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 ?