This comes from http://ocw.mit.edu/NR/rdonlyres/Physics/8-04Quantum-Physics-ISpring2003/44AEFEB2-BD59-4647-9B54-3F2C57C2B57C/0/ps7.pdf" [Broken] of the MIT coursework online. This problem seems straightforward to me and I believe I'm making a stupid math mistake of one kind or another, though its possible I'm (once again) missing something more important.(adsbygoogle = window.adsbygoogle || []).push({});

The problem asks you to normalize a superposition of two states and show that the normalization is good for any time > 0.

[tex]\Psi(x,t) = \sqrt{\frac{1}{a}}sin(\frac{\pi x}{a})e^{-i \omega_1 t} + \sqrt{\frac{1}{a}}sin(\frac{2\pi x}{a})e^{-i \omega_2 t}[/tex]

[tex]1 = \int_l^{l+a} <\psi^*|\psi> \,dx = \int_l^{l+a} \Psi(x,t) = \sqrt{\frac{1}{a}}sin(\frac{\pi x}{a})e^{i \omega_1 t} + \sqrt{\frac{1}{a}}sin(\frac{2\pi x}{a})e^{i \omega_2 t} +[/tex]

[tex] \sqrt{\frac{1}{a}}sin(\frac{\pi x}{a})e^{-i \omega_1 t} + \sqrt{\frac{1}{a}}sin(\frac{2\pi x}{a})e^{-i \omega_2 t}\,dx[/tex]

Taking the complex square I get:

[tex]= \int_l^{l+a} \frac{1}{a}sin^2(\frac{\pi x}{a}) + \frac{1}{a}sin^2(\frac{2 \pi x}{a}) + \left(e^{it(\omega_1 - \omega_2)} + e^{-it(\omega_1 - \omega_2)}\right)\left(\frac{1}{a}sin(\frac{\pi x}{a})sin(\frac{2\pi x}{a})\right)\,dx [/tex]

And integrating the first two pieces (but not the third yet)

[tex] =\left[\frac{x}{2a} - \frac{1}{4\pi}sin(\frac{2\pi x}{a})\right]_l^{l+a} + \left[\frac{x}{2a} - \frac{1}{4\pi}sin(\frac{4\pi x}{a})\right]_l^{l+a} [/tex]

[tex]+ \int_l^{l+a} (e^{it(\omega_1 - \omega_2)}+e^{-it(\omega_1 - \omega_2)})\left(\frac{1}{a}sin(\frac{\pi x}{a})sin(\frac{2\pi x}{a})\right)\,dx[/tex]

Now working out those first two results in a 1/2 for each, so there's my entire probability right there. It would seem to follow that the third will be zero, making the normalization independant to time. That's how I thought this would work out, but either I'm wrong about that logic or my math isn't working - or both.

Looking at just that third part,

[tex] \int_l^{l+a} (e^{it(\omega_1 - \omega_2)}+e^{-it(\omega_1 - \omega_2)})\left(\frac{1}{a}sin(\frac{\pi x}{a})sin(\frac{2\pi x}{a})\right)\,dx[/tex]

[tex](e^{it(\omega_1 - \omega_2)}+e^{-it(\omega_1 - \omega_2)})\int_l^{l+a} \frac{2}{a} sin^2(\frac{\pi x}{a}) cos(\frac{\pi x}{a})\,dx[/tex]

[tex]= (e^{it(\omega_1 - \omega_2)}+e^{-it(\omega_1 - \omega_2)})\left[\frac{1}{3\pi}sin^3(\frac{\pi x}{a})\right]_l^{l+a}[/tex]

If I need to post any of the identities I'm using please let me know.

[tex]= (e^{it(\omega_1 - \omega_2)}+e^{-it(\omega_1 - \omega_2)})\left[\frac{1}{3\pi}sin^3(\frac{\pi (l+a)}{a})\right] - \left[\frac{1}{3\pi}sin^3(\frac{\pi l}{a})\right][/tex]

[tex]= (e^{it(\omega_1 - \omega_2)}+e^{-it(\omega_1 - \omega_2)})\left(\frac{1}{3\pi}\left[sin \frac{\pi l}{a}cos\pi + cos\frac{\pi l}{a}sin\pi\right]^3 - \left[\frac{1}{3\pi}sin^3(\frac{\pi l}{a})\right]\right)[/tex]

[tex]= (e^{it(\omega_1 - \omega_2)}+e^{-it(\omega_1 - \omega_2)})\left[\frac{1}{3\pi}\left(-sin^3(\frac{\pi l}{a})\right) - \frac{1}{3\pi}sin^3(\frac{\pi l}{a})\right][/tex]

Which is not zero by any means, nor would the sum be 1.

Please let me know if you see any errors, and thanks for your time reading.

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# Homework Help: Normalizing in Square Well

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