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I want to find eigenstates of $\hat{x} = a + a^\dagger$ operator, I've started by considering a linear combination like $\sum_{m=0} c_m (a^\dagger)^m |0 \rangle$ and match $c_m$ such that linear combination to be an eigenstate of $\hat{x}$.

But it leads to a recursive relation on $c_m$ which seems hard to evaluate, and I happen to know that eigenstates should be labeled by a complex number $z$, so I don't know how to enter it in my analysis.

Could anyone help me or mention a reference?

Connor Behan
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hossein mohammadi
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  • There's no reason to label eigenstates by a complex number... $\hat{x}$ is Hermitian so its eigenvalues are real. – Connor Behan Dec 14 '21 at 16:51
  • Duplicate. i.e. $\frac{e^{x^2/2}}{\pi^{1/4}} e^{-(a^\dagger-\sqrt{2} x)^2/2} |0\rangle$. – Cosmas Zachos Dec 14 '21 at 16:58
  • @Cosmas Zachos: Is this related to the inverse Weierstrass transfoms $ \exp\left{+\tau\frac {d^2}{dx^2}\right} H_n(x/\sqrt{4\tau}) = \frac 1{\sqrt {4 \pi\tau }} \int_{-\infty}^{\infty} H_n(y/\sqrt{4\tau}) e^{-(x-y)^2/4\tau} dy = \tau^{-n/2}x^n? $ – mike stone Dec 16 '21 at 13:27
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    @mike stone. I'm sure it is... I believe it's all in the Cnudde & DeBie paper cited in my answer, but I did not bother with the fine details of the Segal-Bargmann transform of the esthetes... I just utilized the seat-of-the-pants shortcut $ \langle n| x \rangle= \frac{1}{\pi^{1/4}\sqrt{2^n~n!}}~ (x-\partial_x)^n~ e^{-x^2/2} $. – Cosmas Zachos Dec 16 '21 at 14:14

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The normalized eigenstates of the frequency $\Omega$ Harmonic oscillator are $$ \langle x|n\rangle=\frac{\varphi_n(\sqrt{\Omega_0} x)}{(\Omega_0)^{1/4}} $$ where $$ \varphi_n(x)\equiv \frac{1}{\sqrt{2^n n! \sqrt{\pi}}} H_n(x) e^{-x^2/2}. $$ We want to expand the $\hat x$ eigenvalue $x_0$ eigenstate wavefunction $\langle x|x_0\rangle=\delta(x-x_0)$ in terms of these, but the completeness relation tells us that $$ \sum_{n=0}^\infty \langle x|n\rangle\langle n|x_0\rangle=\delta(x-x_0) $$ and so $$ | x_0\rangle= \sum_n|n\rangle\langle n|x_0\rangle=\sum_n |n\rangle \frac{\varphi_n(\sqrt{\Omega_0} x_0)}{(\Omega_0)^{1/4}} $$

I don't know where complex numbers are coming in. Are you sure that you do not want to introduce coherent states $$ |z\rangle =e^{a^\dagger z}|0\rangle? $$ We do have $$ z\in {\mathbb C}, $$ but these are not eigenstates of $\hat x$, but rather of $\hat a$.

mike stone
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