Eric-- I found the following item which may or may represent fact:
https://uk.answers.yahoo.com/question/index?qid=20091129063508AAEROpC A copy of this link follows: <<<<<<<"The deuteron has a binding energy of 2.2245 +/- 0.0002 MeV and has no stable excited states. It has an angular momentum of 1ħ thus a spin +1 ("triplet") and is a boson. The NMR frequency of deuterium is significantly different from common light hydrogen. The singlet deuteron is a virtual state, with a negative binding energy of ~ 60 keV. There is no such stable particle, but this virtual particle transiently exists during neutron-proton inelastic scattering, accounting for the unusually large neutron scattering cross-section of the proton. Due to the similarity in mass and nuclear properties between the proton and neutron, they are sometimes considered as two symmetric types of the same object, a nucleon. While only the proton has an electric charge, this is often negligible due of the weakness of the electromagnetic interaction relative to the strong nuclear interaction. The symmetry relating the proton and neutron is known as isospin and denoted I (or sometimes T). Isospin is an SU(2) symmetry, like ordinary spin, so is completely analogous to it. The proton and neutron form an isospin doublet, with a "down" state (↓) being a neutron, and an "up" state (↑) being a proton. A pair of nucleons can either be in an antisymmetric state of isospin called singlet, or in a symmetric state called triplet. In terms of the "down" state and "up" state, the singlet is (Dirac Notation): - (1/√2).(|↑↓> - |↓↑<) This is a nucleus with one proton and one neutron, i.e. a deuterium nucleus. The triplet is ↑↑ (1/√2).(|↑↓ + ↓↑<) ↓↓ and thus consists of three types of nuclei, which are supposed to be symmetric: a deuterium nucleus (actually a highly excited state of it), a nucleus with two protons, and a nucleus with two neutrons. The latter two nuclei are not stable or nearly stable, and therefore so is this type of deuterium (meaning that it is indeed a highly excited state of deuterium). The deuteron wavefunction must be antisymmetric if the isospin representation is used (since a proton and a neutron are not identical particles, the wavefunction need not be antisymmetric in general). Apart from their isospin, the two nucleons also have spin and spatial distributions of their wavefunction. The latter is symmetric if the deuteron is symmetric under parity (i.e. have an "even" or "positive" parity) , and antisymmetric if the deuteron is antisymmetric under parity (i.e. have an "odd" or "negative" parity). The parity is fully determined by the total orbital angular momentum of the two nucleons: if it is even then the parity is even (positive), and if it is odd then the parity is odd (negative). The deuteron, being an isospin singlet, is antisymmetric under nucleons exchange due to isospin, and therefore must be symmetric under the double exchange of their spin and location. Therefore it can be in either of the following two different states: Symmetric spin and symmetric under parity. In this case, the exchange of the two nucleons will multiply the deuterium wavefunction by (-1) from isospin exchange, (+1) from spin exchange and (+1) from parity (location exchange), for a total of (-1) as needed for antisymmetry. Antisymmetric spin and antisymmetric under parity. In this case, the exchange of the two nucleons will multiply the deuterium wavefunction by (-1) from isospin exchange, (-1) from spin exchange and (-1) from parity (location exchange), again for a total of (-1) as needed for antisymmetry. In the first case the deuteron is a spin triplet, so that its total spin s is 1. It also has an even parity and therefore even orbital angular momentum l ; The lower its orbital angular momentum, the lower its energy. Therefore the lowest possible energy state has s = 1, l = 0. In the second case the deuteron is a spin singlet, so that its total spin s is 0. It also has an odd parity and therefore odd orbital angular momentum l . Therefore the lowest possible energy state has s = 0, l = 1. Since s = 1 gives a stronger nuclear attraction, the deuterium ground state is in the s =1, l = 0 state. The same considerations lead to the possible states of an isospin triplet having s = 0, l = even or s = 1, l = odd. Thus the state of lowest energy has s = 1, l = 1, higher than that of the isospin singlet. The analysis just given is in fact only approximate, both because isospin is not an exact symmetry, and more importantly because the strong nuclear interaction between the two nucleons is related to angular momentum in spin-orbit interaction that mixes different s and l states. That is, s and l are not constant in time (they do not commute with the Hamiltonian), and over time a state such as s = 1, l = 0 may become a state of s = 1, l = 2. Parity is still constant in time so these do not mix with odd l states (such as s = 0, l = 1). Therefore the quantum state of the deuterium is a superposition (a linear combination) of the s = 1, l = 0 state and the s = 1, l = 2 state, even though the first component is much bigger. Since the total angular momentum j is also a good quantum number (it is a constant in time), both components must have the same j, and therefore j = 1. This is the total spin of the deuterium nucleus. To summarize, the deuterium nucleus is antisymmetric in terms of isospin, and has spin 1 and even (+1) parity. The relative angular momentum of its nucleons l is not well defined, and the deuteron is a superposition of mostly l = 0 with some l = 2. The measured value of the deuterium magnetic dipole moment, is 0.857 μN. This suggests that the state of the deuterium is indeed only approximately s = 1, l = 0 state, and is actually a linear combination of (mostly) this state with s = 1, l = 2 state. The electric dipole is zero as usual. The measured electric quadropole of the deuterium is 0.2859 e·fm2. While the order of magnitude is reasonable, since the deuterium radius is of order of 1 femtometer (see below) and its electric charge is e, the above model does not suffice for its computation. More specifically, the electric quadrupole does not get a contribution from the l =0 state (which is the dominant one) and does get a contribution from a term mixing the l =0 and the l =2 states, because the electric quadrupole operator does not commute with angular momentum. The latter contribution is dominant in the absence of a pure l = 0 contribution, but cannot be calculated without knowing the exact spatial form of the nucleons wavefunction inside the deuterium. Source: 1. Elements of Nuclear Physics, W. E Burcham, Longman,1979. 2. http://en.wikipedia.org/wiki/Hydrogen-2 ">>>>>>> Note the last paragraph and the conclusion about electric quadrupoles not commuting with angular momentum. It suggests that there may be a way to stimulate the D via an electric quadrupole input signal. Also with a magnetic moment the D must respond to a magnetic field and fine tuning of an oscillating magnetic field may very well excite the D to flip up and down in the field. The composite particles of the D should have sligltly different magnetic moments that can respond and create an "excited" state IMHO on a transient short lived time frame. However in a coherent system such a transient may be enough to cause other transitions of similar energy states to occur with mass energy being changed to angular momentum energy. Bob Cook ----- Original Message ----- From: Eric Walker To: [email protected] Sent: Friday, October 17, 2014 9:54 PM Subject: Re: [Vo]:Mizuno, Rossi & copper transmutation On Fri, Oct 17, 2014 at 10:41 AM, Bob Cook <[email protected]> wrote: Do you know if the experiments looked at excited spin energy states that may be possible at higher spin quanta? Unfortunately I don't have any other details and don't know of a particular experiment to refer to. Here is the quote from a textbook I recently finished reading: For nuclear physicists, the deuteron should be what the hydrogen atom is for atomic physicists. Just as the measured Balmer series of electromagnetic transitions between the excited states of hydrogen led to an understanding of the structure of hydrogen, so should the electromagnetic transitions between the excited states of the deuteron lead to an understanding of its structure. Unfortunately, there are no excited states of the deuteron—it is such a weakly bound system that the only "excited states" are unbound systems consisting of a free proton and neutron. [1] Eric [1] Kenneth S. Krane, Introductory Nuclear Physics, pp. 80-81; author's emphasis.
