Grassmann number: a more rigorous treatment?

[taishizhiqiu]

TL;DR Summary

Most textbooks on fermionic path integral only briefly introduce Grassmann numbers. I would want to understand Grassmann numbers a bit more rigorously, especially in the context of quantum mechanics.

Most textbooks on fermionic path integral only briefly introduce Grassmann numbers. However, I want a more systematic treatment to feel comfortable about this approach. For illustration, I have several examples here.

Example 1:

Consider a system with only one state, how to calculate $\langle 0 | \eta | 0 \rangle$? Here $| 0 \rangle$ is the empty state and $\eta$ is a Grassmann number. My understanding is that $\langle 0| \eta | 0 \rangle$ needs to be defined as $\eta$, and this is not trivial. For example, with the above definition, $\langle 1| \eta | 1 \rangle = \langle 0| a \eta | 1 \rangle = -\langle 0| \eta a | 1 \rangle = -\langle 0| \eta | 0 \rangle = -\eta$. Edit: $a$ is the annihilation operator of the state: $a|1\rangle=|0\rangle$, $\langle 0 | a = \langle 1 |$.

Example 2:
To avoid confusion, I will try to avoid use Dirac symbol and denote inner product as $(\phi, \psi)$. For common quantum mechanics, we have $(a \phi, b \psi) = a^* b (\phi, \psi)$. However, what is $(\eta \phi, \nu\psi)$ ($\eta$ and $\nu$ are Grassmann numbers)? Since the complex conjugate of Grassmann numbers are not well defined (see footnote 4 on page 163 of "Condensed Matter Field Theory" second edition), I have no idea how to calculate $(\eta \phi, \nu\psi)$.

As I said, is there a book on a more systematic treatment of Grassmann numbers? It would also be very helpful if anyone can address the above specific questions.

 

[PeterDonis]

Consider a system with only one state, how to calculate $\langle 0 | \eta | 0 \rangle$? Here $| 0 \rangle$ is the empty state and $\eta$ is a Grassmann number.

A Grassmann number is not an operator, it's a number, so I would think you should be able to just pull it out of the bracket: $\bra{0} \eta \ket{0} = \eta \braket{0 | 0}$. If states are normalized in the usual way, then $\braket{0 | 0} = 1$, so $\bra{0} \eta \ket{0} = \eta$.

with the above definition, $\langle 1| \eta | 1 \rangle = \langle 0| a \eta | 1 \rangle = -\langle 0| \eta a | 1 \rangle = -\langle 0| \eta | 0 \rangle = -\eta$.

Where does the $a$ come from? Is it supposed to be an operator? If so, I don't think it can be multiplied with a Grassmann number.

To evaluate $\bra{1} \eta \ket{1}$, It seems to me you would just pull $\eta$ out of the bracket, as was done above, so $\bra{1} \eta \ket{1} = \eta \braket{1 | 1} = \eta$.

If you want to evaluate $\bra{0} a \eta \ket{1}$, that would be $\eta \bra{0} a \ket{1}$. Then you just evaluate $\bra{0} a \ket{1}$ normally.

 

[taishizhiqiu]

Where does the $a$ come from? Is it supposed to be an operator? If so, I don't think it can be multiplied with a Grassmann number.

Thanks for the reply.

$a$ is the annihilation operator of the state: $a|1\rangle=|0\rangle$, $\langle 0 | a = \langle 1 |$ (I have already edited the original post for clarification).

It is important to notice that Grassmann numbers are supposed to anticommute with annihilation and creation operators, which is why $\langle 1| \eta | 1 \rangle = \langle 0| a \eta | 1 \rangle = -\langle 0| \eta a | 1 \rangle = -\langle 0| \eta | 0 \rangle = -\eta$. So $\langle 0| \eta | 0 \rangle = \eta$ is only compatible with $\langle 1| \eta | 1 \rangle = -\eta$.

 

[Demystifier]

see footnote 4 on page 163 of "Condensed Matter Field Theory" second edition

When you refer to a book, always say the the second name of the author.

 

[Demystifier]

It is important to notice that Grassmann numbers are supposed to anticommute with annihilation and creation operators, which is why $\langle 1| \eta | 1 \rangle = \langle 0| a \eta | 1 \rangle = -\langle 0| \eta a | 1 \rangle = -\langle 0| \eta | 0 \rangle = -\eta$. So $\langle 0| \eta | 0 \rangle = \eta$ is only compatible with $\langle 1| \eta | 1 \rangle = -\eta$.

This in fact shows that Grassmann number cannot anticommute with creation and annihilation operators.

In fact, that's easy to understand from analogy with standard QM in terms of c-numbers. In the Schrodinger representation, the position $x$ is an ordinary commutative number, but $x$ and the operator $a$ (represented as a linear combination of $x$ and the derivative operator $p$) don't commute. Similarly, for fermions, $a$ is a linear combination of $\eta$ and the derivative with respect to $\eta$, so $a$ and $\eta$ don't anticommute.

 

[PeterDonis]

It is important to notice that Grassmann numbers are supposed to anticommute with annihilation and creation operators

Why do you think so?

 

[PeterDonis]

is there a book on a more systematic treatment of Grassmann numbers?

This might be helpful:

https://irfu.cea.fr/Phocea/file.php?class=page&file=678/QFT-IRFU9.pdf

Particularly the part starting at section 10.7, which develops the use of Grassman numbers in the context of fermions in QM.

 

[protonsarecool]

Why do you think so?

It's in the book he is citing, Altland & Simons "Condensed Matter Field Theory". On p. 160, they start by defining a fermionic coherent state by requiring that it is an eigenstate of the annihilation operators, just as in the bosonic case.
$$ a_i \ket{\eta} = \eta_i \ket{\eta} $$
Then they quickly notice that these "eigenvalues" will have to anti-commute, so they are part of a Grassmann algebra and not real numbers. $\eta_i \eta_j = - \eta_j \eta_i$. Some more historic stuff and details about Grassmann "derivatives and integrals" follow. Then on p. 163 they get to the points OP is making. They now define the fermionic coherent states as
$$ \ket{\eta} = \exp \left( -\sum_i \eta_i a_i^\dagger \right) \ket{0} $$
Right before that definition, they remark:

With this background, let us now proceed to apply the Grassmann algebra to the con-
struction of fermion coherent states. To this end we need to enlarge the algebra so as to
allow for a multiplication of Grassmann numbers by fermion operators. In order to be con-
sistent with the anti-commutation relations, we need to require that fermion operators and
Grassmann generators anti-commute,

$$[\eta_i, a_j]_+ = 0. $$Now to OP: Note that this anti-commutation relation is only given for different(!) i and j. In your example, you have only 1 "particle type" or lattice site or whatever. So the only different Grassmann numbers are $\eta$ and $\bar \eta$, there are no more different Grassmann numbers in this model. So the only thing you can say is that $\bar \eta$ and $a$ anti-commute, and they use this to write
$$ \bra{\eta} = \bra{0} \exp \left( -\sum_i a_i \bar \eta_i \right) \\
= \bra{0} \exp \left( \sum_i \bar \eta_i a_i \right). $$
But this is because $\eta$ and $\bar \eta$ are formally unrelated elements of the algebra. No complex conjugation or Hermitian conjugate or anything is going on, just two unrelated Grassmann numbers. As far as the commuting or anti-commuting of $a_i$ and $\eta_i$ is concerned: They make no statement about that, they don't need it for the construction of the coherent state, and as you saw with your own example, they have to commute in fact.Edit: Actually, I was wrong, they do refer to $$ a_i \eta_i = - \eta_i a_i $$ in footnote 3 on p. 163, trying to proof in one line that the coherent state is in fact a coherent state. However, I think there is a mistake in their one-line proof anyway. I'll first quote their version in full:
$a_i \ket{\eta} = a_i \exp(-\eta_i a_i^\dagger) = a_i (1 - \eta_i a_i^\dagger) \ket{0} = \eta_i a_i a_i^\dagger \ket{0} = \eta_i \ket{0} = \eta_i (1 - \eta_i a_i^\dagger) \ket{0} = \eta_i \exp(- \eta_i a_i^\dagger) \ket{0} = \eta_i \ket{\eta}$
As you can see, in the third equality, they do commute $$ \eta_i $$ and $$ a_i $$, but actually after that their proof makes no sense anymore. They then get just the ground state, but then somehow introduce the exponential again (like $$ \eta_i $$ would actually work as an annihilation operator instead of just an eigenvalue). I'll have to double-check with other resources because I'm not convinced yet how to fix their proof idea.

 

[vanhees71]

This might be helpful:

https://irfu.cea.fr/Phocea/file.php?class=page&file=678/QFT-IRFU9.pdf

Particularly the part starting at section 10.7, which develops the use of Grassman numbers in the context of fermions in QM.

This is an extraordinarily clear exposition. Can you quote the link to the rest of these lectures and particularly who is the author?

[EDIT:] After some googling I figured it out. It's a lecture by Zinn-Justin (no wonder that's it's an extraordinarily clear exposition ;-)):

https://irfu.cea.fr/Phocea/Page/index.php?id=678

 

[protonsarecool]

I've thought about my post #8 some more, and Altland&Simons treatment of fermionic coherent states is consistent with other textbooks, and the annihilation operators do have to anti-commute with the Grassmann eigenvalues. See for example Eduardo Fradkin "Field Theories of Condensed Matter Physics", p. 44. Essentially, since the coherent state is defined as
$\ket{\eta} = \exp (-\eta a^\dagger) \ket{0} = \ket{0} - \eta \ket{1},$
the "action" (although its not an operator) of \eta on this is
$\eta \ket{\eta} = \eta \ket{0} - \eta^2 \ket{1} = \eta \ket{0}$.
And then a has to anticommute with eta to make this consistent with the action of a:
$a \ket{\eta} = a \ket{0} - a \eta \ket{1} = - a \eta \ket{1} = + \eta a \ket{1} = \eta \ket{0}$

Coming back to OPs question: If anything, this just shows, if you can't just treat eta like an operator in the matrix element <1|eta|1>. You should just consistently write eta outside of the overlap, or you treat it like a full blown operator, but then you can never pull it outside of an overlap, and your last equality in your example is also invalid. Going the route of just treating it as something to be pulled out of overlaps is consistent though. Notice that $\eta \ket{1} = \ket{0} - \ket{\eta},$
so
$\bra{1} \eta \ket{1} = \bra{1} \ket{0} - \bra{1} \ket{\eta} = - (- \eta) \bra{1} \ket{1} = \eta$.

Edit: Regarding the second question on post #1:
$(\eta \phi, \nu \psi) = \bar \eta \nu (\phi, \psi)$.
Don't be too confused by Altlands remark that eta and eta-bar are not related by conjugation, it is just a note of caution that the viewpoint get's problematic in supersymmetric theories (without further elaboration). Note that you have to treat eta and eta-bar independently, but for all practical purposes can use the bar just like a conjugation. Even for normal complex numbers, you can choose to see c and c* as independent, and then the real and imaginary parts are determined(!) by giving the two independent complex numbers c and c* (Re c = (c + c*)/2, I am c = (c - c*)/2i).

 

[taishizhiqiu]

I've thought about my post #8 some more, and Altland&Simons treatment of fermionic coherent states is consistent with other textbooks, and the annihilation operators do have to anti-commute with the Grassmann eigenvalues. See for example Eduardo Fradkin "Field Theories of Condensed Matter Physics", p. 44. Essentially, since the coherent state is defined as
$\ket{\eta} = \exp (-\eta a^\dagger) \ket{0} = \ket{0} - \eta \ket{1},$
the "action" (although its not an operator) of \eta on this is
$\eta \ket{\eta} = \eta \ket{0} - \eta^2 \ket{1} = \eta \ket{0}$.
And then a has to anticommute with eta to make this consistent with the action of a:
$a \ket{\eta} = a \ket{0} - a \eta \ket{1} = - a \eta \ket{1} = + \eta a \ket{1} = \eta \ket{0}$

Coming back to OPs question: If anything, this just shows, if you can't just treat eta like an operator in the matrix element <1|eta|1>. You should just consistently write eta outside of the overlap, or you treat it like a full blown operator, but then you can never pull it outside of an overlap, and your last equality in your example is also invalid. Going the route of just treating it as something to be pulled out of overlaps is consistent though. Notice that $\eta \ket{1} = \ket{0} - \ket{\eta},$
so
$\bra{1} \eta \ket{1} = \bra{1} \ket{0} - \bra{1} \ket{\eta} = - (- \eta) \bra{1} \ket{1} = \eta$.

Edit: Regarding the second question on post #1:
$(\eta \phi, \nu \psi) = \bar \eta \nu (\phi, \psi)$.
Don't be too confused by Altlands remark that eta and eta-bar are not related by conjugation, it is just a note of caution that the viewpoint get's problematic in supersymmetric theories (without further elaboration). Note that you have to treat eta and eta-bar independently, but for all practical purposes can use the bar just like a conjugation. Even for normal complex numbers, you can choose to see c and c* as independent, and then the real and imaginary parts are determined(!) by giving the two independent complex numbers c and c* (Re c = (c + c*)/2, I am c = (c - c*)/2i).

@protonsarecool Thanks for explaining $[a, \eta]_+ = 0$.

However, I think it is not the correct way to just write $\eta$ out of the overlap in, for example,
$\bra{1} \eta \ket{1} = \bra{1} \ket{0} - \bra{1} \ket{\eta} = - (- \eta) \bra{1} \ket{1} = \eta$.
I am not sure I can draw a contradiction from this practice, but in "Quantum Field Theory and Condensed Matter: An Introduction" by Shankar, it is stated that $\eta \ket{1} = - \ket{1} \eta$ (above Eq. 6.53).

 

[protonsarecool]

@protonsarecool Thanks for explaining $[a, \eta]_+ = 0$.

However, I think it is not the correct way to just write $\eta$ out of the overlap in, for example,
$\bra{1} \eta \ket{1} = \bra{1} \ket{0} - \bra{1} \ket{\eta} = - (- \eta) \bra{1} \ket{1} = \eta$.
I am not sure I can draw a contradiction from this practice, but in "Quantum Field Theory and Condensed Matter: An Introduction" by Shankar, it is stated that $\eta \ket{1} = - \ket{1} \eta$ (above Eq. 6.53).

Actually I made a mistake in that line.
$-\bra{1}\ket{\eta} = (-1)^2 \bra{1}\eta\ket{1}$,
so that line of reasoning didn't actually work. But now the picture is clear, and consistent with your quote of Shankar:
$-\eta = - \eta \bra{1}\ket{1} = \bra{1} \eta \ket{1} = - \bra{1}\ket{1} \eta = -\eta$
So, moving η out of the braket to either side is punished with a factor of (-1), because we need to anti-commute with a creation/annihilation operator when moving to either side. This is then also consistent with
$\bra{1} \eta \ket{1} = -\bra{0} \eta \ket{0} = - \eta \bra{0} \ket{0} = -\eta$
We can pull η out of the ground state expectation value just fine, no punishment by signs, because there are no hidden creation/annihilation operators around.