Physics, asked by ibrahimk6137, 1 year ago

How to derive the simplest 1D Superpotential Hamiltonian?

Answers

Answered by vrrunda
0
In the superpotential wiki article there are definitions of 2 supersymmetric operators:

Q1=12[(p−iW)b+(p+iW)b†]Q2=i2[(p−iW)b−(p+iW)b†]Q1=12[(p−iW)b+(p+iW)b†]Q2=i2[(p−iW)b−(p+iW)b†]

Then the following Hamiltonian is defined and presented:

H={Q1,Q1}={Q2,Q2}=p22+W22+W′2(bb†−b†b)H={Q1,Q1}={Q2,Q2}=p22+W22+W′2(bb†−b†b)

Where W′=dW(x)dxW′=dW(x)dx and {b,b†}=1{b,b†}=1 and [b,b†]=0[b,b†]=0.

Why does ={Q1,Q1}={Q2,Q2}=p22+W22+W′2(bb†−b†b)={Q1,Q1}={Q2,Q2}=p22+W22+W′2(bb†−b†b)? I can't see how they got this expression.

What's the motivation/justification to define the superpotential as such? Also why do Q1,Q2Q1,Q2 map "bosonic" states into "fermionic" states and vice versa?

Lastly, why does it take the form: H=p22+W22±W′2H=p22+W22±W′2?

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