Schneibster said:
You know, I have a question. I had thought that all of the talk about electrons being waves, and the phase angle, and all of that, linked right up with classical wave mechanics, and that that was one of the links between QM and CM. From what I'm seeing here, though, it sounds like the phase being used in the probability calculations is a highly abstract entity that has nothing to do with the classical phase of the wave.
There are several issues here, and it all comes down to different aspects of quantum field theory. But no panic, you don't need to study that in all detail with all the renormalization and everything to come clear of it.
First, there is the "non-relativistic" quantum particle, with mass (it is trickier to treat a zero-mass particle non-relativistically, and that's where I got myself in a mess by giving that laser example ! But I promise you that I will try to put my "money" (time and posts) where my mouth is :-), such as an electron, or a neutron.
You have to choose a basis, such as the position basis or the momentum basis, to work in. If you work in the "position" basis, there is ONE possible quantum state corresponding to each position (x,y,z) ; and we denote that state by |x,y,z> (or by |x> or something, doesn't matter).
Applying the Born rule in that basis means that you assign a probability to find the particle to each position.
If you choose to work in the momentum basis, then to each momentum vector, (kx, ky, kz) there corresponds ONE quantum state |kx,ky,kz> or for short |k>.
Note that in quantum mechanics, a state |u> and exp(i a) |u> represent exactly the same state.
It turns out that the position state |x> is a superposition of momentum states, namely the integral over all k of exp(i k x) |k> and vice versa:
A momentum state |k> is a superposition of position states:
Integral over all x of exp(- i k x) |x>.
When you write a momentum state IN THE POSITION BASE, then you CALL THE WAVE FUNCTION this exp(- i k x) : they are the coefficients in the superposition of the position state |x> in the considered state, namely the momentum state |k>.
The absolute square of that coefficient gives you the probability, at that moment, if you measure the position, to find the particle in state |x> (in position x) ; this is an application of the Born rule at that point.
You also see that, because |k> and exp(i a) |k> represent the same state, that the wave function exp(- i k x) or the wave function exp( - i k x + i a) describe the same physical state.
So there is no meaning attached to "the phase of the electron wave function".
What can happen, however, such as in a double slit experiment, is that we reason "as if it were a classical wave" (because the unitary evolution equations will be very similar), and we calculate "optical path differences" with "partial waves that interfere". That's a shortcut, which is in fact meant to calculate the final wave function (on the screen) of the electron. Indeed, "interference effects" will then cause you to have a wave function of which the amplitude will not be a constant. This is, as I said, using "classical wave theory" to calculate THE QUANTUMMECHANICAL UNITARY EVOLUTION as expressed in superpositions of position states.
So what seems to be a "classical calculation" when you use a classical field to do quantum mechanics of particles, comes actually down to applying unitary quantum mechanical evolution. But it is a mathematical trick, that can only be applied when talking about the same particle.
If you apply the Born rule somewhere, you do not "switch to the classical wave" but you would switch to "the classical intensities" and your wave is dead. You only do that when you project on a screen (and it is for all practical purposes - thanks to decoherence theory).
However, working with the classical electromagnetic field is a lot trickier. In fact, the mapping between the classical EM field and the QM representation requires the full machinery of QED. The EM field is not really "the field of the photon", although you MAY use it that way if you work with a one-photon state, in the same way as we did above. But then this is just a mathematical trick, not a correspondence with a real EM field.
The reason is exactly the one you are struggling with:
a real EM field has a definite phase, while a one-photon state hasn't.
So if you want to relate to the "real phase" of a classical EM field, you need to construct, what is called "coherent states".
QED sees the quantum EM field in several states, but now the number of photons (it is in fact the DEFINITION of what is a photon...) is variable. So the different possible quantum states of the quantum EM field are:
|0> Nothing, the vacuum
|k> One - photon state with momentum k (for all vectors k)
|k1,k2> 2 - photon states ; one with momentum k1 and one with momentum k2.
|k1,k2,k3>
|k1,k2,k3,k4>
...
|n-photon state>
...
(I dropped the 2 possible polarisations for each photon).
Note again that there is no "phase" attached to each photon, or to each state. Each state is just a bucket saying that there are 7 photons, one with momentum k1, one with momentum k2... No position, no phase.
It is only if we limit ourselves to one-photon states that we can play the trick with the "wave function".
What corresponds now to a classical EM field WITH phase ?
It is the state, described by the following superposition:
|\alpha,k> = N \sum_{n} \frac{\alpha^n}{n!}|k,k,k..(n)..k>
Here, we have the coherent state which corresponds to a plane wave with wave vector k, intensity given by |alpha|^2 and phase (this time the real, classical phase of the corresponding classical EM field) by the phase of alpha.
You see that we can, as usual, change the quantum phase of the state, it doesn't change the "classical phase" which is encoded in the relative phases of the terms in the superposition.
You also see that a well-defined classical wave consists of superpositions of quantum states with different photon numbers. In fact, this state can also be shown to give rise to the Poisson statistics if we will count photons with a photon detector (applying Born's rule to this state in this basis).
I was trying to work out the effect of stimulated emission, in which I tried to show that this transforms a coherent state in another coherent state with slightly more amplitude, but my calculation screwed up and I have to find out where it did.
cheers,
patrick.
, but I got stuck trying to work it out, and I'm now reading up on it in Mandel and Wolf.
