1 (Contents) (4) Why Has the Superstring Theory Collapsed? Noboru NAKANISHI 2 2. A Periodic Potential Problem

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2 1 (Contents) (4) Why Has the Superstring Theory Collapsed? Noboru NAKANISHI 2 2. A Periodic Potential Problem in Quantum Mechanics (4) Kenji SETO Anti-commutativity among Linearly Independent Imaginary Units Katsusada MORITA The Parity Operator Minoru YONEZAWA The Parity Operator A Comment Noboru NAKANISHI Editorial Comments Shozo NIIZEKI, Tadashi YANO 29

3 Why Has the Superstring Theory Collapsed? * 1 Noboru NAKANISHI * *3 30 *4 72, 345 (1986), ( , 48, 44 (1993). superalgebra graded algebra indecomposable extension SUSY 1 x µ ( ) 1 ( ) *5 SUSY *1 *2 nbr-nak@trio.plala.or.jp *3 *4 3 woit/wordpress P. Woit Some Early Criticism of String Theory, October 30, 2006 *5 3 2

4 SUSY 2 SUSY 2 2 SU(3) SU(2) U(1) *6 S SUSY SUSY SUSY S 2 SUSY SUSY SUSY *7 SUSY SUSY SUSY 1/2 SUSY *6 *7 SUSY SUSY SUSY 3

5 S (NG) SUSY 1/2 NG SUSY NG NG SUSY SUSY (SUGRA) NG SUSY S SUSY SUSY SUSY 1 SUSY 2 1 SUSY CERN LHC SUSY is dead SUSY SUSY 4

6 1/2 *8 BRS *9 x µ x µ NG 2 SU(2) U(1) SU(2) U(1) U(1) U(1) 20 SUSY SUSY SUSY ( *8 *9 FP BRS 5

7 x µ SUSY x µ SUSY SUSY SUSY 3 S 1 2 S S s t 2 2 s s S * 10 l * 11 l s l = α(s) α(s) * 12 α(s) = α 0 + α s *10 *11 *12 6

8 α s 1 t t t α(s) (s 0) log t 1 α 0 1 α 0 = 1 1 s < 0 s * 13 N 2 2 N 4 N = 4 B( α(s), α(t)) = Γ( α(s))γ( α(t)) Γ( α(s) α(t)) = 1 0 dx x α(s) 1 (1 x) α(t) 1 * 14 N N 3 * 15 3 SU(3) *13 *14 N = *15 N 3 7

9 2 2 D 2 * 16 D 2 * 17 2 * 18 D = D * 19 4 SUSY 2 α 0 = 2 2 c h 1/2π ħ 1 20 *16 *17 S *18 *19 D = 26 8

10 α 2 s [l = ] 1 α * /2 * n + 2 (n = 0, 1, 2, ) n = SO(32) E 8 E 8 SU(3) SU(2) U(1) SU(5) SO(10) * 22 * *21 2 *22 9

11 p + 1 Dp 11 M * 23 4 * (2010) 10

12 * * 25 * 26 6 a s (a = (1/x) α > 1) s = p µ p µ = p 2 0 p 2 p µ p 0 p 2 0 *24 D 3 * (2012) *26 L. Smolin The Troubles with Physics (2006) p.279-p.282 S. Mandelstam Mandelstam 1 11

13 2 T T T T T* T T* T T* * T T* 2 * 28 5 The only game in town * 29 The only Game in Town K. Vonnegut * 30 A guy with the gambling sickness loses his shirt every night in a poker game. Somebody tells him that the game is crooked, rigged to send him to the poorhouse. And he says, haggardly, I know, I know. But it s the only game in town. *27 Φ = 0 Φ 1 T Φ T* T* *28 M. Abe and N. Nakanishi, Prog. Theor. Phys. 115, 1151 (2006) 113, 76 (2006) *29 Not Even Wrong (2006) P. Woit *30 B. Schroer, String theory and the crisis in particle physics, special volume of I. J. M. P. D (2006) 12

14 (4) A Periodic Potential Problem in Quantum Mechanics (4) Kenji Seto 1 Kronig- Penney Schrödinger [ ħ2 d 2 ] 2m dx 2 + V (x) Ψ = EΨ (1.1) V (x) 2l V (x + 2l) = V (x) l x < l V (x) = V 0 ( x l ) 2 (1.2) V 0 x, V 0, E x l x, 2ml 2 ħ 2 V 0 V 0, 2ml 2 ħ 2 E E (1.3) [ d 2 ] dx 2 + E V (x) Ψ = 0 (1.4) V (x) 2 V (x + 2) = V (x) 1 x < 1 V (x) = V 0 x 2 (1.5) 2 µ µ = (4V 0 ) 1/4 (2.1) seto@pony.ocn.ne.jp 13

15 x z E κ z = µx, κ = E µ (2.2) 1 x < 1 [ d 2 dz 2 + κ z2 4 ] Ψ = 0 (2.3) Weber (Weber ) D κ (z) 1 F 1 D κ (z) = 2 κ/2 [ π e z2 /4 1 ( Γ ((1 κ)/2) 1 F 1 κ 2, 1 2 ; z 2 ) 2 2 z ( 1 κ Γ ( κ/2) 1 F 1, ; z 2 )] 2 (2.4) *1 Weber Hermite H n (z) κ n D n (z) = e z2 /4 H n (z) Hermite Weber 1 D κ 1 (iz) = 2 (κ+1)/2 [ π e z2 /4 1 ( κ + 1 Γ ((κ + 2)/2) 1 F 1, 2 1 ) 2 ; z2 2 2 iz ( κ )] Γ ((κ + 1)/2) 1 F 1, 2 2 ; z2 2 (2.5) Floquet (2.3) ( S 1 (z) =e z2 /4 κ F 1, 2 ( S 2 (z) =ze z2 /4 κ F 1, 2 1 ) 2 ; z ; z2 2 (2.3) 2 *2 S 1 (z) S 2 (z) S 1, S 2 Wronskian ) (2.6) W (z) = S 1 (z)s 2(z) S 1(z)S 2 (z) (2.7) z S 1, S 2 (2.3) Wronskian z = 0 W (z) 1 (2.8) (1.4) 1 x < 1 A, B Ψ(x) = AS 1 (µx) + BS 2 (µx) (2.9) *1 3 ( ) p λ κ *2 Kummer ( p.67) 1 F 1 (α, γ; z) = e z 1F 1 (γ α, γ; z) S 1, S 2 (2.4)

16 Floquet 1 K (0 K π) e ik e ik 1 x < 3 Ψ(x) = e ik[ ( ) ( )] AS 1 µ(x 2) + BS2 µ(x 2), or Ψ(x) = e ik [ ( ) ( )] AS 1 µ(x 2) + BS2 µ(x 2) (2.10) K 2 1 K = 0, π 2 K 1 K 0 < K < π (2.9) (2.10) x = 1 AS 1 (µ) + BS 2 (µ) = e ik[ AS 1 (µ) BS 2 (µ) ] AS 1(µ) + BS 2(µ) = e ik[ AS 1(µ) + BS 2(µ) ] (2.11) S 1, S 2 S 1, S 2 ( ) ( ) ( ) (1 e ik )S 1 (µ) (1 + e ik )S 2 (µ) A 0 (1 + e ik )S 1(µ) (1 e ik )S 2(µ) = B 0 (2.12) A, B cos(k) = S 1 (µ)s 2(µ) + S 1(µ)S 2 (µ) (2.13) (2.8) Wronskian 1 K 1 S 1, S 2 κ, E A, B (2.11) 1 A = (1 + e ik )S 2 (µ), B = (1 e ik )S 1 (µ) (2.14) x n 2n 1 x < 2n + 1 Ψ(x) = e ikn[ (1 + e ik ( ) )S 2 (µ)s 1 µ(x 2n) (1 e ik ( )] )S 1 (µ)s 2 µ(x 2n) (2.15) 3 2 E, E κ, K κ, κ, K, K S 1, S 2 Ψ(x, E), S i (z, κ), i = 1, 2 15

17 (2.15) Ψ(x, E)Ψ(x, E )dx = n= 2n+1 2n 1 ( Ψ(x, E)Ψ(x, E )dx = n= lim M M n= M ) 1 e i(k K )n Ψ(x, E)Ψ(x, E )dx 1 (3.1) e i(k K )n = 2πδ(K K ), 0 < K, K < π (3.2) E, E (1.4) [ d 2 ] dx 2 + E V (x) Ψ(x, E) = 0, [ d 2 ] dx 2 + E V (x) Ψ(x, E ) = 0 (3.3) E 1 Ψ(x, E ) 2 Ψ(x, E) d [ Ψ(x, E) dψ(x, E ) dx dx Ψ(x, E)Ψ(x, E )dx = dψ(x, E) ] Ψ(x, E ) (E E )Ψ(x, E)Ψ(x, E ) = 0 (3.4) dx 1 [ E E Ψ(x, E) dψ(x, E ) dx dψ(x, E) ] 1 Ψ(x, E ) dx 1 n = 0 (2.15) 16 S 1, S 2 Wronskian (2.8) (2.13) 1 1 Ψ(x, E)Ψ(x, E )dx = 2µ E E [ S 1 (µ, κ)s 2 (µ, κ) [ (1 + e i(k K ) ) cos(k ) (e ik + e ik ) ] (3.5) S 1 (µ, κ )S 2 (µ, κ ) [ (1 + e i(k K ) ) cos(k) (e ik + e ik ) ]] (3.6) (3.2) (3.6) (3.1) E, E (2.2) κ, κ Ψ(x, E)Ψ(x, E )dx = 4π [S µ(κ κ 1 (µ, κ)s 2 (µ, κ) [ (1 + e i(k K ) ) cos(k ) (e ik + e ik ) ] ) S 1 (µ, κ )S 2 (µ, κ ) [ (1 + e i(k K ) ) cos(k) (e ik + e ik ) ]] δ(k K ) (3.7) 1 κ (2.13) K K κ κ κ κ K = K κ κ κ κ 0/0 l Hôpital κ κ κ Ψ(x, E)Ψ(x, E )dx = 8π µ S 1(µ, κ)s 2 (µ, κ) sin(k) dk dκ δ(k K ) (3.8) 16

18 0 < K < π sin(k) S 1 (µ, κ)s 2 (µ, κ) dk/dκ E Ψ(x, E)Ψ(x, E )dx = N 2 (E)δ(E E ), N 2 (E) = 8πµ S 1 (µ, κ)s 2 (µ, κ) sin(k) (3.9) N(E) Ψ(x, E)/N(E) K (2.10) 2 sin(k), dk/dκ 4 (2.6) S 1 (z, κ), S 2 (z, κ) κ S 1 (z, 0) = e z2 /4, S 2 (z, 1) = ze z2 /4 (4.1) κ 1-1, 1-2 z κ S 1 ( 1-1) S 2 ( 1-2) 0 z 5, 0 κ 10 S 1 κ 1.6 z S 2 0 κ < 1 z κ 1 2 z S 1, S 2 κ 4 z S S 1 (z, κ) 1-2 S 2 (z, κ) 17

19 (2.13) V 0 E K V 0 = 50 2 K (3.8) S 1 (µ, κ)s 2 (µ, κ) S 1 (µ, κ)s 2 (µ, κ) dk/de = dk/µ 2 dκ 2 K E (V 0 = 50) 3 V 0 V 0 - E (2.13) V 0 - E 5 4 (2.6) S 1 (z), S 2 (z) 50 18

20 100 z 5.5 z z z z 5.5 V 0 (2.1) µ (2.2) z, κ V 0 < 0 [ ] 19

21 1 1) Anti-commutativity among Linearly Independent Imaginary Units Katsusada Morita 2) n a 1 a n a 2 n a 2 = a 2 z z 2 = z 2 1 x, y x, y xy = x y 2 z = x + iy z 2 = x 2 + y 2 3) z 1 z 2 = z 1 z 2 3 [1] a = a 0 + ia 1 + ja 2, b = b 0 + ib 1 + jb 2 {1, i, j} 1 4) ( ) ab : ab = (ab) 0 + i(ab) 1 + j(ab) 2 + ija 1 b 2 + jia 2 b 1 (ab) 0 = a 0 b 0 a 1 b 1 a 2 b 2 (ab) 1 = a 0 b 1 + a 1 b 0 (ab) 2 = a 0 b 2 + a 2 b 0 (1.1) i 2 = j 2 = 1 ij + ji = 0 (1.2) 5) (1.1) ij(a 1 b 2 a 2 b 1 ) ij(ab) 3 ab 2 = (ab) (ab) (ab) (ab) 2 3 = a 2 b 2 a 2 z 2 = x 2 + y 2 a 2 = a a a 2 2 ab 2 4 ( {1, i, j, ij} ) ij {1, i, j} ij = α + βi + γj, 1) 2) kmorita@cello.ocn.ne.jp 3) z = 0 z 2 = 0 z = 0 x = y = 0. {1, i} a = a 0 + ia 1 + ja 2 = 0 a 2 = a a2 1 + a2 2 = 0 a = 0 a 0 = a 1 = a 2 = 0 {1, i, j} 1 4) 1 Hamilton Dickson [2] 1 {i, j, k} Frobenius 5) ij = ji 0 = i 2 j 2 = (i j)(i+j) j = ±i a a = a 0 +i(a 1 ±a 2 ) ij ji ij = αji, α( +1) R (ij) 2 = (ji) 2 = α α 2 = 1 α 1 α = 1 (1.2) Hamilton [3] 20

22 α, β, γ (ij) 2 = 1 6) i(ij) = (ij)i, j(ij) = (ij)j ij {i, j} {i, j, ij} 1 7) ab 2 ab {1, i, j, ij} 3 a 2 b 2 ab 2 = a 2 b 2 Hurwitz a = b (1.2) (a 2 ) 3 = 0 a 2 2 = a 4 {i, j} (1.2) (1.2) {1, i, j} 1 a a 2 2 = a 4 {1, i, j} 1 2 (1.2) {1, i, j} 1 3 n (n 1) 1 n a a 2 = a 2 4 Hamilton [1] 2 1 Hamilton a [1] a = a 0 + ia 1 + ja 2, i 2 = j 2 = 1, a 0, a 1, a 2 R (2.1) b = b 0 + ib 1 + jb 2, b 0, b 1, b 2 R (1.1) ij ji 2 ( ) z 1 z 2 = z 1 z 2 a = b a µ = b µ (µ = 0, 1, 2) a 2 a 2 = a 2 0 a 2 1 a ia 0 a 1 + 2ja 0 a 2 + (ij + ji)a 1 a 2 (2.2) (1.2) a 2 a 2 2 = a 4 : a 2 2 = (a 2 0 a 2 1 a 2 2) 2 + (2a 0 a 1 ) 2 + (2a 0 a 2 ) 2 = (a a a 2 2) 2 = a 4 (2.3) (2.3) {i, j} (1.2) Hamilton [1] Hamilton ) Hamilton [1] [3] a = a 0 + ia 1 + ja 2 a = (a 0, a 1, a 2 ) 1 = (1, 0, 0), i = (0, 1, 0), j = (0, 0, 1) {1, i, j} α 0 + iα 1 + jα 2 = 0, α 0, α 1, α 2 R (2.4) 6) Hamilton k = ij, k 2 = 1 7)

23 α 0 = α 1 = α 2 = 0 {i, j} 1 iα 1 + jα 2 = 0, α 1, α 2 R (2.5) α 1 = α 2 = 0 (2.5) 2 (α α 2 2) + (ij + ji)α 1 α 2 = 0, ij + ji R (2.6) {i, j} (1.2) α 1 = α 2 = 0 {i, j} 1 {1, i, j} 1 (i ± j) 2 < 0 (i ± j) w = 1 w = i ± j {1, i, j} 1 (i ± j) 2 < 0 ij + ji = 2α R, α < 1 8) α 0 [2] {i, j} α < 1 a, b I = i, J = ai + bj, a = ±α/ 1 α 2, b = ±1/ 1 α 2 (2.7) I 2 = J 2 = 1, IJ + JI = 0 (2.8) I = i, J = j ( (2.1) a 1,2 ) (1.2) 1 {i, j} 3 (n 1) 1 [4] a, b {1, e i } i=1,2,3 l a = a 0 e 0 + a i e i + a 4 l a 0, a i, a 4 R b = b 0 e 0 + b i e i + b 4 l b 0, b i, b 4 R e 2 0 = e 0, l 2 = 1, le i = e i l (i = 1, 2, 3) ab = (ab) 0 e 0 + (ab) i e i + (ab) (4i) e i l + (ab) 4 l (i, j, k = 1, 2, 3 l [4] {e i, le i, l} {e A } A=1,2,3,,7 ): (ab) 0 = a 0 b 0 a i b i a 4 b 4 (ab) i = (a 0 b i + a i b 0 ) + ϵ jki a j b k (ab) 4i = a i b 4 a 4 b i (ab) 4 = a 0 b 4 + a 4 b 0 (2.9) (2.10) ab 2 = (ab) (ab) 2 i + (ab)2 4 + (ab) 2 4i ab 2 = a 2 b 2 a = b (a 2 ) 4i = 0 a 2 a 2 2 = a 4 8) ij + ji = 2β R, β < 1 (2.6) α 1,2 α 1 = α 2 = 0 22

24 3 n a = a 0 + a 1 e 1 + a 2 e a n 1 e n 1, a µ (µ = 0, 1,, n 1) R (3.1) 1 {e i } i=1,2,,n 1 e i e j + e j e i = 2δ ij (3.2) e i e j = δ ij + c ijk e k, c jik = c ijk {e i } {E k } e i e j = δ ij + c ijk E k, c jik = c ijk, E k E k = δ kk a 9) 1 1 e 1 = (0, 1, 0, 0,, 0, 0) e 2 = (0, 0, 1, 0,, 0, 0). e n 1 = (0, 0, 0, 0,, 0, 1) (3.3) α 1 e 1 + α 2 e α n 1 e n 1 = (0, α 1, α 2,, α n 1 ) = (0, 0, 0,, 0) α 1 = α 2 = = α n 1 = 0 e i e 2 1 = e 2 2 = = e 2 n 1 = ( 1, 0,, 0) e 0 ( ) (α 1 e 1 + α 2 e α n 1 e n 1 ) 2 = i i j αi α i α j (e i e j + e j e i ) = 0 2 i j e i e j + e j e i = 0 (i j ) (3.4) i α i = 0 {e i } i=1,2,,n ) 1 (3.4) 11) i, j(i j) {1, e i, e j } 1 (e i ± e j ) 2 < 0 e i e j + e j e i = 2α ij, α ij < 1 {e i, e j } {I i, I j } {I i, I j } (2.7), (2.8) {I i, I j } = {e i, e j } (3.4) {e i } i=1,2,,n 1 1 (3.4) (3.4) ( Einstein ) a 2 2 = (a 0 + a 1 e 1 + a 2 e a n 1 e n 1 ) 2 2 = (a 0 + a i e i ) 2 2 = a a 0 a i e i + a i a j e i e j 2 = a a 0 a i e i + a i a j (e i e j + e j e i )/2 2 = (a 2 0 a 2 i ) 2 + (2a 0 a i ) 2 = (a a 2 i ) 2 = a 4 (3.5) 9) e 1 = i, e 2 = j E 3 = e 1 e 2, E3 2 = 1 10) j 1 e 1 e j = e j e 1 n a = a 0 + a i e i a = a 0 ± i(a a n), i e 1 11) [4] A 23

25 a 2 = a 2 (3.6) (2.3) n 1 a = (a 0, a 1,, a n 1 ) 2 a 2 ab = a b a = b (3.6) (3.6) (3.6) n a 2 a 2 n (3.6) (3.4) a {1, e 1,, e n 1 } 1 ( ) 4 Hamilton [1] ( {i, j}(i 2 = j 2 = 1) ) {1, i, j} 3 {1, i, j} 1 Hamilton a a 2 = a 2 {i, j} a a aa Hamilton Dickson [2] a 1 a 2 = a 2 n a 2 a 2 n n a {e 0, e 1,, e n 1 } 1 Cayley-Dickson a 2 = a 2 ab = a b R, C, H, O Hurwitz, ( ) [1] W. R. Hamilton, Quaternions, Proc. Roy. Irish Acad. vol. L(1945), [2] L. E. Dickson, Linear Algebras, Cambridge: at the University Press, [3] ( 2014). [4] K. Morita, Quasi-Associativity and Cayley-Dickson Algebras, PTEP, 2014, 013A03 (19 pages). 24

26 1 The Parity Operator Minoru Yonezawa 2 1 (Parity operator) P 3 2 [ ] [ 1 n P = ( 2x)n n! x n n=0 m=0 ] [ 1 m ( 2y)m m! y m l=0 ] 1 l ( 2z)l l! z l (2.1) 3 1 (Taylor expansion) f(x + a) = n=0 ( f( x) = exp a d ) f(x) dx = a= 2x ( 1 dn an f(x) = exp a d ) f(x) (2.2) n! dxn dx n=0 1 n! dn ( 2x)n f(x) (2.3) dxn (2.3) (2.3) a f(x) x a = 2x 4 P f(x) = f( x), 1 dn P = ( 2x)n n! dx n (2.4) n=0 1 2 m-yonezawa@mtc.biglobe.ne.jp

27 3 p x = iħ d dx x x + a (unitary operator) U = e i a ħ p x f(x + a) = Uf(x) (2.5) 3 x x pp (translation operator) 2 1 (2.2) a = 2x n 2 f( x) = n=0 ( 1 dn ( 2x)n f(x) = exp 2x d ) f(x) (3.1) n! dxn dx ( exp 2x d ) = dx (3.1) n=0 ( 1 2x d ) n (3.2) n! dx ( ( 2x) n dn dx n 2x d ) n (3.3) dx (2.2) (2.2) a = 2x f(x + a) = f( x) a = a = 2x 26

28 (2.3) 1 (2.3) 2 Taylor Taylor f(x+h) x+h x f(x) f (n) (x), (n = 1, 2, 3, ) f(x + h) h f(x + h) = a 0 + a 1 h + a 2 h 2 + a 3 h 3 + a 4 h a n h n + (3.4) h a 0, a 1, a 2, a 3, a 4, (3.4) h = 0 a 0 = f(x) (3.4) h f (x + h) = a 1 + 2a 2 h + 3a 3 h 2 + 4a 4 h na n h n 1 + (3.5) (3.5) h = 0 a 1 = f (x) (3.5) h f (x + h) = 2a a 3 h + 3 4a 4 h (n 1)na n h n 2 + (3.6) (3.6) h = 0 a 2 = 1 2! f (x) (3.6) h f (x + h) = 1 2 3a a 4 h + + (n 2)(n 1)na n h n 3 + (3.7) (3.7) h = 0 a 3 = 1 3! f (x) a n = 1 n! f (n) (x) (3.4) f(x + h) = f(x) + hf (x) + h2 2! f (x) + h3 3! f (x) + + hn n! f n (x) + (3.8) ( ) 27

29 The Parity Operator A Comment 1 Noboru NAKANISHI 2 P P f(x) = d dx P a x dyf(y) (1) dyk(x, y)f(y) (2) K(x, y) = δ(x + y) (3) ( ) 1 2 nbr-nak@trio.plala.or.jp 28

1 (Contents) (1) Beginning of the Universe, Dark Energy and Dark Matter Noboru NAKANISHI 2 2. Problem of Heat Exchanger (1) Kenji

1 (Contents) (1) Beginning of the Universe, Dark Energy and Dark Matter Noboru NAKANISHI 2 2. Problem of Heat Exchanger (1) Kenji 8 4 2018 6 2018 6 7 1 (Contents) 1. 2 2. (1) 22 3. 31 1. Beginning of the Universe, Dark Energy and Dark Matter Noboru NAKANISHI 2 2. Problem of Heat Exchanger (1) Kenji SETO 22 3. Editorial Comments Tadashi