by Firecracker182 July 6, 2005
Get the susiemug. Susie she really love to laugh has a huge booty oh and hot as (.)(.) she is really nice to be with attracts all the guys and get little moody sometimes
oooo I need that susie
by slay123 October 9, 2018
Get the susiemug. by wildawv April 20, 2009
Get the susiemug. a girl who is very k00L. uber fobby. and is one of the bestest friends in the world. one who you can count on to make you smile.
susie is the k00Lest!
by laLALa October 20, 2004
Get the susiemug. by Quel November 15, 2005
Get the susiemug. a susie is a person who is muchos exotic and tropical and is KOOL FOR KITTENS AND i have love for her, she makes guys wanna ding dong
by NIC-O-LA February 19, 2005
Get the susiemug. An abbreviation for Super-Symmetrical Quantum mechanics.
Consider a QM system with Hamiltonian H and potential V(x) such that H |Ψ> = E |Ψ>
We define a new Hamiltonian H1 in terms of potential V1(x) which is offset by the zero point energy so that:
H1 |0> = 0 ie the enegy of the ground state of H1 is zero.
We define this Hamiltonian in terms of generalized raising and lowering operators A and A dagger such that:
H1 = A_dag A = (p^2/2m) + V1(x)
A = (ip/root(2m)) + W'(x)
Where W(x) is the super potential.
The potential V1(x) can be constructed from the superpotential:
V1(x) = W'(x)^2 - (ћ/root(2m))W"(x)
If we know the H1 ground state Ψ_0(x) then we can derive:
Ψ_0(x) ~ exp(-root(2m)W(x)/ћ)
Which can be used to find the superpotential W(x)
From this superpotential we can derive the partnerpotential V2(x) where:
V2(x) = W'(x)^2 - (ћ/root(2m))W"(x)
which has associated Hamiltonian H2 = A A_dag = (p^2/2m) + V2(x)
This partner potential may allow H2 to have an eigenspectrum which is easier to find. Once this is found we can go from the nth energy level of H2 to the (n+1)th level of H1 by simply applying the A_dag operator. This means we can find the first excited state of H1 by applying A_dag to the ground state of H2.
Note: I wrote this while in class and the prof was talking about some really complex shit I wasn't paying attention to so now I'm fucked for the exam next week.
But sugondese amirite?
At least we still got us a woodshed!
Consider a QM system with Hamiltonian H and potential V(x) such that H |Ψ> = E |Ψ>
We define a new Hamiltonian H1 in terms of potential V1(x) which is offset by the zero point energy so that:
H1 |0> = 0 ie the enegy of the ground state of H1 is zero.
We define this Hamiltonian in terms of generalized raising and lowering operators A and A dagger such that:
H1 = A_dag A = (p^2/2m) + V1(x)
A = (ip/root(2m)) + W'(x)
Where W(x) is the super potential.
The potential V1(x) can be constructed from the superpotential:
V1(x) = W'(x)^2 - (ћ/root(2m))W"(x)
If we know the H1 ground state Ψ_0(x) then we can derive:
Ψ_0(x) ~ exp(-root(2m)W(x)/ћ)
Which can be used to find the superpotential W(x)
From this superpotential we can derive the partnerpotential V2(x) where:
V2(x) = W'(x)^2 - (ћ/root(2m))W"(x)
which has associated Hamiltonian H2 = A A_dag = (p^2/2m) + V2(x)
This partner potential may allow H2 to have an eigenspectrum which is easier to find. Once this is found we can go from the nth energy level of H2 to the (n+1)th level of H1 by simply applying the A_dag operator. This means we can find the first excited state of H1 by applying A_dag to the ground state of H2.
Note: I wrote this while in class and the prof was talking about some really complex shit I wasn't paying attention to so now I'm fucked for the exam next week.
But sugondese amirite?
At least we still got us a woodshed!
Guy one: Look at that physicist fuck over there doing SUSY.
Guy two: Damn straight brother I hear SUSY sucks a lot.
Guy one: Hell yeah she does!
Guy two: Damn straight brother I hear SUSY sucks a lot.
Guy one: Hell yeah she does!
by Migdal Kadanoff April 26, 2019
Get the SUSYmug.