Fock Space and fermionic annihilation & creation operators

Question

I have been trying very hard to understand, I am reading Ballentine's book on this topic, but I need help:

I realized that I don't understand how many particle states work with the creation & annihilation operators $ C_a $ and $C_a^\dagger $ while trying to calculate $\{C_a,C_b^\dagger\}$.

I will illustrate my problem starting with $C_a C_b^\dagger |a...(\sim b)\rangle $ where Ballentine uses $ \sim b$ to mean the state b is not occupied.

Here is my confusion. If I do what seems sensible: $C_a C_b^\dagger |a... (\sim b)\rangle =C_a |a... b\rangle=|(\sim a)... b\rangle $ but $C_b^\dagger C_a |a... (\sim b)\rangle= C_b^\dagger |(\sim a)... (\sim b)\rangle = |(\sim a)... b\rangle $

This is obviously wrong but from the definition I don't get what to do in the above case: $ C_a^\dagger C_b^\dagger |0\rangle=|ab \rangle $

Can someone explain how exactly one can relate a general fock state to the nice but confusing: $ | a,b,c...\rangle$. And how formally one can make sense of just a row of operators $C_a^\dagger C_b$ so I can transfer this to other situations?

I would be really glad for help. If my problem is unclear please comment.

Answer 1

So, the problem is that you've got to enforce Fermionic antisymmetry, but Fock space tries to make things easier by making that invisible.

So if we've got two electrons in a box in a definite Fock state, the electrons definitively occupy some single-particle states which we can just call $1, 2$. The actual state that is being occupied is therefore:

$|\psi\rangle = |12\rangle - |21\rangle$

where the "first electron" (arbitrarily chosen) is in the first numbered state, etc.

Looking at your $C_a^\dagger$ and $C_a$ operators, it is somewhat clear that they are not capturing this distinction completely. Let us say that we're looking at $C_3^\dagger$ and $C_1$. Perhaps the action of $C_3^\dagger$ will look like:

$|123\rangle - |213\rangle - |132\rangle + |231\rangle - |321\rangle + |312\rangle$

Here I am associating the $+$ sign with appending onto the end, a $-$ sign with appending one before that. This means that $C_1$ should probably have a + sign for deleting from the end, a $-$ sign for deleting from the one before that, etc. This sign convention leads to the state:

$|23\rangle - |32\rangle $

But if we reverse these for $C^\dagger_3 C_1$ then the very same sign convention would force us first into the state $ -|2\rangle $ thus generating $ -|23\rangle + |32\rangle$. So you see that the results you get are negatives of each other, but this result is hidden by a naive Fock space solution.

We can focus on the orders which are associated with a + sign and phrase all of this simply as:

1. For $C_1 C^\dagger_3$ I started with [12], prepended a 3 to get [312], swapped 1 to the front to get -[132], then removed the 1 from the front to get -[32].
2. For $C^\dagger_3 C_1$ I started with [12], removed the 1 from the front, prepended with 3, got +[32].

Similarly with a starting point of three states, you start with [123] having a + sign associated with it:

1. For $C_3 C^\dagger_4$ I started with [123], prepended a 4 to get [4123], swapped 3 to the front with 3 swaps to get -[3412], then removed it from the front to get -[412].
2. For $C^\dagger_4 C_3$ I started with [123], swapped the 3 to the front with 2 swaps to get [312], removed the 3 from the front, prepended with 4, got +[412].

Now you can maybe see why they will always be negatives of each other: in the first case you will do $k$ swaps to get that number to the start of the permutation. In the second case you will do $k - 1$ because the 4 will not be there. So you'll do an odd number of swaps total.

Answer 2

The anticommutation rules for creation/annihilation fermionic operators are what defines these operators. The "proof" that they are correct is that they produce a theory that is compatible with the antisymmetric nature of fermions (and with all the other experimental results of course). For example you can check that they produce the expected result for average values of one particle operators, with the minus signs expected for fermions states (see for example the second section of this other Phys.SE answer).

About the particular calculation you refer to: the problem is in the way you "neglect" the order of the creation operators defining your initial state. Starting from the definition of the state $|a,\dots,\sim \! b\rangle$ in terms of creation operators: $$ \tag 1 |a,\dots,\sim \! b\rangle = c_a^\dagger |\sim \! a,\dots,\sim \! b\rangle, $$ you have, using $\{ a^\dagger,b^\dagger \} = 0$, the following: $$ \tag 2 c_a c_b^\dagger| a,\dots,\sim \! b \rangle = c_a c_b^\dagger c_a^\dagger |\sim \! a,\dots,\sim \! b\rangle = - c_a c_a^\dagger c_b^\dagger |\sim \! a,\dots,\sim \! b\rangle \\= -c_ac_a^\dagger |\sim \!a,\dots,b\rangle = - |\sim \!a,\dots,b\rangle. $$ On the other hand, you have $$ \tag 3 c_b^\dagger c_a | a,\dots,\sim \! b \rangle = c_b^\dagger c_a c_a^\dagger | \sim \!a,\dots,\sim \! b \rangle = c_b^\dagger | \sim \!a,\dots,\sim \! b \rangle = | \sim \! a, \dots, b \rangle. $$

This is compatible with the (anti)commutation rules, as you can see summing (2) and (3), and remembering that for $a \neq b$ you have $\{ c_a,c_b^\dagger \} = 0$. The case $a=b$ is also easily verified, but your notation seem to implicitly assume $a \neq b$ so I wan't address it here.