Simple question about anticommutator and spinors

  • Thread starter CGH
  • Start date
  • Tags
    Spinors
In summary: Yes, because the result is not a number.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)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]
  • #1
CGH
7
0
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 [itex]\psi[/itex] and its hermitian timex [itex]\gamma^0[/itex], [itex]\bar \psi[/itex].

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

[tex][\psi,\bar \psi]_+=\psi\bar \psi+\bar\psi \psi[/tex]

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

I tried writting

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

in that case, the first term of the anticommutator gives something like [itex]u\bar u[/itex] (a matrix) and the second [itex]\bar u u[/itex] (a number). The problem is still there, my question is: what did i do wrong?
 
Physics news on Phys.org
  • #2
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?
 
  • #3
SpectraCat said:
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).

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

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

Saludos!
 
  • #4
The Dirac field [tex]\psi[/tex] is really a collection of 4 operators; to remember this you can include an index [tex]\alpha[/tex] that runs from 1 to 4 and write the Dirac field as [tex]\psi_\alpha[/tex]. So the commutator of two Dirac fields, [tex][\psi_\alpha, \psi_\beta][/tex] is a collection of 16 operators, labeled by [tex]\alpha[/tex] and [tex]\beta[/tex], specifying the results of commuting any of the 16 pairs of operators. In this instance I think it's best to stop thinking of [tex]\psi[/tex] as a column vector and [tex]\overline{\psi}[/tex] as a row vector and just use indices; then there is no trouble interpreting an expression like [tex][\psi_\alpha, \overline{\psi}_\beta][/tex]
 
  • #5
CGH said:
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 [itex]\psi[/itex] and its hermitian timex [itex]\gamma^0[/itex], [itex]\bar \psi[/itex].

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

[tex][\psi,\bar \psi]_+=\psi\bar \psi+\bar\psi \psi[/tex]

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:

[tex]\{\psi_i (x,t),\bar{\psi}_j(y,t)\}=\gamma^{0}_{ij}\delta^{3}(x-y)[/tex]
 
  • #6
The_Duck said:
In this instance I think it's best to stop thinking of [tex]\psi[/tex] as a column vector and [tex]\overline{\psi}[/tex] as a row vector and just use indices; then there is no trouble interpreting an expression like [tex][\psi_\alpha, \overline{\psi}_\beta][/tex]

Good advise, many thanks,

Saludos!
 
  • #7
I think it's one of the limits in representation by Dirac equation.

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

[tex] \bar{u}(p) u(p) = 1 \neq u(p)\bar{u}(p)[/tex]

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

[tex] \bar{u}(p) u(p) = 1 = u(p)\bar{u}(p) [/tex]

multiplying by another u(p) from right,

[tex] \bar{u}(p) u(p) u(p)= 1 \times u(p), \quad u(p)\bar{u}(p) u(p) = u(p) \times 1 [/tex]

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

1. What is the anticommutator of two spinors?

The anticommutator of two spinors is a mathematical operation that combines two spinors to produce a scalar quantity. It is defined as the sum of the products of the two spinors, with the order of the spinors reversed and a negative sign in front. In other words, the anticommutator of two spinors, A and B, is given by {A, B} = AB + BA.

2. How is the anticommutator related to the commutator?

The anticommutator and commutator are both mathematical operations that involve two quantities. However, they differ in their order of operations. The commutator is defined as the difference between the products of the two quantities in their original order, while the anticommutator involves reversing the order of the quantities. In addition, the commutator results in a vector or tensor quantity, while the anticommutator produces a scalar quantity.

3. What is the significance of the anticommutator in quantum mechanics?

In quantum mechanics, the anticommutator plays a crucial role in the formulation of the fermionic creation and annihilation operators. These operators are used to describe the behavior of fermions, such as electrons, in quantum systems. The anticommutator of these operators is used to calculate the probabilities of different quantum states and to determine the properties of fermionic systems.

4. Can the anticommutator be used to calculate the expectation value of a quantity?

Yes, the anticommutator can be used to calculate the expectation value of a quantity. The expectation value is a measure of the average value of a physical quantity in a given quantum state. The anticommutator can be used in the derivation of the general expression for the expectation value of an operator in quantum mechanics.

5. Are there any general properties of the anticommutator?

Yes, there are several general properties of the anticommutator. These include the fact that it is anti-symmetric, meaning that switching the order of the spinors results in a negative sign, and that it satisfies the Jacobi identity, which states that the anticommutator of three quantities is equal to the sum of the anticommutators of their pairwise combinations. It also has implications for the uncertainty principle in quantum mechanics, as the anticommutator of two operators is related to their commutator and the expectation value of their product.

Similar threads

Replies
1
Views
591
Replies
2
Views
861
  • Quantum Physics
Replies
1
Views
1K
  • Quantum Physics
Replies
6
Views
959
  • Quantum Physics
Replies
5
Views
2K
Replies
16
Views
2K
Replies
1
Views
1K
  • Differential Geometry
Replies
1
Views
1K
  • Quantum Physics
Replies
6
Views
750
Replies
1
Views
1K
Back
Top