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I'm reading from Landau's book about second quantization and I confused about the bra-ket notation for the creation and annhilation operators. From the book, annhilation oparator defined as $$ a_i|N_i\rangle = \sqrt{N_i}|N_i-1\rangle, \tag{eq. 64.5} $$ and it can be represented in the form of a matrix whose only non-zero element is $$ \langle N_i-1|a_i|N_i\rangle=\sqrt{N_i}. \tag{eq. 64.6} $$

The creation operator can be represented by a matrix whose only non-zero element is (eq. 64.7) $$ \langle N_i|a_i^\dagger|N_i-1\rangle=\sqrt{N_i}. \tag{eq. 64.7} $$

From here, it says that by direct multiplication of the matrices (eqs. 64.6 and 64.7) it can be proved that $$ a_i^\dagger a_i = N_i. $$

The questions are:

  1. What is the form of the matrix from eq. 64.6? Is it correct (for $i = 1,2$)?

$$ \begin{bmatrix}\sqrt{N_1}&0\\0&\sqrt{N_2}\end{bmatrix} $$

  1. How do I represent the multiplication of 64.6 and 64.7 using bra-ket notation? I'm thinking about this but I don't know if it's correct...

$$ \langle N_i-1|a_i|N\rangle \otimes \langle N_i|a_i^\dagger|N_i-1\rangle = \sqrt{N_i}\cdot\sqrt{N_i}=N_i = \begin{bmatrix}N_1&0\\0&N_2\end{bmatrix}. $$

Thanks for help!

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    $\begingroup$ The number operator's matrix representation is diagonal; the ladder operators' matrix representations are off-diagonal. $\endgroup$
    – J.G.
    Commented Nov 6, 2022 at 8:09
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    $\begingroup$ 1. No and 2. No. You can come up with basically infinitely many "is this correct?" attempts - so please explain why do you think this might be the correct way. This enables others to really help you. $\endgroup$ Commented Nov 6, 2022 at 10:58

1 Answer 1

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Both things are wrong:

A bosonic operator has more than one non-zero element in the matrix representation: the $N$ "part" of the $N_i$ index runs from 0 to $\infty$.

So a "stand-alone" annihilation operator would have a matrix representation like so: $$ \hat{a} = \begin{bmatrix} 0 & 1 & 0 & 0 & ...\\ 0 & 0 & \sqrt{2} & 0 & ...\\ 0 & 0 & 0 & \sqrt{3} &... \\ 0 &0 & 0 & 0 & ...\\ \vdots & \vdots & \vdots & \vdots & \ddots \end{bmatrix} $$ Or in bra-ket notation: $$ \hat{a}=\sum_{N>0} \sqrt{N} |N-1\rangle\langle N| $$ And the number operator would be : $$ \hat{a}^\dagger\hat{a} = \begin{bmatrix} 0 & 0 & 0 & 0 & ...\\ 0 & 1 & 0 & 0 & ...\\ 0 & 0 & 2 & 0 & ... \\ 0 & 0 & 0 & 3 & ... \\ \vdots & \vdots & \vdots & \vdots & \ddots \end{bmatrix} $$ Additionally, the $i$ "part" of the $N_i$ index indicates the subspace in which you're acting on, loosely speaking it's only a label. It can be an integer if you're looking at the EM excitations of a box, for example, in which case it will label the mode number of the box. Or it can be a real number indexing the momenta of phonons for example in a (infinite) crystal.

So what we mean when we write down $\hat{a}_i$ (for an integer index) is a tensor product: $$ \hat{a}_i = 1 \otimes ... \otimes 1 \otimes \hat{a} \otimes 1 ... $$

Where the annihilation operator is at the $i$-th place. In principle there's a way to write down the matrix form of tensor product of multiple finite matrices (see Kronecker product) but for bosonic operators you can see it would not be possible as they are represented by infinite matrices.

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  • $\begingroup$ Thanks!! You really helped! $\endgroup$ Commented Nov 6, 2022 at 17:19

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