any value

so sp*sx can take any value

Yes, what it's showing is how the HUP for a quantum SHO is related to a similar concept for the classical SHO.

which is a clear failure of the uncertainty principle.

i can make E arbitrary small.

NB for a quantum SHM $E \ge \tfrac12\hbar\omega$, not less than as you wrote.

ah yes, oops.

anyway, it seems that the HUP is a QM result, and not a general one? So when can we know that the the system is "quantum enough", so as to apply the HUP ?

The quantum effect is the zero point energy $\tfrac12\hbar\omega$

i can construct a classical oscillator with $E= 1/4 \hbar \omega$ for example.

The HUP applies to a quantum SHO because it has a lowest possible energy i.e the zero point energy.

@JohnRennie oh hmm

okay , so the HUP applies for systems with quantized energy?

That's the difference between the quantum and classical systems. The ZPE means there is a lower limit to the uncertainty that doesn't exist in the classical system.

@JohnRennie hmm

so in general, you would need to know something about the system before trying to use HUP

as in, wether there is something as a ZPE or not

Hang on.

The uncertainty exists in classical and quantum systems. It's just a statistical property. But in a classical system the uncertainty can be made arbitrarily low. The difference is that in a quantum system the uncertainty cannot be made arbitrarily low.

how do I look at a system and confirm if its classical or quantum.

@JohnRennie does hup says that photon in a em wave does not travel in straight line?

@satan29 a classical system obeys Newton's laws while a quantum system obeys the Schrodinger equation.

hmmmmmmm

I mean I always have a dilemma related to path travel by photon when Em wave propogat e

Does each photon actually travel with speed C

@JackRod quantum particles don't really have a path. They don't have a well defined position due to the uncertainty principle, so they can't have a well define trajectory. They have a sort of average trajectory.

Sir when we measure light we assume each photon travel in straight line

?

@JohnRennie

Well, we assume the photon has a momentum. Travelling in a straight line is the same as saying it has a constant momentum.

But due the the uncertainty principle there is always some uncertainty in the momentum.

Okay sir last can hup predict the velocity of light as C

Which is a fact

@JackRod no

@JohnRennie but the uncertainty principle comments on the magnitude of p and not the direction right?

Ordinary QM is non-relativistic so the speed of light does not enter into it.

it may have an uncertain magnitude, but whats stopping it from travelling in a straight line?

Let's leave this for now ...

@JohnRennie hi sir

@satan29 hi :-)

sir can you open Griffiths

OK I have it open ...

1.16 c)

problem

ehrenfest theorem

why doesnt it apply ?

Err, I don't know ...

You'll have to ask in the h bar

I need to go now. I'll be around later, or failing that tomorrow as usual.