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I have got a problem understanding the meaning of the Laplace operator in the Schrödinger equation for the hydrogen atom.

$$\Big(-\frac{\hbar^2}{2m_e} \Delta_{r_e} - \frac{\hbar^2}{2M_P} \Delta_{r_p} +V(r) \Big)\Psi(r_e,r_p) = E \Psi(r_e,r_p)$$

What exactly does this $\Delta_{r_e}$ mean? I only know $\Delta = \frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2}$. So I am kind of confused by the index of the $\Delta$. And what does $\Psi(r_e,r_p)$ look like?

Once the barycenter and relative coordinates are introduced (not sure if this translation is correct) the Schrödinger equation is written as:

$$\Big(-\frac{\hbar^2}{2(m_e+M_p)} \Delta_{_{R}} - \frac{\hbar^2}{2\mu} \Delta_{r} +V(r) \Big)\Psi(r,R) = E \Psi(r,R)$$ With: $$R=\frac{m_e \overrightarrow{r_e} + M_e \overrightarrow{r_p}}{m_e + M_p}$$ $$\mu = \frac{m_eM_p}{m_e+M_p}$$

Same problem, what does this $\Delta_{_{R}}$ mean exactly, and what does the $\Psi(r,R)$ look like? And what happens if I use $\Delta_{_{R}}$ on $\Psi(r,R)$?

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2 Answers 2

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What exactly does $\Delta_{r_e}$ mean ?

Your wave function isn't a field in space, it is a field on configuration space, i.e. it assigns complex numbers to a configuration. If your electron is at $(x_e,y_e,z_e)$ and your proton is at $(x_p,y_p,z_p)$ then the configuration is the point $(x_e,y_e,z_e,x_p,y_p,z_p)$ in a 6d space. A point in that 6d space tells you where both the particles are, it tells you the configuration.

So since it is a 6d space there are six directions you can take derivatives in: $\partial/\partial x_e,$ $\partial/\partial y_e,$ $\partial/\partial z_e,$ $\partial/\partial x_p$ $\partial/\partial y_p,$ and $\partial/\partial z_p.$

$\Delta_{r_e}$ equals $$ \frac{\partial^2}{\partial x_e^2}+\frac{\partial^2}{\partial y_e^2}+\frac{\partial^2}{\partial z_e^2}.$$

Similarly $$\Delta_{r_p}=\frac{\partial^2}{\partial x_p^2}+\frac{\partial^2}{\partial y_p^2}+\frac{\partial^2}{\partial z_p^2}.$$

So each one takes derivatives in just three of those six directions.

If you switch coordinates (such as to $\vec r,$ $\vec R$ then you still have six coordinates because it is still a six dimensional space and you can break them into two groups of three and take the Laplacian of each group.

What does the $\Psi(\vec r,\vec R)$ look like ? Separation of variables give a free particle in $R$ and a regular hydrogen atom solution (with reduced mass) in $r$. So, for instance, if there is a plane wave of fixed momentum for the free particle $e^{i\vec P \cdot \vec R}$ and a let $\Phi_n(\vec r)$ be an energy eigenstate of the hydrogen atom (with reduced mass). Then we can consider $\Psi(\vec r, \vec R)=e^{i\vec P \cdot \vec R}\Phi_n(\vec r).$

What does $\Psi(r_e,r_p)$ look like ?

$\Psi(r_e,r_p)=e^{i\vec P \cdot (m_e\vec r_e+M_p\vec r_p)/(m_e+M_p)}\Phi_n(\vec x_e-\vec x_p).$

If you want it to be normalizable you need to take combinations with different $\vec P$ to get a wave packet motion for the center of mass.

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The index of the Laplacian tells you which of the coordinates it acts on, that is, if you write $r = (r_x,r_y,r_z)^T$ and $R = (R_x,R_y,R_z)^T$ as Cartesian coordinates, then \begin{align} \Delta_r & := \frac{\partial^2}{\partial {r_x}^2} + \frac{\partial^2}{\partial {r_y}^2} + \frac{\partial^2}{\partial {r_z}^2} \\ \Delta_R & := \frac{\partial^2}{\partial {R_x}^2} + \frac{\partial^2}{\partial {R_y}^2} + \frac{\partial^2}{\partial {R_z}^2} \end{align}

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  • $\begingroup$ Thanks, that's what i wanted to know. Could u also tell me how the corresponding $\Psi(r,R)$ function looks like ? $\endgroup$
    – user85397
    Commented Aug 23, 2015 at 0:10
  • $\begingroup$ @Mareck: You mean the solution? No, I have not memorized the solution. $\endgroup$
    – ACuriousMind
    Commented Aug 23, 2015 at 0:13

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