I Translating the harmonic oscillator

ergospherical
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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?
 
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ergospherical said:
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 \psi_n(x + a) as a linear combination of the \psi_n, then look for one. Set <br /> \psi_n(x + a) = \sum_{m} M_{nm} \psi_m(x) and take an appropriate inner product with \psi_k(x) (ideally one with repect to which the \psi_k are orthogonal) to determine the M_{nm}, <br /> \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.
 
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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
$$\langle x|\exp(+\mathrm{i} \hat{p} a) \psi_n \rangle = \int_{\mathbb{R}} \mathrm{d} p \langle x | p \rangle \langle p |\exp(\mathrm{i} \hat{p} a) \psi_n \rangle = \int_{\mathbb{R}} \mathrm{d} p \frac{1}{\sqrt{2 \pi}} \exp(\mathrm{i} p x) \exp(\mathrm{i} p a) \langle p |\psi_n \rangle = \int_{\mathbb{R}} \mathrm{d} p \frac{1}{\sqrt{2 \pi}} \exp[\mathrm{i} p(x+a)] \langle p|\psi_n \rangle= \int_{\mathbb{R}} \mathrm{d} p \langle x+a|p \rangle \langle p|\psi_n \rangle= \psi_n(x+a).$$
 
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Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. Towards the end of the first lecture for the Qiskit Global Summer School 2025, Foundations of Quantum Mechanics, Olivia Lanes (Global Lead, Content and Education IBM) stated... Source: https://www.physicsforums.com/insights/quantum-entanglement-is-a-kinematic-fact-not-a-dynamical-effect/ by @RUTA
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