- #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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