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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.

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.

The anticommutation rules for creation/annihilation fermionic operators are what defines them. The "proof" that they are correct is that they produce a theory that is compatible with the wanted 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 (see for example the second section of this other answer of mine), and the minus sign that must be there for the antisymmetry of fermions.

About the particular calculation you refer to: the problem is in the way you "neglect" the order of the creation operators defining the state $|a,b\rangle$. Take as an example a state with quantum numbers $a$ and $b$, $| a,b \rangle$. In terms of creation operators this would be $$ \tag 1 | a,b \rangle = c_a^\dagger c_b^\dagger | 0 \rangle, $$ where $| 0 \rangle$ is the vacuum state. Why did I put $c_a^\dagger$ before $c_b^\dagger$? For no particular reason: it's just a convention, and I could have equivalently defined the state as $$ \tag 2 | a,b \rangle = c_b^\dagger c_a^\dagger | 0 \rangle. $$ What's important is that once you make a choice (e.g. put the creation operators in the same order of the labels of the states) you must be consistent with it. This means that the corresponding bra will have the annihilation operators in inverse order, e.g. for the convention in (1) you would have: $$ \tag 3 \langle a,b | = \langle 0 | c_b c_a. $$

If you now try to apply your reasoning to the state $|a,b\rangle$ expressed as in (1) or (2), you discover that the order of the operators does matter, because you have to make your creation/annihilation operators cross over the creation operators defining the state $| a,b \rangle$.

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.

The trick is in how you define multi-particle states in terms of creation operators.

Take as an example a state with quantum numbers $a$ and $b$, $| a,b \rangle$. In terms of creation operators this would be $$ \tag 1 | a,b \rangle = c_a^\dagger c_b^\dagger | 0 \rangle, $$ where $| 0 \rangle$ is the vacuum state. Why did I put $c_a^\dagger$ before $c_b^\dagger$? For no particular reason: it's just a convention, and I could have equivalently defined the state as $$ \tag 2 | a,b \rangle = c_b^\dagger c_a^\dagger | 0 \rangle. $$ What's important is that once you make a choice (e.g. put the creation operators in the same order of the labels of the states) you must be consistent with it. This means that the corresponding bra will have the annihilation operators in inverse order, e.g. for the convention in (1) you would have: $$ \tag 3 \langle a,b | = \langle 0 | c_b c_a. $$

If you now try to apply your reasoning to the state $|a,b\rangle$ expressed as in (1) or (2), you discover that the order of the operators does matter, because you have to make your creation/annihilation operators cross over the creation operators defining the state $| a,b \rangle$.

The anticommutation rules for creation/annihilation fermionic operators are what defines them. The "proof" that they are correct is that they produce a theory that is compatible with the wanted 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 (see for example the second section of this other answer of mine), and the minus sign that must be there for the antisymmetry of fermions.

About the particular calculation you refer to: the problem is in the way you "neglect" the order of the creation operators defining the state $|a,b\rangle$. Take as an example a state with quantum numbers $a$ and $b$, $| a,b \rangle$. In terms of creation operators this would be $$ \tag 1 | a,b \rangle = c_a^\dagger c_b^\dagger | 0 \rangle, $$ where $| 0 \rangle$ is the vacuum state. Why did I put $c_a^\dagger$ before $c_b^\dagger$? For no particular reason: it's just a convention, and I could have equivalently defined the state as $$ \tag 2 | a,b \rangle = c_b^\dagger c_a^\dagger | 0 \rangle. $$ What's important is that once you make a choice (e.g. put the creation operators in the same order of the labels of the states) you must be consistent with it. This means that the corresponding bra will have the annihilation operators in inverse order, e.g. for the convention in (1) you would have: $$ \tag 3 \langle a,b | = \langle 0 | c_b c_a. $$

If you now try to apply your reasoning to the state $|a,b\rangle$ expressed as in (1) or (2), you discover that the order of the operators does matter, because you have to make your creation/annihilation operators cross over the creation operators defining the state $| a,b \rangle$.