Same theory as Schroeder's Cat, if you're blindfolded and fisting. You do not know whether you are fisting a vagina or a rectum until you remove the blindfold.
Have you heard of Schroedinger's Cat?
No, but have you ever heard of Quantum Fisting? Great time at my stag party.... I think.
No, but have you ever heard of Quantum Fisting? Great time at my stag party.... I think.
by Quantosexual December 7, 2018
Get the Quantum Fisting mug.When you have a boner and lay down flat and start to spin on your peepee head and eventually you spin so fast that quantum mechanics break down and spacetime disintegrates.
Bill: "Hey man, I've been doing the quantum headspin alot lately. It sure does hurt!"
John: "What the hell is quantum headspin?"
John: "What the hell is quantum headspin?"
by jj the large jet plane December 30, 2022
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Get the Quantum Loop mug.iℏ⋅(∂Ψ(x,t))/(∂t)=-ℏ^2/2m⋅(∂^2Ψ(x,t))/(∂x^2)
(This is how the average conversation about quantum mechanics generally goes:)
Person 1: Omg like literally iℏ⋅(∂Ψ(x,t))/(∂t)=-ℏ^2/2m⋅(∂^2Ψ(x,t))/(∂x^2)
Person 2: Right! Lmao it's just ∂Ψ(x,t)=Ae^(i(kx-ωt))
Person 1: Exactly! And iℏ⋅(∂Ψ(x,t))/(∂t)=iℏ⋅(∂Ae^(i(kx-ωt)))/(∂t)=ℏωAe^(i(kx-ωt))
Person 2: Hahaha! -ℏ^2/2m⋅(∂^2Ψ(x,t))/(∂x^2)=-ℏ^2/2m⋅(∂^2Ae^(i(kx-ωt)))/(∂x^2)
Person 1: Yeah and it's like -(ikℏ^2)/2m⋅(∂Ae^(i(kx-ωt)))/(∂x)=-ℏ^2/2m⋅(∂^2Ψ(x,t))/(∂x^2)=-ℏ^2/2m⋅(∂^2Ae^(i(kx-ωt)))/(∂x^2)
Person 2: Bro I was thinking the same thing, as soon as she walked in my mind just went (ℏ^2⋅k^2)/2m⋅Ae^(i(kx-ωt))= -(ikℏ^2)/2m⋅(∂Ae^(i(kx-ωt)))/(∂x)=-ℏ^2/2m⋅(∂^2Ψ(x,t))/(∂x^2)=-ℏ^2/2m⋅(∂^2Ae^(i(kx-ωt)))/(∂x^2)
Person 1: Pfffftttt!!! (ℏ^2⋅k^2)/2m=ℏω
Person 2: Lol yeah man, great shit right there... So anyway, how's Jack?
Person 1: Omg like literally iℏ⋅(∂Ψ(x,t))/(∂t)=-ℏ^2/2m⋅(∂^2Ψ(x,t))/(∂x^2)
Person 2: Right! Lmao it's just ∂Ψ(x,t)=Ae^(i(kx-ωt))
Person 1: Exactly! And iℏ⋅(∂Ψ(x,t))/(∂t)=iℏ⋅(∂Ae^(i(kx-ωt)))/(∂t)=ℏωAe^(i(kx-ωt))
Person 2: Hahaha! -ℏ^2/2m⋅(∂^2Ψ(x,t))/(∂x^2)=-ℏ^2/2m⋅(∂^2Ae^(i(kx-ωt)))/(∂x^2)
Person 1: Yeah and it's like -(ikℏ^2)/2m⋅(∂Ae^(i(kx-ωt)))/(∂x)=-ℏ^2/2m⋅(∂^2Ψ(x,t))/(∂x^2)=-ℏ^2/2m⋅(∂^2Ae^(i(kx-ωt)))/(∂x^2)
Person 2: Bro I was thinking the same thing, as soon as she walked in my mind just went (ℏ^2⋅k^2)/2m⋅Ae^(i(kx-ωt))= -(ikℏ^2)/2m⋅(∂Ae^(i(kx-ωt)))/(∂x)=-ℏ^2/2m⋅(∂^2Ψ(x,t))/(∂x^2)=-ℏ^2/2m⋅(∂^2Ae^(i(kx-ωt)))/(∂x^2)
Person 1: Pfffftttt!!! (ℏ^2⋅k^2)/2m=ℏω
Person 2: Lol yeah man, great shit right there... So anyway, how's Jack?
by wugugugug May 6, 2021
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