# Second Quantization-Kinetic operator

1. Jan 26, 2014

### ChrisVer

I am feeling a little stupid tonight... So let me build the problem...
For a single particle operator $O$, we have in the basis $|i>$ we have that:
$O= \sum_{ij} o_{ij} |i><j|$ with $o_{ij}=<i|O|j>$
Then for N particles we have that:
$T=\sum_{a}O_{a}= \sum_{ij} o_{ij} \sum_{a} |i>_{a}<j|_{a}$ with $a=1,2...,N$

How can we write this afterwards (for bosons) in respect to the creation and annihilation operators as:
$T= \sum_{ij} o_{ij} c_{i}^{t} c_{j}$

From what I suspect, $\sum_{a} |i>_{a}<j|_{a}$ must be equal to the number operator... But for some reason I'm unable to see it...Also I tried taking this:
$O= \sum_{ij} o_{ij} |i><j|$
$O= \sum_{ij} o_{ij} c_{i}^{t}|0><0|c_{j}$
And then summing over the particles:
$O= \sum_{ij} o_{ij} c_{i}^{t}c_{j} \sum_{a}|0>_{a}<0|_{a}$
but I see no reason why the sum must be equal to unity $\sum_{a}|0>_{a}<0|_{a} =1$

Last edited: Jan 26, 2014
2. Jan 29, 2014

### Staff: Mentor

Consider a $N$-particle operator $O$ as a sum of one-particle operators
$$O = \sum_{i=1}^{N} O_i$$
First assume that you can write it as
$$O = \sum_{\alpha, \beta} \langle \alpha | O_1 | \beta \rangle c_\alpha^\dagger c_\beta$$
If you change the basis, you can write
$$O' = \sum_{\gamma, \delta} \langle \gamma | O_1 | \delta \rangle b_\gamma^\dagger b_\delta$$
The creation and annihiliation operators satisfy the linear transformations
\begin{align} b_\gamma^\dagger &= \sum_{\alpha} c_\alpha^\dagger \langle \alpha | \gamma \rangle \\ b_\gamma &= \sum_{\alpha} \langle \gamma | \alpha \rangle c_\alpha \end{align}
Therefore,
\begin{align} O' &= \sum_{\gamma, \delta} \sum_{\alpha, \beta} c_\alpha^\dagger \langle \alpha | \gamma \rangle \langle \gamma | O_1 | \delta \rangle \langle \delta | \beta \rangle c_\beta \\ &= \sum_{\alpha, \beta} \langle \alpha | O_1 | \beta \rangle c_\alpha^\dagger c_\beta \\ &= O \end{align}
This means that the assumed representation of $O$ is independent of the basis. If you choose a basis $| \gamma \rangle$ that diagonalizes $O_1$, then
$$O = \sum_\gamma o_\gamma b_\gamma^\dagger b_\gamma$$
where $o_\gamma$ is the eigenvalue associated with the ket $| \gamma \rangle$, which is what we expected from the first formula above. Therefore, the assumption for the form of $O$ in any basis was correct.