Recovering Fermion States in New Formalism?

  • #1

Main Question or Discussion Point

Hi, I just started a book on QFT and one of the first things that was done was switch from labeling states with their individual particles and instead label states by the number of particles in each momentum eigenstate.

In addition, some "algebras" (not sure if they qualify by the mathematical definition) were defined for creation and annihilation operators as follows:

For creation and annihilation operators for states with definite momenta p and q (working in 3 dimensions), it was defined that $$[a_p,a_q^{\dagger}]=\delta^{(3)}(p-q);\quad [a_p,a_q]=0;\quad [a_p^{\dagger},a_q^{\dagger}]=0$$ for bosons and $$\{a_p,a^{\dagger}_q\}=\delta^{(3)}(p-q);\quad \{a_p,a_q\}=0;\quad \{a_p^{\dagger},a_q^{\dagger}\}=0$$ for fermions, where the square brackets and curly braces denote commutator and anti-commutator, respectively.

The book proceeds to (reverting back to the old notation of denoting each particle in a state individually, I suppose) state that $$\langle p'q'|qp\rangle = \langle 0|a_{p'}a_{q'}a_q^{\dagger}a_p^{\dagger}|0\rangle=\delta^{(3)}(p'-p)\delta^{(3)}(q'-q)\pm\delta^{(3)}(p'-q)\delta^{(3)}(q'-p).$$ On the RHS, the positive sign is taken for bosons and the negative sign for fermions.

I believe the first of these equalities, but for some reason I keep getting a plus instead of a plus/minus on the RHS. Could anyone explain what is wrong with my derivation?

For ease of typing, I let p' = w, q' = x, q = y, p = z and I use a's and b's (with the appropriate subscripts) for the creation and annihilation operators, respectively.

Then we wish to compute $$\langle 0|a_wa_xb_yb_z|0\rangle.$$ By simply applying the commutator and anti-commutator relations, we find $$a_wa_xb_y=a_w(b_ya_x\pm\delta^{(3)}(x-y))=a_wb_ya_x\pm\delta^{(3)}(x-y)a_w.$$ Then $$a_wa_xb_yb_z=a_wb_ya_xb_z\pm\delta^{(3)}(x-y)a_wb_z=a_wb_y(b_za_x\pm\delta^{(3)}(x-z))\pm\delta^{(3)}(x-y)(b_za_w\pm\delta^{(3)}(w-z)).$$

Since we want to compute the result when this expression is sandwiched in between the vacuum state, we can ignore the terms that have an annihilation as the rightmost operator (since those will hit the vacuum state to produce zero). This leaves us with $$\pm\delta^{(3)}(x-z)a_wb_y+\delta^{(3)}(x-y)\delta^{(3)}(w-z)=\pm\delta^{(3)}(x-z)(b_ya_w\pm\delta^{(3)}(w-y))+\delta^{(3)}(x-y)\delta^{(3)}(w-z).$$

Once again, we may remove the terms that will annihilate the vacuum state, leaving us with $$\delta^{(3)}(x-z)\delta^{(3)}(w-y)+\delta^{(3)}(x-y)\delta^{(3)}(w-z).$$ Exactly what I wanted, but with + instead of +/-. What's wrong with my logic?

Thanks in advance!
 

Answers and Replies

  • #2
stevendaryl
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By simply applying the commutator and anti-commutator relations, we find $$a_wa_xb_y=a_w(b_ya_x\pm\delta^{(3)}(x-y))=a_wb_ya_x\pm\delta^{(3)}(x-y)a_w.$$
You have the ##\pm## on the wrong term:

$$a_wa_xb_y=a_w(\pm b_ya_x + \delta^{(3)}(x-y))$$
 
  • #3
You have the ##\pm## on the wrong term:

$$a_wa_xb_y=a_w(\pm b_ya_x + \delta^{(3)}(x-y))$$
OMG. Whoops. Haha, thank you, I spent the last four hours trying to figure this thing out!
 

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