The website I got the derivation from pretty much spelled it out the way I have it so I don’t think it will be much more help. The lines and parentheses are my limited graphics way of trying to reproduce the typical quantum representations of things.
The symbol |x) that I have [as in |state) or |up)] represents the state of the system or particle. The usual representation is actually |x> and is called a ket, but as you can see the graphics don’t really come out that good. Heh, actually they only look like crap in my word program. They actually look pretty good here. Therefore, I’m going to switch notations on you and use things like |state> and |up> instead. These kets are closely related to the wavefunctions of the system. In fact, for our purposes, we can just consider them to be the wavefunctions for the system. With that in mind
|state> — This is the total wavefunction for our two photon system.
|1> and |2> — These would be the wavefunctions of our two photons, photon 1 and photon 2.
However, since we have to consider each photon as traveling along all possible paths each photon actually has 2 wavefunctions, one for the up path and one for the down path. I have these represented like so
|1,up> — Wavefunction for photon 1 along the up path.
|1,down> — Wavefunction for photon 1 along the down path.
|2,up> — Wavefunction for photon 2 along the up path.
|2,down> — Wavefunction for photon 2 along the up path.
The total wavefunction |state> is made up of a linear superposition (this means we just add them together) of all possible states of the system. This gives
|state> = 1/sqrt(2) {|1,up>|2,down> + |1,down>|1,up>}
The factor 1/sqrt(2) (the inverse of the square root of 2) is just a normalization factor which ensures that the total probability for the system to be found in some state is equal to one.
Inside the brackets notice we do not have a states such as |1,up>|2,up> or |1,down>|2,down>. This represents the fact that one of the photons must travel along the upper path and one of the photons must travel along the lower path (in other words |1,up>|2,up> = |1,down>|2,down> = 0). This leaves the only non-zero possible states |1,up>|2,down> and |1,down>|2,up>.
So what is the reality of the situation? Did photon 1 travel along the upper path and photon 2 along the lower? Or did photon 2 travel along the upper path and photon 1 along the lower?
What quantum mechanics shows us is that it is not correct to consider each photon as going along a specific path. Instead we must consider each photon as going along all possible paths simultaneously. This, of course, seems absurd. However, it makes a physical difference whether the photons travel along a specific path or all possible paths and so the question of what path(s) the photons take is not a metaphysical one but an experimental one.
That said, the total wavefunction for our photons must include the possibilities for each photon to travel along both the upper and lower path. This is why both the states (wavefunctions) |1,up>|2,down> and |1,down>|2,up> are included.
The beam splitter gives us two possible paths to the detectors for photons traveling along the upper and lower paths. This means that the up and down states (wavefunctions) (ie. |up> and |down>) are actually comprised of a linear superposition of two other states (wavefunctions), one leading to detector d1 and one leading to detector d2.
The next question is then, how do we represent the states |up> and |down> in terms of the states leading to the two detectors |d1> and |d2>? This is where the question of phase comes in.
You asked how you should try to visualize these wavefuntions along the possible paths and I said to just consider them as sine functions. A better (at least easier mathematically) way to visualize them is in terms of the exponential function such as
|x> = exp[i{(kx-wt) + phase}] = e^i[(kx-wt) + phase] = exp[i(phase)] * exp[i(kx-wt)].
These are known as plane waves.
i is the imaginary number sqrt(-1)
k is the wavenumber and is related to the momentum of the photon.
x is the photon position.
w is the photon frequency which is related to its energy
t is time.
A few things about the experimental setup are important to note.
1) The photons are identical. Therefore, their wavefunctions have identical values for k and w.
2) The photons always travel the same distance to reach a given detector whether they travel along the upper or lower path. Therefore, at the detectors the wavefunctions have identical values for x and t.
This implies that at the detectors the wavefunctions for the photons are identical except for the phase difference exp[i(phase)]. The difference in phase is caused by a phase shift in the wavefunction upon reflection of the photon from the beam splitter. Dr. Rioux sets the value of this phase shift at 90 deg or pi/2. For transmission through the beam splitter there is no change in phase.
Since the only difference in the wavefunctions at the detectors is the difference in phase, there is no sense in writing out the factor exp[i(kx-wt)] everywhere. Therefore, it can just be symbolized |d1> and |d2>. Keep in mind though the actual wavefunctions at the detectors are exp[i(phase)]|d>.
Ok, for transmission through the beam splitter there is no change in phase and we have exp[0] = 1 (transmission)
For reflection we have exp[i(pi/2)] = i (reflection)
Now we are in a position to construct the wavefunctions |up> and |down> in terms of the wavefunctions leading to detectors d1 and d2.
For a photon traveling along the upper path it must be reflected to reach detector d1 and transmitted to reach detector d2. A superposition of the wavefunctions then gives
|up> = 1/sqrt(2) {i|d1> + |d2>}.
For a photon traveling along the lower path it must be reflected to reach detector d2 and transmitted to reach detector d1. A superposition of the wavefunctions then gives
|down> = 1/sqrt(2) {|d1> +i |d2>}.
Once again, the factor 1/sqrt(2) is just there to ensure that the probability that a photon arrives at either d1 or d2 is equal to 1.
Obviously we have these up and down wavefunctions for both photon 1 and photon 2. This is where I get things like
|1,up> = 1/sqrt(2) {i|1,d1> + |1,d2>} and so forth.
Now just substitute these expanded wave functions back into our total wavefunction equation
|state> = 1/sqrt(2) {|1,up>|2,down> + |1,down>|1,up>}
and solve. Take the magnitude square of the coefficient in front of each state [for example |1,d1>|2,d1>] to get the probabilities for the various photon detector combinations.
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