What causes the formation of multiple sub-beams in the Stern Gerlach Experiment?

[fog37]

Hello Everyone,

I am reviewing the SG experiment. I think the experiment was set out to demonstrate that the orbital angular momentum $L$ is quantized producing a quantized magnetic momentum. But silver and hydrogen atoms have no net orbital angular momentum. Still two sub-beams, due to two different intrinsic spin magnetic moments, on which two different forces applied, were produced on the screen. Stern and Gerlarch didn't immediately know that spin was involved...

Here my question: if both orbital angular momentum and spin angular momentum are present, how many sub-beams would be present on the collecting screen? In that case, the total magnetic moment is $\mu_{total} = \mu_{orbital} + \mu_{spin}$.

For example, if $\ell=1$, $m_{\ell} =-1,0,1$ so there are three different orbital magnetic moments and three different forces (one of which is zero). Spin provides two different values: $\frac {1}{2}$ and $- \frac {1}{2}$ hence two different spin magnetic moment.
If we considered all possible six combinations $(m_{\ell} , m_{s})$: $(0,\frac {1}{2})$, $(1,\frac {1}{2})$, $(-1, \frac {1}{2})$,$(0, -\frac {1}{2})$, $(1, - \frac {1}{2})$, $(-1, - \frac {1}{2})$

there would seem to be 6 different possible total magnetic moments , six different forces and 6 different sub-beams generated on the screen. Or does each different orbital magnetic moment have only a single spin value? For example, are the combinations $(1,\frac {1}{2})$ and $(1, - \frac {1}{2})$ no possible together? Why?

Thanks.

 

[DrClaude]

If both orbital and spin angular momentum are present, then you have to look at the total angular momentum $j$. For $l=1$ and $s=1/2$, the possible values of $j$ are $j = 1/2, 3/2$. These are then sub-divided into 2 and 4 values of $m_j$, respectively.

How many spots will be observed in the SG experiment then depends on the atom and how it was prepared. Generally speaking, the $j=1/2$ level will be lower in energy, so one could see only two spots. But it is possible that there is a significant population of the $j=3/2$ level, so a total of 6 spots could be observed. The splitting of theses 6 spots will be unequal, due to the fact that the splitting of the $m_j$'s of the $j=1/2$ and $3/2$ levels are different (different Landé g-factors).

 

[drvrm]

there would seem to be 6 different possible total magnetic moments , six different forces and 6 different sub-beams generated on the screen. Or does each different orbital magnetic moment have only a single spin value? For example, are the combinations (1,12)(1,12)(1,\frac {1}{2}) and (1,−12)(1,−12)(1, - \frac {1}{2}) no possible together? Why?

In the experiment...
The force experienced by the atoms in the beam due to inhomogeneous magnetic field was
F= - g , magnetic moment. x m(j) x (dB/dz) ... where g= gyromagnetic ratio , m(j) = -j...0...+j in steps of 1

So, even if L=1 and S= 1/2 J= 1/2 , and 3/2

For J=1/2 m(j)= +1/2 and -1/2 g = 2/3 gm(j)= +/- 1/3

For J = 3/2 m(j) = +/- 3/2 and +/- 1/2 but g=4/3 so gm(j) = +/-2 , +/- 2/3

so the splittings will depend on F.I do not think one can couple m(l) and m(s) separately for the states ...L and S may be combined first and deflection will be guided by
j- values i.e m(j) and g values. and off course the inhomogeneity of the field...

 

[fog37]

I see. Thank you. I will process the information and learn better about the $j$ quantum number.

It is clear tha if orbital angular momentum is not present, i.e. $\ell=0$, there are always two beams due to spin and two forces. This is what happens with silver atoms since all shells and subshells are filled and the overall orbital angular momentum is zero. The last subshell is $4s$ which has no angular momentum at all since $\ell=0$ implying that the two deflections are purely due to spin.

Would it be possible, for a single atom, to have nonzero orbital angular momentum, i.e. $\ell \ne 0$, and zero overall spin so the sub-beams are only due to the orbital magnetic moments? I know spin is always present so I do not think so. Or would the net spin be zero when an atom has each of the outermost electrons with spin $-\frac {1}{2}$ and $\frac {1}{2}$ so the net spin is zero?

 

[DrClaude]

Would it be possible, for a single atom, to have nonzero orbital angular momentum, i.e. $\ell \ne 0$, and zero overall spin so the sub-beams are only due to the orbital magnetic moments? I know spin is always present so I do not think so. Or would the net spin be zero when an atom has each of the outermost electrons with spin $-\frac {1}{2}$ and $\frac {1}{2}$ so the net spin is zero?

You could prepare an atom in a state with $S = 0$ and $L \neq 0$, but it won't be the ground state of the atom: if the spins are paired because of the Pauli exclusion principle, then it follows that $L=0$; if the spins are paired but equal-energy orbitals are free, then its an exited state (states with parallel spins are lower in energy).

 

[drvrm]

Would it be possible, for a single atom, to have nonzero orbital angular momentum, i.e. ℓ≠0ℓ≠0\ell \ne 0, and zero overall spin so the sub-beams are only due to the orbital magnetic moments? I know spin is always present so I do not think so.

Alkali metals (S=1/2) form doublets.
Ions with 2 electrons in the outer shell,
like He, Ca I or Mg I, may form singlets or triplets. but as you go to complex systems L-S coupling may not hold...

 

[fog37]

Thanks you. I just started reading/studying about $L$ and $S$ considered together into the spin-orbit coupling. Is it possible for these two vectors not to couple at all? I think there are always coupled but in some cases produce spectral line splitting and in some cases they don't.

Both are vectors (vector operators) that produce $$J=L+S$$ The vector $J$ has its own quantum number $m_j$ which assumes discrete values such that $J_z = m_j \hbar$. The possible discrete values of $m_j$ are $-j, -(j-1),...+(j+1), +j$ where $j = \ell \pm \frac {1}{2}$, which means that for every value of $j$ there are $2j+1$ values of $m_j$. Different values of $m_j$ correspond to different orientations of $J$.

Examples: if $\ell = 0$ then #J=S# since $L=0$. For states having $\ell=1$, $j=\frac {3}{2}$ and $\frac {1}{2}$. Books talk about two sub-states, the $p_{3/2}$ and a state $p_{1/2}$. the first state has four values of $J_z$ and the second two values. Why are there exactly two states from this spin-orbit coupling? Why not 4 or 5, etc.? Vectorially, I see how $S$ and $L$ are not (never) aligned. Why? there are two vector arrangements to produce the vector $J$.

We now have two alternative sets of quantum numbers: $(n,\ell,m_{\ell}, m_s)$ for when there is no interaction and $(n,\ell,m_{\ell}, j, m_j)$ for when there is interaction. The internal magnetic field due to the proton motion is parallel or antiparallel to the electron magnetic moment due to spin giving rise to two states of different energy.

For sodium, which has a single valence electron, the electron is in $3s_{1/2}$ which means $n=3$, $\ell=0$, $j=\frac {1}{2}$. This state is not split by spin orbit coupling because orbital angular momentum is zero. However, the state $3p$ splits into two states: $3p_{1/2}$ and $3p_{3/2}$ of different energy giving rise two different energy levels and two different spectral line, the two sodium D lines. If we place sodium inside a B field, these two lines split up into 3 lines (the first) and 4 more lines (the second D line). This should be the anomalous Zeeman effect.

Hope this review helps some other student trying to learn these concepts :)

If what I wrote is correct, what is exactly a singlet? And what about a triplet? It sound like it is three different states. I am still confused about the definition of these two terms.

 

[drvrm]

If what I wrote is correct, what is exactly a singlet? And what about a triplet? It sound like it is three different states. I am still confused about the definition of these two terms.

actually when one applies coupling of angular momentum the quantization rules limits their possible states.

these vectors are not ordinary vectors ,though many a time they are depicted like classical vectors.

when one writes the schrodinger equation for these atoms the equation gets
separable in radial and angular coordinates and the angular part of the equation

leads to angular momentum operators like L^2 and Lz as these two commute with hamiltonian
and can be defined simultaneously or their eigen values are the expected result of measurement .

the other components Lx or Ly are not well defined.

The the angular functions which are eigen functions of these two operators (called spherical harmonics)
lead to eigenvalue equations.

so particular values of L^2 and Lz operators in the space of orbital angular momentum are allowed.
To cut the story short the spin space is very much separate and spin wave functions are not usual spinning top description
so the S^2 and Sz are only well defined.

If one's Hamiltonian has L.S interaction then and then only they get coupled and those states are degenerate..Many a time the degeneracy gets lifted by external magnetic field usually taken in z-direction

i think if you treat H-atom QMechanically then the picture can get cleared.

 

[DrClaude]

To add to the good answers from @drvrm:

Vectorially, I see how $S$ and $L$ are not (never) aligned. Why? there are two vector arrangements to produce the vector $J$.

That's one of the nice peculiarities of quantum mechanics. Whereas a classical angular momentum $\mathbf{J}$ is a vector of length $J = | \mathbf{J} |$, a quantum mechanical angular momentum eigenstate with an eigenvalue of $J$ has $\langle \mathbf{\hat{J}} \rangle = \sqrt{J (J+1)}$. For an eigenstate which is also an eigenstate of $\hat{J}_z$, since the allowed eigenvalues $J_z$ are integer multiples of $J$, in the classical representation the vector can never be aligned with the z axis. Likewise, when summing $\mathbf{L}$ and $\mathbf{S}$ into $\mathbf{J}$, they can never be exactly aligned, since it is impossible to satisfy the equation
$$
\sqrt{L(L+1)} + \sqrt{S(S+1)} = \sqrt{J(J+1)}
$$
for valid values of $J$ (except if $L=0$ or $S=0$).

If what I wrote is correct, what is exactly a singlet? And what about a triplet? It sound like it is three different states. I am still confused about the definition of these two terms.

The multiplicity is given by $2S+1$, with $2S+1 = 1$ a singlet, $2S+1 = 2$ a doublet, etc. The name indeed reflects the number of states the level will split into. The easiest case to illustrate is the one for 2 electrons:
$$ \frac{1}{\sqrt{2}} \left[ | \uparrow \downarrow \rangle - |\downarrow \uparrow \rangle \right] $$
is the singlet spin state, while the triplet is made up of the spin states
$$
| \uparrow \uparrow \rangle \\
\frac{1}{\sqrt{2}} \left[ | \uparrow \downarrow \rangle + |\downarrow \uparrow \rangle \right] \\
| \downarrow \downarrow \rangle
$$

 

[vanhees71]

Of course, what's finally relevant for the SG experiment is the total magnetic moment of the particle/atom/molecule. It's related to angular momentum, but there are the gyro-factors involved. For orbital angular momentum the gyrofactor is 1 for elementary spin 1/2 its close to 2. The orbital part of this relation is understandable heuristically from the classical picture of an angular momentum due to a ring current, while the gyro-factor 2 for the spin part is not understandable from the point of view of classical physics and is due to the heuristic principle of minimal coupling. The corresponding piece in the Hamiltonian of the Pauli equation is
$$\hat{H}_{\text{mag-moment}}=-\mu_{\text{B}} \vec{A} \cdot (\hat{\vec{L}}+g_s \hat{\vec{S}}), \quad g_s \simeq 2.$$
Dabei ist $\mu_{\text{B}}$ das Bohrsche Magneton. In SI-Einheiten: $\mu_{\text{B}}=e/(2 m_e) \simeq 5.8 \cdot 10^{-11} \text{MeV}/\text{T}$.