Simple question about anticommutator and spinors

[CGH]

Hi there,

i have a very simple question, but still, i don't know what the answer is, her it goes.

I havew Dirac spinor $\psi$ and its hermitian timex $\gamma^0$, $\bar \psi$.

My question is the following:
we can think of $\psi$ as a vector and $\bar \psi$ as a row vector, then, if i take

$$[\psi,\bar \psi]_+=\psi\bar \psi+\bar\psi \psi$$

the first term is a matrix, and the second one is a number! What did i do wrong?

I tried writting

$\psi=\int (\tex{something})(b u e^{-ipx}+d^\dagger v e^{ipx})$

in that case, the first term of the anticommutator gives something like $u\bar u$ (a matrix) and the second $\bar u u$ (a number). The problem is still there, my question is: what did i do wrong?

 

[SpectraCat]

Well first off, anti-commutators are usually defined for operators, not states or vectors. Since operators can be represented as matrices, the commutator or anti-commutator is always a sum of two matrix products.

What are you trying to derive the anti-commutator of exactly?

 

[CGH]

Well first off, anti-commutators are usually defined for operators, not states or vectors. Since operators can be represented as matrices, the commutator or anti-commutator is always a sum of two matrix products.

I though about that, but i wasn0t sure, i needed a fourth opinion (me, myself and i already agreed).

What are you trying to derive the anti-commutator of exactly?

Nothing special, I'm just computing anything that comes in my path,

Saludos!

 

[The_Duck]

The Dirac field $$\psi$$ is really a collection of 4 operators; to remember this you can include an index $$\alpha$$ that runs from 1 to 4 and write the Dirac field as $$\psi_\alpha$$. So the commutator of two Dirac fields, $$[\psi_\alpha, \psi_\beta]$$ is a collection of 16 operators, labeled by $$\alpha$$ and $$\beta$$, specifying the results of commuting any of the 16 pairs of operators. In this instance I think it's best to stop thinking of $$\psi$$ as a column vector and $$\overline{\psi}$$ as a row vector and just use indices; then there is no trouble interpreting an expression like $$[\psi_\alpha, \overline{\psi}_\beta]$$

 

[samalkhaiat]

Hi there,

i have a very simple question, but still, i don't know what the answer is, her it goes.

I havew Dirac spinor $\psi$ and its hermitian timex $\gamma^0$, $\bar \psi$.

My question is the following:
we can think of $\psi$ as a vector and $\bar \psi$ as a row vector, then, if i take

$$[\psi,\bar \psi]_+=\psi\bar \psi+\bar\psi \psi$$

the first term is a matrix, and the second one is a number! What did i do wrong?

when you treat 4-spinors as field operators, the anticommutation relations between them become 4 by 4 matrices:

$$\{\psi_i (x,t),\bar{\psi}_j(y,t)\}=\gamma^{0}_{ij}\delta^{3}(x-y)$$

 

[CGH]

In this instance I think it's best to stop thinking of $$\psi$$ as a column vector and $$\overline{\psi}$$ as a row vector and just use indices; then there is no trouble interpreting an expression like $$[\psi_\alpha, \overline{\psi}_\beta]$$

Good advise, many thanks,

Saludos!

 

[ytuab]

I think it's one of the limits in representation by Dirac equation.

$$u(p)$$ is 4 x 1 matirix (column), and $$\bar{u}(p)$$ is 1 x 4 matrix (row).
To be presice, the following things are different, considering matrices ?

$$\bar{u}(p) u(p) = 1 \neq u(p)\bar{u}(p)$$

Because the former is not matrix, and the latter is 4 x 4 matrix.
But if we suppose both are the same,

$$\bar{u}(p) u(p) = 1 = u(p)\bar{u}(p)$$

multiplying by another u(p) from right,

$$\bar{u}(p) u(p) u(p)= 1 \times u(p), \quad u(p)\bar{u}(p) u(p) = u(p) \times 1$$

Both are the same.
So without another u(p), both are different things ? (Because they are "matrices", not usual numbers)
Is it perimissible ?