hey tomoko
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hey tomoko
f~(p)=∫e−2πipxf(x)dxf(x)=∫f~(p)e2πipxdp
Ψ~(p)=∫e−2πipxδ(x)d3x=1δ(x)=∫Ψ~(p)e2πipxd3p=1(2πℏ)3∫Ψ~(p)ei(2πℏp)x/ℏd3(2πℏp)=1(2πℏ)3∫eipx/ℏd3p
Ψ~(p2πℏ)≡Φ~(p)=∫e−2πi(p2πℏ)xΨ(x)d3x=∫e−ipx/ℏΨ(x)d3x=(2πℏ)3/2Ψ(p)Ψ(x)=∫Ψ~(p2πℏ)e2πipxd3p=1(2πℏ)3∫Ψ~(p2πℏ)e2πi(p2πℏ)xd3p=1(2πℏ)3∫Φ~(p)eipx/ℏd3p=1(2πℏ)3/2∫Ψ(p)eipx/ℏd3p
Ψ(p)=1(2πℏ)3/2∫e−ipx/ℏΨ(x)d3xΨ(x)=1(2πℏ)3/2∫Ψ(p)eipx/ℏd3p
Φ~(p,E)=∫e−2πi(px−Et)/2πℏΨ(x,t)dxdt=∫e−i(px−Et)/ℏΨ(x,t)dxdtΨ(x,t)=1(2πℏ)2∫Φ~(p,E)e2πi(px−Et)/2πℏdpdE=1(2πℏ)2∫Φ~(p,E)ei(px−Et)/ℏdpdE
K~(p,E)=−iℏp22m+V−(E+iη)=iℏE−[(p22m+V)−iη]
K(x−x′,t−t′)=1(2πℏ)2∫K~(p,E)ei[p(x−x′)−E(t−t′)]/ℏdpdE=1(2πℏ)2∫(−2πi)(iℏ)ei[p(x−x′)−(p22m+V)(t−t′)]/ℏdp=1(2πℏ)∫ei[p(x−x′)−(p22m+V)(t−t′)]/ℏdp]
H(x′′)K(x′′,t;x′,t0)=∑a′Ea′<x′′|a′><a′|x′>exp[−iEa′(t−t0)ℏ]θ(t−t0)
∫∞−∞dtEa′<x′′|a′><a′|x′>exp[−i(Ea′−iη)(t−t0)ℏ]θ(t−t0)=∫∞0dtEa′<x′′|a′><a′|x′>exp[−i(Ea′−iη)tℏ]=∫∞0ℏEa′−i(Ea′−iη)d(−i(Ea′−iη)tℏ)<x′′|a′><a′|x′>exp[−i(Ea′−iη)tℏ]=ℏEa′−i(Ea′−iη)<x′′|a′><a′|x′>exp[−i(Ea′−iη)tℏ]|t=+∞t=0=−iℏEa′(Ea′−iη)<x′′|a′><a′|x′>=−iℏ<x′′|a′><a′|x′
∑a′−iℏ<x′′|a′><a′|x′>=−iℏδ(x′′−x′)
∫∞−∞dt(−iℏ∂∂t)K(x′′,t;x′,t0)=−iℏK(x′′,t;x′,t0)|+∞−∞=−iℏ∑a′Ea′<x′′|a′><a′|x′>exp[−i(Ea′−iη)(∞−t0)ℏ]−0=0−0=0
∫∞−∞dt(H−iℏ∂∂t)K(x′′,t;x′,t0)=−iℏδ(x′′−x′)
limN→∞(1+aN)N=exp(ln(limN→∞(1+aN)N))=exp(limN→∞ln(1+aN)N)=exp(limN→∞Nln(1+aN))=exp(limN→∞N(aN+O(1N2)))=ea
iℏc¨2=−i(ω−ω21)c˙2−iγ2ℏc2
4π2ℏ2k2+2π(ω−ω21)ℏ2k−γ2=0
2πk(1,2)=−ω−ω212±(ω−ω212)2+(γℏ)2−−−−−−−−−−−−−−−√
|c1(t)|2+|c2(t)|2m2=(ω−ω21)2+2(γℏ)2+(γℏ)22cos(2π(k(1)−k(2))(γℏ)2+2−2cos(2π(k(1)−k(2))=⎛⎝⎜⎜2(ω−ω21)24+(γℏ)2(γℏ)2−−−−−−−−−−−−−⎷⎞⎠⎟⎟2=1
look!
Ψ~(p)=∫e−2πipxδ(x)d3x=1δ(x)=∫Ψ~(p)e2πipxd3p=1(2πℏ)3∫Ψ~(p)ei(2πℏp)x/ℏd3(2πℏp)=1(2πℏ)3∫eipx/ℏd3p
Ψ~(p2πℏ)≡Φ~(p)=∫e−2πi(p2πℏ)xΨ(x)d3x=∫e−ipx/ℏΨ(x)d3x=(2πℏ)3/2Ψ(p)Ψ(x)=∫Ψ~(p2πℏ)e2πipxd3p=1(2πℏ)3∫Ψ~(p2πℏ)e2πi(p2πℏ)xd3p=1(2πℏ)3∫Φ~(p)eipx/ℏd3p=1(2πℏ)3/2∫Ψ(p)eipx/ℏd3p
Ψ(p)=1(2πℏ)3/2∫e−ipx/ℏΨ(x)d3xΨ(x)=1(2πℏ)3/2∫Ψ(p)eipx/ℏd3p
Φ~(p,E)=∫e−2πi(px−Et)/2πℏΨ(x,t)dxdt=∫e−i(px−Et)/ℏΨ(x,t)dxdtΨ(x,t)=1(2πℏ)2∫Φ~(p,E)e2πi(px−Et)/2πℏdpdE=1(2πℏ)2∫Φ~(p,E)ei(px−Et)/ℏdpdE
K~(p,E)=−iℏp22m+V−(E+iη)=iℏE−[(p22m+V)−iη]
K(x−x′,t−t′)=1(2πℏ)2∫K~(p,E)ei[p(x−x′)−E(t−t′)]/ℏdpdE=1(2πℏ)2∫(−2πi)(iℏ)ei[p(x−x′)−(p22m+V)(t−t′)]/ℏdp=1(2πℏ)∫ei[p(x−x′)−(p22m+V)(t−t′)]/ℏdp]
H(x′′)K(x′′,t;x′,t0)=∑a′Ea′<x′′|a′><a′|x′>exp[−iEa′(t−t0)ℏ]θ(t−t0)
∫∞−∞dtEa′<x′′|a′><a′|x′>exp[−i(Ea′−iη)(t−t0)ℏ]θ(t−t0)=∫∞0dtEa′<x′′|a′><a′|x′>exp[−i(Ea′−iη)tℏ]=∫∞0ℏEa′−i(Ea′−iη)d(−i(Ea′−iη)tℏ)<x′′|a′><a′|x′>exp[−i(Ea′−iη)tℏ]=ℏEa′−i(Ea′−iη)<x′′|a′><a′|x′>exp[−i(Ea′−iη)tℏ]|t=+∞t=0=−iℏEa′(Ea′−iη)<x′′|a′><a′|x′>=−iℏ<x′′|a′><a′|x′
∑a′−iℏ<x′′|a′><a′|x′>=−iℏδ(x′′−x′)
∫∞−∞dt(−iℏ∂∂t)K(x′′,t;x′,t0)=−iℏK(x′′,t;x′,t0)|+∞−∞=−iℏ∑a′Ea′<x′′|a′><a′|x′>exp[−i(Ea′−iη)(∞−t0)ℏ]−0=0−0=0
∫∞−∞dt(H−iℏ∂∂t)K(x′′,t;x′,t0)=−iℏδ(x′′−x′)
limN→∞(1+aN)N=exp(ln(limN→∞(1+aN)N))=exp(limN→∞ln(1+aN)N)=exp(limN→∞Nln(1+aN))=exp(limN→∞N(aN+O(1N2)))=ea
iℏc¨2=−i(ω−ω21)c˙2−iγ2ℏc2
4π2ℏ2k2+2π(ω−ω21)ℏ2k−γ2=0
2πk(1,2)=−ω−ω212±(ω−ω212)2+(γℏ)2−−−−−−−−−−−−−−−√
|c1(t)|2+|c2(t)|2m2=(ω−ω21)2+2(γℏ)2+(γℏ)22cos(2π(k(1)−k(2))(γℏ)2+2−2cos(2π(k(1)−k(2))=⎛⎝⎜⎜2(ω−ω21)24+(γℏ)2(γℏ)2−−−−−−−−−−−−−⎷⎞⎠⎟⎟2=1
look!
Last edited by c. robin, muffin man on November 24th 2014, 3:50 pm; edited 1 time in total
the 4th disciple- hello?
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Re: hey tomoko
actually here tomoko i'll make it easier for you
Last edited by c. robin, muffin man on November 24th 2014, 3:51 pm; edited 1 time in total
Re: hey tomoko
explain this shitc. robin, muffin man wrote:actually here tomoko i'll make it easier for you
the 4th disciple- hello?
- Posts : 5821
volume of testosterone : 245730
Join date : 2014-09-01
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Re: hey tomoko
Last edited by c. robin, muffin man on November 24th 2014, 3:51 pm; edited 1 time in total
froggy- tomak and the power of jew jew
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Re: hey tomoko
salty wrote:And titties
now im interested
froggy- tomak and the power of jew jew
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