a super sexy, most likely columbian babe that is awsome to be around and has a great butt. she is most likely a great kisser and is very laid back.
by the tyler j. G. April 1, 2011
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SUSI is a term created in Norway originated by a group of yuths in Fredrikstad , the word went from one person to another thats how it became popular. The term means look at your clothes like they look nice hella drippy.
Boy susi. Peak that susi. Or just susi
by Susi master May 26, 2021
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Cute, sometimes a loser, likes to cuddle, enjoys playing Xbox
Man, that girl is so Susi.
Yee, i know man.
by cccc-cccc February 9, 2014
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The beastliest man alive who protects this great town. His elbows are made of steel and they are his most deadly form of attack. He is not only is great at what he does, but he warns the kids about dangers such as drugs and alcohol. He also loves to play xbox with the kids. All in all, a real people person. He is rumored to have picked up a burning school bus full of children and put it back down again. When he was 7 he failed the swim test at the local pool. To prove to his teacher that he deserved to pass it, he swam across the Atlantic Ocean, then took a giant dump that we now know as asia, europe, and africa. God created the universe, but Susi's elbows created god.
Class - "Thanks Susi, now we know!

Susi - "And knowing is half the battle."
by Darth Sidius February 11, 2009
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Et joku on iha susi
Äijjjä on iha susi
by Mursu123 October 24, 2018
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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!
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!
by Migdal Kadanoff April 26, 2019
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