The Energy Expectation Value for a Moving Hydrogen Atom

In summary, the author is summarizing the content of a conversation. The author states that the integral over ##X## collapses into a delta function, and that ##\psi_0## is normalized. The author then calculates the expectation value of ##\psi_0## and finds that it is equal to ##E_0 + \frac{\hbar^2 |K|^2}{2m}\right). The author also calculates the total energy of the system as\left(\frac{\hbar^2 |K|^2}{2m} + E_0\right)
  • #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)##?
 
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  • #2
I observe time t is missing which should appear in the phase factor. I do not catch "relative" Hamiltonian relative to what, and what is small k.
 
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  • #3
For clarification: If you treat the hydrogen atom as a (non-relativistic) two-body problem with the proton and the electron, then you can introduce new coordinates, the center of mass and relative coordinates,
$$\vec{X}=\frac{1}{M}(m_e \vec{x}_e + m_p \vec{x}_p), \quad \vec{x}=\vec{x}_e-\vec{x}_p.$$
Since there are no operator-ordering problems in writing down the Hamiltonian, working in Gaussian electromagnetic units,
$$\hat{H}=\frac{\vec{p}_e^2}{2m_e} + \frac{\vec{p}_p^2}{2m_p} -\frac{e^2}{|\vec{x}_e-\vec{x}_p|}$$
in terms of the new coordinates and canonical momenta: The canonically conjugated momentum to ##\vec{X}## is ##\vec{P}=\vec{p}_e+\vec{p}_p##, i.e., the total momentum of the entire hydrogen atom, and the canonical momentum to ##\vec{x}## is given by ##\vec{p}=\mu \mathring{\vec{x}}##, where ##\mu=m_e m_p/M## is the reduced mass of the electron-proton system. The Hamiltonian expressed in these coordinates and their conjugate momenta reads
$$\hat{H}=\frac{1}{2M} \vec{P}^2 + \frac{\vec{p}^2}{2 \mu} -\frac{e^2}{r}=\hat{H}\, \quad r=|\vec{x}|.$$
The center of mass thus moves as a free particle, and the energy eigenfunctions are the product of these free-particle energy-eigenfunctions with the usual hydrogensolutions ##\psi_{n \ell m}(\vec{x})##. For the center-of-mass energy eigenfunctions it's most convenient to choose the corresponding momentum eigenfunctions, i.e., the plane waves,
$$u_{\vec{P}}(\vec{X})=\frac{1}{(2 \pi \hbar)^{3/2}} \exp(\mathrm{i} \vec{X} \cdot \vec{P}).$$
With these you build your wave packet for the center-of-mass motion as given as "Relevant Equation".

For the OP: Your reasoning is almost right. You only forgot the normalization factor ##1/(2 \pi \hbar)^{3/2}##. Also I don't understand, why you first calculate the position representation for the center-mass wave function. It's much simpler to just stay in the momentum-representation, i.e., work with the two-body function
$$\Psi(\vec{P},\vec{x})=g(\vec{P}) \psi_0(\vec{x}).$$
Then you don't need to do any calculations at all ;-)).
 
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  • #4
@vanhees71 Thank you so much for the insight. I didn't mention this originally, but the question asked me to solve it in position space as a way to build my mathematical stamina. I agree that it would have been nicer in momentum space. @anuttarasammyak Sorry, that I forgot the factor of ##t## while I was typing. ##k## was also supposed to be ##K##. I just joined the site a week ago, and I still don't know how to preview my latex. Is there a button I should be pressing? Thanks again.
 
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  • #5
1676160368436.png

Preview button is on the right side up.
 
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