Eigenvalues of fermionic field operators - Physics Stack Exchange most recent 30 from physics.stackexchange.com 2019-10-22T21:21:49Z https://physics.stackexchange.com/feeds/question/314170 https://creativecommons.org/licenses/by-sa/4.0/rdf https://physics.stackexchange.com/q/314170 1 Eigenvalues of fermionic field operators StrangeField https://physics.stackexchange.com/users/142765 2017-02-23T08:53:39Z 2017-02-23T12:22:40Z <p>Consider the fermionic field operators $\psi_a(x), \psi^{\dagger}_b(y)$ with the canonical anti-commutation relations $$\{\psi_a(x),\psi_b(y)\} = 0$$ and $$\{\psi^{\dagger}_b(t,\vec{x}),\psi_a(t,\vec{y})\} = \delta_{ab} \delta(\vec{x}-\vec{y}).$$ </p> <p>What can we say about their eigenvalues? Are they real or Grassmann-numbers? </p> <p>I'm a bit confused about this at first I would guess they are Grassmann-numbers since $$\psi_a(x)\psi_a(x) = -\psi_a(x)\psi_a(x) = 0$$ but I'm not sure if this conclusion is true. </p> https://physics.stackexchange.com/questions/314170/-/314193#314193 2 Answer by Qmechanic for Eigenvalues of fermionic field operators Qmechanic https://physics.stackexchange.com/users/2451 2017-02-23T12:21:18Z 2017-02-23T12:22:40Z <ol> <li><p>An eigenvalue $\lambda$ of an operator $\hat{A}$ (with definite Grassmann-parity $|\hat{A}|$) is a complex <a href="http://planetmath.org/supernumber" rel="nofollow noreferrer">supernumber</a> of the same Grassmann-parity. See also <a href="https://physics.stackexchange.com/q/40746/2451">this</a> Phys.SE post and links therein.</p></li> <li><p>Note however that an annihilation operator $\hat{a}$ and a creation operator $\hat{a}^{\dagger}$ of definite Grassmann-parity $|\hat{a}|$ do not <a href="http://ncatlab.org/nlab/show/graded+commutator" rel="nofollow noreferrer">supercommute</a> $$[\hat{a}, \hat{a}^{\dagger}]~:=~\hat{a}\hat{a}^{\dagger}-(-1)^{|\hat{a}|}\hat{a}^{\dagger} \hat{a}~=~ \hbar~\hat{\bf 1}~\neq~0,$$ and are therefore not <a href="http://en.wikipedia.org/wiki/Normal_operator" rel="nofollow noreferrer">supernormal</a>, and hence not diagonalizable, cf. e.g. <a href="https://physics.stackexchange.com/q/82746/2451">this</a> Phys.SE post. OP's fermionic fields are field-theoretic versions of Grassmann-odd annihilation &amp; creation operators.</p></li> </ol>