In second quantization, creation and annihilation operators are defined on Fock space as follows: \begin{align} \begin{cases}a_j^\dagger|n_1,n_2,...,n_j,...\rangle=\xi^{s_j}\sqrt{n_j+1}|n_1,n_2,...,n_j+1,...\rangle, \\ a_j|n_1,n_2,...,n_j,...\rangle=\xi^{s_j}\sqrt{n_j}|n_1,n_2,...,n_j-1,...\rangle.\end{cases} \end{align} where $\xi=\pm 1$ respectively for bosons and fermions and $s_j=\sum_{k=1}^{j-1}n_k$. This should guarantee that the canonical commutation/anticommutation relations are satisfied; however I'm having a little trouble proving that for fermions $\{a_j, a_k^\dagger\}=\delta_{jk}$ in particular. The case $j\ne k$ is fine: assuming $j<k$, we have $$a_ja_k^\dagger|n_1,...,n_j,...,n_k,...\rangle=(-1)^{s_j+s_k}\sqrt{(n_k+1)n_j}|n_1,...,n_j-1,...,n_k+1,...\rangle$$ while $$a_k^\dagger a_j|n_1,...,n_j,...,n_k,...\rangle=(-1)^{s_j+s_k-1}\sqrt{n_j(n_k+1)}|n_1,...,n_j-1,...,n_k+1,...\rangle $$ which means that $a_j a_k^\dagger+a_k^\dagger a_j=0$ as expected. However, when $j=k$ I find $$a_ja_j^\dagger|n_1,...,n_j,...\rangle=(-1)^{2s_j}(n_j+1)|n_1,...,n_j,...\rangle $$ and likewise $$a_j^\dagger a_j|n_1,...,n_j,...\rangle=(-1)^{2s_j}n_j|n_1,...,n_j,...\rangle $$ as $s_j$ is the same in both situations. Clearly this is wrong as $a_ja_j^\dagger+a_j^\dagger a_j\ne 1$. But where?

  • $\begingroup$ @AndyChen Why though? $s_j$ does not see the $j$-th site, it only sums the occupation numbers up to $n_{j-1}$, right? That means applying $a_j$ or $a_j^\dagger$ first shouldn't make a difference to the value of $s_j$. Am I missing something obvious? $\endgroup$
    – Brown Hole
    Jul 2, 2022 at 16:09
  • $\begingroup$ You are right, and I made a mistake here. $a^{\dagger}_ja_j$ should just be $\hat{n}_j$ and should give something always larger or equal to $0$. $\endgroup$
    – Andy Chen
    Jul 2, 2022 at 16:16
  • $\begingroup$ Then the main reason why $\{a_j,a^{\dagger}_j\}=1$ should be as follows: we first notice $n_j$ cannot be larger than $1$ for any state $|n_1,\cdots,n_j,\cdots\rangle$ since $a^{\dagger}_ja^{\dagger}_j=-a^{\dagger}_ja^{\dagger}_j=0$. Therefore, for either case $n_j=0$ or $n_j=1$, $\{a_j,a^{\dagger}_j\}=1$. The anticommutation relation is not as obvious as other cases for the same site $j$. $\endgroup$
    – Andy Chen
    Jul 2, 2022 at 16:22
  • $\begingroup$ @AndyChen That can work as a proof. However, I'd really like to show that simply by employing the definitions of these operators. In other words, instead of taking Pauli's principle for granted and then use it to derive the relations I'd like to derive the relations first so that Pauli's principle is already built in them. $\endgroup$
    – Brown Hole
    Jul 2, 2022 at 16:25
  • 1
    $\begingroup$ I now know where you have been wrong: at the first step, the creation and annihilation operators of fermions are not defined as what you wrote. Instead, we should have: $a^{\dagger}_j|n_1,\cdots,n_j,\cdots\rangle=(-1)^{s_j}\sqrt{1-n_j}|n_1,\cdots,n_j+1,\cdots\rangle$. $\endgroup$
    – Andy Chen
    Jul 2, 2022 at 16:51

1 Answer 1


The wiki page is a reference you can check up the definition of creation and annihilation operators. It shows fermions follow $a^{\dagger}_ja^{\dagger}_j=0$ (the Pauli's exclusion principle) and $a^{\dagger}_j|n_1,\cdots,n_j,\cdots,\rangle=(-1)^{s_j}\sqrt{1-n_j}|n_1,\cdots,n_j+1,\cdots\rangle$ and $a_j|n_1,\cdots,n_j,\cdots\rangle=(-1)^{s_j}\sqrt{n_j}|n_1,\cdots,n_j-1,\cdots\rangle$. However, I believe that your definition can be correct, too, but it requires careful justification. Since $$a_ja^{\dagger}_j|n_1,\cdots,n_j,\cdots\rangle=(-1)^{2s_j}\sqrt{n_j+1}|n_1,\cdots,n_j,\cdots\rangle$$ works if and only if $n_j=0$ (if $n_j=1$, the first application of $a^{\dagger}_j$ on the state would lead to $0$), while $$a^{\dagger}_ja_j|n_1,\cdots,n_j,\cdots\rangle=(-1)^{2s_j}\sqrt{n_j}|n_1,\cdots,n_j,\cdots\rangle$$ works for both $n_j=0$ and $n_j=1$ ($n_j=0$ gives us trivial result), we cannot just simply add them up, obtaining the contradicting result $$(a_ja^{\dagger}_j+a^{\dagger}_ja_j)|n_1,\cdots,n_j,\cdots\rangle=(2n_j+1)|n_1,\cdots,n_j,\cdots\rangle$$ We have to separate the cases apart, and derive $$(a_ja^{\dagger}_j+a^{\dagger}_ja_j)|n_1,\cdots,n_j,\cdots\rangle=|n_1,\cdots,n_j,\cdots\rangle$$

Supplementary on Defining Creation and Annihilation Operators :

While the proof from wiki page is valid, it introduces another two operators, insertion and deletion operators, in the first place to define the creation and annihilation operators. I prefer another equivalent formalism using slater determinant. It is more straightforward (at least to me), and when I forget the exact values of the factors which appear when these operators acting on the basis of the Fock space (like this time), I would have some standard way to rebuild them.

Starting Point: The Fock Space

Consider a basis for single-particle wavefunction $\{|\psi_j \rangle\}$, the vacuum state $|\cdot \rangle$, and a list of integers $[n]=\{n_j|n_j \geq 0\}$. Most of $n_j$'s are $0$ except for $n_{j_k}$'s where $j_k \geq 1$ is increasing with $k$, $m \geq k \geq 1$ and $\sum_{k=1}^{m}{n_{j_k}}=N$. We define the basis of the Fock space \begin{align} |[n]\rangle & =|0,\cdots,0,n_{j_1},0,\cdots,n_{j_2},\cdots,n_{j_m},\cdots\rangle \\ & = {1 \over \sqrt{n_{j_1}!n_{j_2}!\cdots n_{j_m}!N!}}\text{det}_{\pm}\big(|\psi_{j_1} \rangle,\cdots,|\psi_{j_1} \rangle,|\psi_{j_2} \rangle,\cdots,|\psi_{j_2} \rangle,\cdots,|\psi_{j_m} \rangle,\cdots,|\psi_{j_m} \rangle\big) \\ & = {1 \over \sqrt{n_{j_1}!n_{j_2}!\cdots n_{j_m}!N!}}\begin{vmatrix} |\psi_{j_1} \rangle & \cdots & |\psi_{j_1} \rangle & |\psi_{j_2} \rangle & \cdots & |\psi_{j_m} \rangle \\ |\psi_{j_1} \rangle & \cdots & |\psi_{j_1} \rangle & |\psi_{j_2} \rangle & \cdots & |\psi_{j_m} \rangle\\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ |\psi_{j_1} \rangle & \cdots & |\psi_{j_1} \rangle & |\psi_{j_2} \rangle & \cdots & |\psi_{j_m} \rangle \end{vmatrix}_{\pm} \end{align} where $\text{det}_{\pm}$ stands for the slater determinant with $+$ for bosons (so there is no alternating sign) and $-$ for fermions (so it works like a usual determinant), and in the second and the third line, $|\psi_{j_k}\rangle$ appears $n_{j_k}$ times. The factor ${1 \over \sqrt{n_{j_1}!\cdots n_{j_m}!N!}}$ is for normalization. For example, if we have fermions, the state $|1,1,0,\cdots \rangle$ is $$|1,1,0,\cdots \rangle = {1 \over \sqrt{2}} \begin{vmatrix} |\psi_1 \rangle & |\psi_2 \rangle \\ |\psi_1 \rangle & |\psi_2 \rangle \end{vmatrix}_{-} ={1 \over \sqrt{2}}(|\psi_1 \rangle|\psi_2 \rangle-|\psi_2 \rangle|\psi_1 \rangle)$$ Let us see another example, if we have bosons, the state $|2,0,\cdots \rangle$ is $$|2,0,\cdots \rangle = {1 \over \sqrt{2\cdot2}} \begin{vmatrix} |\psi_1 \rangle & |\psi_1 \rangle \\ |\psi_1 \rangle & |\psi_1 \rangle \end{vmatrix}_{+} ={1 \over 2}(|\psi_1 \rangle|\psi_1 \rangle+|\psi_1 \rangle|\psi_1 \rangle)=|\psi_1 \rangle|\psi_1 \rangle$$ There are some key messages we can take away from this definition:

  1. The slater determinant gives symmetric wavefunctions for bosons and antisymmetric ones for fermions as required.
  2. Also, if we have fermions, $n_j \leq 1$ for any $j$, which is Pauli's exclusion principle. This can be seen from the slater determinant that any repeating column would lead to $0$ as the overall result.

Define Creation and Annihilation Operators

Since we are hoping to build some operators which can add particles or remove them from the wavefunction, it is natural for us to start from the vacuum state, and define the creation annihilators with \begin{align} a^{\dagger}_{j_1}a^{\dagger}_{j_2}a^{\dagger}_{j_3} \cdots a^{\dagger}_{j_N}|\cdot \rangle & = {1 \over \sqrt{N!}}\text{det}_{\pm}\big(|\psi_{j_1} \rangle,|\psi_{j_2} \rangle,|\psi_{j_3} \rangle,\cdots,|\psi_{j_N} \rangle\big) \\ & = {1 \over \sqrt{N!}} \begin{vmatrix} |\psi_{j_1} \rangle & |\psi_{j_2} \rangle & \cdots & |\psi_{j_N} \rangle \\ |\psi_{j_1} \rangle & |\psi_{j_2} \rangle & \cdots & |\psi_{j_N} \rangle\\ \vdots & \vdots & \ddots & \vdots \\ |\psi_{j_1} \rangle & |\psi_{j_2} & \cdots & |\psi_{j_N} \rangle \end{vmatrix}_{\pm} \end{align} From the property of slater determinant, we can already see $a^{\dagger}_ja^{\dagger}_{j'}=a^{\dagger}_{j'}a^{\dagger}_j$ for bosons and $a^{\dagger}_ja^{\dagger}_{j'}=-a^{\dagger}_{j'}a^{\dagger}_j$ for fermions. Moreover, if we compare the expression with the basis of the Fock space, we would immediately recover $$a^{\dagger}_j|n_1,n_2,\cdots,n_j,\cdots\rangle=(\pm1)^{s_j}\sqrt{n_j+1}|n_1,n_2,\cdots,n_j+1,\cdots\rangle$$ as given by your definition (if we have fermions here, the equality works only if $n_j=0$). To define the annihilation operators, we express $a^{\dagger}_j$ with $$a^{\dagger}_j=\sum_{[n]}{(-1)^{s_j}\sqrt{n_j+1}|n_1,n_2,\cdots,n_j+1,\cdots\rangle \langle n_1,n_2,\cdots,n_j,\cdots|}$$ and we take the Hermitian conjugate of $a^{\dagger}_j$, having $$a_j=\sum_{[n]}{(-1)^{s_j}\sqrt{n_j}|n_1,n_2,\cdots,n_j-1,\cdots\rangle \langle n_1,n_2,\cdots,n_j,\cdots|}$$ which implies $$a_j|n_1,n_2,\cdots,n_j,\cdots\rangle=(-1)^{s_j}\sqrt{n_j}|n_1,n_2,\cdots,n_j-1,\cdots\rangle$$ Another useful application through defining the creation and annihilation operators using slater determinant is to show that under some unitary transformation $\hat{U}=(U)^i_j$ on the single-particle basis $\{\psi_j\}$ (so we have the new basis $\{\tilde{\psi}_i=\sum_{j}{U^i_j\psi_j}\}$), we have the new creation operators $\tilde{a}^{\dagger}_i=\sum_{j}{U^i_j\tilde{a}^{\dagger}_j}$. This is intuitively true, but can be tricky if we try to show it with the definition given by wiki.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.