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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.

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    $\begingroup$ The action of the $C_a$ operators on the states $|a\ldots b\rangle$ includes a phase of $\pm 1$, which depends on an arbitrary choice of ordering the possible quantum numbers $a,b,\ldots$. It is simpler to take $\{C_a,C_b^\dagger\} = \delta_{ab}$ as a definition and then always write many-particle states like $C_1^\dagger C_2^\dagger \ldots C_m^\dagger \lvert 0 \rangle$ etc. $\endgroup$ Feb 18, 2015 at 16:51
  • $\begingroup$ Possible duplicate: physics.stackexchange.com/q/62604/2451 $\endgroup$
    – Qmechanic
    Feb 18, 2015 at 17:31
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    $\begingroup$ the answer presupposes what i was trying to prove, so thats not very satisfying. $\endgroup$
    – Kuhlambo
    Feb 18, 2015 at 18:17
  • $\begingroup$ @pindakaas I don't know the complete answer, but you make a mistake. Let's deal only with two fermions, "old" and "new", identified by their places, the newly added fermion on the 1st position. The operation $C_b^{\dagger}|\sim b,a⟩$ introduces a new fermion, but one gets an antisymmetrical state, $(|b,a⟩ - |a,b⟩)/ \sqrt {2}$. Whether the old fermion remains with the value $a$ and the new one gets $b$ it is not known. Applying $C_a$ you get $(|b,\sim a⟩ - |\sim a,b⟩)/\sqrt {2}$, so, the old fermion or the new one was destroyed? You see, the position in the list is not fixed. (I continue) $\endgroup$
    – Sofia
    Feb 18, 2015 at 22:05
  • $\begingroup$ @pindakaas The other operation, $C_b^{\dagger}C_a|\sim b,a⟩$, first of all leaves vacuum, $|\sim b,\sim a⟩$. You know for sure that the old fermion was removed. After that, you create a single fermion with spin-projection $b$, $|b, \sim a⟩$. $\endgroup$
    – Sofia
    Feb 18, 2015 at 22:05

2 Answers 2

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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.

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  • $\begingroup$ Thanks a lot to everyone. I think this is at least the way to solving the problem of prooving the commutator relation here. The Operators just always act on the first state (by convention probably) so you have to swap and collect a minus sign so you get the 0 case. $\endgroup$
    – Kuhlambo
    Feb 19, 2015 at 8:01
  • $\begingroup$ @Chris Drost In my opinion this proof is not correct. Instead of this proof I could use the Ballentine's proof. You can't reverse the order of the particles as you please. The position of the particles in the list identifies the particle as I said in my comment. $\endgroup$
    – Sofia
    Feb 19, 2015 at 16:51
  • $\begingroup$ @Sofia: I did not reverse the order of the particles. Perhaps you're misunderstanding. I was defining a new notation for a set of antisymmetric states, $[abc\dots] = |abc\dots\rangle - |bac\dots\rangle + |cab\dots\rangle - |cba\dots\rangle + \dots$. When I am swapping those indexes, I am exploiting equality of the underlying wavefunctions. If you have this notation, and if you have a fixed convention on "which particle is created" (the one at the end of the ket or the one at the beginning are good ones) by a creation/annihilation operator, then you can see directly why the C/A-ops anticommute. $\endgroup$
    – CR Drost
    Feb 20, 2015 at 17:02
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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.

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  • $\begingroup$ did you see his comment to Mark? The user doesn't want to start from vacuum. $\endgroup$
    – Sofia
    Feb 18, 2015 at 20:42
  • $\begingroup$ Your answer doesn't address the question posed by the user. He wants that somebody explain him what is wrong with the operations he does, on the initial state that he chose. I also looked in Ballentine and I disagree with the proof there. Besides that, I am not so sure that $C_a ^{\dagger} C_b ^{\dagger} = C_b ^{\dagger} C_a ^{\dagger}$ because the state created by the second variant is opposite in sign to the state created by the former variant. But let's discuss the issue tomorrow, maybe we'll clarify it. $\endgroup$
    – Sofia
    Feb 18, 2015 at 22:39
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    $\begingroup$ @Sofia indeed, that is not true, and I didn't say that. What I said is that the way you define your initial state is a manner of convention, and you can define it as (1) or as (2). Once you chose one or the other, you must stick with that convention. And of course we are dealing with fermionic operators so $$c_a^\dagger c_b^\dagger = - c_b^\dagger c_a^\dagger, \quad \text{for }a \neq b.$$ Concerning the question of the OP, you are right I didn't read that comment. I edited the question trying to more directly address the question. $\endgroup$
    – glS
    Feb 18, 2015 at 22:43
  • $\begingroup$ I have no pleasure to give minus to someone. I just marked by that, the fact that this is not the answer. Let's see the situation tomorrow (now it's late in my country) and I'd be glad to remove the minus. $\endgroup$
    – Sofia
    Feb 18, 2015 at 22:48
  • $\begingroup$ it also displeases me that Ballentine wrote a superficial proof, very questionable. So, let's talk tomorrow. Mark's comment is also not very useful. $\endgroup$
    – Sofia
    Feb 18, 2015 at 22:50

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