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@xianrenb
Last active November 28, 2020 08:56
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Spatial Probability Density
===========================
假設 ρ 是 spatial probability density
∫∫∫ ρ dx dy dz = 1 ...(1)
∫∫∫ ρ dx dy dz = e^(i 2 π n), n = integer
∫∫∫ ρ dx dy dz = e^(i 2 π n n_t), n_t = integer
∫∫∫ ρ dx dy dz = e^(i 2 π n 1/τ n_t τ), τ = period
設 f = 1/τ, t = n_t τ
∫∫∫ ρ dx dy dz = e^(i 2 π n f t)
∫∫∫ ρ dx dy dz = e^(i 2 π n h f/h t)
∫∫∫ ρ dx dy dz = e^(i 2 π n E/h t)
∫∫∫ ρ dx dy dz = e^(i n E/(h/2 π) t)
∫∫∫ ρ dx dy dz = e^(i n E/h_bar t)
ρ = d^3/(dx dy dz) e^(i n E/h_bar t)
ρ = (d^3/dt^3 e^(i n E/h_bar t)) dt/dx dt/dy dt/dz
ρ = (d^3/dt^3 e^(i n E/h_bar t)) 1/(dx/dt) 1/(dy/dt) 1/(dz/dt)
ρ = 1/(v_x v_y v_z) (d^3/dt^3 e^(i n E/h_bar t))
ρ = 1/(v_x v_y v_z) (d^3/dt^3 e^(i n 1/h_bar E t)) ...(2)
按等效座標理論
p x = E t ...(3)
3D 效果
p . r = E t ...(4)
(2) & (4):
ρ = 1/(v_x v_y v_z) (d^3/dt^3 e^(i n 1/h_bar (p . r)))
ρ = 1/(v_x v_y v_z) d^2/dt^2 (e^(i n 1/h_bar (p . r)) (i n 1/h_bar (dp/dt . r + p . dr/dt)))
ρ = 1/(v_x v_y v_z) d^2/dt^2 (e^(i n 1/h_bar (p . r)) (i n 1/h_bar (F . r + p . v))) ...(5)
F = m E_g + m v × B_g + q E + q v × B ...(6)
(5) & (6):
ρ = 1/(v_x v_y v_z) d^2/dt^2 (e^(i n 1/h_bar (p . r)) (i n 1/h_bar ((m E_g + m v × B_g + q E + q v × B) . r + p . v))) ...(7)
(4):
d/dt (p . r) = d/dt (E t)
dp/dt . r + p . dr/dt = dE/dt t + E
F . r + p . v = t d/dt (h f) + E
F . r + p . v = h t df/dt + E
E = F . r + p . v - h t df/dt ...(8)
(6) & (8):
E = (m E_g + m v × B_g + q E + q v × B) . r + p . v - h t df/dt ...(9)
(2) & (9):
ρ = 1/(v_x v_y v_z) (d^3/dt^3 e^(i n 1/h_bar ((m E_g + m v × B_g + q E + q v × B) . r + p . v - h t df/dt) t)) ...(10)
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