Question

New Tutorial!!
Going from Classical Mechanics to Quantum Mechanics: the example of the Deuteron

Answer 1

here is the PDF file:

Answer 2

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{\mathbf{THE}}\;\;{\mathbf{SCHROEDINGER}}\;\;{\mathbf{EQUATION}}\;\;{\mathbf{FOR}}\;\; \hfill \\
{\mathbf{GROUND}}\;\;{\mathbf{STATE}}\;\;{\mathbf{OF}}\;\;{\mathbf{THE}}\;\;{\mathbf{DEUTERON: }}\; \hfill \\
{\mathbf{EXAMPLE}}\;\;{\mathbf{OF}}\;\;{\mathbf{TRANSITION}}\;\;{\mathbf{FROM}}\;\; \hfill \\
{\mathbf{CLASSICAL}}\;\;{\mathbf{MECHANICS}}\;\;{\mathbf{TO}}\;\;{\mathbf{QUANTUM}}\;\;{\mathbf{MECHANICS}} \hfill \\
\end{gathered} \]

\(By\;\;Michele \; Laino\)

\({\mathbf{ABSTRACT}}\)
In this paper I start with the studying of a system of two interacting particles, with the same mass \(M\), using the framework of classical mechanics, subsequently, I derive the equation of Schroedinger for such system, finally, I update such differential equation, adding the binding energy of the deuteron in its ground state and setting \(M\) equals to the nucleon mass, namely \(M \approx 940\;MeV\).
Really, it is an advanced tutorial, since for its fully comprehension, it is required the knowledge of the Analytical Mechanics and the basics of the Quantum Mechanics as well.

\({\mathbf{1)\;Introduction}}\)

Let's consider a system composed by two particles, of the same mass $M$ and which are interacting with a spring which connects one particle to the other (see figure 1 below).


Within certain limits, we can say that our mechanical system possesses six degrees of freedom, so in order to identify, uniquely, its position, it is suffice to know the six coordinates (\(x,\;y,\;z\)) of both points.
Then, we can conclude that the total kinetic energy \(T\) of such system, is given by the subsequent formula:
\[T = \frac{1}{2}M\left( {\dot x_1^2 + \dot y_1^2 + \dot z_1^2} \right) + \frac{1}{2}M\left( {\dot x_2^2 + \dot y_2^2 + \dot z_2^2} \right)\]
whereas the corresponding \(Lagrangian\; function\) \(\mathcal{L}\), is:
\begin{equation}
\mathcal{L} = T - V = \frac{1}{2}M\left( {\dot x_1^2 + \dot y_1^2 + \dot z_1^2} \right) + \frac{1}{2}M\left( {\dot x_2^2 + \dot y_2^2 + \dot z_2^2} \right) - V\left( {\left| {{{\mathbf{r}}_2} - {{\mathbf{r}}_1}} \right|} \right)
\end{equation}

wherein the function \(V\left( {\left| {{{\mathbf{r}}_2} - {{\mathbf{r}}_1}} \right|} \right)\) takes account of the potential energy of interaction between the two particles of our mechanical system, as a function of the distance between them.

Next we compute the so called \(conjugate\; momentum\) for each degree of freedom, for example, if we consider the coordinate \(x_1\), then the corresponding conjugate momentum \(p_{{x_1}}\), is:
\begin{equation}
{p_{{x_1}}} = \frac{{\partial L}}{{\partial {{\dot x}_1}}} = M{\dot x_1} \Rightarrow {\dot x_1} = \frac{{{p_{{x_1}}}}}{M}
\end{equation}
and similarly for other Lagrangian coordinates: \(y_1,\;z_1,\;x_2,\;y_2,\;z_2\).
Next I write the \emph{Hamiltonian function} of our mechanical system, which is defined by the subsequent formula:
\begin{equation}
\mathcal{H} = \sum\limits_i {{q_i}{p_i}} - \mathcal{\hat L}
\end{equation}

where \( q_i\) are the Lagrangian coordinates, namely \(x_i,\;y_i,\;z_i\), and \(p_i\) are the corresponding conjugate momentum, furthermore, the index \(i\) of summation, runs over all the degrees of freedom. Finally the function \(\mathcal{\hat L}\) is the same Lagrangian function, which is rewritten using the substitutions \((1.2)\).

\[{{\mathbf{r}}_G} = \frac{{M{{\mathbf{r}}_1} + M{{\mathbf{r}}_2}}}{{2M}},\quad {\mathbf{r}} = {{\mathbf{r}}_1} - {{\mathbf{r}}_2}\]
which are the coordinate of the center of mass of the system, and the relative coordinate of the two points, respectively.

From the definitions above, we can write the position vectors \({{\mathbf{r}}_1},\;{{\mathbf{r}}_2}\) , like below:
\[{{\mathbf{r}}_1} = {{\mathbf{r}}_G} + \frac{{\mathbf{r}}}{2},\quad \;{{\mathbf{r}}_2} = {{\mathbf{r}}_G} - \frac{{\mathbf{r}}}{2}\]
As we can see such equations are vector equations, so if we rewrite them using the components of their respective vectors, we get:

\[\begin{gathered}
{x_1} = {x_G} + \frac{x}{2},\quad {y_1} = {y_G} + \frac{y}{2},\quad {z_1} = {z_G} + \frac{z}{2}, \hfill \\
\hfill \\
{x_2} = {x_G} - \frac{x}{2},\quad {y_2} = {y_G} - \frac{y}{2},\quad {z_2} = {z_G} - \frac{z}{2} \hfill \\
\end{gathered} \]
Now, let's consider the first scalar equation, and let's multiply both sides by the mass \(M\), here is what we get:
\[M{x_1} = M{x_G} + \frac{{Mx}}{2}\]
which can be rewritten as follows:
\begin{equation}
M{x_1} = \frac{{{M_{TOT}}}}{2}{x_G} + \mu x
\end{equation}
where \(M_{TOT}\) is the total mass of our mechanical system, and \(\mu\) is the \(reduced\; mass\) of such system, which, in turn, is defined as follows:
\[\frac{1}{\mu } = \frac{1}{M} + \frac{1}{M}\]
Of course, with the aid of the same procedure, we can write similar equations for the remaining degrees of freedom.

If we compute the first derivative of equation \((1.4)\) , we get:
\begin{equation}
{p_{{x_1}}} = \frac{{{p_{{x_G}}}}}{2} + p
\end{equation}
As we can see, equation \((1.5)\) expresses the momentum of particle \(\#1\), along the \(x-\)axis, as a sum between the momentum \(p_G\) of the center of mass of the system, and the momentum \(p\) of the relative motion along the same \(x-\)axis. Again, similar equations can be written for the remaining degrees of freedom.

Finally, I substitute the equations $(1.5)$ into the Hamiltonian $(1.3)$, so I can write this:
\begin{equation}
\mathcal{H} = \frac{{p_G^2}}{{2{M_{TOT}}}} + \frac{{{p^2}}}{{2\mu }} + V\left( r \right) = {\mathcal{H}_G} + {\mathcal{H}_{rel}}
\end{equation}
where:
\[{\mathcal{H}_G} = \frac{{p_G^2}}{{2{M_{TOT}}}},\quad {\mathcal{H}_{rel}} = \frac{{{p^2}}}{{2\mu }} + V\left( r \right)\]

As we can see, we have broken the full Hamiltonian function in two parts, the former \(\mathcal{H}_G\), which describes the motion of the center of mass of the system, and the latter \(\mathcal{H}_{rel}\) , which describes the relative motion of the two particles.