Recent content by cipher42

Is this a two-dimensional problem (i.e. what system are you looking at)? If so, then my guess is that r=\sqrt{x^2+y^2}. From this, you could invert the equations and then recast your Lagrangian (or Hamiltonian) in these two new coordinates (since you should know what L=T-V looks like in...

If you're looking for the probability that it tunnels, that should be |T|^2, not just T (remember, all of the physical significance of the wave function has to do with it's absolute magnitude)

I was just trying to make sure I understood your argument and then state what I thought could be drawn from your premises. You say So your argument is that because T(2,5,7)=(2,5,7), the map is onto R^3, right? I'm just saying that all you have shown is that Tv=v for one single v (namely...

First attempt: 80 wpm w/3 mistakes 2nd: 83 3rd: 89 I love speed drills - especially when it just nice easy words like this one!

Yup, that's dead-on with the proof (though you might want to throw the word 'positive-definite' in there somewhere for clarity). Though I'm still not sure that no such operator exists :smile:

Why not? Is your argument that if, for some v, Tv=v, then T must map onto R3? I'm not sure I see that. In our case, v = (2,5,7) and Tv=v, but all this gives us is that the one-dimensional subspace spanned by v (namely all vectors of the form av, for some scalar a) gets mapped to 0 (and...

I'm not quite sure what you mean by this. Just because it maps to R3 doesn't mean that it has to map onto R3. Or maybe you mean something else, but still, knowing that one vector gets mapped to itself doesn't mean that no other vector can be mapped to zero (unless I'm missing something)...

I'm not sure where you went wrong; I can't find a simple inverse for your first term - which is the one that doesn't match the answers - so I'm guessing that's where your problem is. However, look at #16 on this table of transforms...

The definition of which charge is "positive" and which is "negative" is completely a matter of convention ("What's in a name?"), so at some point, if you want to know which has the positive charge and which has the negative, you're going to have to consult some reference to find out what the...

I wouldn't think so. The length and width are independent directions, so changing one will not affect the other. But with things such as volume or resistance that depend on the the diameter, this will not be the case.

You seem to have this one down. That should definitely do it. To make sure your answer is right, you can always just compute the potential of such a configuration and then plug in the constraints for the given planes (x=0, y=0, and z=0) and see that the potential is zero under these conditions.

Any time your using the method of images to solve for a potential, a uniqueness theorem is exactly what let's you solve the problem. Since the uniqueness theorem states that there can only be one potential that satisfies both Poisson's Equation and the given boundary conditions in the specified...

Right. That's exactly what the question is asking you to find out. You cannot assume you know the answer, you either have to find a counterexample to show that it is not true in some specific case, or make a logical argument (aka find a proof) to show that it must be true in all situations.

Underneath the integral, that's perfectly fine. Since the wavefunction is normalizable, it (and it's derivatives) will go to 0 at infinity and -infinity (the implied limits of integration), so integration by parts let's you transfer the derivative at the cost of a sign without having to worry...

Could you use the standard template for homework help problems and give us what you think are the relevant equations and your attempt at a solution as well? You'll have a much better chance at getting a helpful response that way.