앞 글들:

3. Fermions

모든 과정이 기본적으로 앞에서 다룬 boson의 경우와 유사하다. 단, boson에서는 하나의 single-particle state가 몇 번 나타나는지를 세는 것이 중요했지만, fermion의 경우 siingle-particle state가 최대 1번만 출현하므로 이것은 별 문제가 되지 않는다. 여기서 우리가 조심해야 할 부분은, 입자 교환에 대한 antisymmetry에서 비롯되는 (-) 부호이다.

(1) N-fermion basis state

앞서와 마찬가지로, single particle Hilbert space gif.latex?\\\\small \\\\inline \\\\mathcal{H}와 그 basis gif.latex?\\\\small \\\\inline \\\\mathfrak{B}를 생각하고, gif.latex?\\\\small \\\\inline \\\\mathfrak{B}에 속하는 single-particle state들을 조합하여 아래와 N-fermion basis state를 만들어 보자.

gif.latex?|s_{1},s_{2},\\\\ldots,s_{N}\\\\rangle_{F}^{(N)} \\\\equiv \\\\frac{1}{\\\\sqrt{N!}} \\\\,\\\\mathcal{A}\\\\Big(|s_{1}\\\\rangle\\\\otimes|s_{2}\\\\rangle\\\\otimes\\\\cdots\\\\otimes|s_{N}\\\\rangle\\\\Big).............(16)

여기서 boson의 경우와는 달리 gif.latex?\\\\small \\\\inline |s_{1}\\\\rangle,|s_{2}\\\\rangle,\\\\ldots,|s_{N}\\\\rangle\\\\in\\\\mathfrak{B} 들은 서로 중복되지 않는다.  gif.latex?\\\\small \\\\inline \\\\mathcal{A} (\\\\ldots)는 antisymmetrization을 의미하는데, 다음과 같이 정의된다.

gif.latex?\\\\mathcal{A}\\\\Big(|s_{1}\\\\rangle\\\\otimes|s_{2}\\\\rangle\\\\otimes\\\\cdots\\\\otimes|s_{N}\\\\rangle\\\\Big) \\\\equiv \\\\sum_{P} (-1)^{P} |s_{P(1)}\\\\rangle\\\\otimes|s_{P(2)}\\\\rangle\\\\otimes\\\\cdots\\\\otimes|s_{P(N)}\\\\rangle

gif.latex?\\\\small \\\\inline (-1)^{P}는 permutation gif.latex?\\\\small \\\\inline P의 부호를 의미하는데, even permutation이면 1, odd permutation이면 -1이다. Antisymmetrization에 의해 만들어진 gif.latex?\\\\small \\\\inline N!개의 항들은 서로 직교한다.

Fermion이기 때문에 두 입자를 교환하면 양자상태에 (-) 부호가 붙는다.

gif.latex?\\\\small \\\\inline |s_{1},\\\\ldots,s_{i},\\\\ldots,s_{j},\\\\ldots,s_{N}\\\\rangle_{F}^{(N)} \\\\,=\\\\, -\\\\, |s_{1},\\\\ldots,s_{j},\\\\ldots,s_{i},\\\\ldots,s_{N}\\\\rangle_{F}^{(N)}.


(2) Operator의 작용 및 creation/annihilation operator의 정의

gif.latex?\\\\small \\\\inline |s_{1},s_{2},\\\\ldots,s_{N}\\\\rangle_{F}^{(N)}에  1-particle operator의 building block  gif.latex?\\\\small \\\\inline \\\\hat{O}_{rs}^{(N)}[식 (1) 참조]를 가해보자. gif.latex?\\\\small \\\\inline s_{1},s_{2},\\\\ldots,s_{N} 중에 gif.latex?\\\\small \\\\inline s와 일치하는것이 없으면 gif.latex?\\\\small \\\\inline \\\\hat{O}_{rs}^{(N)}|s_{1},s_{2},\\\\ldots,s_{N}\\\\rangle_{F}^{(N)}=0 이다. Fermion인 만큼 gif.latex?\\\\small \\\\inline s가 여러 번 등장할 가능성은 없으니, 아래와 같이 gif.latex?\\\\small \\\\inline s_{1},s_{2},\\\\ldots,s_{N} 중 하나가 gif.latex?\\\\small \\\\inline s인 상황만을 고려하면 된다. 

gif.latex?\\\\small \\\\\\\\ \\\\hat{O}_{rs}^{(N)}|s_{1},\\\\ldots,s,\\\\ldots,s_{N}\\\\rangle_{F}^{(N)}\\\\\\\\ =\\\\Bigg(\\\\sum_{i=1}^{N} |r\\\\rangle\\\\langle s|_{i}\\\\otimes\\\\bigotimes_{\\\\substack{j=1\\\\\\\\j\\\\neq i}}^{N}\\\\textbf{1}_{j}\\\\Bigg ) |s_{1},\\\\ldots,s,\\\\ldots,s_{N}\\\\rangle_{F}^{(N)}\\\\\\\\ =\\\\frac{\\\\hat{O}_{rs}^{(N)}}{\\\\sqrt{N!}}\\\\,\\\\mathcal{A}\\\\Big(|s_{1}\\\\rangle\\\\otimes\\\\cdots\\\\otimes|s\\\\rangle\\\\otimes\\\\cdots\\\\otimes|s_{N}\\\\rangle\\\\Big)\\\\\\\\\\\\\\\\ =\\\\frac{1}{\\\\sqrt{N!}}\\\\,\\\\mathcal{A}\\\\Big(|s_{1}\\\\rangle\\\\otimes\\\\cdots\\\\otimes|r\\\\rangle\\\\otimes\\\\cdots\\\\otimes|s_{N}\\\\rangle\\\\Big)\\\\\\\\ =|s_{1},\\\\ldots,r,\\\\ldots,s_{N}\\\\rangle_{F}^{(N)}..........................(17)

여기서 한 가지 주의할 점이 있는데, 기존의 gif.latex?\\\\small \\\\inline s_{1},s_{2},\\\\ldots,s_{N} 중에 gif.latex?\\\\small \\\\inline r과 일치하는 것이 있는 경우 아래와 같이 0이 나온다는 점이다.

gif.latex?\\\\\\\\|s_{1},\\\\ldots,r,\\\\ldots,r,\\\\ldots,s_{N}\\\\rangle_{F}^{(N)} \\\\\\\\=\\\\frac{1}{\\\\sqrt{N!}}\\\\,\\\\mathcal{A}\\\\Big(|s_{1}\\\\rangle\\\\otimes\\\\cdots\\\\otimes|r\\\\rangle\\\\otimes\\\\cdots\\\\otimes|r\\\\rangle\\\\otimes\\\\cdots\\\\otimes|s_{N}\\\\rangle\\\\Big)=0
(두  gif.latex?\\\\small \\\\inline |r\\\\rangle의 교환에 대해 대칭이므로, 이것을 antisymmetrize하면 0)

이제 아래와 같이 creation operator를 정의해보자.

gif.latex?c_{s}^{\\\\dagger}|s_{1},s_{2},\\\\ldots,s_{N}\\\\rangle_{F}^{(N)} = |s, s_{1},s_{2},\\\\ldots,s_{N}\\\\rangle_{F}^{(N+1)}...........(18-1)

gif.latex?\\\\small \\\\inline s_{1},s_{2},\\\\ldots,s_{N} 중 gif.latex?\\\\small \\\\inline s와 일치하는 것이 있으면 우변이 0이 됨에 유의하자. Annihilation operator는 위의 adjoint에 해당하므로 다음과 같이 주어진다.

gif.latex?c_{s}|s,s_{1},s_{2},\\\\ldots,s_{N-1}\\\\rangle_{F}^{(N)} = |s_{1},s_{2},\\\\ldots,s_{N-1}\\\\rangle_{F}^{(N-1)}.............(18-2)

기존의 양자상태에 gif.latex?\\\\small \\\\inline |s\\\\rangle가 존재하지 않는 경우, gif.latex?\\\\small \\\\inline c_{s}를 가하면 단순히 0을 얻는다.
gif.latex?c_{s}|s_{1},s_{2},\\\\ldots,s_{N}\\\\rangle_{F}^{(N)}=0\\\\qquad\\\\Big(s\\\\notin \\\\{s_{1},s_{2},\\\\ldots,s_{N}\\\\}\\\\Big) ............(18-3)

또한, 식 (18-1) 및 (18-2)와 동등하게, 다음이 성립한다.

gif.latex?c_{s}^{\\\\dagger} \\\\,\\\\mathcal{A}\\\\Big(|s_{1}\\\\rangle\\\\otimes|s_{2}\\\\rangle\\\\otimes\\\\cdots\\\\otimes|s_{N}\\\\rangle\\\\Big) = \\\\frac{1}{\\\\sqrt{N+1}}\\\\,\\\\mathcal{A}\\\\Big(|s\\\\rangle\\\\otimes|s_{1}\\\\rangle\\\\otimes|s_{2}\\\\rangle\\\\otimes\\\\cdots\\\\otimes|s_{N}\\\\rangle\\\\Big)............(19-1)
gif.latex?c_{s} \\\\,\\\\mathcal{A}\\\\Big(|s\\\\rangle\\\\otimes|s_{1}\\\\rangle\\\\otimes|s_{2}\\\\rangle\\\\otimes\\\\cdots\\\\otimes|s_{N-1}\\\\rangle\\\\Big) = }{\\\\sqrt{N}\\\\,\\\\mathcal{A}\\\\Big(|s_{1}\\\\rangle\\\\otimes|s_{2}\\\\rangle\\\\otimes\\\\cdots\\\\otimes|s_{N-1}\\\\rangle\\\\Big)...........(19-2)

우리는 gif.latex?\\\\small \\\\inline |s\\\\rangle가 미리 존재하지 않는 양자상태에  gif.latex?\\\\small \\\\inline c_{s}를 가한 경우와 gif.latex?\\\\small \\\\inline |s\\\\rangle가 이미 존재하는 상황에서 gif.latex?\\\\small \\\\inline c_{s}^{\\\\dagger}를 가한 경우 모두 0을 얻음을 확인하였다. 이것은 gif.latex?\\\\small \\\\inline c_{s} 또는 gif.latex?\\\\small \\\\inline c_{s}^{\\\\dagger} 를 두 번 가하면 무조건 0이 됨을 의미한다. 즉, 다음이 만족된다.

gif.latex?c_{s}^{2} = (c_{s}^{\\\\dagger})^{2} = 0.

식 (18-1)~(19-2)에서 한 가지 유의할 점은, 우리가 굳이 gif.latex?\\\\small \\\\inline |s\\\\rangle 를 순서 상 제일 앞쪽에 놓았다는 것이다. gif.latex?\\\\small \\\\inline |s\\\\rangle가 첫번째가 아닌 i번째에 놓여 있으면 어떻게 될까? 앞에서부터 인접한 것끼리 i-1차례 교환함으로써 gif.latex?\\\\small \\\\inline |s\\\\rangle를 첫번째에서 i번째 위치로 옮길 수 있기 때문에, 우선 다음이 성립한다.

gif.latex?\\\\\\\\|s_{1},\\\\ldots,s_{i-1},\\\\underline{s},s_{i},\\\\ldots,s_{N-1}\\\\rangle_{F}^{(N)} = (-1)^{i-1} |s,s_{1},s_{2},\\\\ldots,s_{N-1}\\\\rangle_{F}^{(N)}\\\\\\\\ \\\\indent\\\\qquad\\\\quad{}^{\\\\textmd{i}^{\\\\textmd{th}} \\\\, \\\\textmd{position}}

이로부터 아래와 같은 결론을 얻는다.

gif.latex?\\\\\\\\ c_{s}^{\\\\dagger}|s_{1},s_{2},\\\\ldots,s_{N}\\\\rangle_{F}^{(N)}=(-1)^{i-1}|s_{1},\\\\ldots,s_{i-1},\\\\underline{s},s_{i},\\\\ldots,s_{N}\\\\rangle_{F}^{(N+1)} \\\\\\\\ \\\\indent\\\\qquad \\\\qquad\\\\qquad\\\\qquad\\\\qquad\\\\qquad\\\\qquad\\\\qquad {}^{\\\\textmd{i}^{\\\\textmd{th}} \\\\, \\\\textmd{position}}.............(18-1-2)
gif.latex?\\\\\\\\c_{s}|s_{1},\\\\ldots,s_{i-1},\\\\underline{s},s_{i},\\\\ldots,s_{N-1}\\\\rangle_{F}^{(N)} = (-1)^{i-1} |s_{1},s_{2},\\\\ldots,s_{N-1}\\\\rangle_{F}^{(N-1)}\\\\\\\\ \\\\indent\\\\quad \\\\ \\\\qquad{}^{\\\\textmd{i}^{\\\\textmd{th}} \\\\, \\\\textmd{position}}.............(18-2-2)

마지막으로, boson의 경우와 마찬가지로 vacuum state gif.latex?\\\\small \\\\inline |0\\\\rangle를 생각할 수 있고, gif.latex?\\\\small \\\\inline |0\\\\rangle과 관련된 creation/annihilation operator의 작용은 다음과 같이 정의된다.

gif.latex?c_{r}|s\\\\rangle =\\\\delta_{rs}|0\\\\rangle...........(20-1)
gif.latex?c_{r}^{\\\\dagger}|0\\\\rangle = |r\\\\rangle................(20-2)
gif.latex?c_{r}|0\\\\rangle = 0....................(20-3)

식 (18-1)~(20-3)을 이용하면 다음을 보일 수 있다.

gif.latex?c_{s}^{\\\\dagger}c_{s}\\\\,|s_{1},s_{2},\\\\ldots,s_{N}\\\\rangle_{F}^{(N)} = n_{s}\\\\,|s_{1},s_{2},\\\\ldots,s_{N}\\\\rangle_{F}^{(N)}........(21-1)
gif.latex?c_{s}c_{s}^{\\\\dagger}\\\\,|s_{1},s_{2},\\\\ldots,s_{N}\\\\rangle_{F}^{(N)} = (1-n_{s})\\\\,|s_{1},s_{2},\\\\ldots,s_{N}\\\\rangle_{F}^{(N)}........(21-2)

Boson의 경우와 유사하게 gif.latex?\\\\small \\\\inline c_{s}^{\\\\dagger}c_{s}는 number operator의 역할을 한다. 또한 여기서 gif.latex?\\\\small \\\\inline n_{s}가 가질 수 있는 값은 0 또는 1 뿐임을 잊지 말자.

여기서 정의한 fermion creation/annihilation operator의 가장 큰 특징은 boson의 경우와는 달리 anticommutation relation을 만족한다는 것이다. 예를 들어, 

gif.latex?c_{r}c_{s}|r,s,s_{1},s_{2},\\\\ldots\\\\rangle_{F}^{(N)} = -|r,s_{1},s_{2},\\\\ldots\\\\rangle_{F}^{(N-1)} = -|s_{1},s_{2},\\\\ldots\\\\rangle_{F}^{(N-2)}
gif.latex?c_{s}c_{r}|r,s,s_{1},s_{2},\\\\ldots\\\\rangle_{F}^{(N)} = |s,s_{1},s_{2},\\\\ldots\\\\rangle_{F}^{(N-1)} = |s_{1},s_{2},\\\\ldots\\\\rangle_{F}^{(N-2)}

gif.latex?c_{r}^{\\\\dagger}c_{s}|s,s_{1},s_{2},\\\\ldots\\\\rangle_{F}^{(N)} = |s_{1},s_{2},\\\\ldots\\\\rangle_{F}^{(N-1)} = |r,s_{1},s_{2},\\\\ldots\\\\rangle_{F}^{(N)}
gif.latex?c_{s} c_{r}^{\\\\dagger}|s,s_{1},s_{2},\\\\ldots\\\\rangle_{F}^{(N)} = |r,s,s_{1},s_{2},\\\\ldots\\\\rangle_{F}^{(N+1)} = -|r,s_{1},s_{2},\\\\ldots\\\\rangle_{F}^{(N)} \\\\quad (r \\\\neq s).

지금까지의 결과를 종합하면, 다음과 같은 anticommutation relation을 적을 수 있다.

gif.latex?\\\\{c_{r},c_{s}^{\\\\dagger}\\\\} = \\\\delta_{rs}..........................(22-1)
gif.latex?\\\\{c_{r},c_{s}\\\\} = \\\\{c_{r}^{\\\\dagger},c_{s}^{\\\\dagger}\\\\} =0...........(22-2)

또한 식 (17)과 (18-1-1) 및 (18-2-2)를 비교하면, gif.latex?\\\\small \\\\inline c_{r}^{\\\\dagger}c_{s}가 1-particle operator의 building block과 같은 작용을 함을 알 수 있다.

gif.latex?\\\\hat{O}_{rs}^{(N)} \\\\,|s_{1},s_{2},\\\\ldots,s_{N}\\\\rangle_{F}^{(N)} = c_{r}^{\\\\dagger}c_{s}\\\\, |s_{1},s_{2},\\\\ldots,s_{N}\\\\rangle_{F}^{(N)}...........(23)

2-particle operator에 대해서도 비슷한 관계식을 구하기 위해 식 (11)을 다시 들여다 보자.
gif.latex?\\\\\\\\ \\\\hat{O}_{r,r^{\\\\prime};s,s^{\\\\prime}}^{(N)} = \\\\sum_{\\\\substack{i,j=1\\\\\\\\i\\\\neq j}}^{N}|r\\\\rangle\\\\langle s|_{i} \\\\!\\\\otimes\\\\! |r^{\\\\prime}\\\\rangle\\\\langle s^{\\\\prime}|_{j} \\\\otimes \\\\bigotimes_{\\\\substack{k=1\\\\\\\\k\\\\neq i,j}}^{N}\\\\textbf{1}_{k}\\\\\\\\\\\\\\\\ = \\\\sum_{\\\\substack{i,j=1\\\\\\\\i\\\\neq j}}^{N}\\\\Big[|r\\\\rangle\\\\langle s|_{i} \\\\otimes \\\\bigotimes_{\\\\substack{k=1\\\\\\\\k\\\\neq i}}^{N}\\\\textbf{1}_{k}\\\\Big]\\\\Big[|r^{\\\\prime}\\\\rangle\\\\langle s^{\\\\prime}|_{j} \\\\otimes \\\\bigotimes_{\\\\substack{k=1\\\\\\\\k\\\\neq j}}^{N}\\\\textbf{1}_{k}\\\\Big]\\\\\\\\\\\\\\\\ =\\\\sum_{i,j=1}^{N}\\\\Big[|r\\\\rangle\\\\langle s|_{i} \\\\otimes \\\\bigotimes_{\\\\substack{k=1\\\\\\\\k\\\\neq i}}^{N}\\\\textbf{1}_{k}\\\\Big]\\\\Big[|r^{\\\\prime}\\\\rangle\\\\langle s^{\\\\prime}|_{j} \\\\otimes \\\\bigotimes_{\\\\substack{k=1\\\\\\\\k\\\\neq j}}^{N}\\\\textbf{1}_{k}\\\\Big] - \\\\sum_{i=1}^{N} \\\\delta_{r^{\\\\prime}s}|r\\\\rangle\\\\langle s^{\\\\prime}|_{i} \\\\otimes \\\\bigotimes_{\\\\substack{k=1\\\\\\\\k\\\\neq i}}^{N}\\\\textbf{1}_{k}\\\\\\\\\\\\\\\\ =\\\\hat{O}_{rs}^{(N)}\\\\hat{O}_{r^{\\\\prime}s^{\\\\prime}}^{(N)} - \\\\delta_{r^{\\\\prime}s}\\\\hat{O}_{r s^{\\\\prime}}^{(N)}............(11)

위 식은 boson이냐 fermion이냐에 상관없이 성립하며, 이로부터 아래와 같은 결과가 유도된다.


gif.latex?\\\\\\\\ \\\\hat{O}_{r,r^{\\\\prime};s,s^{\\\\prime}}^{(N)} |s_{1},s_{2},\\\\ldots,s_{N}\\\\rangle_{F}^{(N)}=\\\\Big(\\\\hat{O}_{rs}^{(N)} \\\\hat{O}_{r^{\\\\prime}s^{\\\\prime}}^{(N)} - \\\\delta_{r^{\\\\prime}s}\\\\hat{O}_{r s^{\\\\prime}}^{(N)}\\\\Big) |s_{1},s_{2},\\\\ldots,s_{N}\\\\rangle_{F}^{(N)}\\\\\\\\ =\\\\Big(c_{r}^{\\\\dagger}c_{s}c_{r^{\\\\prime}}^{\\\\dagger}c_{s^{\\\\prime}}-\\\\delta_{r^{\\\\prime}s}c_{r}^{\\\\dagger}c_{s^{\\\\prime}}\\\\Big)|s_{1},s_{2},\\\\ldots,s_{N}\\\\rangle_{F}^{(N)}\\\\\\\\=\\\\Big(c_{r}^{\\\\dagger}\\\\big(\\\\{c_{s},c_{r^{\\\\prime}}^{\\\\dagger}\\\\}-c_{r^{\\\\prime}}^{\\\\dagger}c_{s}\\\\big)c_{s^{\\\\prime}}-\\\\delta_{r^{\\\\prime}s}c_{r}^{\\\\dagger}c_{s^{\\\\prime}}\\\\Big)|s_{1},s_{2},\\\\ldots,s_{N}\\\\rangle_{F}^{(N)}\\\\\\\\\\\\\\\\= - c_{r}^{\\\\dagger}c_{r^{\\\\prime}}^{\\\\dagger}c_{s}c_{s^{\\\\prime}}|s_{1},s_{2},\\\\ldots,s_{N}\\\\rangle_{F}^{(N)} \\\\\\\\\\\\\\\\ =c_{r}^{\\\\dagger}c_{r^{\\\\prime}}^{\\\\dagger}c_{s^{\\\\prime}}c_{s}|s_{1},s_{2},\\\\ldots,s_{N}\\\\rangle_{F}^{(N)}\\\\\\\\..............(24)

3-particle 이상의 경우에도 상응하는 관계가 성립함을 보일 수 있다.


(3) Change of basis

1-(4)에서 다룬 boson의 경우와 모든 중간과정이 동일하며, 단지 creation/annihilation operator를 fermion의 그것으로 바꾸고, gif.latex?\\\\small \\\\inline \\\\mathcal{S}(\\\\ldots) 대신 gif.latex?\\\\small \\\\inline \\\\mathcal{A}(\\\\ldots)를 사용하기만 하면 된다. 즉, single-particle Hilbert space gif.latex?\\\\small \\\\inline \\\\mathcal{H}에 대해, gif.latex?\\\\small \\\\inline \\\\mathcal{H}의 두 서로 다른 orthonormal basis gif.latex?\\\\small \\\\inline \\\\mathcal{B}와 gif.latex?\\\\small \\\\inline \\\\mathcal{B}^{\\\\prime}를 기준으로 정의된 두 벌의 fermion creation/annihilation operator들이 있을 때, 이들 사이의 변환 관계는 다음과 같다.

gif.latex?c_{\\\\alpha}^{\\\\dagger} = \\\\sum_{r} \\\\langle r|\\\\alpha\\\\rangle \\\\, c_{r}^{\\\\dagger} .............(25-1)

gif.latex?c_{\\\\alpha} = \\\\sum_{r} \\\\langle \\\\alpha|r\\\\rangle \\\\, c_{r}.............(25-2)

(gif.latex?\\\\small \\\\inline |r\\\\rangle\\\\in\\\\mathcal{B}gif.latex?\\\\small \\\\inline |\\\\alpha\\\\rangle\\\\in\\\\mathcal{B}^{\\\\prime})


(4) Fock space

Fock space 역시 boson의 경우와 유사하다. 단, 이 경우에는 N-particle Hilbert space gif.latex?\\\\small \\\\inline \\\\mathcal{H}^{(N)}에 속하는 state들 중 입자의 교환에 대해 antisymmetric한 것들로 이루어진 subspace gif.latex?\\\\small \\\\inline \\\\mathcal{F}^{(N)}을 고려하여야 한다. (N>0.) 이때 gif.latex?\\\\small \\\\inline \\\\mathcal{H}^{(N)}에서 gif.latex?\\\\small \\\\inline \\\\mathcal{F}^{(N)}으로의 projection operator  gif.latex?\\\\small \\\\inline \\\\mathcal{P}_{\\\\mathcal{F}}^{(N)}을 생각한다. (gif.latex?\\\\small \\\\inline \\\\mathcal{F}^{(N)}=\\\\mathcal{P}_{\\\\mathcal{F}}^{(N)}\\\\mathcal{H}^{(N)}.)  gif.latex?\\\\small \\\\inline \\\\mathcal{F}^{(0)}는 앞서와 마찬가지로 vacuum state만으로 이루어진 1차원 vector space이다. 이제 fermion의 Fock space는 다음과 같이 정의된다.

gif.latex?\\\\small \\\\mathcal{F}\\\\equiv\\\\mathcal{F}^{(0)}\\\\oplus \\\\mathcal{F}^{(1)}\\\\oplus\\\\mathcal{F}^{(2)}\\\\oplus\\\\cdots = \\\\bigoplus_{N=0}^{\\\\infty}\\\\mathcal{F}^{(N)}(Fermion의 Fock space)

이때 1-particle 및 2-particle operator들은 다음과 같은 Fock space representation을 가진다.

gif.latex?c_{r}^{\\\\dagger}c_{s} = \\\\bigoplus_{N=1}^{\\\\infty} \\\\mathcal{P}_{\\\\mathcal{F}}^{(N)}\\\\hat{O}_{rs}^{(N)}\\\\mathcal{P}_{\\\\mathcal{F}}^{(N)} =\\\\bigoplus_{N=1}^{\\\\infty}\\\\, \\\\mathcal{P}_{\\\\mathcal{F}}^{(N)}\\\\!\\\\left[\\\\sum_{i=1}^{N}|r\\\\rangle\\\\langle s|_{i}\\\\otimes\\\\bigotimes_{\\\\substack{j=1\\\\\\\\j\\\\neq i}}^{N} \\\\textbf{1}_{j} \\\\right ]\\\\!\\\\mathcal{P}_{\\\\mathcal{F}}^{(N)}=\\\\bigoplus_{N=1}^{\\\\infty}\\\\, \\\\mathcal{P}_{\\\\mathcal{F}}^{(N)}\\\\!\\\\left[\\\\sum_{i=1}^{N}|r\\\\rangle\\\\langle s|_{i} \\\\right ]\\\\!\\\\mathcal{P}_{\\\\mathcal{F}}^{(N)}.........(26-1)

gif.latex?\\\\\\\\c_{r}^{\\\\dagger}c_{r^{\\\\prime}}^{\\\\dagger}c_{s^{\\\\prime}}c_{s} = \\\\bigoplus_{N=2}^{\\\\infty} \\\\mathcal{P}_{\\\\mathcal{F}}^{(N)}\\\\hat{O}_{r,r^{\\\\prime};s,s^{\\\\prime}}^{(N)}\\\\mathcal{P}_{\\\\mathcal{F}}^{(N)} \\\\\\\\=\\\\bigoplus_{N=2}^{\\\\infty}\\\\, \\\\mathcal{P}_{\\\\mathcal{F}}^{(N)}\\\\!\\\\left[\\\\sum_{\\\\substack{i,j=1\\\\\\\\i\\\\neq j}}^{N}|r\\\\rangle\\\\langle s|_{i}\\\\otimes|r^{\\\\prime}\\\\rangle\\\\langle s^{\\\\prime}|_{j}\\\\otimes\\\\bigotimes_{\\\\substack{k=1\\\\\\\\k\\\\neq i,j}}^{N} \\\\textbf{1}_{k} \\\\right ]\\\\!\\\\mathcal{P}_{\\\\mathcal{F}}^{(N)} =\\\\bigoplus_{N=2}^{\\\\infty}\\\\, \\\\mathcal{P}_{\\\\mathcal{F}}^{(N)}\\\\!\\\\left[\\\\sum_{\\\\substack{i,j=1\\\\\\\\i\\\\neq j}}^{N}|r,r^{\\\\prime}\\\\rangle\\\\langle s,s^{\\\\prime}|_{ij} \\\\right ]\\\\!\\\\mathcal{P}_{\\\\mathcal{F}}^{(N)}...........(26-2)

(3-particle 이상은 역시 생략.)


(5) State의 표현

식 (19-1)을 이용하여 vacuum state에서 하나씩 입자의 수를 늘려 나가 보면, 다음이 만족됨을 알 수 있다.
gif.latex?\\\\mathcal{A}\\\\Big(|s_{1}\\\\rangle\\\\otimes|s_{2}\\\\rangle\\\\otimes\\\\cdots\\\\otimes|s_{N}\\\\rangle\\\\Big) = \\\\sqrt{N!}\\\\,\\\\,c_{s_{1}}^{\\\\dagger}c_{s_{2}}^{\\\\dagger}\\\\ldots c_{s_{N}}^{\\\\dagger}|0\\\\rangle

따라서 다음이 성립한다.
gif.latex?\\\\\\\\|s_{1},s_{2},\\\\ldots,s_{N}\\\\rangle_{F}^{(N)}\\\\\\\\=\\\\frac{1}{\\\\sqrt{N!}}\\\\,\\\\mathcal{A}\\\\Big(|s_{1}\\\\rangle\\\\otimes|s_{2}\\\\rangle\\\\otimes\\\\cdots\\\\otimes|s_{N}\\\\rangle\\\\Big)\\\\\\\\= c_{s_{1}}^{\\\\dagger}c_{s_{2}}^{\\\\dagger}\\\\ldots c_{s_{N}}^{\\\\dagger}|0\\\\rangle..................(27)

위 식을 바탕으로 occupation number representation을 정의할 수 있다. Boson의 경우와 마찬가지로, single-particle state들을 1, 2, 3, ....으로 표시하고, 이들의 occupation이 gif.latex?\\\\small \\\\inline n_{1},n_{2},n_{3},\\\\ldots일 때(fermion이라서 각각의 occupation은 0 또는 1의 값만을 가짐에 유의), 이를 gif.latex?\\\\small \\\\inline |n_{1},n_{2},n_{3},\\\\ldots\\\\rangle으로 나타내는 것이다.  단, fermion의 creation operator들은 작용 순서에 따라 부호가 달라지기 때문에 문제가 발생한다. gif.latex?\\\\small \\\\inline |0\\\\rangle에서부터 gif.latex?\\\\small \\\\inline |n_{1},n_{2},n_{3},\\\\ldots\\\\rangle를 만들려면 gif.latex?\\\\small \\\\inline n_{m}=1인 모든 gif.latex?\\\\small \\\\inline m에 대한  gif.latex?\\\\small \\\\inline c_{m}^{\\\\dagger}를 가해 주어야 하는데, 이들을 특정한 순서로 gif.latex?\\\\small \\\\inline |0\\\\rangle에 작용시킨 것이 것이 gif.latex?\\\\small \\\\inline \\\\pm |n_{1},n_{2},n_{3},\\\\ldots\\\\rangle 중 어느 쪽인지 정해 줄 필요가 있다. 가장 자연스러운 정의는 gif.latex?\\\\small \\\\inline c_{m}^{\\\\dagger}들의 배열이 왼쪽에서부터 gif.latex?\\\\small \\\\inline m에 대한 오름차순(gif.latex?\\\\small \\\\inline n_{1},n_{2},n_{3},\\\\ldots와 순서가 일치)이 되는 것을 gif.latex?\\\\small \\\\inline +|n_{1},n_{2},n_{3},\\\\ldots\\\\rangle로 정의하는 것이다. 예를 들어 다음과 같다.

gif.latex?|n_{1}\\\\!=\\\\!1,n_{2}\\\\!=\\\\!0,n_{3}\\\\!=\\\\!1,n_{4}\\\\!=\\\\!0,n_{5}\\\\!=\\\\!1,\\\\ldots\\\\rangle \\\\equiv c_{1}^{\\\\dagger}c_{3}^{\\\\dagger}c_{5}^{\\\\dagger}\\\\cdots|0\\\\rangle

한편, 위의 양자상태에 gif.latex?\\\\small \\\\inline c_{4}^{\\\\dagger}를 가하는 경우를 생각해 보자.

gif.latex?\\\\\\\\c_{4}^{\\\\dagger}\\\\,|n_{1}\\\\!=\\\\!1,n_{2}\\\\!=\\\\!0,n_{3}\\\\!=\\\\!1,n_{4}\\\\!=\\\\!0,n_{5}\\\\!=\\\\!1,\\\\ldots\\\\rangle\\\\\\\\ =c_{4}^{\\\\dagger}c_{1}^{\\\\dagger}c_{3}^{\\\\dagger}c_{5}^{\\\\dagger}\\\\cdots|0\\\\rangle\\\\\\\\ =(-1)^{2}c_{1}^{\\\\dagger}c_{3}^{\\\\dagger}c_{4}^{\\\\dagger}c_{5}^{\\\\dagger}\\\\cdots|0\\\\rangle\\\\\\\\ =c_{1}^{\\\\dagger}c_{3}^{\\\\dagger}c_{4}^{\\\\dagger}c_{5}^{\\\\dagger}\\\\cdots|0\\\\rangle\\\\\\\\ =|n_{1}\\\\!=\\\\!1,n_{2}\\\\!=\\\\!0,n_{3}\\\\!=\\\\!1,n_{4}\\\\!=\\\\!1,n_{5}\\\\!=\\\\!1,\\\\ldots\\\\rangle

두 번째 줄에서 세 번째 줄로 넘어가면서 붙은 gif.latex?\\\\small \\\\inline (-1)^{2}는, 오름차순으로 정렬하기 위해 gif.latex?\\\\small \\\\inline c_{4}^{\\\\dagger}의 위치를 두 칸 오른쪽으로 옮기면서 생겼다. 같은 이유로 임의의 경우에는 다음이 성립한다.

gif.latex?c_{s}^{\\\\dagger}|n_{1},n_{2},\\\\ldots,n_{s}\\\\!=\\\\!0,\\\\ldots\\\\rangle = (-1)^{\\\\sum_{rs}\\\\!n_{r}} |n_{1},n_{2},\\\\ldots,n_{s}\\\\!=\\\\!1,\\\\ldots\\\\rangle...........(28-1)

또한 위 식의 adjoint를 취하면 다음과 같다.
gif.latex?c_{s}|n_{1},n_{2},\\\\ldots,n_{s}\\\\!=\\\\!1,\\\\ldots\\\\rangle = (-1)^{\\\\sum_{rs}\\\\!n_{r}} |n_{1},n_{2},\\\\ldots,n_{s}\\\\!=\\\\!0,\\\\ldots\\\\rangle............(28-2)