Recent content by kehler

I found the uncertainty between delta x (position) and delta H (Hamiltonian) to be greater or equal to (h_bar*<p>)/ 2m. Does this mean for stationary states, where <p>=0, the uncertainty can be zero? ie we can precisely measure the position and energy?

Ahh ok I see :). Thanks. Gosh it seems really easy now.

I've done the first two steps. It's the third that I'm having trouble with. What do I do from here? :s

What's a resolution of the identity? I don't have the Shankar text. We use Griffiths in class... No unfortunately I'm not too familiar with Dirac Notation :(. I know roughly what it is but I'm not confident in using it just yet. So far we've just been taught the integral form. I think we're just...

Homework Statement Homework Equations The Attempt at a Solution I've gone round in circles doing this! I started of by writing it as an integral of (psi* x A_hat2 x psi) w.r.t dx, then using the equation above but I keep coming back at my original equation after flipping it...

Is it possible though to derive the whole of classical mechanics from quantum mechanics?

^ Thanks :). That makes it clear

I don't know. I was just wondering what the theorem does...

Is there any physical significance of this theorem? Can we make some kind of conclusion about space and time because the derivative of the expectation value of momentum with respect to time is equal to the negative of the expectation value of the derivative of potential energy w.r.t. space...

Homework Statement Show whether the functions psi_I = A cos(kx - wt) psi_II = A sin(kx - wt) are solutions of Schrodinger equation for a free particleHomework Equations Schrodinger equation The Attempt at a Solution For psi_I = A cos(kx - wt), d2psi_I/dx2 = -Ak2psi[/SUB]I[/SUB] dpsi_I/dt =...

Thanks :). That helped.

Just something like on page 3, fig 1.2 where it's a continuous graph..

I was referring to a graph of the square of the wavefunction vs position. That's what the textbook that I'm reading (Griffiths) uses to depict the probability of where a particle associated with some wavefunction will collapse.. It's only taking 1-D into account I think.

I don't quite get it :S. From my understanding, the probability distribution graph depicts the probability of where the particle will collapse. But you're saying that it actually represents the probability of a particle, currently at a particular position on the graph, collapsing onto an eigenstate?

I just started learning QM. I was wondering, if a wavefunction can only collapse onto a few eigenstates, how come the probability distribution graph is a usually continuous one? :S