Understanding the Dirac Commutation Relations in QFT

[Silviu]

Hello! I am reading Peskin's book on QFT and at a point he wants to show that the Dirac field can't be quantified using this commutation relations: $[\psi_a(x),\psi_b^\dagger(x)]=\delta^3(x-y)\delta_{ab}$ (where $\psi$ is the solution to Dirac equation). I am not sure I understand the math behind the commutation relation (I understand why physically it is wrong) as you have a column and a raw vector, so doing the commutation you have the difference between a number and a 4x4 matrix and I am not sure how does this work. Can someone explain it to me? Thank you!

 

[ChrisVer]

Just to solve your misconception about "physically wrong", what you have on the left-hand-side is again a 4x4 matrix (has indices a,b running from 1 to 4), and similarily for the right-hand-side.

 

[Silviu]

The in

Just to solve your misconception about "physically wrong", what you have on the left-hand-side is again a 4x4 matrix (has indices a,b running from 1 to 4), and similarily for the right-hand-side.

The indeces a and b are fixed, they don't run from 1 to 4. They just specify the spin state.

 

[ChrisVer]

The indeces a and b are fixed, they don't run from 1 to 4. They just specify the spin state.

what do you mean by fixed? the equation has 16 fixed numbers in the left-hand-side (like a 4x4 matrix) and 16 fixed numbers in the right hand side (again like a 4x4 matrix).
The indices indicate one of the Dirac 4-spinor components:
\psi = \begin{pmatrix} \psi_1 \\ \psi_2 \\ \psi_3 \\ \psi_4 \end{pmatrix}
Why would you have vectors then for \psi_a ?