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秋の分科会講演予稿集

1a-D-3 物理的状態はベクトルか行列か?

豊 豊沢

Published Sep 13, 1985• 0 citations
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Flux/Flow Equations and Momentum as p=Ev Versus v=E/p Part II

Francesco R. Ruggeri

In Part I we argued that two flow equations for a free particle, namely: d/dx (partial) A(x,t) = p ((1a)) and d/dt (partial) A(x,t) = -E ((1a)) may be applied to both particles with rest mass and those without. In fact we argue that there are at least three scenarios associated with ((1a)) and ((1b)). The first is classical mechanics for a particle with rest mass mo and constant velocity. Using ((1a)) and ((1b)) with p=E(v)v this scenario yields special relativity. A second solution is to write ((1a)) and ((1b)) as eigenvalue equations: -id/dx (partial) exp(iA) = p exp(iA) and id/dt (partial) exp(iA)= E exp(iA) where A=-Et+px. This yields a quantum mechanical solution for a particle with rest mo if one assumes p=Ev. In such a case one may use the special relativistic result EE = pp + momo and establish the Klein-Gordon equation. A third solution pertains to objects with zero rest mass. As argued in Part I, these do not follow p=E(v)v, but rather the classical wave result: v=E/p. In this note we examine massless particles, namely waves on a string, photons and sound, in more detail. For the zero mass case, dA=0 so there is motion in such a way as to keep A constant. If one considers periodic motion and uses ((1a)) and ((1b)), a solution which is real and periodic is: d/dt d/dt cos(A) = vv d/dx d/dx cos(A) ((2)) A second order partial differential equation is needed as one insists the solution be real. This differs from the quantum mechanical solution of exp(-iEt+ipx) which is complex for a reason. The modulus must be linked to a probability of 1 for a free particle. The zero mass solution is one of a smeared-out physical wave. We note that ((1a)) and ((1b)) deal strictly with E energy and p momentum, but cos(A)=cos(-Et+px) is associated with a period proportional to 1/E and a wavelength proportional to 1/p. This raises two questions. First, what physical entity is periodic? Secondly, how does the energy appear in the problem as energy and not just 2*3.14 frequency of the physical entity? We try to consider this issue for a wave on a string, photon and sound wave.

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3a-D-8 1次元ハバード模型の励起スペクトルの構造

勉 桃井

日本物理学会講演概要集. 秋の分科会•Aug 16, 1994•0 citations

3a-D5-3 2次元ハバード模型のランジュバンシミュレーション

和夫 斉籐, 慧 高田

秋の分科会講演予稿集•Sep 16, 1988•0 citations
In Part I we argued that two flow equations for a free particle, namely: d/dx (partial) A(x,t) = p ((1a)) and d/dt (partial) A(x,t) = -E ((1a)) may be applied to both particles with rest mass and those without. In fact we argue that there are at least three scenarios associated with ((1a)) and ((1b)). The first is classical mechanics for a particle with rest mass mo and constant velocity. Using ((1a)) and ((1b)) with p=E(v)v this scenario yields special relativity. A second solution is to write ((1a)) and ((1b)) as eigenvalue equations: -id/dx (partial) exp(iA) = p exp(iA) and id/dt (partial) exp(iA)= E exp(iA) where A=-Et+px. This yields a quantum mechanical solution for a particle with rest mo if one assumes p=Ev. In such a case one may use the special relativistic result EE = pp + momo and establish the Klein-Gordon equation. A third solution pertains to objects with zero rest mass. As argued in Part I, these do not follow p=E(v)v, but rather the classical wave result: v=E/p. In this note we examine massless particles, namely waves on a string, photons and sound, in more detail. For the zero mass case, dA=0 so there is motion in such a way as to keep A constant. If one considers periodic motion and uses ((1a)) and ((1b)), a solution which is real and periodic is: d/dt d/dt cos(A) = vv d/dx d/dx cos(A) ((2)) A second order partial differential equation is needed as one insists the solution be real. This differs from the quantum mechanical solution of exp(-iEt+ipx) which is complex for a reason. The modulus must be linked to a probability of 1 for a free particle. The zero mass solution is one of a smeared-out physical wave. We note that ((1a)) and ((1b)) deal strictly with E energy and p momentum, but cos(A)=cos(-Et+px) is associated with a period proportional to 1/E and a wavelength proportional to 1/p. This raises two questions. First, what physical entity is periodic? Secondly, how does the energy appear in the problem as energy and not just 2*3.14 frequency of the physical entity? We try to consider this issue for a wave on a string, photon and sound wave.