Eigenvalues of Fermionic field operator

[Neutrinos02]

Hello,

I'm a bit confused about the eigenvalues of the second quantized fermionic field operators $\psi(x)_a$. Since these operators satisfy the condition $\{\psi(x)_a, \psi(y)_b\} = 0$ the eigenvalues should also anti-commute? Does this mean that the eigenvalues of $\psi(x)_a$ are Grassmann-numbers?

Thanks for your help

 

[DrDu]

The Pauli matrices $\sigma_x$ and $\sigma_y$ also anti-commute. Would you also conclude that they have Grassmann valued eigenvalues?

 

[Neutrinos02]

No, but the eigenvalues are not matrices? And in the path integral formalism we use Grassmann valued fields.

 

[Demystifier]

The Pauli matrices $\sigma_x$ and $\sigma_y$ also anti-commute. Would you also conclude that they have Grassmann valued eigenvalues?

There is one important difference:
$$ \psi_a(x)\psi_a(x)=0$$
but
$$\sigma_x\sigma_x \neq 0$$
A fermionic field operator cannot be diagonalized and it does not have eigenvalues. A Pauli matrix can be diagonalized and it does have eigenvalues. Fermionic field operator is not hermitian, Pauli matrix is hermitian.

 

[Neutrinos02]

But isn't there a self-adjoint extension with generalized eigenstates like for the QM operators $x,p$?

 

[Demystifier]

But isn't there a self-adjoint extension with generalized eigenstates like for the QM operators $x,p$?

No.

 

[DrDu]

I just wanted to give a counter example of anti-commuting operators which have eigenvalues. Even more field operator like are the matrices $\sigma_\pm=\sigma_x\pm i\sigma_y$. They anticommute and are non-hermitian. They are not diagonisable.

 

[Demystifier]

I just wanted to give a counter example of anti-commuting operators which have eigenvalues. Even more field operator like are the matrices $\sigma_\pm=\sigma_x\pm i\sigma_y$. They anticommute and are non-hermitian. They are not diagonisable.

Yes, $\sigma_\pm$ is a much better analog of fermionic field. $\sigma_\pm$ does not have (non-zero) eigenvalues.

 

[stevendaryl]

In some developments of the quantum mechanics of fermions (an example is "Quantum Field Theory of Point Particles and Strings" by Brian Hatfield), Feynman's path integral is extended to include integration over non-commuting Grassman variables. I personally find it hard to understand, because I'm used to thinking of noncommutation as being a property of operators, not variables.

 

[Demystifier]

Feynman's path integral is extended to include integration over non-commuting Grassman variables.

A long time ago I have tried to overcome this
https://arxiv.org/abs/hep-th/0210307
but a smart referee found an error for the case of many fermionic degrees, which I couldn't repair. (If someone is interested, Eq. (21), which would be needed in a consistent approach, is not really satisfied by my construction. A more complicated construction is needed to satisfy (21), but then the formalism is too complicated to be useful.)

But in the case of one fermionic degree of freedom my approach is perfectly fine. The final result in this case is the functional integral (14). It's nice from a conceptual point of view, but probably not too useful.

 

[Neutrinos02]

So the only possibility to get hermitian operators are the number-operators? But could the Lagrangian (it should also be hermitian) be rewriten in terms of number-operators?

 

[DrDu]

So the only possibility to get hermitian operators are the number-operators? But could the Lagrangian (it should also be hermitian) be rewriten in terms of number-operators?

No, you can also express your Hamiltonian in terms of e.g. $\psi+\psi^\dagger$ and $i(\psi-\psi^\dagger)$ which are hermitian.

 

[Neutrinos02]

No, you can also express your Hamiltonian in terms of e.g. $\psi+\psi^\dagger$ and $i(\psi-\psi^\dagger)$ which are hermitian.

Thanks. So the eigenvalues of this operators are real?

But shouldn't the hermitian operators $\psi + \psi^\dagger$ and $i(\psi - \psi^\dagger)$ you gave, satisfy the commutation relations if and only if $\psi, \psi^\dagger$ satisfy the anti-commutation relations?

 

[Demystifier]

But shouldn't the hermitian operators $\psi + \psi^\dagger$ and $i(\psi - \psi^\dagger)$ you gave, satisfy the commutation relations if and only if $\psi, \psi^\dagger$ satisfy the anti-commutation relations?

No. See Eq. (4) in the paper I linked above.