 #1
uxioq99
 11
 4
 Homework Statement:

Let ##\Psi(X, x, t)## be given by
##\Psi(X, x, t) = \int_{\mathbb R^3} g(K) \psi_0(x) e^{iK\cdot X + \left(\frac{\hbar^2 K^2}{2m} + E_0\right)} d^3 K##
where ##X## is the center of mass coordinate, ##x## is the relative coordinate, and ##t## is the time. ##\psi_0(x)## is a normalized eigenfunction of the relative Hamiltonian ##H_{\text{Rel}}## such that ##H_{\text{Rel}} \psi_0 (x) = E_0 \psi(x)##. ##g(k)## is a function that peaks in the neighborhood ##K \approx K_0##.
 Relevant Equations:
 ##\Psi(X, x, t) = \int_{\mathbb R^3} g(K) \psi_0(x) e^{iK\cdot X + \frac{i}{\hbar} \left(\frac{\hbar^2 K^2}{2m} + E_0\right)} d^3 K##
##\begin{align}
\langle E \rangle &=
\int_{\mathbb R^3}
\int_{\mathbb R^3}
\int_{\mathbb R^3}
\int_{\mathbb R^3}
g^\dagger (\tilde K) g(K) \psi_0(x)^2
\left(E_0 +\frac{\hbar^2 K^2}{2m}\right)
e^{i(K\tilde K)\cdot X \frac{i}{\hbar} \left(\frac{\hbar^2 K^2}{2m}\frac{\hbar^2 K^2}{2m}\right)}
d^3 K d^3 \tilde K d^3 x d^3 X \\
&=
\int_{\mathbb R^3} \psi_0(x)^2 d^3 x
\int_{\mathbb R^3}
\int_{\mathbb R^3}
\int_{\mathbb R^3}
e^{i(K\tilde K)\cdot X} d^3 X
g^\dagger (\tilde K) g(K)
\left(E_0 +\frac{\hbar^2 K^2}{2m}\right)
e^{\frac{i}{\hbar} \left(\frac{\hbar^2 K^2}{2m}\frac{\hbar^2 K^2}{2m}\right)}
d^3 K
d^3 \tilde K
\\ &=
\left(\frac{\hbar^2 K^2}{2m} + E_0\right)
\end{align}##
as the integral over ##X## collapses into a delta function, ##\psi_0## is normalized, and both integrals peak at ##K_0##. Is my reasoning correct? I apologize in advance if my math is poorly formatted. I am still new to the site.
Do I need a factor of ##2 \pi## for the delta function? The expectation value must real because it is observed right? So, I believe that the phase must cancel. Originally, I was curious if one could argue that
##\frac{\hbar^2 K^2  \hbar^2 \tilde K^2}{2m} \approx A \frac{\hbar^2 2K}{2m}##
when ##K\tilde K\le\epsilon## and
##A = \frac{1}{\frac{4}{3}\pi\epsilon^3} \int_{\mathcal B(K, \epsilon)} \tilde K  K d^3 \tilde K##
fixing ##K## and considering ##\tilde K## to be free. This strengthening of the approximation would have introduced a phase. Is that why ##X## has to introduce the ##\delta## function to cancel it out. What are the purpose of ##g(K)## and ##g^\dagger (\tilde K)##?
\langle E \rangle &=
\int_{\mathbb R^3}
\int_{\mathbb R^3}
\int_{\mathbb R^3}
\int_{\mathbb R^3}
g^\dagger (\tilde K) g(K) \psi_0(x)^2
\left(E_0 +\frac{\hbar^2 K^2}{2m}\right)
e^{i(K\tilde K)\cdot X \frac{i}{\hbar} \left(\frac{\hbar^2 K^2}{2m}\frac{\hbar^2 K^2}{2m}\right)}
d^3 K d^3 \tilde K d^3 x d^3 X \\
&=
\int_{\mathbb R^3} \psi_0(x)^2 d^3 x
\int_{\mathbb R^3}
\int_{\mathbb R^3}
\int_{\mathbb R^3}
e^{i(K\tilde K)\cdot X} d^3 X
g^\dagger (\tilde K) g(K)
\left(E_0 +\frac{\hbar^2 K^2}{2m}\right)
e^{\frac{i}{\hbar} \left(\frac{\hbar^2 K^2}{2m}\frac{\hbar^2 K^2}{2m}\right)}
d^3 K
d^3 \tilde K
\\ &=
\left(\frac{\hbar^2 K^2}{2m} + E_0\right)
\end{align}##
as the integral over ##X## collapses into a delta function, ##\psi_0## is normalized, and both integrals peak at ##K_0##. Is my reasoning correct? I apologize in advance if my math is poorly formatted. I am still new to the site.
Do I need a factor of ##2 \pi## for the delta function? The expectation value must real because it is observed right? So, I believe that the phase must cancel. Originally, I was curious if one could argue that
##\frac{\hbar^2 K^2  \hbar^2 \tilde K^2}{2m} \approx A \frac{\hbar^2 2K}{2m}##
when ##K\tilde K\le\epsilon## and
##A = \frac{1}{\frac{4}{3}\pi\epsilon^3} \int_{\mathcal B(K, \epsilon)} \tilde K  K d^3 \tilde K##
fixing ##K## and considering ##\tilde K## to be free. This strengthening of the approximation would have introduced a phase. Is that why ##X## has to introduce the ##\delta## function to cancel it out. What are the purpose of ##g(K)## and ##g^\dagger (\tilde K)##?
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