Grassmann variables and orthogonality of coherent fermionic states

Question

Let a coherent fermionic state $$ \left|\phi\right> := \left|0\right> + \left|1\right> \phi,\tag{0} $$ where $\phi$ is a Grassmann number (i.e. it anticommutes with other Grassmann numbers). Now, I wish to see if it's orthogonal to another state $\left|\phi'\right>$: $$ \left<\phi|\phi'\right> = \left[ \left<0\right| + \phi \left<1\right| \right] \left[ \left|0\right> + \left|1\right> \phi' \right] = 1 + \phi\phi' \equiv e^{\phi\phi'},\tag{1} $$ where I've used $\left<n|m \right>=\delta_{n,m}$ and the Taylor series for the exponential. Now, this should be a delta function, so \begin{align} \int d\phi\, e^{\phi\phi'} f(\phi) =& \int d\phi\, e^{\phi\phi'} (a + b\phi)\\ =& \int d\phi\, (1 + \phi\phi') (a + b\phi) \\ =& \int d\phi\, (a + b\phi + a\phi\phi')\\ =& b + a\phi' \neq f(\phi').\tag{2} \end{align} Why do I find that (1) does not satisfy the definition of a delta function?

Also, in my lecture notes I've seen that I should get $$ \left<\phi|\phi' \right> = \phi - \phi',\tag{3} $$ which turns out to be a nicely behaving delta function. But in some other reference (eq 28.16) I've seen that (1) is correct! How can I derive (3)?

Answer 1

1. Bosonic (Grassmann-even) and fermionic (Grassmann-odd) coherent states are overcomplete bases, that are not orthogonal.

2. One may show that the fermionic definition of coherent states$$ |\eta \rangle~:=~e^{\hat{c}^{\dagger}\eta}|0 \rangle~=~|0 \rangle+|1 \rangle\eta, \qquad |1 \rangle~:=~\hat{c}^{\dagger}|0 \rangle, \tag{0}$$ does satisfy the completeness relation $$\int_{\mathbb{C}^{0|1}} d\bar{\eta}~d\eta ~e^{-\bar{\eta}\eta} |\eta \rangle\langle \bar{\eta} |~=~\mathbb{1}.$$

3. If we introduce a picture-changing operator $$\hat{P}~:=~|1 \rangle\langle 0|-|0 \rangle\langle 1| ,$$ we can write $$ \langle \bar{\eta} |\hat{P}|\eta \rangle~=~\bar{\eta}-\eta~=~\delta(\bar{\eta}\!-\!\eta), $$ OP's eq. (3).

References:

1. A. Altland & B. Simons, Condensed matter field theory, 2nd ed., 2010; p. 160-164.

2. T. Lancaster & S.J. Blundell, QFT for the Gifted Amateur, 2014; section 28.2.

Answer 2

The reason that some lecture notes say that the inner product of those two states is $1 + \phi \phi'$ and others say it is $\phi - \phi'$ is because they are using different conventions. From memory, I think two results are just the difference between using the definition $|\phi \rangle = |0 \rangle + \phi | 1 \rangle$ and $|\phi \rangle = |1 \rangle + \phi | 0 \rangle$, where here $|0 \rangle$ and $|1 \rangle$ are of course represented by $1$ and $\theta$, respectively, where $\theta$ is a dummy Grassmann number. (Or it might come from defining the bras and kets differently.) In any case, with either convention, it all comes out in the wash at the end of the day.