Question:
Given
$\phi\left(x, t \right) = \sum_{n=0}^{\infty}a_n \psi_n\left(x\right)\exp{\left(-i E_n t/\hbar \right)} = \exp{\left(-i \omega t/2 \right)} \Phi\left(x,t\right) $
What is true about $ \Phi\left(x,t\right) $?
Solution:
First, let's look at the energies. $ |\phi\left(x, t \right)|^2 = |\exp{\left(-i \omega t/2 \right)} \Phi\left(x,t\right)|^2 = |\exp{\left(-i \omega t/2 \right)}|^2 |\Phi\left(x,t\right)|^2 = |\Phi\left(x,t\right)|^2 $, so the energies are unchanged while pulling a phase term out.
Also, the Schrödinger equation guarantees that the wavefunction stay normalized, so if the wavefunction was normalized, it will still be at any time.
To look at the periodicity of the term, we simply need to follow what we have done in homework 3 bonus question. The exponential terms looks like $ e^{-\frac{i\hbar E_n}{\hbar}} $, and we also have $ E_n = \hbar\omega(n + \frac{1}{2}) $. With the phase term pulled out, we are left with $ e^{-i\omega n} $, that means we have the periods to be $ \frac{2\pi}{\omega n} $, with the largest one being $ \frac{2\pi}{\omega} $, that would be the same as question 1.
Given
$\phi\left(x, t \right) = \sum_{n=0}^{\infty}a_n \psi_n\left(x\right)\exp{\left(-i E_n t/\hbar \right)} = \exp{\left(-i \omega t/2 \right)} \Phi\left(x,t\right) $
What is true about $ \Phi\left(x,t\right) $?
Solution:
First, let's look at the energies. $ |\phi\left(x, t \right)|^2 = |\exp{\left(-i \omega t/2 \right)} \Phi\left(x,t\right)|^2 = |\exp{\left(-i \omega t/2 \right)}|^2 |\Phi\left(x,t\right)|^2 = |\Phi\left(x,t\right)|^2 $, so the energies are unchanged while pulling a phase term out.
Also, the Schrödinger equation guarantees that the wavefunction stay normalized, so if the wavefunction was normalized, it will still be at any time.
To look at the periodicity of the term, we simply need to follow what we have done in homework 3 bonus question. The exponential terms looks like $ e^{-\frac{i\hbar E_n}{\hbar}} $, and we also have $ E_n = \hbar\omega(n + \frac{1}{2}) $. With the phase term pulled out, we are left with $ e^{-i\omega n} $, that means we have the periods to be $ \frac{2\pi}{\omega n} $, with the largest one being $ \frac{2\pi}{\omega} $, that would be the same as question 1.
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