I am trying formally derive the projection of the energy eiegenstates of the 1D quantum harmonic oscillator into the $x$ basis $$ \phi_n(x) = \langle x | n \rangle = \langle x | \frac{{a^{\dagger}}^{n}}{\sqrt{n!}} | 0 \rangle = \frac{{a^{\dagger}}^{n}}{\sqrt{n!}} \, \phi_0(x) $$

I'm not sure how to make the last step

$$ \langle x |\frac{{a^{\dagger}}^{n}}{\sqrt{n!}} | 0 \rangle = \frac{{a^{\dagger}}^{n}(x)}{\sqrt{n!}} \langle x | 0 \rangle $$

where I use ${a^{\dagger}}^{n}(x)$ to clarify that the operator is in the $x$ basis.

I feel I need to use the resolution of identity

$$ \int \langle x | \frac{{a^{\dagger}}^{n}}{\sqrt{n!}}| x' \rangle \langle x'| 0 \rangle \: \mathrm{d} x'$$

I can do this integral for a single creation operator

$$ \int \langle x | a^{\dagger}| x' \rangle \langle x'| 0 \rangle \: \mathrm{d} x' = \int \langle x | \gamma X - i \epsilon P| x' \rangle \langle x'| 0 \rangle \: \mathrm{d} x' $$

where $ \gamma = \sqrt{\frac{m \omega}{2 \hbar}}$ and $ \epsilon = \sqrt{\frac{1}{2m \omega \hbar}}$. Using the matrix representations of the $X$ and $P$ operators in the $x$ basis

$$ \langle x |\gamma X - i \epsilon P| x' \rangle = x' \gamma \delta(x - x') + \hbar \epsilon\delta^{(1)}(x - x')$$

where $\delta^{(1)}(x)$ is the first derivative of the Dirac delta function. So the integral is

$$ \phi_1(x) = \int \left[ x' \gamma \delta(x - x') + \hbar \epsilon\delta^{(1)}(x - x') \right] \phi_0 (x') \: \mathrm{d} x' = \left[ \gamma x - \hbar \epsilon\frac{\mathrm{d}}{\mathrm{d} x} \right]\phi_0 (x)$$

So how do I continue with this derivation? Do I make some sort of inductive step? Is this even the right path to go down, is there a much simpler method?

  • 1
    $\begingroup$ You've already done it! What do you even want help with? $\endgroup$
    – Prahar
    Commented Apr 20, 2017 at 13:15
  • $\begingroup$ To check the recursive generation of Hermite polynomials, consult WP. $\endgroup$ Commented Apr 20, 2017 at 14:35
  • $\begingroup$ That is, your already know that $\phi_{n+1}(x)\propto \left ( x- {d\over dx}\right ) \phi_n(x)$. $\endgroup$ Commented Aug 1, 2020 at 15:29

2 Answers 2


I have not checked the details and this is too long for a comment but it looks right and yes it is induction. Instead of starting with $\phi_n(x)=\langle x\vert \frac{(\hat a^\dagger)^n}{\sqrt{n!}}\vert 0\rangle$, start with \begin{align} \phi_{n+1}(x)&=\frac{1}{\sqrt{n+1}}\langle x\vert \hat a^\dagger\vert n\rangle\, ,\\ &= \frac{1}{\sqrt{n+1}}\left[\gamma x-\hbar \epsilon \frac{d}{dx}\right] \phi_n(x) \tag{1} \end{align} where you obtain (1) by repeating your steps starting with $\vert n\rangle$ rather than $\vert 0\rangle$. You can then start the induction.


There is another way to perform the calculation, which differs from the standard approach. I've provided this alternative approach in [Phys. Rev. A 98, 043841 (2018); arXiv]. The alternative approach is based on generating functions.

First, we simplify the problem by replacing the position and momentum operators by quadrature operator into which all dimensional parameters are absorbed, so that the ladder operator can be expressed in terms of the quadrature operators $\hat{q}$ and $\hat{p}$ by $$ \hat{a} = \frac{1}{\sqrt{2}}(\hat{q}+i\hat{p}) , $$ and its Hermitian conjugate.

The question is now to compute the expression for an arbitrary Fock state as a function of $q$, which is the eigenvalues of the quadrature operator associated with (or analogous to) the position operator. In other words, we want to compute $$ \psi_n(q)=\langle q|n\rangle = \langle q|\frac{1}{\sqrt{n!}} (\hat{a}^{\dagger})^n|\text{vac}\rangle. $$ Here we introduce the generating function: we multiply the right-handside by $J^n/\sqrt{2^n n!}$ and sum over $n$. It gives $$ \mathcal{G}(J) = \langle q|\exp\left(\frac{J\hat{a}^{\dagger}}{\sqrt{2}}\right)|\text{vac}\rangle . $$ Since $$ \exp\left(K\hat{a}\right)|\text{vac}\rangle=|\text{vac}\rangle , $$ we can insert such an extra exponential to get $$ \mathcal{G}(J) = \langle q|\exp\left(\frac{J\hat{a}^{\dagger}}{\sqrt{2}}\right) \exp\left(\frac{J\hat{a}}{\sqrt{2}}\right)|\text{vac}\rangle . $$ Using the Baker-Campbell-Hausdorff formula $$ \exp(\hat{A})\exp(\hat{B})=\exp(\hat{A}+\hat{B}+[\hat{A},\hat{B}]/2) , $$ we combine the two exponents: $$ \mathcal{G}(J) = \langle q|\exp\left(\frac{J\hat{a}^{\dagger}}{\sqrt{2}} +\frac{J\hat{a}}{\sqrt{2}}-\frac{1}{4}J^2\right)|\text{vac}\rangle . $$ The ladder operators can now be combined into the quadrature operator $$ \mathcal{G}(J) = \langle q|\exp\left(J\hat{q}-\frac{1}{4}J^2\right)|\text{vac}\rangle . $$ The resulting operator only depends on $\hat{q}$. Therefore we can pull it out of the inner product by replacing $\hat{q}$ by its eigenvalue $q$. Thus we get $$ \mathcal{G}(J) = \psi_0(q)\exp\left(Jq-\frac{1}{4}J^2\right) , $$ where $$ \psi_0(q)=\langle q|\text{vac}\rangle . $$ You may recognize the resulting generating function as having the same form as the generating function for the Hermite polynomials: $$ \exp\left(2x\nu-\nu^2\right) =\sum_{n=0}^{\infty} \frac{\nu^n}{n!} H_n(x) . $$ So the individual wavefunctions have the form $$\psi_n(q)= \frac{\psi_0(q)}{\sqrt{2^n n!}} H_n(q) . $$ It now only remains to determine $\psi_0(q)$, which we can find with the aid of the orthogonality condition. In the end we have $$\psi_n(q)= \frac{H_n(q)}{\pi^{1/4}\sqrt{2^n n!}}\exp\left(-\frac{q^2}{2}\right) . $$

Hope that was not too painful.


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