Magnetic sublevels of a random atom

[Malamala]

Hello! Assume I create an atom by some non-state-selective method (e.g. laser ablation, or hitting a proton on a target) and let's say that the atom is in a $J=1$ state. In the absence of magnetic fields, the $m_J = 0, \pm 1$ levels are degenerate. If I am to define arbitrary a z-axis (say along the beamline in which I will study the atom), will the wavefunction of this atom be:

$$a|m_J=-1>+b|m_J=0>+c|m_J=1>$$
i.e. a coherent superposition of different magnetic sublevels? And will different atoms (usually I generate several in the $J=1$ state) have different a, b and c values?

I know that on average, I expect 1/3 of the atoms in the $J=1$ state in each magnetic substate. So if I am to rotate my frame of reference such that the wavefunction above becomes, say, $|m_J=0>$ will in this frame of reference all the atoms be exactly in one of the states $|m_J=-1>$, $|m_J=0>$ or $|m_J=1>$ or can there still be some of them in linear superpositions of these basis?

Also, if I am in the initial frame, where I have a linear superposition of substates and I apply a magnetic field (adiabatically such that I don't transfer population) along the current defined z-axis, will the atom still have the same population in each of the (now non-degenerate) magnetic substates? Thank you!

 

[Twigg]

For a process like ablation, you can safely assume that the resulting atoms are in a completely incoherent statistical mixture of spin states.

So if I am to rotate my frame of reference such that the wavefunction above becomes, say, |mJ=0> will in this frame of reference all the atoms be exactly in one of the states |mJ=−1>, |mJ=0> or |mJ=1> or can there still be some of them in linear superpositions of these basis?

Since there is no anisotropic Hamiltonian (e.g. Zeeman shift or Stark shift) to define a quantization axis, you should expect each atom could be in any superposition of the mJ sublevels. If you found that the atoms were exactly in one of the basis states in one given frame of reference, then you would have strong reason to believe that the atoms are strongly interacting. Atoms coming off an ablation target are waaaay waaaaay too hot and not dense enough to have strong dipole-dipole interactions.

Also, if I am in the initial frame, where I have a linear superposition of substates and I apply a magnetic field (adiabatically such that I don't transfer population) along the current defined z-axis, will the atom still have the same population in each of the (now non-degenerate) magnetic substates? Thank you!

Each atom's magnetic moment will start Larmor precessing around the z-axis. I believe the overall statistical distribution over spin states will be identical if you adiabatically turn the B-field on, but the spin states of individual atoms will not be the same. This is because the mJ = +1 part of the spin wavefunction and the mJ = -1 part of the spin wavefunction will precess in opposite directions (and the mJ = 0 part won't precess at all). The reason I say the statistical distribution over spin states will be the same is Larmor precession is a reversible process and it shouldn't be able to reduce the entropy of the ensemble (by turning a completely random distribution into a less random one). You would need a process that would also create some entropy to compensate for the entropy loss of the atoms, such as optical pumping.

 

[Malamala]

For a process like ablation, you can safely assume that the resulting atoms are in a completely incoherent statistical mixture of spin states.

Since there is no anisotropic Hamiltonian (e.g. Zeeman shift or Stark shift) to define a quantization axis, you should expect each atom could be in any superposition of the mJ sublevels. If you found that the atoms were exactly in one of the basis states in one given frame of reference, then you would have strong reason to believe that the atoms are strongly interacting. Atoms coming off an ablation target are waaaay waaaaay too hot and not dense enough to have strong dipole-dipole interactions.Each atom's magnetic moment will start Larmor precessing around the z-axis. I believe the overall statistical distribution over spin states will be identical if you adiabatically turn the B-field on, but the spin states of individual atoms will not be the same. This is because the mJ = +1 part of the spin wavefunction and the mJ = -1 part of the spin wavefunction will precess in opposite directions (and the mJ = 0 part won't precess at all). The reason I say the statistical distribution over spin states will be the same is Larmor precession is a reversible process and it shouldn't be able to reduce the entropy of the ensemble (by turning a completely random distribution into a less random one). You would need a process that would also create some entropy to compensate for the entropy loss of the atoms, such as optical pumping.

Thanks for the reply! So if I have a linearly polarized laser, and say that I want to excite a given atom from the $J=1$ state to a $J=0$ state, what would happen? Should I assume that the atom is in a given $m_J$ level (hence it can be excited only if $m_J=0$)? Should I assume it is in an equal quantum (or classical?) superposition of all $m_J$ values? Intuitively I imagine that only $1/3$ of the atoms in a large ensemble will be excited, but I am not sure what happens to a single atom.

 

[Twigg]

Thanks for the reply! So if I have a linearly polarized laser, and say that I want to excite a given atom from the J=1 state to a J=0 state, what would happen?

Did you mean a circularly polarized ($\sigma_+$) laser? A linearly polarized laser would excite J=1 to J=1 (if polarized along the z axis aka $\pi$-polarized) or J=0 to J=1 and J=0 to J=-1 with equal probability (if polarized perpendicular to the z axis).

If you meant a $\sigma_+$ polarized laser, and if we assume that the electronic excited state is also $J=1$, then you'll end up optically pumping and the atoms will accumulate in the $m_J = +1$ state. Suppose your atom starts in the state $\frac{1}{\sqrt{2}} \left( |-1\rangle + |0 \rangle \right)$. The $\sigma_+$ excitation will drive it to an electronic excite manifold in a higher spin state: $\frac{1}{\sqrt{2}} \left( |0\rangle + |+1 \rangle \right)$. Then this state will return to the ground manifold by spontaneous emission, changing it's spin state randomly by any of three possibilities $\Delta m_J = 0, \pm1$. The state $m_J = +1$ is dark because it cannot be excited by $\sigma_+$ light since the excited state is also $J=1$ and doesn't have a $m_J = 2$ sublevel. Thus, population will accumulate in the dark state, and you'll wind up with polarized atoms in the pure stretched state $m_J = +1$.

For a bit of mathematical perspective, the density matrix for an atom in this ensemble will look like $\rho = \left( \begin{array}{ccc} \frac{1}{3} & 0 & 0 \\ 0 & \frac{1}{3} & 0 \\ 0 & 0 & \frac{1}{3} \end{array} \right)$. This is a mixed state since $\mathrm{tr}(\rho^2) = \frac{1}{3} < 1$. This does not mean that the atoms are always in an eigenstate of the $S_z$ (the z-axis spin component).

 

[Malamala]

I think there is a bit of a confusion (sorry about that). What I meant is: assume, after the ablation, I want to test a given atom, and drive a (known) transition from a ground electronic state with $J=1$ to an excited electronic state with $J=0$. As my laser is linearly polarized (and here I assume just one excitation, without caring about the decay of the atom, or multiple excitations), I can only drive this transition from the $J=1, m_J=0$ state. So if my atom is in a $m_J=1$ state, the transition won't happen (at least in the electric dipole approximation). If I think of a large ensemble of atoms, given that $m_J = 0,\pm 1$ all have equal populations (on average), as they have the same energy in the absence of a magnetic field, with a linearly polarized I can excite only 1/3 of them (the ones in the $m_J = 0$ state). However, I am not sure what happens when I think about one atom at a time. When the laser interacts with an atom, will the atom be in a well defined $m_J$ state, such that if it is in, say, $m_J = 1$ I know that I won't be able to excite that atom no matter how long I keep the laser on, or will the atom be in a linear combination of $m_J$ states, such that if I keep the laser on long enough, I can excite it?

 

[Twigg]

Let's say your single atom starts in state $a|1,0\rangle + b|1,+1 \rangle$. It interacts with the exciting laser for some time before it spontaneously emits, accumulating a phase on the Bloch sphere (where the south pole is the $m_J = 0$ ground state and the north pole is the $J=0$ excited state). For the sake of argument, let's suppose the atom is fully excited (reaches the north pole of the Bloch sphere) because it spontaneously emits. The atom's state is now $a|1,0 \rangle + b|0,0\rangle$. When the atom spontaneously emits, the atom's state will be projected to the excited state $|0,0\rangle$ before emitting a photon and coming back down to the ground manifold. I see no reason why the atom would necessarily fall into only one $m_J$ state (as opposed to a superposition) after spontaneous emission, since the quantum vacuum is isotropic so the photon emitted can have any polarization.

 

[Malamala]

Let's say your single atom starts in state $a|1,0\rangle + b|1,+1 \rangle$. It interacts with the exciting laser for some time before it spontaneously emits, accumulating a phase on the Bloch sphere (where the south pole is the $m_J = 0$ ground state and the north pole is the $J=0$ excited state). For the sake of argument, let's suppose the atom is fully excited (reaches the north pole of the Bloch sphere) because it spontaneously emits. The atom's state is now $a|1,0 \rangle + b|0,0\rangle$. When the atom spontaneously emits, the atom's state will be projected to the excited state $|0,0\rangle$ before emitting a photon and coming back down to the ground manifold. I see no reason why the atom would necessarily fall into only one $m_J$ state (as opposed to a superposition) after spontaneous emission, since the quantum vacuum is isotropic so the photon emitted can have any polarization.

But my question is for the state before the atom spontaneously emit. My question is simply what is the state of the atom when the first photon of the laser interacts with it.

 

[Malamala]

But my question is for the state before the atom spontaneously emit. My question is simply what is the state of the atom when the first photon of the laser interacts with it.

Actually let me put the problem a bit more practical (i.e. what I actually need in practice, although I would like to understand the theory behind it better). Say I create my species (atom or molecule) at a temperature T (and it is thermalized). Let's say that based on the Boltzman distribution at T, I know that the expected population in my $J=1$ state is N. If I want to drive a transition from this $J=1$ level to an excited one with $J=0$, using a linearly polarized laser, will I be able to excite only N/3 of these atoms?

Going one step further (based on your replies), assuming the atom-laser interaction is long enough, such that I have many excitations and spontaneous emissions. Given that the excited atom (in $J=0, m_J=0$) can decay to any of the $m_J=0,\pm 1$ levels of the $J=1$ state, but the laser can only excite from the $m_J=0$ state (as it is linearly polarized), does this mean that after a long enough time, I won't be able to excite any atom anymore, as once they decay to $m_J = \pm 1$ they can't be excited anymore?

 

[Twigg]

Say I create my species (atom or molecule) at a temperature T (and it is thermalized). Let's say that based on the Boltzman distribution at T, I know that the expected population in my J=1 state is N. If I want to drive a transition from this J=1 level to an excited one with J=0, using a linearly polarized laser, will I be able to excite only N/3 of these atoms?

Yes. If your laser is applying a coherent pulse, then you can excite a maximum of N/3 atoms (for a time much less than a lifetime).

However, that does *not* mean that 1/3rd of the atoms are initially in the $m_J = 0$ state. It means that an atom will project into the $m_J = 0$ state with probability 1/3. In principle, you could concoct a situation in which none of the atoms occupy the $|J=1, m_J = 0 \rangle$ state but still get 1/3rd excited by a pi polarized laser.

But my question is for the state before the atom spontaneously emit. My question is simply what is the state of the atom when the first photon of the laser interacts with it.

To answer your question, the atoms are in a mixed state before excitation. There is no single ket that can describe a mixed state. The mixed state is a statistical ensemble over pure states. Specifically, the atoms are in a uniform statistical distribution over all possible $J=1$ spin states. This mixed state is mathematically described by the density matrix I gave in post #4.

This is why I started with an arbitrary ket in post #6. The arbitrary ket is just a stand-in for any ket in the statistical ensemble.