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Given an harmonic oscillator I need to calculate the eigenvector $|x=0\rangle$. Knowing that $$x|x=0\rangle = 0 \quad \Rightarrow \quad (a + a^\dagger) | x = 0 \rangle = 0 $$ I started to plug in the matrix notation $$\begin{pmatrix} 0 & 1 & 0 & \cdots & \cdots \\ 1 & 0 & \sqrt 2 & 0 & \cdots \\ 0 & \sqrt 2 & 0 & \sqrt 3 & \cdots \\ \cdots & \cdots & \cdots & \cdots & \cdots \end{pmatrix} \begin{pmatrix} c_0 \\ c_1 \\ c_2 \\ \cdots \end{pmatrix} = 0 $$ Now I want to get the $c_i$ values to get an expression for the ket, but I am kind of stuck. I started to normalize $$ \frac{c_0}{N} = 1 $$ and got some expression for the eigenket $$ |x=0\rangle = N \begin{pmatrix}1 \\ 0 \\ - \frac{1}{\sqrt 2} \\ 0 \\ \left(- \frac{1}{\sqrt 2} \right) \left(- \frac{\sqrt 3}{\sqrt 4} \right) \\ \cdots \end{pmatrix}$$

Now I would like to know how to continue? Or if there is a different way to solve this problem?

I am also a little bit insecure about the normalization: Regarding $$N^2 \cdot \vec c^2 = N^2 \cdot \left( 1 + \frac{1}{2} + \frac{3}{8} + \cdots \rightarrow \infty\right)= 1 \quad \Rightarrow \quad N \to 0$$ So it seems that I cannot normalize the function probperly.

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    $\begingroup$ Yes. I don't think $x$ eigenkets are normalizable. You can check this by looking at $\langle x' | {\hat x} | x \rangle$ and simplifying it in two ways which then implies $(x-x') \langle x' | x \rangle = 0$. This implies that $\langle x' | x \rangle \propto \delta(x-x')$. Thus, they are $\delta$-function normalizable. $\endgroup$ – Prahar Feb 4 '16 at 15:59

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