Question
... snip ...
Suppose $ \psi(x, \frac{T}{2}) = \alpha\psi(\beta x + \gamma L, 0) $, we are the values of $ \alpha $, $ \beta $ and $ \gamma $?
Solution:
We have $\phi(x, t) = \sum_{n=1}^{\infty} a_n \psi_n(x) \exp \left(-iE_n t/\hbar \right) $. Notice that the value of $ t $ only impacts the exponential term, so let's focus there.
Substitute in $ E_n = \frac{\pi^2 \hbar^2 n^2}{2mL^2} $ and $ T = \frac{8mL^2}{h} $, we have the exponential term at $ t = \frac{T}{2} $ equals to $ \exp \left(-i\frac{\pi^2 \hbar^2 n^2}{2mL^2} \frac{8mL^2}{h}/\hbar \right) = \exp \left(-in^2\pi \right) = (-1)^n $.
The last equality comes from nowhere, really, it is just an observation on those values, now we know that by shifting time by $ \frac{T}{2} $, we are inverting the odd terms and keeping the even terms untouched.
Next we wanted to derive the values for $ \alpha $, $ \beta $ and $ \gamma $ to achieve the effect of inverting the odd terms and leaving the even terms untouched. The identity $ \sin(x + n\pi) = (-1)^n \sin(x) $ comes in really handy. It is really telling us that set $ \gamma $ to 1.
Thanks for the discussion forum's reminder, I should make sure the range I passed into $ \psi $ lies between $ 0 $ and $ L $, so it leaves me no choice but set $ \beta $ to -1 to maintain that constraint.
Last but not least, setting $ \beta $ to -1 inverted the terms unconditionally by 1, so we need to set $ \alpha $ to -1 to flip it back, so the answer is -1,-1,1
Substitute in $ E_n = \frac{\pi^2 \hbar^2 n^2}{2mL^2} $ and $ T = \frac{8mL^2}{h} $, we have the exponential term at $ t = \frac{T}{2} $ equals to $ \exp \left(-i\frac{\pi^2 \hbar^2 n^2}{2mL^2} \frac{8mL^2}{h}/\hbar \right) = \exp \left(-in^2\pi \right) = (-1)^n $.
The last equality comes from nowhere, really, it is just an observation on those values, now we know that by shifting time by $ \frac{T}{2} $, we are inverting the odd terms and keeping the even terms untouched.
Next we wanted to derive the values for $ \alpha $, $ \beta $ and $ \gamma $ to achieve the effect of inverting the odd terms and leaving the even terms untouched. The identity $ \sin(x + n\pi) = (-1)^n \sin(x) $ comes in really handy. It is really telling us that set $ \gamma $ to 1.
Thanks for the discussion forum's reminder, I should make sure the range I passed into $ \psi $ lies between $ 0 $ and $ L $, so it leaves me no choice but set $ \beta $ to -1 to maintain that constraint.
Last but not least, setting $ \beta $ to -1 inverted the terms unconditionally by 1, so we need to set $ \alpha $ to -1 to flip it back, so the answer is -1,-1,1
No comments:
Post a Comment