Question:
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Given $\phi(x,T/4)= \exp{(-i\theta)}\left(\cos{(\theta)}\phi(x,0) + i \, \sin{(\theta)} \phi(x,T/2)\right)$ where $ \theta = \pi/k$ Find $ k $.
Solution:
Using the same approach we did in question 9, we found out that the exponential term becomes $ \exp(-in^2\frac{\pi}{2}) $ when $ t = \frac{T}{4} $.
Inspecting the values, we see what the term really do is that it brings odd terms to the imaginary part and even terms to the real parts.
Next, we look at what $ \theta $ can do for us on the left hand side.
For simplicity, let $ A = \phi(x,0) $ and $ B = \phi(x,\frac{T}{2}) $, the expression is now simplified to
$ \exp{(-i\theta)}(\cos{(\theta)}A + i \, \sin{(\theta)} B) = (\cos(\theta) - i\sin(\theta))(\cos{(\theta)}A + i \, \sin{(\theta)} B) = cos^2(\theta)A + \sin^2(\theta)B + i(\cos(\theta)\sin(\theta)B - \cos(\theta)\sin(\theta)A) $.
We already knew from question 9 that applying $ t = \frac{T}{2} $ inverting the odd terms and leaving the even terms alone.
Here is the key intuition, to get rid of odd terms in the real part, we make $ \cos^2(\theta) = \sin^2(\theta) = \frac{1}{2} $. That makes the real part simply $ \frac{1}{2}(A + B) $, which eliminates the odd terms and leave the even terms alone. So $ \theta = \frac{\pi}{4} $. The same $ \theta $ works for the imaginary part as well and therefore it is the answer!
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Given $\phi(x,T/4)= \exp{(-i\theta)}\left(\cos{(\theta)}\phi(x,0) + i \, \sin{(\theta)} \phi(x,T/2)\right)$ where $ \theta = \pi/k$ Find $ k $.
Solution:
Using the same approach we did in question 9, we found out that the exponential term becomes $ \exp(-in^2\frac{\pi}{2}) $ when $ t = \frac{T}{4} $.
Inspecting the values, we see what the term really do is that it brings odd terms to the imaginary part and even terms to the real parts.
Next, we look at what $ \theta $ can do for us on the left hand side.
For simplicity, let $ A = \phi(x,0) $ and $ B = \phi(x,\frac{T}{2}) $, the expression is now simplified to
$ \exp{(-i\theta)}(\cos{(\theta)}A + i \, \sin{(\theta)} B) = (\cos(\theta) - i\sin(\theta))(\cos{(\theta)}A + i \, \sin{(\theta)} B) = cos^2(\theta)A + \sin^2(\theta)B + i(\cos(\theta)\sin(\theta)B - \cos(\theta)\sin(\theta)A) $.
We already knew from question 9 that applying $ t = \frac{T}{2} $ inverting the odd terms and leaving the even terms alone.
Here is the key intuition, to get rid of odd terms in the real part, we make $ \cos^2(\theta) = \sin^2(\theta) = \frac{1}{2} $. That makes the real part simply $ \frac{1}{2}(A + B) $, which eliminates the odd terms and leave the even terms alone. So $ \theta = \frac{\pi}{4} $. The same $ \theta $ works for the imaginary part as well and therefore it is the answer!
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