A 4 4 diagonal matrix Schrödinger equation from relativistic total energy with a 2 2 Lorentz invariant solution.

Transcription

1 A 4 4 diagonal matrix Shrödinger equation from relativisti total energy with a 2 2 Lorentz invariant solution. Han Geurdes 1 and Koji Nagata 2 1 Geurdes datasiene, 2593 NN, 164, Den Haag, Netherlands han.geurdes@gmail.om 2 Department of Physis, Korea Advaned Institute of Siene and Tehnology, Daejeon 34141, Korea ko mi na@yahoo.o.jp Dated: Otober 13, 2017 In this paper an algebrai method is presented to derive a non-hermitian Shrödinger equation from E = V + m p e A2 with E i and p i. In the derivation no use is t made of Dira s method of four vetors and the root operator isn t squared either. In this paper use is made of the algebra of operators to derive a matrix Shrödinger equation. It is demonstrated that the obtained equation is Lorentz invariant. PACS numbers: 03.65Aa, w, LxQuantum omputer

2 2 I. INTRODUCTION In many textbooks, treatises or leture notes, e.g. [1, Chapter 2 page 40], [3, Chapter 6] or [4], one an read that Dira s road to relativisti quantum mehanis is the way to quantize the total relativisti energy. In quantum field theory the Dira spinor is the key objet of researh [2]. In some textbooks, for instane, [5, Chapter 2, setion 2.4], Kramer s work on the relativisti quantum mehanis or Weyls s equation is mentioned as a uriosity. Weyl s work e.g. refers to neutrinos. The main diffiulty with the quantization of the total relativisti energy is that a diret more or less straightforward Shrödinger equation appears to be impossible beause of the square root term E = V + m p e A 2 1 In this equation, V is the potential energy, m the mass of the quantum, p the momentum, e the unit of harge, the veloity of light in vauum and A the eletromagneti field vetor. In the following setion we present a method with operator algebra to takle the operator form in the square root term of 1. II. OPERATOR ALGEBRA Let us firstly rewrite equation 1 and write, H V = E V /. Introdue a vetor of operators, p A = p e A. We suppose A 0. This results into the equivalent. H V = m p A 2 1/2 Let us subsequently observe that beause of p i the operator p 2 A R e p 2 A and an imaginary, I m p 2 A part. Given this format it follows that 2 = i e A 2 will ontain a real, m p A 2 = m R e p A 2 + ii m p A 2 3 If we then introdue two real operator 4-vetor funtions H 1 and H 2, the operator in 3 an be equal to H 1 + ih 2 2. So we look at m p A 2 = H 1 + ih Similar to p A 2 = p A p A, we have H 1 + ih 2 2 = H 1 + ih 2 H 1 + ih 2. Combining 3 with 4 the following two equations an be obtained β = m R e p A 2 = H 2 1 H 2 2 γ = I m p A 2 = H 1 H 2 + H 2 H 1 5 The H 1 and H 2 operators are spanned by {ê µ } 4 and for κ = 1, 2, 3, 4, we have ê µ κ = δ µ,κ. This defines the four unit base vetors. The δ µ,ν, is the Kroneker delta, µ, ν = 1, 2, 3, 4. The produt of basis vetors {ê µ } 4 therefore shows, ê µ ê ν = δ µ,ν. III. DEFINITION OF THE H 1 AND H 2 OPERATOR Let us define the operators that are used in 5. We have H 1 = ê 4 σm + e ê k A k x, t H 2 = ê k 6 x k

3 We will demonstrate that for σ { 1, 1} H 1 and H 2 an be employed in 5. Beause the {ê µ } 4 are orthonormal it is found that, noting σ 2 = 1, e 2 H1 2 = m A 2, 3 H 1 H 2 + H 2 H 1 = e H2 2 = 2 2, A k + A k x k x k Then, the operators in the previous equation math the definitions of β and γ in 5. 7 A. Four-vetor root terms Subsequently it must be noted that 4 has a = on the salar level. Hene, we annot flat out take the square root on both sides of 4 and have H 1 + ih 2, a 1 4 form, on the right hand and H V, defined in 2 a 1 1 form on the left hand. However, let us define a 1 4 form E V as E V = ê µ H V,µ 8 If, H V,µ = µ H V Seondly, then the following an be observed. In the first plae we have from 2, 4 and 5 that E 2 V = E V E V = Suppose we have, 4 2 µ = 1. Then, we an have H 2 V = H 1 + ih {H V,µ } 2 = HV 2 2 µ. E V = H 1 + ih 2 10 as a solution of 9. In 8-10 E i t. Hene, H V,µ = µ i t V x, t for µ = 1, 2, 3, 4. Bearing in mind that we work in the omplex wave funtions, we may take, 1 = 1, 2 = 3 = i and 4 = 2 and note that 4 2 µ = = 1. B. Quantisation equation The result 10 an now be employed where the inner produt of left and right hand side result in a Shrödinger equation. Let us define ψ = 4 ν=1 êνψ ν x, t, and E V ê ν ψ ν = H 1 + ih 2 ê ν ψ ν 11 with ν = 1, 2, 3, 4. If, e.g. ν = 4, looking at 8 and 10, then on the left hand of 11 we will find, 4 i t V x, t ψ 4. On the right hand side, looking at 10 and the definitions in 6 we see, σ { 1, 1}, Hene, a Shrödinger equation σmψ 4 i t ψ 4x, t = σb 4 m 2 ψ 4 x, t + V x, tψ 4 x, t 12 is found. If ν = k = 1, 2, 3, then it is found from 10 and the operator definitions in 6 that Here b µ = 1/ µ, for, µ = 1, 2, 3, 4. i t ψ kx, t = b k i x k ψ k x, t + {V x, t + b k ea k x, t} ψ k x, t. 13

4 4 C. 2 2 Lorentz invariane Suppose we take, 1 = 1, 2 = 3 = i and 4 = 2 together with the ollapse ψ 2 = ψ 3 0. In the Lorentz transformation we look at ψ 4 = ψ 4 x, t and ψ 1 = ψ 1 x, t and take for brevity, x = x 1. There is no transformation along x 2 and x 3. The resulting equations then an be written as, i t ψ 4x, t = σ 2 m2 ψ 4 x, t + V x, tψ 4 x, t i t ψ 1x, t = i x ψ 1x, t + {V x, t + ea 1 x, t} ψ 1 x, t 14 The theoretial phenomenon we are looking at has, hene, only one important spatial diretion. The Lorentz transformations for an observer with onstant veloity v along the x-axis, related to the x, t system, are, x = γx vt t = γ t vx 2 with γ = 1/ 1 v/ 2. The inverse transformation is equal to x = γx + vt t = γ t + vx 2 For ease of argument, let us take in equations 12, 13 and 14, V = V 0 onstant in x, t, suppressing for onveniene, the x 2 and x 3. We start the Lorentz transformation exerise by looking at the transformation rule of ψ 4 x, t. Suppose Here, ψ 0 4 is a onstant in x, t. The onstant λ 0 in this equation is defined by ψ 4 x, t = ψ 0 4 exp [λ 0 x t] 17 λ 0 = V 0 i σm i 2 From the definitions one an derive that the Shrödinger equation for, ψ 4, 12 for onstant V = V 0 applies. Moreover, it is found for ψ 4 that Subsequently, the Lorentz transformations of t and x are i t ψ 4x, t = i x ψ 4x, t 18 x = γ x γ v 2 t t = γv x + γ t 19 This implies that equation 12, with V = V 0 onstant in x, t, is Lorentz invariant and entails the transformation ψ 4x, t = γ 1 v ψ 4 [xx, t, tx, t ] 20 Here equation 16 is observed on the rhs. Hene, if λ 0 = γ 1 v λ0, then ψ 4x, t = ψ 0 4 exp [λ 0x t ] 21 Looking at 17, ψ4 0 = ψ4γ 0 1 v. Looking at the definition of λ0 we also find V 0 = V 0 γ 1 v. The latter has to be in aordane with the Lorentz transformation of ψ 1. Let us take a loser look at this transformation. From 19 we find i γ t vγ x ψ 1 = i γ x γ v 2 t ψ 1 + {V 0 + ea 1 } ψ 1 22

5 This leads us to Lorentz transformed i t ψ 1 = i x ψ 1 + {V 0 + ea 1} ψ 1 23 with ψ 1 = ψ 1 γ 1 + v and A 1 = A 1 γ 1 v suh that, together with V 0 = V 0 γ 1 v, we have beause γ 1 v γ 1 + v = 1. {V 0 + ea 1} ψ 1 = {V 0 + ea 1 } ψ 1 5 D. 2 2 Parity & Time The 2 2 form whih transforms with the Lorentz transformations is Here, ψx, t = ψ 1 x, t, ψ 2 x, t and H = H 0 + U, with, i ψx, t = H ψx, t 24 t U = diag ea 1 + V 0, V 0 H 0 = diag i x, σ 2 m 2 Employing the operators P and T 25 and noting, P 2 = T 2 = 1, [8, Equation 7, Page 3], it is easy to aknowledge, H0 T P = PT H 0 T P = PT diag i x, σ 2 m 2 T P = H IV. CONCLUSION & DISCUSSION In the present paper a 4 4 Shrödinger equation was diretly derived from the total relativisti energy equation. The present authors already have established a 4 4, but not Lorentz invariant, equation [7]. The found equations are different from Dira s quantization of the total relativisti energy. In our derivation, no use was made of Clifford algebra. We also did not square the root term of the total energy. Instead, use was made of operator algebra in a R 4 Eulidean spae. It is noted that the total relativisti energy is also the starting point for Dira s treatment of relativisti quantum mehanis. The obtained 2 2 set of equations represents a non-hermitian system. The advaned operator algebra is opening a door to study different, possibly also Lorentz invariant, alternatives of quantization of relativisti total energy. Looking at the definition of the 2 2 diagonal H 0 in equation 25 it is easy to aknowledge that indeed H0 T P = H 0. So, if it is assumed that U T P = U, then the Hamiltonian is T P symmetri. Given the 2 2 Lorentz invariane, possibly physial states an be assoiated to the 2 2 Shrödinger equation. Note also [6]. In the ase that there is no physial state assoiated to our set of equations, one an do mathematial experiments to see if the outome of omputations with Lagrangians of real physial states hange under the addition of the found equations. We also refer to [3, Page 148]. The adding terms to the Lagrangian method resembles somewhat the speulative and beyond standard model addition of e.g. the axion equation to the Lagrangian. In the latter ase the axion was introdued to solve the strong CP problem, see e.g. [9, Chapter 10]. [1] R. Clarkson and D. MKeon, Quantum field theory, leture notes [2] R.F. Streater and A.S. Wightman, PCT, spin statistis and all that, Walter Benjamin, NY, [3] J. Ziman, Elements of advaned quantum theory, Camb Univ. Press, NY, [4] T. Das, Relativisti quatum mehanis of eletrons, Harper & Row, NY, [5] H. Pilkunh, Relativisti quantum mehanis, Springer, N.Y [6] A.K. Pati, U. Singh, U. Sinha, Measuring non-hermitian operators via weak values, arxiv: quant-ph/ [7] H. Geurdes and K Nagata, Relativisti invariane in diret derivation of a 4 4 diagonal matrix Shrödinger equation from relativisti total energy, submitted High Eneergy Density Physis, [8] C.M. Bender, Making sense of non-hermitian Hamiltonians, arxiv: hep-th/ v1. [9] E.W. Kolb and M.S. Turner, The Early Universe, Addison-Wesley, Redwood ity, 1989.