User:Delayedchoice/sandbox

change eqn 4 to eqn 3???

Professor Feynman (Lect. on Phys., v. 3, 1965, ps. 3-7 to 3-9) maintained in his neutron scattering off a crystal experiment that which-way information can exist even if one does not perform a measurement. This interaction involves the possibility of a spin flip for both the neutron and nucleus that the neutron scatters off. With the flip, the spin of the nucleus that the neutron scattered off becomes different than the spin direction of the other nuclei in the crystal that the neutron could have scattered off but did not. The spins of the other nuclei are the same. It may be possible to eliminate the which-way information as long as particle detections have not been made, either through spin-lattice relaxation after the neutron-nucleus interaction occurs or through shutting off the uniform, strong, external magnetic field B that initially aligns all of the spins of the crystal nuclei along the same axis. A delayed choice allows for obtaining either a probability distribution showing interference or instead one showing which-way information.

Which-Way Information vs No Which-Way Information (Interference)
Professor Feynman summarized the difference between which-way information and the absence of which-way information (interference)in a distribution of particles in what he called "First principles of quantum mechanics". Vol. 3, Sect.1-7,The Feynman Lectures on Physics - Vol. 3 - Quantum Mechanics. These principles are: 1) The probability of an event in an ideal experiment is given by the square of the absolute value of a complex number ϕ which is called the probability amplitude. 2) When an event can occur in several alternative ways, the probability amplitude for the event is the sum of the probability amplitudes for each way considered separately. There is interference. 3) If an experiment is performed which is capable of determining whether one or another alternative is actually taken, the probability of the event is the sum of the probabilities for each alternative. The interference is lost.

Given the same experimental setup, the probabilities of the same events in a number of experimental runs is determined by 1). If these probabilities are determined subject to 2) there is interference in the distribution. If these probabilities are determined subject to 3) there is no interference and instead there is which-way information in the distribution.

Experiments
Professor Feynman in vol. 3 of his lectures (quantum mechanics) discussed the following experiments, the overall structure of which follows: Neutrons are sent one at a time against a crystal. The concern is with the nuclei of the atoms comprising the crystal that are lined up basically in a periodic array. The neutrons scatter off the nuclei in the crystal into a detector (Fig. 1). The nuclei in the crystal either do not have spin or instead spin 1/2. The neutrons have spin 1/2. Professor Feynman wrote: “We have here a large number of apparently indistinguishable routes. They are indistinguishable because a low-energy neutron is scattered from a nucleus without knocking the atom out of its place in the crystal—no ‘record’ is left of the scattering [except if there is spin flip in the nucleus the neutron scattered off].”

Experiment 1 - All of the Nuclei Don't Have Spin
If all of the nuclei of the crystal do not have spin, the distribution of the neutrons at the detector exhibit interference since it is not known which atomic nucleus the neutron scattered off. The distribution of the neutrons is given by Eqn. 1:


 * $$P = |\psi_|^2 = |[\frac{1}\sqrt[2]{a}[\psi_\psi_ + \psi_\psi_ + ... +\psi_\psi_]]|^2$$[1]

The distribution of the neutrons is depicted in Fig. 2. Table 1 indicates the initial states of the neutrons and nuclei in Experiment 1, and Table 2 indicates the final states of the neutrons and nuclei in Experiment 1.

Experiment 2 - Nuclei Have Spin and Their Spin Direction is Same as the Neutron
If the nuclei have spin one half and if the spin of both the neutron and the nuclei are in the same direction, interference is exhibited in the distribution of the neutrons since there is no change in spin of the neutron or the nucleus that it impacts. The nucleus that the neutron impacted cannot be distinguished from the other nuclei in the crystal that the nuclei could have impacted, but did not, impact. The distribution of the neutrons is given by:


 * $$P = |\psi_|^2 = |[\frac{1}\sqrt[2]{a}[\psi_\psi_ + \psi_\psi_ + ... +\psi_\psi_]]|^2$$[1]

The distribution of the neutrons is depicted in Fig. 2. Table 3 indicates the initial states of the neutrons and nuclei in Experiment 2, and Table 4 indicates the final states of the neutrons and nuclei in Experiment 2.

Experiment 3 - Nuclei Have Spin and Their Spin Direction in Opposite Direction as the Neutron
If the spin of the neutron and the spins of the nuclei are initially in opposite directions, one up and the other down, two outcomes are possible. The distribution of neutrons in this Experiment, as in all of the Experiments, is developed over a number of runs. In Experiment 3 we do not know whether or not we will have a spin flip in the neutron and the nucleus it impacts.

Possibility 1 - Spin Flip in Interaction between Neutron and Specific Nucleus
If there is spin flip when the neutron interacts with one of the nuclei, we obtain a which-way pattern since it can be known which nucleus the neutron scattered off (possibility 1). In possibility 1 there is which way information since we have spin flip. The neutron can scatter off any one of the nuclei in the crystal. The probability distribution of the neutrons is depicted in Fig. 3. Professor Feynman wrote: “Well, if we can tell which atom did the scattering, what have the other atoms got to do with it? Nothing, of course. The scattering is exactly the same as that from a single atom.” The probability distribution of the neutrons is given by:


 * $$P_ = [1/a[|(\psi_| crystal_\rangle)|^2 + |(\psi_| crystal_ \rangle)|^2 +... + |(\psi_| crystal_\rangle)|^2]$$ [2]

where
 * $$\langle crystal_ | crystal_\rangle = 1$$

and
 * $$\langle crystal_ | crystal_\rangle = 0$$

The probability distribution of the neutrons is depicted in Fig. 3. Table 5 indicates the initial states of the neutrons and nuclei in Experiment 3 – possibility 1 or possibility 2, and Table 6 indicates the final states of the neutrons and nuclei in a particular run in this scenario for possibility 1. The probability distribution of neutrons in this Experiment, as in all of the Experiments, is developed over a number of runs. Now in Experiment 3 we do not know whether or not we will have a spin flip in the neutron and the nucleus it impacts. Figure 3 and the Table 6 are concerned with those runs in Experiment 3 where we have a spin flip (possibility 1).

In possibility 1, Professor Feynman concludes we don’t have to observe “which atom [which atomic nucleus] is up" (that is flipped) to get a which-way distribution from these neutrons (where there is spin flip) over a series of runs. The information is in the nuclei and that is enough to provide which-way information to the neutron. This which-way information is found in the distribution of the neutrons over a series of runs where they are detected in the case where there is spin flip in the interaction between a neutron and the nucleus it impacts. Professor Feynman wrote: “You may argue, ‘I don’t care which atom is up.’ Perhaps you don’t, but nature knows; and the probability is, in fact, what we gave above—there isn’t any interference.”  Section 3–3 Scattering from a crystal.

The implication appears to be that once this interaction that produces the which-way information occurs, this which-way information appears irrevocable since it does not matter whether the spin of the nucleus that the neutron scattered off is measured. Contrary to this implication, it may be possible to eliminate the which-way information as long as particle detections have not been made. Eliminating the which-way information before any particle detections are made is found in Experiment 4 below.

Possibility 2 - No Spin Flip in Interaction between Neutron and Specific Nucleus
If there is no spin flip when the neutron interacts with one of the nuclei, we obtain the interference pattern since we still cannot tell which nucleus the neutron scattered off (possibility 2). Since we do not have spin flip, the interference pattern is still exhibited since we cannot tell which nucleus the neutron scattered off. The relevant eqn. is:


 * $$P = |\psi_|^2 = |[\frac{1}\sqrt[2]{a}[\psi_\psi_ + \psi_\psi_ + ... +\psi_\psi_]]|^2$$[1]

The probability distribution of the neutrons is depicted in Fig. 2. Table 5 indicates the initial states of the neutrons and nuclei in Experiment 3 – possibility 1 or possibility 2, and Table 7 indicates the final states of the neutrons and nuclei in a particular run in this scenario for possibility 2. Figure 2 depicts data for a series of runs in Experiment 3 where we do not have a spin flip (possibility 2).



The Difference Between Possibilities 1 and 2
Professor Feynman wrote that the difference between possibilities 1 and 2 is the following. In possibility, 1 the phases disappear since we first take the absolute square of each possible neutron-nucleus interaction and then sum the probabilities for each possible interaction with the result that the phases disappear when the absolute squares are calculated. In possibility 2, we first sum the amplitudes for each possible interaction, which now means summing the phases for each amplitude, and then take the absolute square of the sum yielding interference since the phase for each amplitude is preserved.

Spin Flip and No Spin Flip in Interaction between Neutrons and Specific Nuclei
In Experiment 3, since we don’t know whether or not there is going to be spin flip (since the neutron and the atomic nuclei have opposite spins to begin with), the resulting overall probability distribution from combining the results for possibilities 1 and 2 over a number of runs with neutrons shows a which-way pattern punctuated with interference (Fig. 4).
 * $$P_ = P_ + P_$$   [3]

the sum of Eqns. 1 and 2:


 * $$P_ = [1/a[|(\psi_| crystal_\rangle)|^2 + |(\psi_| crystal_ \rangle)|^2 +... + |(\psi_| crystal_\rangle)|^2] +$$


 * $$     |[\frac{1}\sqrt[2]{a}[\psi_\psi_ + \psi_\psi_ + ... +\psi_\psi_]]|^2$$    [4]

Experiment 4
Through the use of a delayed choice, it may be possible to eliminate the which-way information as long as particle detections have not been made. A delayed choice occurs where the choice of whether an experiment will yield which-way information or instead no which-way information (interference) in a series of experimental runs is made after a particle is already in route to a detection device (i.e., after the experiment has begun). Perhaps through spin-lattice relaxation or a similar process after the neutron-nucleus interaction occurs, a spin flip, if it occurred in a nucleus as a result of its interaction with the incoming neutron, would reverse before any detection is made. It would no longer be possible to determine which nucleus in the crystal the neutron scattered off (possibility 1). The result is interference in the distribution of the neutrons where there would have been which-way information exhibited if the spin flip was not reversed before any detections are made. The distribution of the neutrons would be affected by a change in information regarding the nuclei in the crystal since there is no physical process whereby the change in the nuclei as a result of the delayed choice can affect the distribution of the neutrons. The phonon created in the relaxation of the spin of the impacted nucleus would be for all practical purposes not possible to detect with the thermal activity in the crystal. (G. Jeschke, Relaxation and slow motion in EPR spectroscopy)

If spin-lattice relaxation occurs before detections are made, even if we have a spin flip in the interaction between a neutron and a nucleus, all the nuclei in the crystal that can interact with the neutron will be found with their spin in the same direction (possibility 1). Which-way information held by the nucleus off of which the neutron is initially to have thought to have scattered is lost, since it can no longer be distinguished from the other nuclei in the crystal. The motion of the neutron after impacting the crystal itself is certainly capable of producing interference since when no spin flip occurs, interference in the distribution of the neutrons in a series of experimental runs occurs. The equation for the situation after spin lattice relaxation and all the nuclei again have their spin in the same original direction is Eqn 1. All the nuclei that the neutron could have scattered off now have the same orientation before any detections are made. The crystal possesses no which-way information before detections are made and the distribution of the neutrons in possibility 1 should show interference.

Possibility 2 is making a neutron detection before spin-lattice relaxation occurs for the nucleus that scattered the neutron so that there is which-way information when the neutron is detected and the distribution of these neutrons where there is spin flip in interaction with specific nuclei in the crystal is the wide round hump characteristic of which way information. Possibility 2 could occur through lengthening the spin relaxation time for the impacted nucleus over the spin relaxation time possibility 1.

There exists a delayed choice for possibilities 1 or 2 whether or not to make a detection of the neutron before spin-lattice relaxation occurs. The two possible distributions of neutrons over a series of runs either exhibit only interference (Figure 2) OR instead a combination of interference and which-way information (Figure 4). 

Possibility 1
The equation is:


 * $$P_ = |\psi_|^2 = |[\frac{1}\sqrt[2]{a}[\psi_\psi_ + \psi_\psi_ + ... +\psi_\psi_]]|^2$$[1]

The relevant figure for the neutron distribution is Figure 2.

Possibility 2
The relevant equation is Eqn. 4 for Experiment 3,


 * $$P_ = P_ + P_$$   [3]

where
 * $$P_ = [1/a[|(\psi_| crystal_\rangle)|^2 + |(\psi_| crystal_ \rangle)|^2 +... + |(\psi_| crystal_\rangle)|^2]$$    [2]

and


 * $$P_ = |[\frac{1}\sqrt[2]{a}[\psi_\psi_ + \psi_\psi_ + ... +\psi_\psi_]]|^2$$    [1]

The relevant figure for the neutron distribution is Figure 4.

Experiment 5
Another experiment would be to shut off the uniform, strong external magnetic field that initially lines up all of the spins of the nuclei in the same direction after the spin flip and before the neutron is detected. An ultrafast switch would be needed in this circumstance that would shut off the external magnetic field after the neutron was scattered off the crystal but before the neutron was detected (and leaving enough time for the spins of the nuclei to randomize before the neutron is detected). Which-way information would be eliminated since the spin directions of all the nuclei would quickly become essentially random. This is very similar to possibility 1 in Experiment 4where which-way information is lost. Possibility 1 is turning off the external magnetic field before a neutron detection is made and with enough time for the nuclei spins to randomize. In this case we obtain interference only as in Equation 1 and Figure 2. Figure 5 shows the experimental design. Possibility 2 is not altering the external magnetic field. In this case we obtain the mix of the which-way and interference distributions discussed in Experiment 4 in possibility 2: Equation 4 and Figure 4. Figure 6 shows the experimental design.

Neutralizing Any Effect of the Neutron Spin after It Interacts with the Crystal on the Neutron Intensity Distribution
It is useful to take into account of the possibility that the spin flip in the neutron, when it occurs, could be a factor. In the intensity distributions of the neutrons. To eliminate this possibility one could “mix up” the spin orientations along the axis of the B field in the following way: 1) extend the B field past the crystal almost to the neutron detector. Apply pulsed oscillating magnetic radiation (B’) to the neutrons  in the area between the crystal and the neutron detector at a 90O angle to the axis of the external homogenous magnetic field B (where the intensity of the oscillating is much less than that of the external homogenous magnetic field B.  This frequency of the oscillating magnetic field would be at the Larmor precession frequency of the neutrons. It can flip neutrons from a low energy state to a high energy state or from a high energy state to a low energy state. If one used an appropriate intensity for B’ where only a certain percentage of the neutron spins were flipped by the oscillating magnetic field (either higher to lower or lower to higher in the same proportion) one could eliminate the neutron spin state itself as a factor in determining whether there is or is not which-way information in the neutron intensity distribution as determined at the neutron detector.


 * $$P_ = [1/a[|(\psi_| crystal_\rangle)|^2 + |(\psi_| crystal_ \rangle)|^2 +... + |(\psi_| crystal_\rangle)|^2]$$ [2]


 * $$P = |\psi_|^2 = |[\frac{1}\sqrt[2]{a}[\psi_\psi_ + \psi_\psi_ + ... +\psi_\psi_]]|^2$$[1]


 * $$P_ = [1/a[|(\psi_| crystal 1\rangle)|^2 + |(\psi_| crystal 2\rangle)|^2 +... + |(\psi_| crystal a\rangle)|^2]$$


 * $$P_ = [1/a[|(\psi_| crystal_\rangle)|^2 + |(\psi_| crystal_ \rangle)|^2 +... + |(\psi_| crystal_\rangle)|^2]$$


 * $$ \langle A | B \rangle = \text{the inner product of ket } | A \rangle \text{ with ket } | B \rangle$$





Σ

$$P = |\psi_{\rm{first}} + \psi_{\rm{second}}|^2 = |\psi_{\rm{first}}|^2 + |\psi_{\rm{second}}|^2 + 2 |\psi_{\rm{first}}| |\psi_{\rm{second}}| \cos (\varphi_1-\varphi_2)$$

from Schrodinger page

For $N$ particles in one dimension, the Hamiltonian is:


 * $$ \hat{H} = \sum_{n=1}^{N}\frac{\hat{p}_n^2}{2m_n} + V(x_1,x_2,\cdots x_N) \,,\quad \hat{p}_n = -i\hbar \frac{\partial}{\partial x_n}$$

where the position of particle $n$ is $x_{n}$.

$$P = |\psi_{\rm{first}} + \psi_{\rm{second}}|^2$$

$$P = |\psi_ + \psi_|^2$$

$$P = |\psi_|^2 = |\psi_ + \psi_|^2$$

\sum_{n=1}^{N} \frac{\hat{p}_n^2}{2m_n} \sqrt[root]{arg}

$$P = |\psi_|^2 = |\psi_ + \psi_|^2$$

$$P = |\psi_|^2 = \frac{\hat{p}_n^2}{2m_n}|\psi_ + \psi_|^2$$

$$P = |\psi_|^2 = \frac{1}{a}|\psi_ + \psi_|^2$$

$$P = |\psi_|^2 = |[\frac{1}\sqrt[2]{a}\psi_ + \psi_]|^2$$

$$P = |\psi_|^2 = |[\frac{1}\sqrt[2]{a}\psi_\psi_ + \psi_\psi_ + ... +\psi_\psi_]|^2$$

dddddd dddd

"fdfgsdfgsgfdgdsg"

Feynman, but in volume 3 I think at the end of the first chapter as I recall that when summarizing when to add amplitudes and then take the absolute square (interference or when you don't know which way something happened) or instead take the absolute square of the amplitudes first and then add them (when you know which way something happened or could determine which way something happened)



FROM NEUTRON SPIN MAGNETIC MOMENT Thumbnail



FROM LARMORPRECESSION



FROM NMR





from neutron spin magnetic moment page

For an elementary particle to have an intrinsic magnetic moment, it must have both spin and electric charge. The neutron has spin 1/2 ħ, but it has no net charge. The existence of the neutron's magnetic moment was puzzling and defied a correct explanation until the quark model for particles was developed in the 1960s. The neutron is composed of three quarks, and the magnetic moments of these elementary particles combine to give the neutron its magnetic moment.

In the quark model for hadrons, the neutron is composed of one up quark (charge +2/3 e) and two down quarks (charge −1/3 e).[43] The magnetic moment of the neutron can be modeled as a sum of the magnetic moments of the constituent quarks,[44] although this simple model belies the complexities of the Standard Model of particle physics.[45] The calculation assumes that the quarks behave like pointlike Dirac particles, each having their own magnetic moment, as computed using an expression similar to the one above for the nuclear magneton:

\mu_\mathrm{q} = {{e_\mathrm{q} \hbar} \over {2 m_\mathrm{q}}},

where the q-subscripted variables refer to quark magnetic moment, charge, or mass. Simplistically, the magnetic moment of the neutron can be viewed as resulting from the vector sum of the three quark magnetic moments, plus the orbital magnetic moments caused by the movement of the three charged quarks within the neutron.

In one of the early successes of the Standard Model (SU(6) theory), in 1964 Mirza A. B. Beg, Benjamin W. Lee, and Abraham Pais theoretically calculated the ratio of proton to neutron magnetic moments to be −3/2, which agrees with the experimental value to within 3%.[46][47][48] The measured value for this ratio is −1.45989806(34).[49] A contradiction of the quantum mechanical basis of this calculation with the Pauli exclusion principle, led to the discovery of the color charge for quarks by Oscar W. Greenberg in 1964.[46]

From the nonrelativistic, quantum mechanical wavefunction for baryons composed of three quarks, a straightforward calculation gives fairly accurate estimates for the magnetic moments of neutrons, protons, and other baryons.[44] For a neutron, the end result of this calculation is that the magnetic moment of the neutron is given by μn = 4/3 μd − 1/3 μu, where μd and μu are the magnetic moments for the down and up quarks, respectively. This result combines the intrinsic magnetic moments of the quarks with their orbital magnetic moments, and assumes the three quarks are in a particular, dominant quantum state.

from nmr page

from magnetic field page



me

As added insurance that which-way information is not conclusively not provided by the neutron that leaves with spin opposite in direction to that which it had before it interacted with the crystal, in the Feynman experiment when there is spin flip between the incoming neutron (with spin up initially) and the specific nucleus (with spin down initially) in the crystal that it impacts, after the neutron leaves the region of the constant magnetic field over the crystal on the neutron's path to the neutron detector, the neutron could be placed in an environment where it very quickly loses its "spin down" along the "axis of quantization" (the direction of the constant magnetic field surrounding the crystal). Such an environment would be, for example, where there is no external constant magnetic field in the neutron's path from the crystal to the neutron detector. The neutron's spin would then not be defined up or down along any axis unless a subsequent measurement of spin for the neutron is made. The same would hold for spin up neutrons for which there is no spin flip when they interact with nuclei in the crystal: these neutrons' spins would then not be defined up or down along any axis unless a subsequent measurement of spin for the neutron is made. Then neutrons which experienced spin flip and those that did not in different runs of the experiment could not be distinguished from one another.


 * Distribution_with_ww_and_no_ww_information_events_edited.jpg
 * One round hill characteristic of which-way information in quantum mechanics.jpg