Electron transitions and spin

Question

Take a single proton and electron. Write $n,\ell,m$ for the quantum numbers of the electron, and $1/2$ or $-1/2$ for its spin.

The space of states of this system is the separable Hilbert space $V$ with orthonormal basis $$\{v_{n,\ell,m,\pm 1/2}\ : \ n,\ell,m \text{ as above}\}.$$ The Hamtiltonian $H$ is a Hermitian operator on $V$.

Question: I'm interested in the probability of one state turning into another, or equivalently in $$\langle v_{n,\ell,m,\pm 1/2}|H|v_{n',\ell',m',\pm' 1/2}\rangle.$$ What is this, as a function of $n,n',\ell,\ell',m,m',\pm, \pm'$?

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I'm happy for the answer to be an integral, a sum etc. as long as it's a function of those variables. Since the state is completely described by $n,\ell,m,\pm 1/2$, no other variables (like $\textbf{S},\textbf{L},\psi,...$) should appear, unless they are some clearly-stated functions of $n,\ell,m,\pm 1/2$.

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Edit: to be clear, $n,\ell,m$ are the principal, azithumal and magnetic quantum numbers.

Secondly, my question is what the real-life entries of $H$ are (because up to a little ambiguity they can be observed). i.e. what are leading few terms after the diagonal, and what physical effects are they caused by?

For instance, some effects which might add non-diagonal terms to $H$: $n$ changing due to absorption/emission of a photon, the spin may effect the energy a bit (causing an "$n$-$\pm 1/2$" off-diagonal term), some "$\ell$-$\pm 1/2$" off-diagonal terms due to the $\mathbf{L}\cdot\mathbf{S}$ terms I've seen before in similar contexts, etc.

Answer 1

The short answer would that this function is determined by what $H$ is, and $\langle v_{n,\ell,m,\pm 1/2}|H|v_{n',\ell',m',\pm' 1/2}\rangle$ is the so-call "matrix-element of $H$ in angular momentum basis".

If we need a general answer, we must look at the symmetry of H. Because the $|n,l,m,s\rangle$ is eigenstates of operator $\mathbf{L}^2$, $L_z$, $S^2$, $S_z$, $V(\mathbf{r})$. That means if $H$ commutes with all of these operators, there wouldn't be any off-diagonal elements. Real life example is the ideal Hydrogen atom without relativity.

If there is terms like $L\cdot S$(spin-orbit coupling of atoms) or $\hat{x}$(atom interaction with plane wave lights) in the Hamiltonian. Then $|n,l,m,s\rangle$ is no longer an eigenstate, since $[H,L_z]\ne 0$. These terms will produce off-diagonal elements and it lets 'neighboring' levels couple to each other. In particular, $L\cdot S$ is caused by the fact that a moving electron can 'feel' magnetic field due to relativity.

For a minimum example, let's write things explicitly in (x,y,z) basis for a spinless Hydrogen wave function, and calculate the first few matrix elements.

The wavefunction of hydrogen atom is $${\displaystyle \psi _{n\ell m}(r,\vartheta ,\varphi )={\sqrt {{\left({\frac {2}{na_{0}^{*}}}\right)}^{3}{\frac {(n-\ell -1)!}{2n(n+\ell )!}}}}e^{-\rho /2}\rho ^{\ell }L_{n-\ell -1}^{2\ell +1}(\rho )Y_{\ell }^{m}(\vartheta ,\varphi )}$$ according to wikipedia: Hydrogen atom

Put the formula in Mathematica

a0 = 0.1;
\[Psi][n_, l_, m_, x_, y_, z_] := 
   Sqrt[4 (n - l - 1)!/(a0^3 *n^3 (n + l)!)]
   E^(-2 Sqrt[x^2 + y^2 + z^2]/(n a0)/2) 2 Sqrt[x^2 + y^2 + z^2]/(n a0)^l  
  LaguerreL[n - l - 1, 2 l + 1,2 Sqrt[x^2 + y^2 + z^2]/(n a0)] 
  (-1)^m SphericalHarmonicY[l, m, ArcTan[z, Sqrt[x^2 + y^2]], ArcTan[x, y]]

Where $a0$ is an arbitrary Bohr radius that I defined.

Visualization:

plist = Flatten[Table[{n, l, m}, {n, 2}, {l, 0, n - 1}, {m, -l, l}], 2];
 Table[
  DensityPlot[
  Evaluate[Abs[\[Psi] @@ (i~Join~{x, 0, z})]]^2, {x, -20. a0, 
  20 a0}, {z, -20 a0, 20 a0}, PlotRange -> {0, 1}],
 {i, plist}]

Which gives

Now let $H_I$ be the light matter interaction term. One of the simplest Hamiltonian is $$ H_I = \hat{V}(x,y,z) = c \hat{x}, $$ where $c$ is a constant and $\hat{x}$ is the $x$ position operator.

For $H_I=x$ the integral is easy without computer, but for a generic $H_I$ you can always calculate it numerically.

data = Table[
NIntegrate[
 Conjugate[\[Psi] @@ (i~Join~{x, y, z})] x (\[Psi] @@ (j~
      Join~{x, y, z})),
 {x, -20 a0, 20 a0}, {y, -20 a0, 20 a0}, {z, -20 a0,  a0}],
{i, plist}, {j, plist}]; // Quiet

It yields the matrix element of the $nlm=100,200,21-1,210,211$

Round[data, 0.0001]
(* {{0., 0., -0.0436, 0., 0.0436}, 
    {0., 0., 0.1882, 0., -0.1882}, 
    {-0.0436, 0.1882, 0., 0., 0.}, 
    {0., 0., 0., 0., 0.}, 
    {0.0436, -0.1882, 0., 0., 0.}} *)

The only non-zero elements are $\Delta l=\pm 1, \Delta m = \pm 1$. For $H_I= \hat{z}$, it is $\Delta m =0$. Combine together, it gives the selection rule $\Delta l=\pm 1, \Delta m = 0,\pm 1$ for electric dipole transition.