Open main menu
Home
Random
Recent changes
Special pages
Community portal
Preferences
About Wikipedia
Disclaimers
Incubator escapee wiki
Search
User menu
Talk
Dark mode
Contributions
Create account
Log in
Editing
Supersymmetric quantum mechanics
(section)
Warning:
You are not logged in. Your IP address will be publicly visible if you make any edits. If you
log in
or
create an account
, your edits will be attributed to your username, along with other benefits.
Anti-spam check. Do
not
fill this in!
== Example == Let's look at the example of a one-dimensional nonrelativistic particle with a 2D (''i.e.,'' two states) internal degree of freedom called "spin" (it's not really spin because "real" spin is a property of 3D particles). Let <math>b</math> be an operator which transforms a "spin up" particle into a "spin down" particle. Its adjoint <math>b^\dagger</math> then transforms a spin down particle into a spin up particle; the operators are normalized such that the anticommutator <math>\{b,b^\dagger\}=1</math>. And <math>b^2=0</math>. Let <math>p</math> be the momentum of the particle and <math>x</math> be its position with <math>[x,p]=i</math>. Let <math>W</math> (the "[[superpotential]]") be an arbitrary complex analytic function of <math>x</math> and define the supersymmetric operators : <math>Q_1=\frac{1}{2}\left[(p-iW)b+(p+iW^\dagger)b^\dagger\right]</math> : <math>Q_2=\frac{i}{2}\left[(p-iW)b-(p+iW^\dagger)b^\dagger\right]</math> Note that <math>Q_1</math> and <math>Q_2</math> are self-adjoint. Let the [[Hamiltonian (quantum mechanics)|Hamiltonian]] : <math>H=\{Q_1,Q_1\}=\{Q_2,Q_2\}=\frac{(p+\Im\{W\})^2}{2}+\frac{{\Re\{W\}}^2}{2}+\frac{\Re\{W\}'}{2}(bb^\dagger-b^\dagger b)</math> where ''W''β² is the derivative of ''W''. Also note that {{mset|''Q''<sub>1</sub>, ''Q''<sub>2</sub>}} = 0. This is nothing other than ''N'' = 2 supersymmetry. Note that <math>\Im\{W\}</math> acts like an electromagnetic [[vector potential]]. Let's also call the spin down state "bosonic" and the spin up state "fermionic". This is only in analogy to quantum field theory and should not be taken literally. Then, ''Q<sub>1</sub>'' and ''Q<sub>2</sub>'' maps "bosonic" states into "fermionic" states and vice versa. Reformulating this a bit: Define : <math>Q=(p-iW)b</math> and, : <math>Q^\dagger=(p+iW^\dagger)b^\dagger</math> : <math>\{Q,Q\}=\{Q^\dagger,Q^\dagger\}=0</math> and : <math>\{Q^\dagger,Q\}=2H</math> An operator is "bosonic" if it maps "bosonic" states to "bosonic" states and "fermionic" states to "fermionic" states. An operator is "fermionic" if it maps "bosonic" states to "fermionic" states and vice versa. Any operator can be expressed uniquely as the sum of a bosonic operator and a fermionic operator. Define the [[supercommutator]] <nowiki>[,}</nowiki> as follows: Between two bosonic operators or a bosonic and a fermionic operator, it is none other than the commutator but between two fermionic operators, it is an [[anticommutator]]. Then, ''x'' and ''p'' are bosonic operators and ''b'', <math>b^\dagger</math>, ''Q'' and <math>Q^\dagger</math> are fermionic operators. Let's work in the [[Heisenberg picture]] where ''x'', ''b'' and <math>b^\dagger</math> are functions of time. Then, : <math>[Q,x\}=-ib</math> : <math>[Q,b\}=0</math> : <math>[Q,b^\dagger\}=\frac{dx}{dt}-i\Re\{W\}</math> : <math>[Q^\dagger,x\}=ib^\dagger</math> : <math>[Q^\dagger,b\}=\frac{dx}{dt}+i\Re\{W\}</math> : <math>[Q^\dagger,b^\dagger\}=0</math> This is nonlinear in general: ''i.e.'', x(t), b(t) and <math>b^\dagger(t)</math> do not form a linear SUSY representation because <math>\Re\{W\}</math> isn't necessarily linear in ''x''. To avoid this problem, define the self-adjoint operator <math>F=\Re\{W\}</math>. Then, : <math>[Q,x\}=-ib</math> : <math>[Q,b\}=0</math> : <math>[Q,b^\dagger\}=\frac{dx}{dt}-iF</math> : <math>[Q,F\}=-\frac{db}{dt}</math> : <math>[Q^\dagger,x\}=ib^\dagger</math> : <math>[Q^\dagger,b\}=\frac{dx}{dt}+iF</math> : <math>[Q^\dagger,b^\dagger\}=0</math> : <math>[Q^\dagger,F\}=\frac{db^\dagger}{dt}</math> and we see that we have a linear SUSY representation. Now let's introduce two "formal" quantities, <math>\theta</math>; and <math>\bar{\theta}</math> with the latter being the adjoint of the former such that : <math>\{\theta,\theta\}=\{\bar{\theta},\bar{\theta}\}=\{\bar{\theta},\theta\}=0</math> and both of them commute with bosonic operators but anticommute with fermionic ones. Next, we define a construct called a [[superfield]]: : <math>f(t,\bar{\theta},\theta)=x(t)-i\theta b(t)-i\bar{\theta}b^\dagger(t)+\bar{\theta}\theta F(t)</math> ''f'' is self-adjoint. Then, : <math>[Q,f\}=\frac{\partial}{\partial\theta}f-i\bar{\theta}\frac{\partial}{\partial t}f,</math> : <math>[Q^\dagger,f\}=\frac{\partial}{\partial \bar{\theta}}f-i\theta \frac{\partial}{\partial t}f.</math> Incidentally, there's also a U(1)<sub>R</sub> symmetry, with ''p'' and ''x'' and ''W'' having zero R-charges and <math>b^\dagger</math> having an R-charge of 1 and ''b'' having an R-charge of β1.
Edit summary
(Briefly describe your changes)
By publishing changes, you agree to the
Terms of Use
, and you irrevocably agree to release your contribution under the
CC BY-SA 4.0 License
and the
GFDL
. You agree that a hyperlink or URL is sufficient attribution under the Creative Commons license.
Cancel
Editing help
(opens in new window)