Translating the harmonic oscillator

[ergospherical]

Let's say I know the position space wavefunctions of the 1d harmonic oscillator ##\psi_n(x)## corresponding to the state ##| n \rangle## are known. I want to write ##\psi_m(x + a)##, for fixed ##m = 1,2,...##, in terms of all of the ##\psi_n(x)##. I know \begin{align*}
\psi_n(x+a) = \langle x | e^{-iaP}| n \rangle &= \int \langle x | e^{-iaP} | p \rangle \langle p | n \rangle dp \\
&= \int e^{-iap} \langle x | p \rangle \bar{\psi}_n(p) dp \\
&= \frac{1}{\sqrt{2\pi}} \int e^{i(x-a)p} \bar{\psi}_n(p) dp
\end{align*}To get it in terms of ##\psi_n(x)## we could Fourier transform, i.e. (?)
\begin{align*}
\psi_n(x+a) = \frac{1}{2\pi} \iint e^{ip(x-x')} e^{-iap} \psi_n(x') dx' dp
\end{align*}It doesn't really look helpful?

 

[pasmith]

Let's say I know the position space wavefunctions of the 1d harmonic oscillator ##\psi_n(x)## corresponding to the state ##| n \rangle## are known. I want to write ##\psi_m(x + a)##, for fixed ##m = 1,2,...##, in terms of all of the ##\psi_n(x)##. I know \begin{align*}
\psi_n(x+a) = \langle x | e^{-iaP}| n \rangle &= \int \langle x | e^{-iaP} | p \rangle \langle p | n \rangle dp \\
&= \int e^{-iap} \langle x | p \rangle \bar{\psi}_n(p) dp \\
&= \frac{1}{\sqrt{2\pi}} \int e^{i(x-a)p} \bar{\psi}_n(p) dp
\end{align*}[/tex]

This appears to be related to the fourier shift theorem (see row 102 of the table here).

If you want [itex]\psi_n(x + a)[/itex] as a linear combination of the [itex]\psi_n[/itex], then look for one. Set [tex]
\psi_n(x + a) = \sum_{m} M_{nm} \psi_m(x)[/tex] and take an appropriate inner product with [itex]\psi_k(x)[/itex] (ideally one with repect to which the [itex]\psi_k[/itex] are orthogonal) to determine the [itex]M_{nm}[/itex], [tex]
\int w(x)\psi_n(x + a)\bar{\psi}_k(x)\,dx = \sum_m M_{nm} \int w(x)\psi_m(x)\bar{\psi}_k(x)\,dx.[/tex]

 

[vanhees71]

As you can easily see yourself, using your Fourier-decomposition method, you simply got the sign wrong in the very first exponential-operator expression, i.e., you have
$$\langle x|\exp(+\mathrm{i} \hat{p} a) \psi_n \rangle = \int_{\mathbb{R}} \mathrm{d} p \langle x | p \rangle \langle p |\exp(\mathrm{i} \hat{p} a) \psi_n \rangle = \int_{\mathbb{R}} \mathrm{d} p \frac{1}{\sqrt{2 \pi}} \exp(\mathrm{i} p x) \exp(\mathrm{i} p a) \langle p |\psi_n \rangle = \int_{\mathbb{R}} \mathrm{d} p \frac{1}{\sqrt{2 \pi}} \exp[\mathrm{i} p(x+a)] \langle p|\psi_n \rangle= \int_{\mathbb{R}} \mathrm{d} p \langle x+a|p \rangle \langle p|\psi_n \rangle= \psi_n(x+a).$$