In an interesting version of the energy-time uncertainty principle Δt = τ/π, where τ is the time it takes Ψ(x,t). to evolve into a state orthogonal to Ψ(x,0). Test this out, using a wave function that is a linear combination of two (orthonormal) stationary states of some (arbitrary) potential: Ψ(x,0

Question

In an interesting version of the energy-time uncertainty principle Δt = τ/π, where τ is the time it takes Ψ(x,t). to evolve into a state orthogonal to Ψ(x,0). Test this out, using a wave function that is a linear combination of two (orthonormal) stationary states of some (arbitrary) potential: [latex] \Psi(x, 0)=(1 / \sqrt{2})\left[\psi_{1}(x)+\psi_{2}(x)\right] [/latex].

Accepted Answer

[latex] \Psi(x, t)=\frac{1}{\sqrt{2}}\left(\psi_{1} e^{-i E_{1} t / \hbar}+\psi_{2} e^{-i E_{2} t / \hbar} ight) ; \quad\langle\Psi(x, t) \mid \Psi(x, 0) angle=0 \Rightarrow [/latex]

[latex] \frac{1}{2}\left(e^{i E_{1} t / \hbar}\left\langle\psi_{1} \mid \psi_{1} ight angle+e^{i E_{1} t / \hbar}\left\langle\psi_{1} \mid \psi_{2} ight angle+e^{i E_{2} t / \hbar}\left\langle\psi_{2} \mid \psi_{1} ight angle+e^{i E_{2} t / \hbar}\left\langle\psi_{2} \mid \psi_{2} ight angle ight) [/latex]

[latex] =\frac{1}{2}\left(e^{i E_{1} t / \hbar}+e^{i E_{2} t / \hbar} ight)=0, ext { or } e^{i E_{2} t / \hbar}=-e^{i E_{1} t / \hbar}, ext { so } e^{i\left(E_{2}-E_{1} ight) t / \hbar}=-1=e^{i \pi} [/latex].

Thus [latex] \left(E_{2}-E_{1} ight) t / \hbar=\pi [/latex] (orthogonality also at 3π; 5π; etc., but this is the first occurrence).

[latex] herefore \Delta t \equiv \frac{t}{\pi}=\frac{\hbar}{E_{2}-E_{1}} . \quad ext { But } \Delta E=\sigma_{H}=\frac{1}{2}\left(E_{2}-E_{1} ight) [/latex]. (Problem 3.20). So [latex] \Delta t \Delta E=\frac{\hbar}{2} [/latex].

Question 3.38: In an interesting version of the energy-time uncertainty pri...

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