Chapter 12

The Spin

12.1  INTRODUCTION

In Chapter 10, we solved Schrodinger equation for an electron in hydrogen atom and found discrete values of energy and orbital angular momentum. When an electron makes transition from higher energy state with energy E_m to a lower energy state with energy E_n, allowed by selection rules (to be discussed in a later chapter), the emitted radiation has wavelength λ = hc /(E_m − E_n). When the predicted spectrum was compared with the observed spectral lines, there was agreement in almost whole spectrum except in finer details of (some) lines. In fact, the theoretical spectrum cannot account for a few additional spectral lines seen experimentally. Most important of these is the fact that many of the observed spectral lines, instead of single lines predicted by Schrodinger equation, consist of two closely spaced lines. For example, this so-called fine-structure is seen in the first line of Balmer series of hydrogen atom (corresponding to the transition from n = 3 to n = 2 level). For this particular case, whereas the theoretical prediction is a single line of wavelength λ = 6563Å, the experimental spectrum, instead, shows two lines separated by ∆λ = 1.4Å. Though it is a small effect, it cries for the explanation.

We may cite another example of the discrepancy between the theoretical results and the corresponding experimental observations. This is seen in the Zeeman Effect. We shall discuss the details of Zeeman Effect in Chapter 15. Suffice here to mention that when an atom is put in a constant homogeneous magnetic field, the interaction of orbital magnetic moment and external magnetic field results in splitting of energy levels and hence in splitting of a spectral line into three lines, termed as ‘Normal Zeeman Effect’. On the other hand, what is generally seen is a single line splitting into four, six, or even more lines in the presence of magnetic field, termed as ‘Anomalous Zeeman Effect’.

Both the above mentioned discrepancies may be sorted perfectly well, if we assume, along with Goudsmit and Uhlenbeck, that the electron has some internal degrees of freedom and as a result has additional angular momentum and magnetic moment along with its orbital angular momentum and orbital magnetic moment. Goudsmit and Uhlenbeck imagined the electron as a classical charged spherical ball spinning around its axis. Though this picture of spinning electron may not be true, as it leads the electron to have angular speed so high that the equatorial velocity of the imagined spherical ball is many times larger than the velocity of light, still it is termed as spin angular momentum and spin magnetic moments, only for historical reasons. We really do not know what type of internal motion in electron causes the spin angular momentum and spin magnetic moment. Not only the experimental results cried for the presence of spin angular momentum and spin magnetic moment, the consistency of relativistic quantum mechanics, developed by Dirac, confirmed the existence of electron spin as a fundamental requirement. In this chapter, we shall have preliminary discussions about the quantum mechanical description of spin, spin operators, and spin dynamics, and so on. Firstly, we start with orbital angular momentum and magnetic moment.

12.2  ORBITAL ANGULAR MOMENTUM AND MAGNETIC MOMENT

Let us consider an electron, as a classical particle, moving in a circular orbit of radius r, with speed v. Let us take the orbit in x−y plane (at z = 0) with its centre at the origin 0, as shown in Figure 12.1.

Figure 12.1 Electron moving in a circular orbit of radius r in x-y plane

The angular momentum vector L points in z-direction (L_x = L_y = 0) and has magnitude

 

 

Now the rotating electron is equivalent to an electric current I which is given by

 

in the direction of electron motion. We know that a current I in a small loop of area A is equivalent to a magnet of dipole moment μ, perpendicular to the plane of the loop (in a sense given by the right handed screw rule with respect to sense of current) given by

 

For the general case, we may write

Here the subscript l signifies that the magnetic moment is present due to orbital motion. It is clear from Eq. (12.4) that as the electron carries negative charge (–e), the orbital magnetic moment vector μ_l points in opposite direction to that of the orbital angular momentum vector L.

Now switching over to quantum mechanical case, it is interesting to note that the same relation [Eq. (12.4)] holds even when the electron is described quantum mechanically. But we know from Chapter 9 that in quantum mechanical description of angular momentum, not only the magnitude of angular momentum vector L is quantized , the z-component of L is also quantized with values L_x = m_l ħ [m₁ having (2l + 1) values –l, –l + 1,... l]. Consequently, the z-component may have any one of the (2l + 1) values given by

where μ_B = (eħ/2m_e) (= 9.27 × 10^–24JT^–1) is known as Bohr magneton and g_l is known as Lande g-factor which for orbital case is unity.

12.3  THE ELECTRON SPIN: SPIN OPERATORS AND SPIN EIGENSTATES

As mentioned in the Introduction, there are experimental evidences (e.g., fine structure and Zeeman effect) and theoretical compulsions (in Dirac’s relativities quantum mechanical theory of electrons) to assume that electrons, in addition to their orbital angular momentum and orbital magnetic moment, have intrinsic spin angular momentum and spin magnetic moment. Further, the results of Stern–Gerlach experiment (which shall be discussed in next section) reinforce in the same direction and help in drawing conclusions about spin angular momentum and spin magnetic moment.

So, based on various facts, for example, (i) results of Section 11.3 with eigenvalues of J_z = m_jħ (m_j = –j, –j + 1, ...,j) and allowing j to have values – odd integer alongwith integer values, (ii) Uhlenbeck and Goudsmit suggestion of explaining experimental results, (iii) Dirac’s relativistic quantum mechanical theory of electrons requiring the presence of spin angular momentum for consistency, and (iv) Stern–Gerlach experimental results [explainable only if we include spin magnetic moment μ_z= − g_s(e/2m_e)S_z], we have the following expressions of spin angular momentum and spin magnetic moment.

 

 

where q_s (≈ 2.0) is known as Lande g-factor for spin.

In fact, we had already obtained the operator form of angular momentum components for the case of quantum number j = (in Section 11.7). This case really corresponds to spin angular momentum of electron. We start writing s and m_s in place of j and m_j (s is called as spin quantum number and m_s as spin magnetic quantum number). The spin operators Ŝ_x, Ŝ_y, Ŝ_z, are written as

where

are the famous Pauli spin matrices. The eigenstates and eigenvalues of the spin angular momentum component operator Ŝ_z are easily obtained as

So the operator Ŝ_z represented by a 2 × 2 matrix has eigenstates and with corresponding eigenvalues and − . Let us call these states as and , respectively.

Thus represents a state with ‘spin up’ and represents a state with ‘spin down’. Let us write a general spin state as

Hence the matrix elements C₁ and C₂ are the complex amplitudes of ‘spin up’ and ‘spin down’ states, respectively. The column matrix in Eq. (12.11) is generally referred to as a spinor and, in fact, represents a state vector in an abstract two-dimensional complex linear vector space. The state vectors and are the two basis vectors, which form the complete orthonormal set, in this two-dimensional linear vector space. If the spin state of an electron is given by the spinor of Eq. (12.11), then the probability of finding the electron with spin up is |C₁|² and that of finding it with spin down is |C₂|². Obviously the normalization of state means

Let us rewrite the spin operator components,

Using notations of Eq(12.10) for up-spin state and down-spin state , we write below, how the components of Ŝ act on and

where is a general spin state. Above equations may be re-written in the notations of and for spin-up and spin-down states as

Let us now explore the algebra of spin operators in details. From Eqs (12.8) we have

from these, we get

and

Using Eq. (12.7), we get from Eqs (12.18)−(12.20)

 

 

 

 

These equations are, in fact, true for any components of spin operator, so we may write

 

 

 

 

where i, j, k = x, y, or z, in cyclic order. These relations are similar to those obtained for the orbital angular momentum operator components.

12.3.1  Physical Significance of above Results

We have seen results of eigenstates and eigenvalues of various spin angular momentum operators. For example, the spin operator Ŝ_z has eigenvalues m_sħ, that is, ħ/2 and – ħ/2 with corresponding spin-up and spin-down states and [that means, if an experiment (say Stern–Gerlach experiment, discussed in Section 12.6) is performed to measure z-component of spin of an electron, the result would be either ħ/2 or –ħ/2]. Also we have eigenvalues of Ŝ² as (3/4)ħ² in both the above mentioned states. So the magnitude of the spin angular momentum vector S is . We show schematically in Figure 12.2a, the z-component S_z and magnitude S of spin angular momentum vector. These two quantities are related through

 

 

giving the value of cosθ as

So, when we say an electron is in spin-up state, its spin vector is making an angle with z-axis. To maintain this angle θ with z-axis, the vector S may point in any direction along the surface of the cone shown in Figure 12.2a. Similarly in spin-down state, the vector S makes same angle θ with negative z-axis and may lie in any direction along the surface of the lower cone in Figure 12.2a. This simply means that as and when S_z is exactly known, the S_x and S_y components are totally uncertain. This is consistent with the uncertainty relation which states that two quantities corresponding to two non-commuting operators cannot be determined accurately simultaneously. And we have seen that no two components of S commute. Figure 12.2b shows a similar situation for spin-up and spin-down states with respect to y-axis.

Figure 12.2 (a) The electron spin in spin-up and spin-down states with respect to z-direction points along the surface of upper (lower) cone (b) Spin-up and spin-down states with respect to y-direction

12.4  TOTAL WAVE FUNCTION OF AN ELECTRON

We know that the wave function of an electron, in a three-dimensional space, is characterized by three quantum numbers n_x, n_y, n_z (or n, l, m_l) and is a function of three continuous space coordinates x, y, z (or r, θ, ϕ). The wave function is generally denoted by ψ_{nx, ny, nz} (x, y, z) or ψ_{n, l, ml} (r, θ, ϕ) This is all about the space part of the wave function. Now the electron, whether it is moving in an atom, in a solid or in field free space, has spin angular momentum. We have learnt in this chapter that the z-component of spin angular momentum. S_z,. has two values and − , corresponding to two different values of spin quantum number m_s ( and −). So, if we want to write the spin wave function of the electron at the same footing as its space wave function, the spin wave function should be a function of spin coordinates (say ξ) and be characterized by the spin quantum number m_s. With this notion, let us write the spin wave function as χ_ms (ξ).

Now, for two different values Of m_s, the two wave functions χ_1/2(ξ) and χ_–(1/2)(ξ) correspond, respectively to spin-up and spin-down states. Let us denote χ_1/2(ξ) by α(ξ) and χ_–(1/2)(ξ) by β(ξ)· So, when an electron is in up spin state α(ξ), the spin operator Ŝ_z has eigenvalues while in down spin state β(ξ), Ŝ_z has eigenvalue − . At this stage, we really do not know the forms of α(ξ) and β(ξ) and also do not know what values the spin coordinate ξ may take in spin-space. And, therefore, if we want to apply orthonormality conditions on α(ξ) and β(ξ), that is,

and

we do not know, how to integrate over ξ. To overcome the difficulty we take help from the eigen-states of spin operator Ŝ_z. The operator Ŝ_z has eigenvalue in eigenstate and in eigenstate . Therefore, we conclude that the state is nothing but the up spin state α(ξ) and is same as β(ξ). If we assign only two values +1 and –1 to the spin coordinate ξ and if we permit the spin functions to have only two values 0 and 1, such that

we see the above mentioned correspondence is fulfilled. Also the integrals (12.29) for orthonormality conditions may be evaluated.

and

The various notations for (orthonormal) spin–eigenstates along z-direction are summarized in Table 12.1.

Table 12.1 Various Notations of Spin Eigenstate

Spin-up state	Spin-down state
χ+	χ−
α(ξ)	β(ξ)

Table 12.2 Spinor and Ket Notations

Description of state	Spinor notation	Ket notation
Spin-up state along z-direction
Spin down (z-direction)
Spin up (x-direction)
Spin down (x-direction)
Spin up (y-direction)
Spin down (y-direction)

Also in Table 12.2, we show various notations for spin-up and spin-down states along x-, y-, and z-directions.

Now, we turn to writing the total wave function of an electron including its space part as well as spin part. If we assume that the Hamiltonian of the electron does not contain a term, involving spin angular momentum operator Ŝ (which is the case when, for example, spin-orbit interaction is neglected in atomic case), the total wave function of electron may be written as the product of the space—as well as the spin wave functions,

 

As discussed above, the spin wave function χ_ms, for two internal degrees of freedom (i.e., for m_s = + and − ) is represented by the spinors and , respectively. So we need two wave functions to describe the total state of an electron at a space-point (x, y, z). We may write these two wave functions in column matrix form as

This is known as the Pauli wave function (designated by P). The spinor wave function ψ_Ρ may also be written as a linear combination of basis spinors and .

Now, normalization of wave function ψ_Ρ gives

As we have

So

The first term in above equation represents the probability of the electron in spin-up state i.e., in state and the second term represents the probability for spin-down state.

If Ĥ represents the Hamiltonian of the electron, its equation of motion may be written as

This is generally called as Pauli equation of motion. Now, just like the spin-dependent (total) wave function of an electron has been written in spinor (i.e., column matrix) form [Eq. (12.33)], the spin-dependent operator may be written in a 2 × 2 matrix form. For example, let us consider spin-orbit interaction part of an electron in a hydrogen-like atom (see Section 15.6) which is given as

where

So

This is the Pauli operator form of Hamiltonian .

12.5  THE STERN-GERLACH EXPERIMENT

In 1922 Stern and Gerlach demonstrated in their experiment that the magnetic moment (and the angular momentum) of an electron in an atom has certain discrete orientations. Thus the results of Schrodinger equation of an electron in an atom that the angular momentum component (say) L_z has only certain discrete values m_l ħ (or magnetic moment component μ_z has values –μ_Β m_l) were verified.

Before discussing the Stern–Gerlach experiment, let us consider a magnet in a constant magnetic field B. The potential energy of interaction is

 

 

The magnet experiences a torque T, given by

 

 

The magnet will also experience a force F given by

 

 

Combining Eqs (12.42) and (12.44), we get

 

 

Now for a constant uniform field ∇B = 0 and, therefore, there is no net force on the magnetic moment. This case is shown in Figure 12.3a for a (classical) bar magnet, where equal and oppose forces work on the two poles of the magnet. So there is a finite torque but not net force on the magnet. For a non-uniform field

 

Figure 12.3 A classical magnet: (a) in a uniform magnetic field, (b) in a non-uniform magnetic field

∇B is non-zero and therefore there is a net force working on the magnetic moment. The case is shown in Figure 12.3b for a bar magnet put in a non-uniform field. The forces at the two magnetic poles of the magnet are of different magnitudes and, therefore, there is a net force on the magnet alongwith a finite torque.

The set up used in Stern–Gerlach experiment is shown in schematic form in Figure 12.4. The first experiment was performed using silver atoms. A piece of silver metal is vapourized in an oven. Silver atoms emerging from a small hole, and collimated by slits enter the magnetic field region in the form of narrow atomic beam. This beam passes between the two poles of a magnet which produce non-uniform magnetic field, as shown in Figure 12.4. The beam is finally detected on the plate P.

From Eq. (12.45) we have

 

 

With the shape of the poles as in Figure 12.4, we have B_y = 0. Moreover the magnetic field is predominantly in z-direction, so we neglect the component B_x also. Therefore

 

 

Now the magnet is symmetrical about the y–z plane and the atomic beam is confined in this plane, therefore, (∂B_z/∂x) = 0. Also, if we neglect edge effects, (∂B_z/∂y) = 0. Therefore, Eq. (12.47) may be written as

As shown in Figure 12.4 the atoms travel a distance l along y-axis while a force F is working on these in z-direction. As a result, the displacement in z-direction, ∆z is given by

Atoms of mass M are taken to enter the field with velocity v in y-direction. After crossing the field and with displacement ∆z in z-direction, the atoms move in straight line and reach the plate.

Figure 12.4 The Stern–Gerlach apparatus

If atomic magnets were behaving like classical magnets, as μ_z=μcosθ, the angle θ between the magnetic moment vector and the direction of magnetic field (z-direction) may have any value between 0 and π. So the magnetic moments of different atoms in the beam might have all possible orientations and after passing through the non-uniform magnetic field, should have spread along z-direction. Instead, to their surprise, Stern and Gerlach found silver atoms striking the plate only in two regions, symmetrically situated about the point of no deflection. The very fact that the silver beam had split into only two components on passing through the inhomogeneous magnetic field, dictates that the magnetic moment vector of silver atoms must have only two orientations with respect to z-axis.

The quantization of the component of the magnetic moment along the direction of the magnetic field is termed as space quantization. We now know that of the 47 electrons in silver atom, 46 electrons form a state with total orbital angular momentum L = 0 and total spin angular momentum S = 0. So the total angular momentum and total magnetic moment of the silver atom in its ground state is due to the spin of last electron.

12.6  SPIN AND ROTATION (SPINOR TRANSFORMATION)

Let us recollect, we studied the transformation of state vector ψ in a coordinate system obtained by rotating it about an axis along unit vector e through an angle ϕ. In this section, we are concerned with the transformation properties of spin wave function under similar rotation. Following Eq. (9.61b), a spin wave function χ_s of a particle of spin s transforms into a new spin wave function in the following way

For a spin particle, we have , so

Here in Eq. (12.50b), the definition of the exponential of an operator is,

Now using the identity (see Exercise 12.3)

We have

and

So collecting even and odd powers in the expansion (12.51), we may write the rotation operator as

or

Here I is 2 × 2 unit matrix. It may be checked from Eq. (12.54) that for ϕ = 2π, we get

 

 

Therefore, for ϕ = 2π, we have

 

 

So a full rotation by 360° about a fixed axis, which, in fact, is equivalent to zero rotation or simply identity operator, as seen in Eq. (12.56) changes the sign of every spinor component. Let us mention here that the usual vectors and tensors in three-dimensional co-ordinate space behave differently under the said rotation operator: these simply return to their original value upon a rotation of 2π. However, it may be mentioned here that for spinors getting their sign changed under 2π-rotation does not de-value their usefulness. It may be remembered that all physical quantities (expectation values) evaluated using these sign changed spinors are same as those obtained by using original spinors.

Let us write the spin wave function χ as

and consider a rotation, about z-axis, through an angle ϕ. Under this rotation, let χ be transformed to χ′, written as

Using Eq. (12.54), we have

or

As a second example, let us consider rotation about x-axis through an angle ϕ. In that case we have,

or

12.7  A MAGNETIC MOMENT IN A UNIFORM MAGNETIC FIELD: THE LARMOR PRECESSION

12.7.1  Classical Treatment

Let us firstly consider a classical system. For example, we consider a classical particle of charge q and mass m moving in a circular orbit and having angular momentum L and magnetic moment μ. L and μ are collinear, related through the equation

Here γ, the ratio of the magnetic moment and the angular momentum is called the gyromagnetic ratio. Let the classical system be put in a constant homogeneous magnetic field.

 

 

As a result, a torque is exerted on the system. The classical equation of motion of L is given by

or

or

 

 

Here θ is the angle that μ makes with Β. Δμ will be at right angles to μ and, therefore, magnetic moment μ, without changing angle θ, precesses about the direction of B, as shown in Figure 12.5. Let at time t = 0, μ be along OA and at t = Δt, along OB. Then

 

 

and

 

 

 

From Eqs (12.63) and (12.64), we have

ω₀ is the frequency of precession of μ around direction of B and is called Larmor frequency. It may be noted that the Larmor frequency is only one-half of the cyclotron frequency, that is, the frequency with which a free charged particle goes around in a circular path in presence of a constant homogeneous magnetic field.

Figure 12.5 Larmor precession of a classical magnetic moment

12.7.2  Quantum Mechanical Treatment: Precession of an Electron Spin in a Magnetic Field

Let us consider an electron at rest in a constant uniform magnetic field pointing in z-direction.

 

 

The Hamiltonian is the interaction energy

In the matrix representation, where Ŝ_z is diagonal, Ĥ becomes

The energy eigenvalues of the electron are given as (see Figure 12.6),

where

with corresponding spin states

Figure 12.6 Splitting of spin energy level in the presence of constant magnetic field B = B₀ e_z

Let the electron be initially (t = 0) in state

To find the state at time t, we start with the time-dependent Schrodinger equation

where

and we have to find quantities C₁(t) and C₂(t).

Substituting Eq. (12.73) into Eq. (12.72) and using Eq. (12.68), we get

or

or

Equations (12.75) have got solutions

 

 

Therefore, the normalized state at time t is

Let us consider time development of some particular initial states.

Case I: Let us take the initial state corresponding to C₁(0) = C₂(0) = 1, that is,

Let

Then from Eq. (12.77), we get

So, if T denotes the period, T = (2π/Ω) then

Here we have used notations mentioned in Table 12.2. From Eqs (12.80) we conclude the following:

1. At t = 0, the electron is in an eigenstate of Ŝ_x with corresponding eigenvalue ħ/2.
2. At t = T/4 it is in an eigenstate of Ŝ_y with eigenvalue ħ/2.
3. At t = T/2 in eigenstate of Ŝ_x with eigenvalue –ħ/2.
4. At t = 3T/4 in eigenstate of Ŝ_y with eigenvalue –ħ/2.

Let us remember the discussion of Section 12.3 and the connected Figure 12.2 where it was mentioned explicitly that when an electron is found in spin-up state of Ŝ_x, the spin is not really pointing along +x direction but lies on the surface of the cone with its axis coinciding with +x-axis. So at t = 0, the spin is lying on the cone surface as shown in Figure 12.7. In the presence of magnetic field B = B₀ e_z, the axis of the cone rotates in x–y plane such that at t = T/4, the axis of the cone is along y-direction. At time t = T/2 the axis of the cone is along negative x-axis. At t = 3T/4 along negative y-axis and at t = T again at starting point, that is, along x-axis. So the cone and, therefore, the spin is precessing around z-axis, with frequency Ω = (eB₀/m_e). This is termed as Larmar’s precession.

Figure 12.7 Precession of a spinning electron in a constant uniform magnetic field B = B₀e_z. Initially the electron is in state

We may reach the same conclusions, as above, about the precession of electron spin (about z-axis), by finding the expectation values and . For example,

Putting from Eq. (12.79), we have

or

Similarly, we get

and

Equations (12.82) imply that the spin angular momentum vector and, therefore, spin magnetic moment vector rotates about z-axis, in presence of magnetic field B = B₀e_z, with angular frequency Ω = (eB₀/m_e).

It may also be checked here that, though the initial (spin) eigenstate of the electron changes as a function of time to the sate in presence of magnetic field B and thus the spin precesses about z-axis with angular frequency Ω, its energy eigenvalue does not change with time. We can find its energy in state as

Case II: Let the initial state correspond to C₁(0) = 1, C₂(0) = 0, that is,

From Eq. (12.77), we may write the state at time t as

It can be easily seen that the energy eigenvalue in state is

Let us now find

So, if an electron is put in magnetic field B = B₀ e_z in spin-up state at time t = 0, its energy eigenvalue is E₁ = (ħΩ/2) and it remains in spin-up state for all time (with energy E₁). Similarly if the electron is in spin-down state with eigenvalue E₂ = –(ħΩ/2), it will remain in spin-down state (with energy E₂) for all time.

12.8  ELECTRON SPIN RESONANCE

We studied in Section 12.7 the effect of constant homogeneous magnetic field B = B₀ e_z on the spin states and energy eigenvalues of an electron. We found that irrespective of the initial spin state of the electron, its energy eigenvalue does not change with time. If initially the electron is in state , with the passage of time the state changes to to to to ; that is, the spin angular momentum vector precesses in x–y plane about z-axis with angular frequency Ω. However, if initially the electron is in state with energy eigenvalue , its state at time t remains (except a multiplicative phase factor e^–i(Ω/2)t). Similarly if initially it is in state with energy eigenvalue , its state remains with a phase factor e^{i(Ω / 2)t}.

In this section, we shall consider, in addition to the uniform magnetic field B = B₀ e_z, a rotating radio frequency (r.f.) field B₁, in x–y plane, given as

 

 

We shall see that the r.f. field would cause transitions between spin-up and spin-down states. The total Hamiltonian of electron in presence of both the fields B and B₁, is

 

 

where

and

We shall now solve the time-dependent Schrodinger equation

Let us express the spin state in terms of basis state and (of Hamiltonian Ĥ₀), as

where, as mentioned above and . Substituting Eq. (12.90) in Eq. (12.89), we get

First two terms on L.H.S. are same as first two terms on R.H.S. So, we get

Taking inner product of Eq. (12.91) with eigen-bra (0 1) (i.e., by pre-multiplying Eq. (12.91) by and noting that and , we get

or

or

where

and

Similarly by pre-multiplying Eq. (12.91) by eigen-bra (1 0), we have

Now differentiating Eq. (12.94) and using Eq. (12.92) for (dC₂/dt), we get

where

If we take a solution of the form

 

 

We get

 

 

with

Therefore, the general solution of Eq. (12.95) may be written as

 

 

Let us take the initial spin state to be

i.e.,

 

 

Then Eq. (12.99) gives

 

 

To find C₂(t), we use Eq. (12.94)

so

and we get

From Eq. (12.101) and Eq. (12.102), we have

and

giving the required condition

Now the initial spin state (at t = 0) is [see Eq. (12.100b)]

Therefore, the quantity | C₁(t) |² is nothing but the probability that the state has changed to at time t; alternatively, P(t) = | C₁(t) |² gives the probability of spin–flip. | C₁(t) |² may be re-written as

We can see from Eq. (12.106) that the transition probability P(t) is small for | ω – Ω | >> ω₁. For ω = Ω, the spin–flip probability is maximum. In fact, spin–flip probability is unity at times t = (π / 2T), (3π / 2Τ), ··· for ω = Ω. This is called the resonance condition. The resonance condition may be achieved either by varying the frequency ω of the r.f. field or by varying Ω (by varying the strength B₀ of the homogeneous field in z-direction).

Figure 12.8 shows the variation of probability P(t) as a function ω/Ω. We get resonance at ω/Ω = 1. We show variation of transition probability P(t) as a function of time in Figure 12.9 for fixed values of ω – Ω; Figure 12.9a for ω − Ω = 6ω₁ and 12.9b for ω = Ω.

Figure 12.8 Variation of transition probability P(t) = |C₁(t)|² as a function of (ω/Ω), for two different values of (ω₁/Ω) = (B₁/B₀)

Figure 12.9 Variation of the transition probability P(t) between and states under the effect of rotating magnetic field B₁(t) (a) shows the case for outside resonance, for example, ω – Ω = 6ω₁, (b) shows the resonance case ω = Ω, where even for small value of B₁ the transition probability P(t) may become 1

EXERCISES

Exercise 12.1

In a Stern–Gerlach experiment, a beam of hydrogen atoms with velocity 3 × 10³ m/s, passes through an inhomogeneous magnetic field of length 50 cm and having a gradient of 200 T/m perpendicular to the direction of the incident beam. Find out the transverse deflection of the atoms of the beam at the point where the beam leaves the field.

Exercise 12.2

Show that Pauli spin matrices satisfy

1. 
2. 
3. 

(∊_ijk is the Levi–Civita symbol:

Exercise 12.3

Show that

where A and B are arbitrary vectors.

Exercise 12.4

Consider an electron at rest in a constant uniform magnetic field in x-direction B = B₀e_x. The corresponding Hamiltonian is

What are the eigenvalues and eigenkets of Ĥ₀?

Exercise 12.5

Spin eigenket of an electron is represented by the spinor . Express this spinor in a co-ordinate system obtained by rotation about y-axis through an angle ϕ.

Exercise 12.6

Express the spinor in a co-ordinate system obtained by

1. rotation about x-axis through an angle 60°
2. rotation about y-axis through an angle 60°
3. rotation about z-axis through an angle 60°

Exercise 12.7

Show that any spinor may be expressed as a linear combination of any of the three pairs of eigenkets.

or

or

Exercise 12.8

In a measurement, the z-component of spin of an electron is found to have value ħ/2.

1. What is the spin state of the electron after the measurement?
2. Show that in this state

Exercise 12.9

Show that it is impossible for a spin particle to be in a state such that

Exercise 12.10

A polarized beam of electrons in state is sent through Stern–Gerlach analyzer which measures S_x. What values (and with what probabilities) will be found?

Exercise 12.11

An assembly of electrons has an isotropic distribution of spin values. If you choose an electron at random, what is the probability of finding this electron with the following spin components?

1. 
2. 
3. 

Exercise 12.12

A neutron is in spin state . What is the probability that a measurement finds each of the following?

1. 
2. 
3. 
4. 
5. 
6. 

Exercise 12.13

Consider a spin particle in the state

What is the probability that a measurement finds the following?

1. 
2. 
3. 
4. 
5. 
6. 

Exercise 12.14

Find the expectation values of Ŝ_x, Ŝ_y, Ŝ_z of spin particle in spinor state given in Exercise 12.13.

Exercise 12.15

Consider an electron in spin state

1. Determine the normalization constant.
2. Find the expectation values of Ŝ_x, Ŝ_y and Ŝ_z.
3. Find the expectation values of , and .
4. Find the uncertainties ∆S_x, ∆S_y and ∆S_z. Here, for example,

Exercise 12.16

For a normalized spinor

Compute

1. 
2. 

Exercise 12.17

Suppose a 2 × 2 matrix is written as

where b₀, b_k(k = 1, 2, 3) are numbers.

1. Find tr(X) and
2. Find b₀ and b_k in terms of matrix elements X_ij(i, j = 1, 2). This shows that any 2 × 2 matrix X may be expressed a linear combination of Pauli spin matrices (written in this exercise as ) and a 2 × 2 unit matrix I.

Exercise 12.18

Consider a two level system whose Hamiltonian operator is given as

Here number a is having dimension of energy. Find the energy eigenvalues and the corresponding eigen-kets (in terms of states and ).

Exercise 12.19

An electron is in spin state Compute

where the expectation value is taken for the state . Check the generalized uncertainty relation.

where

Exercise 12.20

A beam of spin atoms goes through three Stern–Gerlach measurements 1, 2, and 3 in succession in following manner:

1. Measurement 1 accepts S_z = ħ/2 atoms and rejects S_z = –ħ/2.
2. Measurement 2 accepts S_n = ħ/2 atoms and rejects S_n = –ħ/2 atoms. Here S_n is the eigenvalue of the operator Ŝ · e_n; unit vector e_n makes an angle ϕ in the xz-plane with respect to z-axis.
3. Measurement 3 accepts S_z = ħ/2 and rejects S_z = -ħ/2.

What is the intensity of the final S_z = ħ/2 beam if the S_z = ħ/2 beam surviving after measurement 1 is normalized to unity? For what orientation ϕ (of measurement 2) shall we get the maximum intensity in measurement 3?

SOLUTIONS

Solution 12.1

 

so

Solution 12.2

1. 

   and so on.

2. We can easily see
   

   so result follows,

3. From (a) and (b), we can easily find result (c).

Solution 12.3

Using relations of Solution 12.2, we get

Solution 12.4

The Hamiltonian Ĥ₀ may be written in matrix form as

We can easily see, Ĥ_o has

eigenvalue = µ_BB₀, corresponding to eigenstate ,

eigenvalue = −µ_BB₀, corresponding to eigenstate .

Solution 12.5

Let denote the spinor in the rotated coordinate system. Using Eq. (12.54), we have

Solution 12.6

1. Using relation (12.59b), we have
   
2. Using relation (12.107), we have
   
3. Using relation (12.58b), we have
   

Solution 12.7

Let us write

so

giving

Therefore,

Similarly, we may get

and

Solution 12.8

1. After the measurement the electron remains in state .
2. It is straight forward to evaluate these expectation values in state .

Solution 12.9

Let us start with operator Ŝ_z and find the state in which

or

Let us set | a | = 1. Then may be written, without loss of generality, as

where θ₁ and θ₂ are the phases of a and b, respectively. Now let us find for this state

Also, we may get

Now there is no value of (θ₁ – θ₂), which gives

 

 

Therefore, there is no state for which expectation values of all the three spin operator components are zero.

Solution 12.10

The state may be written as

So, if S_x is measured on electrons in state , we will get value ħ/2 with probability and –ħ/2 with probability .

Solution 12.11

Suppose the spin component S_x is measured. Then obviously one of two values ħ/2 or –ħ/2 will be found. The spin is isotropic, so these two values are equally likely. Hence, probability is for both values S_x = ħ/2 and S_x = ħ/2. Similar is the case for S_y and S_z.

Solution 12.12

1. 1
2. 0
3. 
4. 
5. 
6. 

Solution 12.13

We have seen in Exercise 12.7 that a state may be written in following forms:

so

From Eq. (12.110a)

1. Probability of
2. Probability of

   From Eq. (12.110b)

3. Probability of
4. Probability of

   From Eq. (12.110c)

5. Probability of
6. Probability of

Solution 12.14

First Method

Expectation value of

Expectation value of

Expectation value of

Second Method

Solution 12.15

1. 

   so

2. 
   
   
3. 
4. Values of standard deviations ∆S_x, and so on, may be easily calculated from the values in (a), (b), and (c) above.

Solution 12.17

1. 
2. We have
   X₁₁ = b₀ + b_z
   X₂₂ = b₀ + b_z
   X₁₂ = b_x – ib_y
   X₂₁ = b_x + ib_y

   So

   

Solution 12.18

Ĥ may be written in matrix form as

in the basis and

where and

It may be easily checked that matrix Ĥ has got eigenstate

eigenstate with eigenvalue

and

eigenstate with eigenvalue

Here normalization constants are

Eigenstates and may be expressed in terms of basis states and as

and

Solution 12.19

so

Similarly, we may get

Now

So

So, we get

Solution 12.20

After passing through measurement 1, the beam has atoms in state

In the co-ordinate system obtained by rotating it about y-axis through an angle ϕ, the state may be written (with the help of the result of Solution 12.5) as

Measurement accepts only those atoms having S_n = ħ/2.

So after second measurement the eigenstate is

Now after second measurement, the spin component is measured along z-direction of the original co-ordinate system, which is obtained by rotating the changed coordinate system about y-axis through an angle –ϕ. So we get

So, the number of atoms having S_z = ħ/2 are having probability cos⁴ϕ/2. This probability is maximum (unity) for ϕ = 0.

REFERENCES

1. Shankar, R. 1980. Principles of Quantum Mechanics. New York, NY: Plenum.
2. Baym, G. 1969. Lectures on Quantum Mechanics. New York, NY: W. A. Benjamin.
3. Ziman, J.M.1969. Elements of Advanced Quantum Mechanics. London, UK: Cambridge University Press.
4. Merzbacher, E. 1999. Quantum Mechanics, 3rd edn., New York, NY: John Wiley & Sons.
5. Tinkham, M. 1964. Group Theory and Quantum Mechanics. New York, NY: McGraw-Hill.
6. Cohen-Tannoudji, C., Diu, B. and Laloe, F. 1977. Quantum Mechanics, Vol. I and II, New York, NY: John Wiley & Sons.
7. Feynman, R.P., Lighton, R.B. and Sands, M. 1965. The Feynman Lectures on Physics, Vol. III, Reading, MA: Addison-Wesley.