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This comes from http://ocw.mit.edu/NR/rdonlyres/Physics/8-04Quantum-Physics-ISpring2003/44AEFEB2-BD59-4647-9B54-3F2C57C2B57C/0/ps7.pdf" 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.
The problem asks you to normalize a superposition of two states and show that the normalization is good for any time > 0.
\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}
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} +
\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
Taking the complex square I get:
= \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
And integrating the first two pieces (but not the third yet)
=\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}
+ \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
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 independent 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,
\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
(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
= (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}
If I need to post any of the identities I'm using please let me know.
= (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]
= (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)
= (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]
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.
The problem asks you to normalize a superposition of two states and show that the normalization is good for any time > 0.
\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}
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} +
\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
Taking the complex square I get:
= \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
And integrating the first two pieces (but not the third yet)
=\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}
+ \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
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 independent 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,
\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
(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
= (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}
If I need to post any of the identities I'm using please let me know.
= (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]
= (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)
= (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]
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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