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This may be a noob question but I've tried searching about this and haven't been able to put things into the context of what I've been studying.

(Dot means the usual derivative w.r.t. time)

If $c$ and $\bar{c}$ are independent anti-commuting variables, I want to confirm a few properties: first, will $\dot{c}$ and $\dot{\bar{c}}$ anti-commute with themselves ($\left\{\dot{c}, \dot{c}\right\}$ = $\left\{\dot{\bar{c}}, \dot{\bar{c}}\right\}$ = 0)?

Can we say that $c$ and $\dot{c}$ (or $\bar{c}$ and $\dot{\bar{c}}$) anti-commute?

And finally, can $\dot{c}$ and $\dot{\bar{c}}$ be considered as mutually anti-commuting variables? In fact, can they be considered anti-commuting variables by themselves?

Thanks in advance.

EDIT: Well at some point later in the text (which I'm referring to), the author "uses" $\left\{\dot{\bar{c}}, \dot{c}\right\} = 0$. So I tried to come up with an explanation (note that square brackets in the following do NOT represent commutators, but curly braces represent anti-commutators):

$\left\{\frac{d}{dt}(c + \bar{c}), \frac{d}{dt}(c + \bar{c})\right\} = 2[\frac{d}{dt}(c + \bar{c})][\frac{d}{dt}(c + \bar{c})]$

$= 2[\frac{dc}{dt}\frac{dc}{dt} + \left\{\frac{dc}{dt},\frac{d \bar{c}}{dt}\right\} + \frac{d \bar{c}}{dt}\frac{d \bar{c}}{dt}]$

$= \left\{\dot{c}, \dot{c}\right\} + 2\left\{\dot{c}, \dot{\bar{c}}\right\} + \left\{\dot{\bar{c}}, \dot{\bar{c}}\right\}.$

Now the result that I mentioned above (which was "used") makes sense to me only if the total derivatives anti-commute with themselves, BUT I'm not too sure about this.

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2 Answers 2

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The "theorem" you want is a definition--- anticommuting variables square to zero, so the anticommutator of a with itself is zero, since the anticommutator of a variable with itself is twice the square.

The other (easy to prove) theorem you should know is that any linear combination of anticommuting variables is again an anticommuting variable, so that it squares to zero and anticommutes with any other linear combination of the anticommuting variables.

The time derivative of an anticommuting variable which depends on time is a linear combination of the variables at adjacent times, you conclude that the time derivative is anticommuting variable, and it has all the ordinary properties of anticommuting variables.

What you should be aware of is that in quantum field theories, when you have an anticommuting field that is time dependent, $c(t)$ is an independent anticommuting variable at each time, so that the time derivative:

$$ {dc\over dt} = {c(t+\epsilon) - c(t)\over \epsilon} $$

is not in any ordinary sense convergent in $\epsilon$. As you make $\epsilon$ smaller, you get completely different anticommuting variables in the difference, the variable at each instant is independent of the variable at the previous instant. The way to understand these derivatives is to make a lattice, interpret the derivatives as lattice derivatives in the appropriate way, and then take the small lattice limit after you do the path integral to get the quantity you want out.

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Well, that provides some perspective. Looks like I need to study a bit of Grassman algebra. So far I was "working backwards" and getting properties of anti-commuting variables from results which used them, rather than vice-versa. Thanks... –  1989189198 Jul 13 '12 at 9:48
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One could ask the same question about Grassmann-even variables $x^i$ (if one replaces anticommutators with commutators):

  1. Is $x^i x^j - x^j x^i~=~0?$

  2. Is $\dot{x}^i \dot{x}^j -\dot{x}^j \dot{x}^i~=~0?$

  3. Is $x^i \dot{x}^j - \dot{x}^jx^i~=~0?$

Do they commute classically (meaning $\hbar=0$)? Yes.

Do they commute quantum mechanically? Well, (i) that depends on the dynamics, i.e., the Lagrangian (or equivalently, the Hamiltonian) of the system, and (ii) it depends on what exactly is meant by the time derivative $\dot{x}^i$ of the operator $x^i$ if regularization is needed.

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Well, in this case I'm talking about quantum-mechanical anti-commutation relations. I believe the time-derivative here is just an ordinary (non-Grassman) derivative. –  1989189198 Jul 12 '12 at 13:03
    
I've edited the question a bit... –  1989189198 Jul 12 '12 at 13:14
    
The answer to v2 is in principle still the same point (i) and (ii). –  Qmechanic Jul 12 '12 at 15:18
    
This doesn't answer the question really, and quantum mechanical matrix commutation is not what is asked about here. –  Ron Maimon Jul 13 '12 at 15:52
    
1. The way I read OP's question, I think my answer does answer it. 2. The answer isn't restricted to quantum mechanical matrix formulation. It could also be inside a path integral, or in some other mathematical setting. 3. When I answer a question, I won't always answer all aspects of a question, in particular if the question is not completely well-defined to begin with. –  Qmechanic Jul 13 '12 at 16:56
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