The Nature of the solutions to the Dirac Equation ans spin

i ħ ∂∂ t = c ħi∑ n⋅nThe Nature of the Dirac Equationby Kevin GibsonMay 18, 2011IntroductionThe Dirac Equation 1 is a staple of relativistic quantum theory and is widely applied toobjects such as electrons and protons.∂∂ x n m 0 c2 ⋅ (1)The α n and β are 4×4 matrices, the Ψ is a 4-component column matrix and m 0 is the restmass. The motivation behind this piece is to explore this equation, what does it say andmean.Reworking the equationIt can be shown that the solutions to (1) must also satisfy the Klein-Gordon equation 2 3 ,which is a scalar equation.∇ 2 − 1 c 2⋅∂2 ∂t =− m 20 c2 ħ (2)In this light the basis set to the solution of (1) can be expressed as the product of a scalarsolution to the Klein-Gordon equation, call it Ψ k , along with a column matrix consistingof constants, call it K.= K k (3)Putting (3) into (1) and utilizing the assumption that K is constant.i ħ ∂ K k= c ħ ∑ ∂ t i n⋅ ∂K kn ∂ x n m 0 c2 ⋅ K k n⋅K ∂ k∂ x n m 0 c2 ⋅K k (4)K⋅i ħ ∂ k= cħ ∑∂ t i nWe can reduce this down to a scalar equation by performing a matrix multiplication withthe transpose of the complex conjugate of K. This is to allow for potentially complexelements.K T ⋅K⋅i ħ ∂ k= c ħ ∑∂t i nK T ⋅ n⋅K ∂ k∂ x n m 0 c2 K T ⋅⋅K ki ħ ∂ k∂ t = ∑n c∣K∣ 2with ∣K∣ 2 ≡ K T ⋅KK T ⋅ n⋅K ⋅ ħ i ⋅∂ k∂ x n m 0 1c2∣K∣ 2⋅ K T⋅⋅K k (5a)(5b)1.

The portions of (5a) contained in the parentheses are in fact scalars. With theappropriate substitutions this can be re-written as:i ħ ∂ k∂ t = V⋅ ħi ⋅ ∇ k m 0 c2 k (6a)V n= c∣K∣ 2⋅ K T ⋅ n⋅K (6b)= 1∣K∣ 2⋅ K T ⋅⋅K (6c)At this point the matrix equation (1) has been transformed into scalar equation (6a). Asstated above Ψ k must be a solution to the Klein-Gordon equation, therefore equation(6a) can be viewed as a boundary condition to be satisfied on (2). But what does Kmean? Thus far there are two things that can be deduced. First, it is needed to interfacethe scalar Ψ k into (1). Second, it influences the boundary condition (6a). Yet there isstill more to be learned.Connecting to relativityBefore exploring K some non-quantum mechanical physics needs to be explored. Themass-shell relationship is 4E 2 =c 2 p 2 m 0 c 2 2 (7)This can be linearized using the standard equations for relativistic energy andmomentum 5 .m 0 c 2 E=c 2 m 0 u ⋅pm 0 2 c 4E=u⋅pm 0c 2 ⋅ 1 For energy, momentum eigenstates states (6a) and (8) are compatible with twoassumptionsV u 1 (8)(9)2.

The nature of the solutionSo what is the significance of the elements of K? To determine this substitute thestandard expressions for α n and β into (6a) and (6b). This givesV 1= 2c∣K∣ 2⋅ R K 1K 4R K 2K 3 (10a)V 2= 2c∣K∣ 2⋅ R K 1K 3−R K 2K 4 (10b)V 3= 2c∣K∣ 2⋅ I K 1K 4−I K 2K 3 (10c)= ∣K 1∣ 2 ∣K 2 ∣ 2 −∣K 3 ∣ 2 −∣K 4 ∣ 2= 1∣K∣ 2(10d)The conclusions are interesting.● Because of (10d), states where K 1 and K 2 are the only nonzero elements representpositive energy while states where K 1 and K 2 are the only nonzero elementsrepresents negative energy.●●An examination of equations (10a) to (10c) shows that states with only one nonzeroelement of K will represent a state at rest.In order for there to be motion one needs to have either K 1 or K 2 as a non-zeroelements and either K 3 or K 4 as a non-zero elements.Because motion in a particular direction, ala (10), does not depend on a single elementof K, it cannot represent a vector, at least not in space-time.3.

The state of the UnionAn interesting consequence arises from the third bullet point above. Consider threescenarios below.K matrix Γ Energies1K =K 1 E = m 0 c 200=00K04-1 E = -m 0 c 20K1K =K 040K-1≤Γ≤1 |E| ≥ m 0 c 2Table 1.A state with both K 1 and K 4 elements will be in motion and so will have a positiveenergy greater than its rest energy or a negative energy less than its negative rest energy.A conceptual conflict arises if we envision each element of K as a separate state. Thereason for the conflict is that if K 1 and K 4 represent states then the third state in theabove table should constitute a mixture of positive and energy states and so having anenergy∣E∣≤m 0 c 2 (11)should be possible. However it is not. It is therefore logically inconsistent to assign astate to each element. All that one can say is that a particular configuration of K leads toa particular state.SpinIn the literature the first and third elements of Ψ are assigned to the spin up states whilethe balance belong to spin down. However, as can be seen from the above arguments,this is a poor interpretation. The most that can be said is that a state with the onlynonzero elements being K 1 and K 3 can be assigned the label of “spin-up.” Likewisestates with only nonzero elements being K 2 and K 4 can be called “spin-down.” Howeversimply classifying other states as a mixture of spin-up and spin down is problematic.The difficulty lies in not being able to separate any Ψ into spin-up and spin-down states.4.

The reason is that while K can be so divided the associated Ψ k from (3) cannot, for whenone divides K the associated Ψ k would have to changeSo what can be said about spin according to Dirac? According to the paradigm, theDirac equation shows spin as an intrinsic property 6 . However the derivation itself doesnot include any information about any of the intrinsic properties of the particle inquestion. Moreover this solution is said to be the solution for matter with two spinstates, called spin ½ particles, while the Klein-Gordon equation is only valid forparticles without spin. Yet as was seen above the elements of Ψ cannot be assigned tospecific states. Thus the question of what (1) says about spin is still left open and vaque.Connecting the Dirac and Klein-Gordon equations via (8) shows the underlying physicsbehind the derivations. Indeed both are extensions of (7). So whereas the Klein-Gordonis a second order scalar equation the Dirac is a first order matrix equation. Neitherequation contains neither extra information nor a deficit of information and so the twotell the same story, thus demonstrating a weakness in the orthodox interpretation of theelements of Ψ.Works Cited1. http://en.wikipedia.org/wiki/Dirac_equation2. http://www.iitg.ernet.in/physics/fac/gsetlur/DiracEqn.pdf3. http://www.physics.ucdavis.edu/~cheng/teaching/230A-s07/rqm2_rev.pdf4. Paul A. Tipler and Ralph A. Llewellyn Modern Physics 4 th ed. p. 915. Kevin Gibson A new statistical view for elementary matter (unpublished)http://www.mesacc.edu/~kevinlg/i256/A_new_statistic.pdf6. G.E. Uhlenbeck and S. Goudsmit, Naturwissenschaften 47 (1925) 953Contact informationkevin.gibson@asu.edukevinlg@mesacc.eduwww.mesacc.edu/~kevinlg/i2565.