The position operator can be expressed in terms of the harmonic oscillator ladder operators

$$\hat x = \hat a + \hat a^\dagger,$$

in natural units. Therefore we have

$$\hat x |n\rangle = \frac{\hat a + \hat a^\dagger}{\sqrt 2} |n\rangle = \sqrt{\frac{n}{2}}|n-1\rangle + \sqrt{\frac{n+1}{2}}|n+1\rangle.$$

My question is if there is a way to write up the state resulting from applying the position operator an arbitrary number of times? Due to the above relation we can write

$$\hat x^m |n_0\rangle = \sum_{n=n_0-m}^{n_0+m} c_n |n\rangle,$$

where only even or odd coefficients will be non-zero, and where we have assumed $n_0>m$. But is there a way we can write a general expression for $c_n$ as a function of $m,n_0$ by using the recurrence relation above?


1 Answer 1


A foolproof approach is doing this in position representation (where it is just multiplication by $x^m$) using your favorite methods of dealing with Hermit polynomials (e.g., Schiff has a clear presentation of using the generating function, and thus reducing everything to Gaussian integrals.)

Another way is by evaluating first $$e^{\lambda(a^\dagger + a)}|n\rangle\propto e^{\lambda a^\dagger}e^{\lambda a}|n\rangle,$$ where the proportionality coefficient is obtained with Baker-Hausdorff formula. One can then expand result in powers of $\lambda$. In my memory, one ends up with Laguerre polynomials or something like that - so don't expect it to be simple, whatever method one uses.

Finally, one could try induction by evaluating $x|n\rangle$, $x^2|n\rangle$, etc. and guessing the general rule.

  • $\begingroup$ See here. $\endgroup$ Commented May 24, 2023 at 22:30
  • $\begingroup$ @CosmasZachos this seems to be the comment to the OP, rather than my answer. I know how to do this - I didn't go into details, because this is homework stuff. $\endgroup$
    – Roger V.
    Commented May 25, 2023 at 7:32

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