Recent content by AntiStrange

Homework Statement Consider the harmonic scillator potential perturbed by a small cubic term, so that V(x) = \frac{1}{2}m\omega^{2} (x^{2} - \frac{x^{3}}{a}) if a is large compared to the characteristic dimension (\hbar /m\omega)^{1/2}, the states will all me metastable, since there can be no...

I'm reading through the Introduction to Quantum Mechanics book by Griffiths (2nd edition) and it is describing delta-function potential wells. When it describes how to find bound states the energy is E < 0 (negative) in the region x < 0 (negative). It says the general solution is: \psi (x)...

How do I know when it is ok to change a derivative into a partial derivative? For example there is something like: (d/dt)∫f(x,t)dx then they simplify it to: ∫[∂f(x,t)/∂t]dx I mean, it sort of makes sense to me, because the integral will be a function of t only so the (d/dt) is fine, but when you...

Sorry my responses take so long, busy schedule. I guess real voltage sources have an upper limit to the current they can produce, but would that really affect the terminal voltage like you say? So I'm guessing that the current is already being overdrawn when there is no resistance through...

I'm sorry but I still don't think that I understand #2. Here is my way of thinking: let's say the resistance of the photoresistor is 20kΩ, and the ammeter has a resistor connected in series with it of 2kΩ and the battery supplies a potential difference of 9V. The total resistance would be R =...

How does this "Dark Meter" work? In our physics lab we used photoresistors to make a simple circuit that is light activated. It was essentially a battery hooked up to a photoresistor which was in series with an ammeter to measure the current. When the photoresistor is blocked there is low...

Ah, ok that is the way you wrote it too :) Nice ty very much.

Does it just require more simplifying? I see how I could get it like this: \psi(x,0) = e^{-ixk_{0}}\int^{\infty}_{-\infty} dq' e^{-\alpha q'^{2} / 2} e^{ixq'} but the final form doesn't have the power of e in front of the integral as negative like I do.

I'm sorry but I don't follow the first step for the variable change. So if we have q' = k-k₀ then: dq' = 1*dk correct? If it is right, then we now have: \psi(x,0) = \int^{\infty}_{-\infty} dq' e^{-\alpha q'^{2} / 2} e^{ikx} which doesn't look right.

I'm reading up on Quantum Mechanics and I don't follow an integration they use. They start with this: \psi(x,t) = \int^{\infty}_{-\infty} dk A(k) e^{i(kx-\omega t)} They begin by considering the wave packet at time t=0: \psi(x,0) = \int^{\infty}_{-\infty} dk A(k) e^{ikx} "and...

Ok that makes perfect sense now. Thank you!

Homework Statement I'm supposed to read this little packet on mechanics, specifically dealing with central forces. and they are explaining a scenario where something is moving due to a central force (say, the Earth rotating around the sun) where all the motion takes place in the xy plane. They...

You don't actually need the shell method to be incorporated into the problem. I'm sure Fightfish brought it up so that you could see that the outer shells cancel out and all you are left with is what is under you. But instead of doing concentric shells, just consider a single shell of arbitrary...

You are very close. Actually I think in it's current form it will give you the correct answer, but for the wrong reasons. But in any case, consider just the portion of your equation: 4\pi \int^{r}_{0}r^{2} dr what is this equal to?

I apologize for not including them, but before I saw your two posts, I edited mine. So just look at it again.