An appropriately symmetrized (and normalized) $N$-particle physical ket can be expressed in the form: $$| \beta_{1}, \beta_{2}, \dots, \beta_{N} \rangle \equiv \frac{1}{\sqrt{N! \prod_{i=1}^{\infty}(n_{\beta_{i}}!)}} \sum_{\mathcal{P} \in S_{N}} \zeta^{[1-\mathrm{sgn}(\mathcal{P})]/2} \, \mathcal{P} (|\beta_{1}\rangle \otimes |\beta_{2} \rangle \otimes \dots \otimes |\beta_{N} \rangle)\,,$$ with $\sum_{i=1}^{\infty} n_{i} = N$. In this equation:

  • $n_{\beta_{i}}$ is the occupation number of the state $|\beta_{i}\rangle$ in the set of states $\{|\beta_{1}\rangle, |\beta_{2}\rangle, \dots, |\beta_{N}\rangle\}$
  • for fermions $\zeta = -1$, while for bosons $\zeta = 1$
  • $\mathcal{P}$ is a permutation operator belonging the the symmetric group on $N$ letters, $S_{N}$, and acting on the tensor product state $|\beta_{1}\rangle \otimes |\beta_{2} \rangle \otimes \dots \otimes |\beta_{N} \rangle$
  • $\mathrm{sgn}(\mathcal{P})$ is the sign of the permutation $\mathcal{P}$, being +1 if $\mathcal{P}$ is even or -1 if it is odd.

In the book by Negele, J.W. and Orland, H., Quantum Many-Particle Systems, the creation operator $a_{\lambda}^{\dagger}$ is defined in a Fock space as: $$a_{\lambda}^{\dagger}| \beta_{1}, \beta_{2}, \dots, \beta_{N} \rangle \equiv \sqrt{n_{\lambda} + 1}\,| \lambda, \beta_{1}, \beta_{2}, \dots, \beta_{N} \rangle\,,$$ where $n_{\lambda}$ is the occupation number of the state $|\lambda\rangle$ in $| \beta_{1}, \beta_{2}, \dots, \beta_{N} \rangle$.

Following this definition, the annihilation operator $a_{\lambda}$ is simply the adjoint of $a_{\lambda}^{\dagger}$, and its action on a many-particle state is determined using the closure relation in the Fock space:

$$|0\rangle\langle0|+\sum_{M=1}^{\infty}\frac{1}{M!}\sum_{\alpha_{1}, \dots, \alpha_{M}} \prod_{i=1}^{\infty} (n_{\beta_{i}}!)\,|\alpha_{1}, \alpha_{2}, \dots, \alpha_{M}\rangle\langle\alpha_{1}, \alpha_{2}, \dots, \alpha_{M}| = \mathbb{1}\,.$$

Now, the problem is that when I try to derive an expression for $a_{\lambda}$, I am not able to obtain the prefactor $\frac{1}{\sqrt{n_{\lambda}}}$, as in equation 1.78b of the aforementioned book, which reads:

$$a_{\lambda} |\beta_{1}, \beta_{2}, \dots, \beta_{N} \rangle = \frac{1}{\sqrt{n_{\lambda}}} \sum_{i = 1}^{N} \zeta^{i-1} \delta_{\lambda, \beta_{i}} |\beta_{1}, \dots, \overline{\beta_{i}}, \dots, \beta_{N}\rangle\,,$$ in which $\delta$ is the Kronecker delta and $\overline{\beta_{i}}$ indicates that the state $|\beta_{i}\rangle$ has been removed from the $N$-particle state $|\beta_{1}, \beta_{2}, \dots, \beta_{N}\rangle$.

Any help? Thanks.


1 Answer 1


The formula for the annihilator is written to be valid for bosons and fermions and assumes that $n_\lambda$ is not zero and (because $\frac{1}{\sqrt{n_\lambda}}$ will be $1/0$ then, but we can define that $0/\sqrt{0} = 0$). This generality causes the complicated structure (for bosons one would usually work with states labelled by occupation numbers, since all permutations of the state labels signify the same state).

In the Fermi case $n_\lambda$ is one or zero and only one term will contribute on the right hand side, the $\zeta^{i-1}$ produces the correct signs. (Note: even if the Fermi state is a zero written in a fancy way like $\left|1,0,1\right>$ or $\left|1,1,0\right>$ the formula will be correct, the two terms on the right hand side will cancel).

Now for bosons all permutations of the labels signify the same state, so we can think in terms of occupation numbers, and derive that $a_\lambda\left|n\right> = \sqrt{n}\left|n-1\right>$ by taking the adjoint of the formula for the creator (and that the occupation numbers of all other states do not matter).

Now there are $n_\lambda$ equal terms on the right hand side (since any permutation of the labels implies the same state) and $a_\lambda$ has to create a factor $\sqrt{n_\lambda}$ overall to be the adjoint to $a_\lambda^\dagger$, therefore you need the $n_\lambda^{-1/2}$ to arrive at the correct factor.


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