Problem-3-11-Find-the-momentum-space-wave-function-p-t-for-a-particle-in-the-ground-state-of-the-harmoic-oscillator-What-is-the-probability-to-two-signficant-digits-that-a-measurement-of-on-a-

Question Number 221380 by SdC355 last updated on 02/Jun/25

$$ \\ $$$$\mathrm{Problem}\:\mathrm{3}.\mathrm{11}\:\mathrm{Find}\:\mathrm{the}\:\mathrm{momentum}\:\mathrm{space}\:\mathrm{wave}\: \\ $$$$\mathrm{function}\:\boldsymbol{\Psi}\left({p},{t}\right)\:\mathrm{for}\:\mathrm{a}\:\mathrm{particle}\:\mathrm{in}\:\mathrm{the}\:\mathrm{ground}\:\mathrm{state}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{harmoic}\:\mathrm{oscillator}.\:\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{probability} \\ $$$$\left(\mathrm{to}\:\mathrm{two}\:\mathrm{signficant}\:\mathrm{digits}\right)\mathrm{that}\:\mathrm{a}\:\mathrm{measurement}\:\mathrm{of}\:\mathrm{on}\:\mathrm{a}\:\mathrm{particle}\: \\ $$$$\mathrm{in}\:\mathrm{this}\:\mathrm{state}\:\mathrm{would}\:\mathrm{yield}\:\mathrm{value}\:\mathrm{outside}\:\mathrm{the}\: \\ $$$$\mathrm{classical}\:\mathrm{range}\left(\mathrm{for}\:\mathrm{the}\:\mathrm{samenergy}\right) \\ $$$$\mathrm{Hint}\:\mathrm{Look}\:\mathrm{in}\:\mathrm{a}\:\mathrm{math}\:\mathrm{table}\:\mathrm{under}\:\mathrm{Normal}\:\mathrm{Distribution} \\ $$$$\mathrm{Error}\:\mathrm{Function}\:\mathrm{for}\:\mathrm{the}\:\mathrm{numerical}\:\mathrm{partor}\:\mathrm{use}\:\mathrm{Mathematica} \\ $$

Answered by MrGaster last updated on 06/Jun/25

$$\boldsymbol{\Psi}\left({p},{t}\right)=\left(\frac{\mathrm{1}}{\pi {m}\omega}\right)^{\mathrm{1}/\mathrm{4}} \mathrm{exp}\left(−\frac{{p}^{\mathrm{2}} }{\mathrm{2} {mu}}\right)\mathrm{exp}\left(−{i}\frac{\omega{t}}{\mathrm{2}}\right) \\ $$$${P}=\int_{\mid{p}\mid>\sqrt{{m} \omega}} \mid\boldsymbol{\Psi}\left({p},{t}\right)\mid^{\mathrm{2}} {dp} \\ $$$$\mid\boldsymbol{\Psi}\left({p},{t}\right)\mid^{\mathrm{2}} =\frac{\mathrm{1}}{\:\sqrt{\pi {m}\omega}}\mathrm{exp}\left(−\frac{{p}^{\mathrm{2}} }{ {m}\omega}\right) \\ $$$$\sigma_{{p}} ^{\mathrm{2}} = {m}\omega \\ $$$${p}_{{max}} =\sqrt{{m} \omega}=\sigma_{{p}} \\ $$$${f}\left({p}\right)=\frac{\mathrm{1}}{\:\sqrt{\pi\sigma_{\rho} ^{\mathrm{2}} }}\mathrm{exp}\left(−\frac{{p}^{\mathrm{2}} }{\sigma_{{p}} ^{\mathrm{2}} }\right) \\ $$$${P}=\mathrm{2}\int_{\sigma_{{p}} } ^{\infty} {f}\left({p}\right){dp} \\ $$$${u}=\frac{{p}}{\sigma} \\ $$$${P}=\frac{\mathrm{2}}{\:\sqrt{\pi}}\int_{\mathrm{1}} ^{\infty} {e}^{−{u}^{\mathrm{2}} } {du} \\ $$$$\int_{\mathrm{1}} ^{\infty} {e}^{−{u}^{\mathrm{2}} } {du}=\frac{\sqrt{\pi}}{\mathrm{2}}\mathrm{erfe}\left(\mathrm{1}\right) \\ $$$${P}=\mathrm{erfe}\left(\mathrm{1}\right) \\ $$$$\mathrm{erfe}\left(\mathrm{1}\right)\approx\mathrm{0}.\mathrm{157} \\ $$$${P}\approx\mathrm{0}.\mathrm{16} \\ $$

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