Recent content by cepheid

Don't give me your indignation, man. You were supposed to use the template for posting homework help threads, the one that requires an *exact* problem statement and an attempt at the solution. You know, the template that you deleted when you made your original post, in spite of being asked in...

It would help if you had actually posted parts b and c of the problem, then :rolleyes:

Welcome to PF Inklings (or Sofia, actually!), There is no link attached. I will make some assumptions about what is going on in your example. If the electric field is given by the function: E(t) = E0sinωt Then, the acceleration is F/m = (qE)/m, where q is the charge that the field is...

The Friedmann equation should be ##H^2 = \frac{8\pi G}{3}\rho##, where rho is the total mass density of the universe (in this case considered to be entirely due to matter). You can ignore the second term with kappa entirely, since it's zero (flat universe). You are correct that you can write...

I don't know what you are saying here. At a given density and temperature, yes. Change these, and you change the equilibrium point of the system.

That is correct. The rest of your work (applying the conservation of energy( looks fine.

Your trig is wrong. D is the hypotenuse of the triangle, and h is the side opposite the angle. Draw a picture.

Yeah, keep in mind that a bunch of gas exists in the disc of our galaxy, permeating the space between the stars: the interstellar medium. You're right that if a particular "cloud" were too sparse, it would be indistinguishable from this background, whose average density is about 1 hydrogen atom...

If the constants are equal (i.e. two distinct states have the same value -- they are said to be degenerate states), then yes, it would work. Their linear combination would also be an eigenstate with that same energy.

Use a more physical argument to motivate your answer: eigenstates (of the Hamiltonian) are states of definite energy. So if the particle is in an eigenstate, a measurement of its energy is sure to yield the same answer always. But if the particle is in a linear combination of two eigenstates...

The answer is indeed no. But why aren't you sure about your answer? This is math after all, there is a definite answer. So don't guess, do the algebra. Take your psi and multiply it by some constant C (or E, or whatever). Is this the same as what you got for H|psi>?

What BruceW is saying is that for psi to satisfy the time-independent Schrodinger equation, it would have to be true that ##H|\psi\rangle = E|\psi\rangle## where E is *some constant.* Look at your above result for ##H|\psi\rangle##. Is it equal to "some constant" multiplied by psi?

Yeah, that's right. That is a linear combination. So what happens when you apply H to this state? Remember that H is a linear operator.

Okay, first of all, by analogy with eigenvectors/values, an energy eigenstate is just a state (wavefunction) for which this is true: when you apply the Hamiltonian operator to it, you just get a constant (the eigenvalue, E) multiplied by it. So, if |E> is an energy eigenstate, then:$$H|E\rangle...

Hi Ascendant78, I replied to your PM regarding this question. Sorry for not getting back to you sooner. For the benefit of other thread goers, the gist of what I said was: Only two forces act on the object: gravity and the normal force. The normal force does no work.