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Jan
14
comment Physical meaning of linear combination of possible states in infinite well
And don't forget that all the states you mix together have a time component...they are multiplied by exp(jwt), where w depends on the energy level. So the pattern is changing with time as the relative phases move with respect to one another.
Jan
14
comment Physical meaning of linear combination of possible states in infinite well
The amplitudes don't remain fixed in time. You need to think about the simple example I gave where you mix just the the ground function and the first excited state. Since the electron is bouncing back and forth, it is radiating. So it loses energy. The amplitude "drains" from the excited state to the ground state until eventually it stops moving. It's the exact same thing that happens to a hydrogen atom in a superposition of the ground (1s) state and first excited (2p) state.
Jan
13
comment Physical meaning of linear combination of possible states in infinite well
You're assuming you add them all in phase at the middle of the box and with equal amplitudes. Yes, you get a wave packet at that instant but I don't think it holds together...I think it explosed all over the box. I'm not sure it's all that easy to get coherent packets that stick together, like the ones you can create in the harmonic oscillator potential.
Jan
13
answered Physical meaning of linear combination of possible states in infinite well
Jan
11
comment Some applications of the Einstein-Podolsky-Rosen (EPR) paradox?
In general, it's a good idea to be suspicious if you find yourself going to extreme lengths to justify how something might be measurable.
Jan
11
comment Some applications of the Einstein-Podolsky-Rosen (EPR) paradox?
@hypnosifl I'm not sure that the very problematical measurement system you propose actually meets the condition of being executable even "in principle". But if it is, you're saying that if you had a properly isolated block of uranium emitting a continuous flux of alpha particles...that you wouldn't be able to set up a double-slit system and measure a diffraction pattern? I don't think that's right. I think you would get a diffraction pattern.
Jan
11
comment Some applications of the Einstein-Podolsky-Rosen (EPR) paradox?
If you have only one uranium atom, then when it decays, the momentum of the thorium must be equal and opposite to the momentum of the alpha particle. If there's a block or uranium, no such restriction. The momentum is divided three ways, so measuring the alph particle tells you nothing about the momentum of the thorium atom.
Jan
11
comment Some applications of the Einstein-Podolsky-Rosen (EPR) paradox?
@hypnosifl your answer is very helpful and I am still trying to work out the consequences; but I see now that the example I gave fails to create faste-than-light communication for a more prosaic reason. You DO get an interference pattern because when I substitued the block of uranium for the single uranium atom, I lost the perfect entanglement of the momenta of the thorium and alpha particles. The momenta were only entangled because there was nowhere else to go when the SINGLE uranium atom decayed. There is no entanglement with a block of uranium.
Jan
10
revised Some applications of the Einstein-Podolsky-Rosen (EPR) paradox?
added 2246 characters in body
Jan
10
comment Some applications of the Einstein-Podolsky-Rosen (EPR) paradox?
I can't comment on your question as to how the thorium atom is linked to the alpha particle. In standard QM, we say they do not each have their own independent wave function, but instead there is a single six-dimensional wave function which contains both of them. (That's the wave function that "collapses" when either one is detected.)
Jan
9
answered Some applications of the Einstein-Podolsky-Rosen (EPR) paradox?
Jan
6
answered Plants and Quantum Mechanics!
Dec
29
comment Why are electromagnetic waves not able to pass through a hole with a diameter smaller than the wavelength?
Oh come on. If you merge the two holes into one, the diameter becomes squrt(2). And the power is 4 times: that is, diameter to the fourth. Like I said in the first place. For a correct explanation of the exponent, re-read what I wrote in my answer.
Dec
27
answered Why are electromagnetic waves not able to pass through a hole with a diameter smaller than the wavelength?
Dec
23
answered On the foundations of quantum physics
Dec
20
answered Why can't unequal current sources be connected in series?
Dec
17
answered Derivation of the speed of light using the integral forms of Maxwell's Equations
Dec
15
comment How does a Wavefunction collapse?
Let's remember your starting position. Here is what you said: "You can measure the energy eigenstate of a hydrogen atom by measuring the magnetic moment of the atom, which will depend on the angular momentum of the electronic state, the energy of the electron, and the alignment of the electron's spin."
Dec
10
comment How does a Wavefunction collapse?
Oh come on. They both have zero orbital energy. And dont' try to squirm out of it by invoking the fine structure constant.
Dec
10
comment How does a Wavefunction collapse?
@jerrySchirmer no you can't measure the energy of a hydrogen atom by measuring its spin state. The 1s and 2s states have the same spin, so a hydrogen atom with a spin of 1/2 can be in any superposition of the 1s and 2s states.