Michele_Laino
  • Michele_Laino
New Tutorial!! Going from Classical Mechanics to Quantum Mechanics: the example of the Deuteron
Physics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Michele_Laino
  • Michele_Laino
here is the PDF file:
Michele_Laino
  • Michele_Laino
\[\begin{gathered} {\mathbf{THE}}\;\;{\mathbf{SCHROEDINGER}}\;\;{\mathbf{EQUATION}}\;\;{\mathbf{FOR}}\;\; \hfill \\ {\mathbf{GROUND}}\;\;{\mathbf{STATE}}\;\;{\mathbf{OF}}\;\;{\mathbf{THE}}\;\;{\mathbf{DEUTERON: }}\; \hfill \\ {\mathbf{EXAMPLE}}\;\;{\mathbf{OF}}\;\;{\mathbf{TRANSITION}}\;\;{\mathbf{FROM}}\;\; \hfill \\ {\mathbf{CLASSICAL}}\;\;{\mathbf{MECHANICS}}\;\;{\mathbf{TO}}\;\;{\mathbf{QUANTUM}}\;\;{\mathbf{MECHANICS}} \hfill \\ \end{gathered} \] \(By\;\;Michele \; Laino\) \({\mathbf{ABSTRACT}}\) 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). |dw:1449657116102:dw| 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.
Michele_Laino
  • Michele_Laino
\({\mathbf{2) The\; Ground\; State\; of\; the\; Deuteron}}\) The deuteron is a bound state of the pair proton-neutron, with a binding energy \(\epsilon = 2.226\;MeV\) and a potential well of a square shape, defined as below: \[V\left( r \right) = \begin{array}{*{20}{c}} { - {V_0} = - 35\;MeV,\quad r \leqslant {r_0} = 1.7 \cdot {{10}^{ - 13}}\;cm} \\ {0,\quad r > {r_0}} \end{array}\] |dw:1449658745865:dw| In order to describe the relative motion of the pair neutron-proton, we can consider only the part $\mathcal{H}_{rel}$ of the full Hamiltonian function: \begin{equation} {\mathcal{H}_{rel}} = \frac{{{p^2}}}{{2\mu }} + V\left( r \right) \end{equation} Subsequently I write the square of the momentum \(p^2\) of the relative motion, using its Cartesian components: \[{p^2} = p_x^2 + p_y^2 + p_z^2\] and after a substitution into the formula of \({\mathcal{H}_{rel}}\), we can write this: \[{\mathcal{H}_{rel}} = \frac{1}{{2\mu }}\left( {p_x^2 + p_y^2 + p_z^2} \right) + V\left( r \right)\] The transition to Quantum Mechanics, can be made simply replacing \(p_x,\;p_y,\;p_z\) with their corresponding operators: \[{p_x} = - i\hbar \frac{\partial }{{\partial x}},\quad {p_y} = - i\hbar \frac{\partial }{{\partial y}},\quad {p_z} = - i\hbar \frac{\partial }{{\partial z}}\] in doing so, I get the corresponding \(Hamiltonian\; operator\) \(\mathcal{\hat H}_{rel}\): \begin{equation} \hat {\mathcal{H}}_{rel} = \frac{{ - {\hbar ^2}}}{{2\mu }}\left( {\frac{{{\partial ^2}}}{{\partial {x^2}}} + \frac{{{\partial ^2}}}{{\partial {y^2}}} + \frac{{{\partial ^2}}}{{\partial {z^2}}}} \right) + V\left( r \right) \end{equation} Such operator, can be used in order to write the \emph{Schr\"odinger Equation} for relative motion of the neutron-proton system: \begin{equation} \left\{ {\frac{{ - {\hbar ^2}}}{{2\mu }}\left( {\frac{{{\partial ^2}}}{{\partial {x^2}}} + \frac{{{\partial ^2}}}{{\partial {y^2}}} + \frac{{{\partial ^2}}}{{\partial {z^2}}}} \right) + V\left( r \right)} \right\}\psi = - \epsilon \; \psi \end{equation} \noindent or in a more symbolic form: \[\left\{ {\frac{{ - {\hbar ^2}}}{{2\mu }}{\Delta ^2} + V\left( r \right)} \right\}\psi = - \epsilon \; \psi \] where the operator \(\Delta^2\) is the \(Laplace\; operator\). \\ We can simplify such equation, recalling that, the angular moment \(l\) of the ground state of the deuteron is zero: \(l=0\). So the Schroedinger equation can be rewritten, like below: \begin{equation} \left\{ {\frac{{ - {\hbar ^2}}}{{2\mu }}\frac{1}{{{r^2}}}\frac{\partial }{{\partial r}}\left( {{r^2}\frac{\partial }{{\partial r}}} \right) + V\left( r \right)} \right\}\psi = - \epsilon \; \psi \end{equation} Of course, the function \(V(r)\) which appears in equation \((2.4)\), can be replaced by the square potential well, defined at beginning of this section.

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Michele_Laino
  • Michele_Laino
@ganeshie8 @iambatman @rvc @Empty @IrishBoy123 @Astrophysics @BAdhi @AlexandervonHumboldt2 @pratima99
rvc
  • rvc
Im sure ill be needing this somewhere in my study life :) You are extremely great! :) Thank you!
Michele_Laino
  • Michele_Laino
:D :D @rvc
Kainui
  • Kainui
Some questions: I have not really used the Lagrangian except a few times, so I can't say I really understand what these 'conjugate momenta' are. It appears as though they just end up giving the same momentum we'd expect, which in some sense is good, but what are cases where the conjugate momenta aren't the same as the regular linear momenta? Also, I was just curious more about the justification for this formula being true, \[\begin{equation} \mathcal{H} = \sum\limits_i {{q_i}{p_i}} - \mathcal{\hat L} \end{equation}\] I am not really looking for a rigorous answer, I am really more interested in an intuitive or heuristic argument of some kind for these questions if possible. Actually everything after this I have seen before, which was nice and relaxing to read. Also, I am kind of curious why you chose this potential.
Kainui
  • Kainui
Overall I really enjoyed reading it, thanks for sharing! Do you have any handy links to all your other tutorials so I can browse through them at all?
rvc
  • rvc
check his profile @Kainui
Michele_Laino
  • Michele_Laino
thanks!! :) @rvc @kainui I don't have a special link so, please refer to my profile the concept of conjugate momenta, is more general concept, namely if the generalized coordinates are the cartesian coordinates, then the conjugate momenta are equal to the ordinary momenta, nevertheless in some cases the generalized coordinates can be the components of the potential vector \(\mathbf{A}\), and also in this case we have to define the momenta associated to theses coordinates, such momenta are given by the conjugate momenta I'm not able to justify the formula for the hamiltonian function, since our teacher of Analytical Mechanics, gave us such formula as a simple definition
Michele_Laino
  • Michele_Laino
I choose such potential, since I have referred to my notes on nuclear physics. Here is another reference which uses the same potential: \[\begin{gathered} {\text{A}}{\text{.G}}{\text{. Sitenko}}{\text{, V}}{\text{.K}}{\text{. Tartakovskij}} \hfill \\ {\text{Lezioni di Teoria del Nucleo}} \hfill \\ \end{gathered} \] please see the chapter 1 of such reference furthermore, please note that it is an italian traslation from russian language @Kainui
Kainui
  • Kainui
Ok, thanks I appreciate it. I know next to nothing about gauge theory, but I am aware that since the divergence of the magnetic field is zero everywhere from Maxwell's equations that we can consider it to be the curl of a vector potential. Is this magnetic field vector potential related to this sort of vector potential like you describe or do you know? I'm pretty curious about all this stuff cause I know there are some things not too far out of my grasp that I'd really like to understand better like Noether's theorem and stuff.
Michele_Laino
  • Michele_Laino
yes! such vector potential \(\mathbf{A}\) is such that \[{\mathbf{B = }}\nabla {\mathbf{ \times A}}\] ok! I can write another tutorial about potentials and quantum mechanics Furthermore, the value of \(V_==35\;MeV\) can be derived starting from the value of \(r_0\), nevertheless we have to solve the Schroedinger equation, which I wrote in my tutorial
Michele_Laino
  • Michele_Laino
oops.. I meant the value of \(V_0\)...
Michele_Laino
  • Michele_Laino
thanks!! for your appreciation to my tutorial!! :) @Kainui
Kainui
  • Kainui
Ahh ok, yeah the reason I asked is cause I think a parabolic potential might be easier to solve and more realistic too, but yeah thanks for the responses I look forward to further stuff! :D
Anaise
  • Anaise
I don't like it.
Michele_Laino
  • Michele_Laino
@CShrix @arindameducationusc @Robert136
Michele_Laino
  • Michele_Laino
@ParthKohli
Astrophysics
  • Astrophysics
This is great, thanks Michele!!
anonymous
  • anonymous
I don't think I have gone through enough courses to understand some of it, but I'll definitely be saving it to read from time to time and see if I understand more and more X) Good work!!!
Michele_Laino
  • Michele_Laino
thanks!! :D @Astrophysics @CShrix
arindameducationusc
  • arindameducationusc
Thanks @Michele_Laino Will go through it
Michele_Laino
  • Michele_Laino
thanks for your appreciation to my work! :) @arindameducationusc

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