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

That moment when you think a bug is crawling on you, and it turns out to be true. LMD

A FML moment. LMD

Pretty much you go, JUST LET ME DIE!

Wish this never happened. LMD

A FML moment. LMD

Pretty much you go, JUST LET ME DIE!

Wish this never happened. LMD

by misspinkyperfectionnnll July 09, 2011

by adrian sosa December 17, 2003

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