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We have learned creation operator $\hat{a}^\dagger_i$ adds a particle in $i^{th}$ state, and annihilation operator $\hat{a}_j$ remove a particle from $j^{th}$ state. They can be interpretated in such ways, and when we act them on number state we can get some very elegant results. But how do we actually express them individually in terms of eigenstates or in matrix form, similarly as we express other observables?

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To express an operator as a matrix, you must first choose a basis. Matrices and vectors are just mathematical objects that group numbers. The basis gives them physical meaning.

Let's choose the basis of the number eigenstates $|n\rangle$. That is, given a column vector, the $n^{\mathrm{th}}$ entry represents the occupation of the $n^{\mathrm{th}}$ number eigenstate. Just to be clear, "number eigenstates" are eigenstates of the number operator $a^\dagger a$, which may count the number of particles or number of energy quanta.

In this basis, the vacuum is $n=0$ and hence $|0\rangle$:

\begin{equation} \left| 0 \right\rangle = \left( \begin{array}{c} 1 \\ 0 \\ \vdots \\ 0 \end{array} \right) \end{equation}

To determine the matrix entries of $a^\dagger$, you need to use the definition of how this operator acts on our chosen basis: \begin{align} a ^\dagger _{ nm} & = \left\langle n \right| a ^\dagger \left| m \right\rangle \\ & = \sqrt{ m + 1 } \delta _{ n , m + 1 }. \end{align}

Then, we have, \begin{equation} \left( \begin{array}{cccc} 0 & 0 & 0 & ...\\ 1 & 0 & 0 &...\\ 0 & \sqrt{2} & 0 &... \\ 0 &0 & \sqrt{3} & \ddots\\ \vdots &... & \ddots & \ddots \end{array} \right). \end{equation}

To show that it works, let's check how $a^\dagger$ acts on some states: $$ a^\dagger |0\rangle = \left( \begin{array}{cccc} 0 & 0 & 0 & ...\\ 1 & 0 & 0 &...\\ 0 & \sqrt{2} & 0 &... \\ 0 &0 & \sqrt{3} & \ddots\\ \vdots &... & \ddots & \ddots \end{array} \right) \cdot \left( \begin{array}{c} 1 \\ 0 \\ \vdots \\ 0 \end{array} \right) = \left( \begin{array}{c} 0 \\ 1 \\ \vdots \\ 0 \end{array} \right) = |1\rangle, $$ as expected.

And: $$ a^\dagger |1\rangle = \left( \begin{array}{cccc} 0 & 0 & 0 & ...\\ 1 & 0 & 0 &...\\ 0 & \sqrt{2} & 0 &... \\ 0 &0 & \sqrt{3} & \ddots\\ \vdots &... & \ddots & \ddots \end{array} \right) \cdot \left( \begin{array}{c} 0 \\ 1 \\ 0 \\ \vdots \\ 0 \end{array} \right) = \left( \begin{array}{c} 0 \\ 0 \\ \sqrt{2} \\ \vdots \\ 0 \end{array} \right) = \sqrt{2}|2\rangle, $$ as expected.

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  • $\begingroup$ Thanks. How do you write in first quantized notation? $\endgroup$
    – Noah Ren
    Jun 19 '20 at 4:18
  • $\begingroup$ It's the same thing. The difference between first/second quantisation is the interpretation of the state $|n\rangle$. $\endgroup$
    – SuperCiocia
    Jun 19 '20 at 4:19
  • $\begingroup$ Thank you very much. $\endgroup$
    – Noah Ren
    Jun 19 '20 at 6:43

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