Anticommutation Example: Solving for Operator p in Relation to Variable q"

[jostpuur]

If I have a variable $$q$$, and think of it as an operator, and want to postulate another operator $$p$$ so that it satisfies a commutation relation $$[q,p]=i\hbar$$, a definition $$p=-i\hbar\partial_q$$ will do this. If I instead wanted an anticommutation relation $$\{q,p\}=i\hbar$$, what would be an example of such $$p$$?

 

[jostpuur]

I just realized that $$p=i\hbar/(2q)$$ would do the job, but it looks horrible. Nobody wants operator like that...

 

[TriTertButoxy]

In elementary quantum mechancs, I believe $\hat{q}=q\times$ and $\hat{p}=-i\hbar\frac{d}{dq}$ is merely a representation (the position representation) of the coordinate and momentum operators that satisfies the commutation relations. There's also the momentum representation which defines the coordinate and momentum differently, but still satisfies the commutation relations. The physics should only depend on the commutation relations; explicit representations provide a way to make calculations and give us an intuitive picture of what's going on.

That didn't answer your question, but helps you formalize your request: you are looking for an explicit representation for $\hat{q}$ and $\hat{p}$ that satisfies anticommutation relations.

 

[Fra]

An illustrative exercise to elaborate on your own is a connection between a bosonic field and a fermionic field, by transforming the klein-gordon equation into a fermionic equation (dirac equation), by trading away higher order dynamics. Try to interpret the dirac equation in terms of klein gordon.

Then you can identify the pauli and gamma matrices that are anticommuting.

/Fredrik

 

[jostpuur]

Another related question: Have you ever seen Klein-Gordon field being quantized so that the Fourier modes of the conjugate field $$\Pi(\boldsymbol{p})$$ get replaced with derivative operators $$-i\hbar(2\pi\hbar)^3\delta^3(0)\partial_{\phi(\boldsymbol{p})}$$? (The strange constant isn't necessarely correct, but that's what I got.) That derivative operator is supposed to operate on the wave functional of the field. I have seen this... so far only in my own notes.

Then you can identify the pauli and gamma matrices that are anticommuting.

Anticommutations of the gamma matrices are not what I was after. I'm struggling with the anticommutations of the dirac field operators $$\psi$$ and $$\psi^\dagger$$ themselves.

I understand that it is possible to have a theory of operators without explicit representations available for them, but these representations would be nice to have.

 

[olgranpappy]

but these representations would be nice to have.

for a single fermionic oscillator the creation and annihilation operators can be written down explicitly as two by two matrices... this is much neater than the matrices of the analogous bosonic oscillator operators which are infinite dimensional...tho also pretty easily written down...

For the fermionic operators we have:
$$a_{11}=a_{21}=a_{22}=0\quad a_{12}=1$$
and
$$a^{\dagger}_{11}=a^{\dagger}_{12}=a^{\dagger}_{22}=0\quad a^{\dagger}_{21}=1\;.$$
these operators satisfy
$$\left\{a,a^{\dagger}\right\}=1$$

 

[jostpuur]

for a single fermionic oscillator the creation and annihilation operators can be written down explicitly as two by two matrices... this is much neater than the matrices of the analogous bosonic oscillator operators which are infinite dimensional...tho also pretty easily written down...

For the fermionic operators we have:
$$a_{11}=a_{21}=a_{22}=0\quad a_{12}=1$$
and
$$a^{\dagger}_{11}=a^{\dagger}_{12}=a^{\dagger}_{22}=0\quad a^{\dagger}_{21}=1\;.$$
these operators satisfy
$$\left\{a,a^{\dagger}\right\}=1$$

This seems to fine for a one dimensional system, but if we want this kind of operators $$a_p$$ and $$a_p^\dagger$$ for each fixed $$p\in\mathbb{R}^3$$, as is the case for fields, and also want to have $$a_p a_{p'}= -a_{p'} a_p$$ (and the same for conjugates) for $$p\neq p'$$, then explicit representation seems to be more difficult again?

In bosonic system large number of dimensions doesn't bring any difficulty because partial derivative operators commute still.