Recent content by gareththegeek

So do you mean that the expectation value is the probability of that measurement regardless of time because the electron is in a fixed pattern about the nucleus?

Nice! Perhaps I'm finally beginning to get the hang of this ere quantum stuff then eh, eh?

I have read that the time dependent wavefunction is related to the Fourier transform of the wavefunction for the angular wavenumber like so \bar{\psi}(k,t) = \frac{1}{\sqrt{2\pi}}\int \psi(x,t)e^{-ikx}dx Can anyone explain why it is relevant to take the Fourier transform of the...

The squared value of the wavefunction represents a particular measurement and a measurement must be taken at a certain time, so wouldn't the time dependence be implied? The measurement shows the state of the system at a given time, reducing the probability of any other (mutually exclusive)...

These two observations seem to be at odds. So what does the phase mean in this case? Many thanks! G

I guess I am confused by how quantum mechanics began (and I have only read and tried to understand early quantum mechanics at this point) by examining electrons &c using wave-particle duality and deriving the equations from treating particles as waves but went on to a point where frequency...

Granted, but if I observe the atom's state when it was stationary and again when it was moving, both times from my own stationary frame, would there be any difference?

So has science determined if there is a concrete difference between mass and energy? If the "relativistic" mass were increased by kinetic energy on an atom rather than a particle then would that affect the internal balance of forces of the atom? Could it alter the orbital radius &c? Thanks...

Fascinating, thanks. That has cleared up some of the material I have read online. Taking the wavefunction model does the first orbital's "average" (or, rather, most probable) radius come out to being comparable to the Bohr radius? As for the wavefunction not having a frequency doesn't it...

Thanks! Yes, I realized I asked my question badly. So to clarify the total mass of the moving particle is equal to the initial mass-energy plus the kinetic energy so the kinetic energy could be thought of as being absorbed into the total mass of the moving particle. Correct?

If a particle is moving through free space (no forces acting upon it) should its kinetic energy equal its mass energy equivalence or am I getting confused. In other words is an object's kinetic energy absorbed within its mass? Is the following true? E_{kinetic} = \frac{p^{2}}{2m} = mc^{2}...

I noticed this thread https://www.physicsforums.com/showthread.php?t=294836" So my mistake is to use equations for a massless particle on a massive particle. \lambda = \frac{h}{mv} \nu = \frac{v}{\lambda} \lambda = \frac{h}{mv} = \frac{v}{\nu} \nu = \frac{mv^{2}}{h} E = mc^{2}...

So isn't the wavefunction just a probability field for where the wave can be found? In a hypothetical world couldn't we take the frequency of an electron following the most probable orbit given by the wavefunction and say that it is comparable to the above calculations? Or am I on the wrong...

Thanks for the response. I was under the impression that de Broglie hypothesised that electron might exhibit both wave and particle nature and used the electron as a wave to explain the discreet energy levels held by electrons. The electron would only orbit in integer multiples of its wave...

I think I see the problem, 13.6eV is the energy required by the electron (in addition to its rest state energy) in order to free the electron from orbit about the nucleus. Taking Bohr radius to be equal to 5.2917720859x10-11 we have E = \frac{\hbar c}{r} E_{rest} \approx 3731eV...