Search results

^ No. And I searched this forum and found out how momentum can be measured - crystal diffraction and similar experiments. So I am answered for the nonce. If I have any further questions about those experiments, I'll start a new thread. Thanks everybody.

I understand what the gradient tells us :) the part I have a problem with is your equating the (vector) function momentum operator maps the wave function to (i.e grad psi) as the momentum of the particle. What we measure is not this direction (which would be a function of space), but...

And what does an ionized track correspond to - a momentum eigenvalue? But then, each interaction with an atom is a position measurement, no? @ Muppet With 'conclusion', what I was trying to say was : this is not the vector we measure. What we measure are eigenvalues, which we...

'The way I think about ordinary momentum [In QM] is that the gradient of a function is a vector that points in the direction in which the function increases fastest; so the momentum operator picks out the direction in space in which you're most likely to find the particle.' Yes, but do we...

In CM, we have a prescription for measuring momentum. Velocity is defined, and we can measure it, and we can find out the momentum. Now, does quantum mechanics have the same prescription for measuring momentum (for a single free particle at least) ? I mean, for a single free particle, can we...

In Modern Quantum Mechanics page 44-46, Sakurai compares two experiments: 1. Here three observables A,B,C are measured. The probability of obtaining the results a,b and c respectively are then :(|<c|b>|^2)(|<b|a>|^2). Summing up over all values of b we'll find...

But this does not mean that the wave function cannot be discontinuous at the initial time t=0, right? So if at t=0 a wavefunction is discontinuous at a point, doesn't that go against the probabilistic interpretation? I mean, probability of finding a particle at that point would be undefined.

No,my question is why do we call it 'momentum'?what has this got to do with momentum as defined in classical mechanics?

ok,so why do we have the operator for momentum that we have?why that particular operator and no other?

so what's the deal here?why do we quantize the classical variables in the prescribed way and no other?

Are the quantization rules for momentum and energy axioms of quantum mechanics? Do we have such quantization rules because the expressions for energy,momentum and angular momentum they furnish are conserved?Or are there any other reason?

ummm...a quantity that remains conserved because the lagrangian is time-independent? can anyone please give me a sort of outline how i can derive the De-broglie relation for energy ?

so the de-broglie relation holds only for constant potential,is that what you are saying?what is the definiton of energy then?

What is the status of the De-Broglie relations then?Is it a derived result or an axiom?does it always hold?

How is energy defined in quantum mechanics?is it defined by the De-Broglie relations or from the eigenvalue equation of H operator?Are the two somehow equivalent?

For a free particle,the one dimensional Schrodinger's equation gives a solution of the form Ae^i(kx - wt).This solution does not meet the normalisation requirement.According to Bransden-Joachain's texr,there are 2 ways out of this difficulty.One is to superpose and form localised wave...