1. Magnetic Moment
2. Spin Angular Momentum Of Electron
3. Spin Angular Momentum Equation
4. Spin Angular Momentum Of Light
5. Spin Angular Momentum Formula
6. Spin Angular Momentum Cones
7. Spin Angular Momentum Pdf

Mar 18, 2020 The important feature of the spinning electron is the spin angular momentum vector, which we label (S ) by analogy with the orbital angular momentum (L ). We define spin angular momentum operators with the same properties that we found for the rotational and orbital angular momentum operators. The total spin momentum has magnitude Square root of √S(S + 1) (ℏ), in which S is an integer or half an odd integer, depending on whether the number of electrons is even or odd. The possible value of the total spin angular momentum can be found from all the possible orientations of electrons within the atom. An angular momentum and a magnetic moment could indeed arise from a spinning sphere of charge, but this classical picture cannot fit the size or quantized nature of the electron spin. The property called electron spin must be considered to be a quantum concept without detailed classical analogy.

Addition of angular momentum

Problem:

You have a system of two electrons whose orbital quantum numbers are l₁ = 2 and l₂ = 4 respectively.
(a) Find the possible values of l (total orbital angular momentum quantum number) for the system.
(b) Find the possible values of s (total spin angular momentum quantum number) for the system.
(c) Find the possible values of j (total angular momentum quantum number) for the system

Solution:

• Concepts:
  Addition of angular momentum
• Reasoning:
  We are supposed to add the orbital and spin angular momentum.
• Details of the calculation:
  (a) The quantum numbers associated with the total orbital angular momentum will range from a maximum value found by adding the individual quantum numbers together to a minimum value found by taking the absolute value of the difference of the two numbers in integer steps.
  l_max = l₁ + l₂ = 2 + 4 = 6
  l_min = l₁ - l₂ = 2 - 4 = 2
  The possible values for L are 2, 3, 4, 5, and 6 .
  (b) The quantum numbers associated with the total spin angular momentum will range from a maximum value found by adding the individual quantum numbers together to a minimum value found by taking the absolute value of the difference of the two numbers in integer steps.
  s_max = s₁ + s₂ = ½ + ½ = 1
  s_min = s₁ - s₂ = ½ - ½ = 0
  The possible values for S are 0 and 1.
  (c) The largest possible value for j will be found by adding together the maximum values for both l and s. The minimum value for j will be found by subtracting the largest possible value of s from the smallest possible value for l.
  j_smax = l_max + s_max = 6 + 1 = 7
  j_min = l_min - s_max = 2 - 1 = 1
  j = 1, 2, 3, 4, 5, 6, 7

Problem:

A hydrogen atom is known to be in a state characterized the quantum numbers n = 3, l = 2.
(a) Give the allowed values of j.
(b) For each of the allowed values of j, calculate the square of the magnitude of the total angular momentum.

Solution:

• Concepts:
  Addition of angular momentum
• Reasoning:
  We are supposed to add the orbital and spin angular momentum of the electron in the hydrogen atom.
• Details of the calculation:
  (b) (a) l = 2, s = ½, j = 5/2, 3/2.
  J² jm> = j(j + 1)ħ² jm>.
  For j = 5/2, j(j + 1)ħ² = (35/4)ħ².
  For j = 3/2, j(j + 1)ħ² = (15/4)ħ².

Problem:

Consider a deuterium atom (composed of a nucleus of spin 1 and an electron). The electronic angular momentum is J = L + S, where L is the orbital angular momentum of the electron and S is its spin. The total angular momentum of the atom is F = J + I, where I is the nuclear spin. The eigenvalues of J² and F² are j(j + 1)ħ² and f(f + 1)ħ² respectively.
(a) What are the possible values of the quantum numbers j and f for a deuterium atom in the 1s ground state?
(b) What are the possible values of the quantum numbers j and f for a deuterium atom in the 2p excited state state?

Solution:

• Concepts:
  Addition of angular momenta
• Reasoning:
  We are asked to add three angular momenta
• Details of the calculation:
  (a) For the 1s ground state we have l = 0, J = L + S = S, j = ½.
  F = J + I, i = 1.
  The possible values for f are f = i + j, i - j; f = 3/2, ½.
  (b) For the 2p state we have l = 1.
  The possible values for j are j = l + s, l - s; j = 3/2, ½.
  If j = 3/2 then the possible values for f are
  f = j + i, j + i - 1, j - i; f = 5/2, 3/2, ½.
  If j = ½ then the possible values for f are
  f = i + j, i - j; f = 3/2, ½.
  So the possible values for f are f = 5/2, 3/2, ½.

Problem:

We are to add two angular momenta characterized by the quantum numbers j₁ = 2 and j₂ = 1.
(a) What are the possible values for j?
(b) Express all eigenkets j₁, j₂; j, m> = 2, 1; 1, m> in terms of j₁, j₂; m₁, m₂>.
(c) Express the ket j₁, j₂; m₁, m₂> = 2, 1; 0, 0> in terms of j₁, j₂; j, m>.
(d) What are the expectation values of J_1z and J_2z in the state j₁, j₂; j, m> = 2, 1; 1, 1>?

Solution:

• Concepts:
  Addition of angular momentum, the Clebsch-Gordan coefficients
• Reasoning:
  Consider two angular momentum operators J₁ and J₂. J₁ operates in E₁ and J₂ operates in E₂.
  Let J = J₁ +J₂. J operates in E = E₁⊗E₂.
  Since the operators J₁², J_1z , J₂², and J_2z all commute, a basis of common eigenvectors for E exists.
  We denote this basis by { j₁, j₂; m₁, m₂>}.
  Since the operators J₁², J₂², J², and J_z all commute, a basis of common eigenvectors for E exists.
  We denote this basis by { j₁, j₂; j, m>}.
  We can write the vectors of one basis as linear combinations of the vectors of the other basis.
  j₁,j₂;j,m> = ∑_m1∑_m2 C^j_m1m2 j₁,j₂;m₁,m₂>,
  j₁,j₂;m₁,m₂> = ∑_j∑_m C^j_m1m2 j₁,j₂;j,m>.
  The C^j_m1m2 = <j₁,j₂;m₁,m₂ j₁,j₂;j,m> are called the Clebsch-Gordon coefficients.
• Details of the calculation:
  (a) j₁ - j₂ ≤ j ≤ j₁ + j₂. j = 3, 2, 1.
  (b) j₁, j₂; j, m>
  Using a table of Clebsch-Gordan coefficients we find:
  2, 1; 1, 1> = (3/5)^½ 2, 1; 2, -1> - (3/10)^½ 2, 1; 1, 0> + (1/10)^½ 2, 1; 0, 1>
  2, 1; 1, 0> = (3/10)^½ 2,1;1 ,-1> - (2/5)^½ 2, 1; 0, 0> + (3/10)^½ 2, 1; -1, 1>
  2, 1; 1, -1> = (1/10)^½ 2,1;0,-1> - (3/10)^½ 2, 1; -1, 0> + (3/5)^½ 2, 1; -2, 1>
  (c) j₁, j₂; m₁, m₂>
  Using a table of Clebsch-Gordan coefficients we find:
  2, 1; 0, 0> = (3/5)^½ 2, 1; 3, 0> - (2/5)^½ 2, 1; 1, 0>
  (d) Using the result from part (b) we have
  <J_1z> = 2ħ(3/5) + ħ(3/10) = 3ħ/2,
  <J_2z> = -ħ(3/5) + ħ(1/10) = -ħ/2.

Problem:

A particle of spin 3/2, at rest in the laboratory, disintegrates into two particles, one of spin ½ and one of spin 0.
(a) What values are possible for the relative orbital angular momentum of the two particles? Show that there is only one possible value if the parity of the relative orbital state is fixed.
(b) Assume the decaying particle is initially in the eigenstate of S_z with eigenvalue mħ. Is it possible to determine the parity of the final state by measuring the probabilities of finding the spin ½ particle either in the +> or -> state?

Solution:

• Concepts:
  Conservation of angular momentum, addition of angular momentum, the Clebsch-Gordan coefficients, orbital angular momentum and parity
• Reasoning:
  The total angular momentum of an isolated system is conserved. The initial state has only spin angular momentum. The spin angular momenta and the orbital angular momentum of the particles in the final state must add to give the angular momentum of the initial state.

• Details of the calculation:
  (a) The initial value of j is j = 3/2. We need the final value to be j = 3/2, since angular momentum of an isolated system is conserved. Let us call the initial particle particle a, the spin ½ particle particle b, and the spin 0 particle particle c. Then the total spin of the two particles in the final state is S = S_b + S_c = S_b. Therefore the spin quantum number is s = ½. The possible values for the orbital angular momentum quantum number are l = 1 and l = 2.
  (j = l + s, ..., l - s; j = l + ½, l - ½, implies l = 1 or l = 2.) The parity of the orbital state is (-1)^l. If the parity is odd, we have l = 1, if the parity is even, we have l = 2.
  (b) Assume that the final state has l = 1.
  It may be written as j₁, j₂; j, m> = 1, ½; 3/2, m>.
  We may write it as a linear combination of j₁, j₂; m₁, m₂> = 1, ½; m_l, m_s >.
  m can take on the values 3/2, ½, -½, -3/2. Using a table of Clebsch-Gordan coefficients we find

  Here P(+) is the probabilities of finding the spin ½ particle in the +> state.

  Now assume that the final state has l = 2.
  It may be written as j₁, j₂; j, m> = 2, ½; 3/2, m>.
  We may write it as a linear combination of j₁, j₂; m₁, m₂> = 2, ½; m_l, m_s >.
  m can take on the values 3/2, ½, -½, -3/2. We find

  Yes, it is possible to determine the parity of the final state by measuring the probabilities of finding the spin ½ particle either in the +> or -> state.

This chapter discusses angular momentum, a quantity involved in the description of rotational motion. Such motions are present for electrons in molecules, and for molecules in the gas phase. The latter was already discussed in a previous chapter. Here we will introduce general concepts of angular momentum and then focus on describing the rotational motion of electrons within molecules, including the magnetic effects that result from this motion.

This chapter also takes a closer look at spin angular momentum, which is another type of angular momentum that is intrinsic to electrons and to certain nuclei.

7.1. Classical angular momentum

Angular momentum, (boldsymbol{l}) is the rotational analog of linear momentum. The angular momentum of a particle is the vector cross product of its position (relative to some origin) (boldsymbol{r}) and its linear momentum (boldsymbol{p} = mboldsymbol{v}).

Magnetic Moment

(boldsymbol{l}) is a vector that points along the rotation axis. This is illustrated in Fig. 7.1. The SI unit of angular momentum is kg m² s^-1 = J s, joule-second.

The angular momentum vector can be written as

where (l_x), (l_y), and (l_z) are the components of (boldsymbol{l}) along the x, y, and z axis. The magnitude-squared of this vector is

There are two ways of writing the kinetic energy of a rotating body, either using the moment of inertia (I) and the angular frequency (omega), or (I) and the angular momentum:

7.2. Quantum angular momentum

To obtain the Hamiltonian for a rotating body, we replace the angular momentum in the above equation for the energy with its operator:

In the chapter on rotating molecules, we have solved the TISE for this already. Therefore, the eigenfunctions of (hat{l}^2) are the spherical harmonics (Y_l^{m_l}(theta,phi)), with eigenvalues (hbar^2 l(l+1)):

with the quantum numbers (l = 0,1,2,dots) and (m_l=0,pm1,dots,pm l). This means that the magnitude of the angular momentum is quantized - only the discrete values (hbarsqrt{l(l+1)}) are possible.

The spherical harmonics are also eigenfunctions of of the operator (hat{l}_z) (the (z) component of (hat{boldsymbol{l}})):

This means that, if one measures the angular momentum along an axis (call it (z)), then the only possible outcomes are multiples of (hbar), (mhbar). Since there are only a discrete allowed projections along (z), the angular-momentum vector cannot point anywhere in space - it is confined to a cone for each (m_l) state. This is called space quantization.

Commutation relations. The components of the angular momentum do not commute:

This means that there exist no quantum states where they are all simultaneously precise (see uncertainty principle). In contrast, the angular-momentum operator (hat{l}^2) and (hat{l}_z) commute:

Fig. 7.2 The vector model for quantum angular momentum. Left: The delocalization cone for a specific ((l,m_l)) state. Right: Cones for all ((l,m_l)) states with (l=2).

Fig. 7.2 (left) depicts the vector model for an angular momentum state with quantum numbers (l) and (m_l). In this model, angular momentum is pictured as a vector of defined length (hbarsqrt{l(l+1)}) and defined projection (m_lhbar) along the direction (z). To indicate the fact that the x and y components are indeterminate, a cone is drawn. The orientation of the angular-momentum vector is “delocalized” over this cone, i.e. the tip of the vector could be anywhere on the cone rim. This delocalization of orientation is analogous to the delocalization of position in the particle-in-a-box and the other model systems we looked at earlier.

Fig. 7.2 (right) shows the vector model for all states with (l=2) (d functions). In this case, the magnitude is (hbarsqrt{6}), and the projections are (-2hbar), (-hbar), 0, (hbar), and (2hbar).

Note

It is incorrect to read into the vector model that the vector is precessing on the cone! The cone represents the possible orientations the vector can be found in when measured. The vector is orientationally delocalized over the cone in the same sense as an electron is delocalized in space.

The angular-momentum operators commute with the Hamiltonian of a hydrogenic atom:

This indicates that, in addition to the energy, also the magnitude of the angular momentum, as well as its projection onto a direction, are conserved quantities (constants of motion). There is one quantum number per operator: (n), (l), (m_l). The wavefunctions of the stationary states of the hydrogen atom are simultaneous eigenfunctions of all three operators.

7.3. Spin angular momentum

In addition to orbital angular momentum due to its motion around the nucleus, an electron has an additional intrinsic angular momentum indicated by (boldsymbol{s}). This is present even if the eletron has no orbital angular momentum (e.g. in a 1s orbital). This intrinsic angular momentum is called spin angular momentum, or simply spin, since it is possible to picture it as being due to the electron spinning around its own axis.

The spin angular momentum of the electron is fixed and a magnitude of (hbar sqrt{3/4}), corresponding to a half-integer quantum number, (s=1/2). It is fixed, meaning that it is not possible to stop or accelerate the spinning. The half-integer value of the quantum number (s) is in contrast to the orbital angular momentum quantum numbers (l), which are intergers. The spin projection quantum number (m_s) can assume two values, (pm1/2). The state with (m_s=+1/2) is called spin-up and indicated by (alpha), and the state with (m_s=-1/2) is called spin-down and indicated by (beta). The vector model representation of these two states is shown in Fig. 7.3.

Fig. 7.3 Vector model for the two stationary states of an electron spin: spin-up ((alpha), (m_s=+1/2)) and spin-down ((beta), (m_s=-1/2)).

Aside from this difference in the quantum numbers, one further difference between orbital and spin angular momentum is that the spin eigenfunctions are not functions of continuous spatial variables (like (theta) and (phi)), but can only be written as abstract symbols ((alpha) and (beta)), with the quantum number (m_s) are the only (discrete) coordinate. Properties of spin and orbital angular momentum are summarized in the following table.

The spin operators commute with the orbital operators and the Hamiltonian, so that there are a total of five commuting operators: (hat{H}), (hat{l}^2), (hat{l}_z), (hat{s}^2), and (hat{s}_z).

Due to the spin of the electron, five quantum numbers are needed to fully specify an eigenstate of a hydrogenic atom: (n), (l), (m_l), (s), and (m_s). Since (s) is fixed, it is usually not explicitly listed, and only the remaining four quantum numbers are given. The total wavefunction (psi_{n,l,m_l,m_s}) is the product of the spatial wavefunction and the spin function:

Spin Angular Momentum Of Electron

The spatial part (psi_{n,l,m_l}) is called a spatial orbital, and the full wavefunction is called a spinorbital. In chemistry, spin orbital symbols are natural extensions of the symbols for spatial orbitals. For example, (1mathrm{s}alpha) indicates (psi_{1,0,0,+1/2}), and (3mathrm{p}_zbeta) indicates (psi_{3,1,0,-1/2}).

Electrons are not the only particles that possess spin. Neutrons and protons (¹H nuclei) are spin-1/2 particles. Since nuclei are composed of protons and neutrons, many nuclei (but not all) have non-zero spin as well. The following table list a few common nuclear isotopes and their nuclear spin. The spin of a nucleus is traditionally indicated by the letter (I).

7.4. Magnetic dipole moments

Both orbital and spin angular momenta give rise to a magnetic field.

Orbital magnetic dipole moment. An electron with non-zero orbital angular momentum can be pictured as rotating around a nucleus. This represents a circulating electric current, which produces a magnetic field. Therefore, an atom with an electron in an (lneq0) state possesses an orbital magnetic dipole moment(boldsymbol{mu}_l) that is proportional to its orbital angular momentum. It is given by

Spin Angular Momentum Equation

where (e) is the (positive) elementary charge and (m_mathrm{e}) is the mass of the electron. The dimension of (boldsymbol{mu}_l) is energy per field, and its SI unit is J T^-1 (joule per tesla). Due to the negative charge, the magnetic dipole moment vector is antiparallel to the angular momentum vector.

Spin magnetic dipole moment. An electron, with its non-zero spin angular momentum, constitutes a spinning negative charge. This is also equivalent to a circular current, and leads to a spin magnetic dipole moment:

The (g_mathrm{e}) factor is about 2.0023.

Protons and other nuclei with non-zero spin also have a spin magnetic dipole moment.

7.5. Zeeman effect

When placed in an external magnetic field (boldsymbol{B}), the energy of a magnetic dipole moment (boldsymbol{mu}) depends on its orientation relative to the field. Mathematically, its energy is the negative scalar product of the two vectors

where (theta) is the angle between the two vectors. The energy is lowest if the magnetic moment is parallel to the field, and highest if it is antiparallel. This is illustrated in Fig. 7.4.

The above equation also shows that the energy depends linearly on the strength of the external magnetic field. This magnetic-field dependence of the energy is called the Zeeman effect.

Note

Magnetic field strengths span a large range: The earth magnetic field is between 25 and 65 μT, depending on location. In NMR (nuclear magnetic resonance) spectroscopy, a 400 MHz spectrometer uses 9.4 T. In clinical MRI (magnetic resonance imaging), 1.5-3 T are used. A neodymium magnet can generate fields up to 1.4 T.

Fig. 7.4 The energy of an electron in a magnetic field depends on the relative orientation between its magnetic moment (boldsymbol{mu}) and the magnetic field (boldsymbol{B}).

If we define the (z) axis along the magnetic field, the Hamiltonian describing the Zeeman interaction between the orbital magnetic moment of the electron and the magnetic field is

The associated eigenvalues are

The factor (mu_mathrm{B} = ehbar/2m_mathrm{e}) is called the Bohr magneton and is an atomic-scale unit for the magnetic moment (similar to the Bohr radius for atomic-scale lengths).

The Hamiltonian and the eigenvalues for the electron spin are analogous:

The Zeeman effect is also operative for magnetic nuclei. The eigenvalues for a nuclear spin with spin angular momentum quantum number (I) (and (m_I = -I,-I+1,...,+I)) are

where (mu_mathrm{N}) is the nuclear Bohr magneton, a universal constant, and (g_mathrm{n}) is the nuclear g factor, which depends on the particular isotope. For example, for a proton it is (g_mathrm{n} = +5.58569).

Note

The Zeeman effect of nuclear spins is the basis of nuclear magnetic resonance (NMR) spectroscopy and magnetic resonance imaging (MRI). The Zeeman effect of unpaired electrons is the basis for electron paramagnetic resonance (EPR) spectroscopy.

Fig. 7.5 shows the energy level diagram of a proton (¹H nucleus) as a function of the strength of an applied magnetic field. To represent energies (E), in NMR and EPR is is common to use the frequency equivalent, (E/h).

Fig. 7.5 The Zeeman energy level diagram for a proton in an external field. In a typical NMR spectrometer with a 9.39 T field, the proton absorbs radiation of frequency 400 MHz.

7.6. Spin-orbit coupling

An electron moving around nucleus sees a moving nuclear charge. From the point of view of the electron, this is equivalent to an electric current! Since an electric current generates a magnetic field, the electron feels a magnetic field. This field is

Spin Angular Momentum Of Light

where (boldsymbol{E}) is the electric field and (boldsymbol{v}) is the velocity of the electron.

This field is purely internal to the atom, a result of the electron moving around a charged nucleus. It does not involve externally applied fields (as in the Zeeman effect).

Since the electron is magnetic as a result of its spin, it interacts with this magnetic field, i.e. its energy depends on the relative orientation between the magnetic field and the magnetic moment of the electron. This is the spin-orbit coupling (SOC). The associated Hamiltonian is

where (xi) is a function of (r). This results in different energies depending on the relative orientation between spin and the orbital-angular-momentum-induced magnetic field. This is illustrated in Fig. 7.6.

Fig. 7.6 Spin-orbit coupling for an electron with orbital angular momentum (boldsymbol{l}) and spin angular momentum (boldsymbol{s}).

An important consequence of the spin-orbit coupling is now that the operators (boldsymbol{l}) and (boldsymbol{s}) don’t commute with the Hamiltonian anymore - they are not conserved quantities. However, the total angular momentum, denoted by (boldsymbol{j}), is conserved. The total angular momentum is the vector sum of the orbital and the spin angular momentum:

The total angular momentum has the same properties as any other angular momentum: The eigenvalue of (hat{j}^2) is (hbar^2 j(j+1)), and the eigenvalues of (hat{j}_z) are (m_j = -j,-j+1,dots,j).

The total number of states with given specific values of (l) and (s) can be determined as follows: for (l), we have (2l+1) states, and for (s), we have (2s+1) states. With (s=1/2), thus, the total is (2(2l+1)).

In order to determine the possible values of (j), we need to examine the possible (m_l) and (m_s), determine the associated (m_j), and then infer which (j) is associated with the (m_j) values.

If (l=0), then (m_l=0) and (m_s=pm1/2), and we have two states total. The possible values for (m_j) are (m_j=m_l+m_s=pm1/2). Such a range of (m_j) can only be due to (j = 1/2), (2 states). All states are accounted for.

If (l=1), then (m_l=-1,0,1) and (m_s=pm1/2), and we have a total of 6 combinations of (m_l) and (m_s). We can classify them according to the value of (m_j), as shown in the following table:

There is one state with (m_j=+3/2). Since this is the largest (m_j), we can conclude that there is a a group of states with (j=3/2), with 4 states ((m_j=+3/2), (+1/2), (-1/2), (-3/2)). If we remove these states, we are left with one (m_j=1/2) state and one (m_j = -1/2) state. These can only be due to (j=1/2) (2 states, (m_j=pm1/2)). Therefore, the coupling of the 6 states with (l=1) and (s=1/2) yields one set of 4 states with (j=3/2) and one set of 2 states with (j=1/2).

The above are special cases of the general rule for the addition of angular momenta. For one electron, the possible values of (j) are

(if (l=0), these two are identical).

Spin Angular Momentum Formula

For several electrons, the orbital and spin angular momenta of the individual are combined first to give (L = sum_k l_k) (orbital) and (S = sum_k s_k) (spin), and then (L) and (S) are combined to give the total angular momentum (J), with the possible values

For each (J), there are (2J+1) states ((m_J = -J,dots,J)). For example, (L=2) and (S=1) will give (J=1), 2, and 3 terms with 3, 5, and 7 states. This type of coupling is called Russell-Saunders coupling. For a one-electron atom, (J=j), (L=l), and (S=s=1/2).

A group of (2J+1) states with the same (J) is called a term and is identified by a term symbol. The general form for a term symbol is

The pre-superscript indicates the spin multiplicity and is called spin singlet for (2S+1=1), spin doublet for (2S+1=2), spin triplet for (2S+1=3), etc. A one-electron atom has (S=s=1/2) and (2S+1=2), so it is a spin doublet. (X) is a letter representing (L): S ((L=0)), P (1), D (2), F (3), etc. (J) is the total angular momentum quantum number.

Note

The term symbols for the one-electron configurations in a hydrogen atom are

• 1s: (^2mathrm{S}_{1/2})

• 2s: (^2mathrm{S}_{1/2})

• 2p: (^2mathrm{P}_{3/2}), (^2mathrm{P}_{1/2})

If the electron is in a 2p state, this indicates that there are two different sets of states with different total angular momenta.

Note

A (^4mathrm{D}_{5/2}) term indicates a group of states with spin angular momentum (S=3/2) (since (2S+1=4)), orbital angular momentum (L=2), and total angular momentum (J=5/2). The term consists of (2J+1=6) states.

7.7. Fine structure of H atom

Fig. 7.7 shows the consequences of spin-orbit coupling (SOC) for the energy levels hydrogen atom. On the left, SOC is neglected, and all states of the 2p electron configuration are degenerate. The SOC coupling causes a splitting between the two terms of 2p configuration, the lower-energy term being (^2mathrm{P}_{1/2}) (2 states), and the higher-energy term being (^2mathrm{P}_{3/2}) (4 states). Instead of a single emission line, the 2p(rightarrow)2s transition consists in reality of two lines that are very close together. If a magnetic field is applied in addition, the states within the two terms split due to the Zeeman effect, and there are many emission lines with (very slightly) different wavelengths.

Spin Angular Momentum Cones

Spin Angular Momentum Pdf

The selection rules for transitions are (Delta L = pm1), (Delta S = 0), (Delta J = 0,pm1), and (Delta m_J = 0,pm1).

The selection rule (Delta S = 0) is of particular importance. It indicates that only transitions between spin singlet states, or between spin triplet states, are possible. Transitions between states of different spin multiplicity (such as from a triplet state to a singlet state) are called spin-forbidden and are very unlikely.