Quantum Mechanics – Exploring 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.