2
$\begingroup$

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.

Improve this question
$\endgroup$
5
  • $\begingroup$ 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. $\endgroup$
    – Autolatry
    Commented Nov 11, 2014 at 15:31
  • $\begingroup$ 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 $k$. $\endgroup$
    – Xin Wang
    Commented Nov 11, 2014 at 15:40
  • $\begingroup$ 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. $\endgroup$
    – Autolatry
    Commented Nov 11, 2014 at 15:44
  • $\begingroup$ so you cannot convert the operators into each other ? $\endgroup$
    – Xin Wang
    Commented Nov 11, 2014 at 15:51
  • $\begingroup$ 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. $\endgroup$
    – mavzolej
    Commented Apr 10, 2016 at 16:19

2 Answers 2

Reset to default
4
$\begingroup$

@Xin Wang's last comment: In the first case you are simply, formally, looking at collection of k_max different, uncoupled oscillators. But you're only doing anything with the k'th one. k is an index in this case, nothing else but giving this specific oscillator a name.

In the second case you only have one oscillator in your notation, so actually you don't need to give the annihilation operator an index, as it is implicitly fixed. It is acutally even clumsy, since you're not giving the corresponding occupation number variable n the same index.

Your question may be a semantic issue, but since you're not doing anything with all other but the k'th oscillator, their particle number will be fixed during the operation. It's just a definition to count the 'total particle number' by adding up all n_m.

Improve this answer
$\endgroup$
1
$\begingroup$

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.

Improve this answer
$\endgroup$
1
  • 1
    $\begingroup$ 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 $\endgroup$
    – Xin Wang
    Commented Nov 11, 2014 at 15:40

Your Answer

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

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