# Second quantization, creation and annihilation operators

I found two notions of states for second quantization.

One representation uses occupation numbers here, for example

Another one creates the n+1 th particle in a collection of n existent states. see for instance here.

Now, the problem is that in the first case the creation operator does $a_k^{\dagger} |N_1,N_2,..\rangle = \sqrt{N_k+1 } |N_1,N_2,..,N_{k}+1,..\rangle$ and in the latter case $a_k^{\dagger} |n\rangle = \sqrt{n+1 } |n+1 \rangle.$

So the action of this operator is very different depending on whether you write down the states in terms of their occupation number or whether you write them in terms of the ensemble of all the existing states.

Unfortunately, I just don't get how these two pictures are related to each other.

If anything is unclear, please let me know.

• It doesn't look to me like the actions are different; in both (creation) cases you list, there is still a additional particle being added to the system. – Autolatry Nov 11 '14 at 15:31
• but in the latter case $n$ is the total number of particles and in the former case ${N_k}$ is the number of particles in state $k$. – Xin Wang Nov 11 '14 at 15:40
• Well, ok but your second example refers specifically to the (gas) particles in state $j$ on a 1-D lattice, your observation is true, the pictures of these two creation examples are different. The 1 D lattice example uses modified creation operators. – Autolatry Nov 11 '14 at 15:44
• so you cannot convert the operators into each other ? – Xin Wang Nov 11 '14 at 15:51
• Here I explain how the formalism of the occupation numbers (the long ket) works for a single harmonic oscillator. The meaning of these two lines is pretty different. In one case you create a new particle in a system of identical particles. Another case is simply changing the energy of a single harmonic oscillator. – mavzolej Apr 10 '16 at 16:19

Instead of $\sqrt{k+1}$ in your first expression, it should say $\sqrt{N_k+1}$. I think this should answer your question.
N.B. I edited your question so it might now say $\sqrt{N_k+1}$ in the first expression.
• well, actually no, cause as I said in the comments: in the latter case $n$ is the total number of particles and in the former case ${N_k}$ is the number of particles in state $k$. So I would say that the actions are somehow different – Xin Wang Nov 11 '14 at 15:40