Quantum Mechanics by Konstantin K. Likharev - HTML preview
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l m l.
(5.162) between
m and l
This relation of the quantum numbers m and l is semi-quantitatively compatible with the classical image of the angular momentum vector L, of the same length L, pointing in various directions, thus affecting the value of its component Lz. In this classical picture, however, L 2 would be equal to the square of ( Lz)max, i.e. to ( l)2; however, in quantum mechanics, this is not so. Indeed, applying both parts of the second of the operator equalities (155) to the top state’s vector l, m max l, l, we get ˆ2
ˆ
ˆ2
ˆ ˆ
L l, l L l, l L l, l L L l, 2
l l l,
2 2
l l l, l 0
z
z
(5.163)
2
l l
1 l, l .
Since by our initial assumption, all eigenvectors l, m correspond to the same eigenvalue of 2
ˆ L , this
result means that all these eigenvalues are equal to 2 l( l + 1). Just as in the case of the spin-½ vector operators discussed in Sec. 4.5, the deviation of this result from 2 l 2 may be interpreted as a result of unavoidable uncertainties (“fluctuations”) of the x- and y-components of the angular momentum, which give non-zero positive contributions to L 2
2
x and Ly , and hence to L 2, even if the angular momentum vector is aligned with the z-axis in the best possible way.46
(For applications, one more relation, in one of its two equivalent forms, may be convenient: ˆ L l, m l l
m m
l m
l m
l m
l m
.
(5.164)
1 11/2 , 1 1
1/2 , 1
This equality, valid to the multiplier ei with an arbitrary real phase , may be readily proved from the above relations in the same way as the parallel Eqs. (89) for the harmonic-oscillator operators (65) were proved in Sec. 4; due to this similarity, the proof is also left for the reader’s exercise.47) 46 Curiously, a similar formula L 2 = 2 l( l + 1) may be also obtained by assuming that all (2 l + 1) values Lz = m of a system with fixed l have equal probability. (Let me leave the proof for the reader’s exercise.) 47 The reader is also challenged to use the commutation relations discussed above to prove one more important property of the common eigenstates of the operators L ând 2
ˆ L :
z
l, m r ˆ l' , m' ,
0
l'
unless l and
1
m'
m
either
m
or
1
.
j
This property gives the selection rule for the orbital electric-dipole quantum transitions, to be discussed later in the course, especially in Sec. 9.3. (The final selection rules at these transitions may be affected by the particle’s spin – see the next section.)
Chapter 5
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Note that the formulas discussed in this section, with the sole exception of Eq. (146), are not conditioned by a particular Hamiltonian of the system under analysis. However, they (as well as those discussed in the next section) are especially important for particles moving in spherically-symmetric potentials, which were discussed in Sec. 3.6. It is easy (and hence is also left for the reader’s exercise) to prove that in this case, the particle’s Hamiltonian operator commutes with that of the angular momentum, so according to Eq. (4.199), in the Heisenberg picture of quantum dynamics, the Cartesian components L âs well as 2
ˆ L do not depend on time, and hence their expectation values are integrals of j
motion.
By using the expression of Cartesian coordinates via the spherical ones exactly as this was done in Eq. (152), we get the following expressions for the ladder operators (153) in the coordinate representation:
i
ˆ L
e
i
cotan
.
(5.165)
Angular
momentum
operators: Now plugging this relation, together with Eq. (152), into any of Eqs. (155), we get coordinate
representation
1
1
2
2
2
ˆ L
sin
.
(5.166)
2
2
sin
sin
But this is exactly the operator (besides its division by the constant parameter 2m R 2) that stands on the left-hand side of Eq. (3.156). Hence that equation, which was explored by the “brute-force” (wave-mechanical) approach in Sec. 3.6, may be understood as the eigenproblem for the operator 2
ˆ L in the
coordinate representation, with the eigenfunctions Y m
l (,) corresponding to the eigenkets l, m, and the
eigenvalues L 2 = 2m R 2 E. As a reminder, the main result of that, rather involved analysis was expressed by Eq. (3.163), which now may be rewritten as
2
L 2
2
2
R
m
E l( l )
1 ,
(5.167)
l
l
in full agreement with Eq. (163), which was obtained by much more efficient means based on the bra-ket formalism. In particular, it is fascinating to see how easy it is to operate with the eigenvectors l, m, while the coordinate representations of these vectors, the spherical harmonics Y m l (,), may be only
expressed by rather complicated functions – please have one more look at Eq. (3.171) and Fig. 3.20.
5.7. Spin and total angular momentum
The theory described in the last section is useful for much more than orbital motion analysis. In particular, it helps to generalize the spin-½ results discussed in Chapter 4 to other values of spin s – the parameter still to be quantitatively defined. For that, let us notice that the commutation relations (4.155) for spin-½, which were derived from the Pauli matrix properties, may be rewritten in exactly the same form as Eqs. (149) and (151) for the orbital momentum:
Spin
operators:
commutation
3
ˆ ˆ
S , S
i
S
S S
(5.168)
j
j'
ˆ
,
j"
jj'j"
ˆ2 ˆ, j 0
relations
j" 1
Chapter 5
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It had been postulated (and then confirmed by numerous experiments) that these relations hold for quantum particles with any spin. Now notice that all the calculations of the last section have been based almost exclusively on such relations – the only exception will be discussed imminently. Hence, we may repeat them for the spin operators, and get the relations similar to Eqs. (158) and (163): Spin
operators:
S ˆ s, m m
s, m
S ˆ
,
2 s, m 2
s( s )
1 s, m , 0 s, s m s ,
(5.169) eigenstates
z
s
s
s
s
s
s
and
eigenvalues
where ms is a quantum number parallel to the orbital magnetic number m, and the non-negative constant s is defined as the maximum value of ms . The c-number s is exactly what is called the particle’s spin.
Now let us return to the only part of our orbital moment calculations that has not been derived from the commutation relations. This was the fact, based on the solution (146) of the orbital motion problems, that the quantum number m (the analog of ms) may be only an integer. For spin, we do not have such a solution, so the spectrum of numbers ms (and hence its limits s) should be found from the more loose requirement that the eigenstate ladder, extending from – s to + s, has an integer number of steps. Hence, 2 s has to be an integer, i.e. the spin s of a quantum particle may be either integer (as it is, for example, for photons, gluons, and massive bosons W and Z0), or half-integer (e.g., for all quarks and leptons, notably including electrons).48 For s = ½, this picture yields all the properties of the spin-½
that were derived in Chapter 4 from Eqs. (4.115)-(4.117). In particular, the operators 2
ˆ S and S ˆ have
z
two common eigenstates ( and ), with Sz = ms = /2, both with S 2= s( s +1)2 = (3/4)2.
Note that this analogy with the angular momentum sheds new light on the symmetry properties of spin-½. Indeed, the fact that m in Eq. (146) is an integer was derived in Sec. 3.5 from the requirement that making a full circle around the z-axis, we should find a similar final value of the wavefunction m, which may differ from the initial one only by an inconsequential factor exp{2 im} = +1. With the replacement m ms = ½, such an operation would multiply the wavefunction by exp{ i} = –1, i.e.
reverse its sign. Of course, spin properties cannot be described by a usual wavefunction, but this odd parity of electrons, shared by all other spin-½ particles, is clearly revealed in properties of multiparticle systems (see Chapter 8 below), and as a result, in their statistics (see, e.g., SM Chapter 2).
Now we are sufficiently equipped to analyze the situations in which a particle has both the orbital momentum and the spin – as an electron inside an atom. In classical mechanics, such an object, with the spin S interpreted as the angular moment of its internal rotation, would be characterized by the total angular momentum vector J = L + S. Following the correspondence principle, we may assume that quantum-mechanical properties of this observable may be described by the similarly defined vector operator:
Jˆ Lˆ Sˆ ,
(5.170) Total
angular
with Cartesian components
momentum
ˆ
ˆ
ˆ
J L S , etc. ,
(5.171)
z
z
z
and the magnitude squared equal to
ˆ 2
ˆ 2
ˆ 2
ˆ 2
J J J J .
(5.172)
x
y
z
48 As a reminder, in the Standard Model of particle physics, such hadrons as mesons and baryons (notably including protons and neutrons) are essentially composite particles. However, at non-relativistic energies, protons and neutrons may be considered fundamental particles with s = ½.
Chapter 5
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Let us examine the key properties of this vector operator. Since its two components (170) describe different degrees of freedom of the particle, i.e. belong to different Hilbert spaces, they have to be completely commuting:
ˆ ˆ
L , S
L S
L S
L S
(5.173)
j
j'
,
0
ˆ2 ˆ,2 ,0 ˆ ˆ,2
j
,0 ˆ2 ˆ, j 0.
The above formulas are sufficient to derive the commutation relations for the operator Jˆ , and unsurprisingly, they turn out to be absolutely similar to those of its orbital and spin components: Total
3
momentum:
commutation
ˆ ˆ
J , J
i
J
J J
.
(5.174)
j
j'
ˆ
,
j"
jj'j"
ˆ2 ˆ, j 0
relations
j" 1
Now by repeating all the arguments of the last section, we may derive the following expressions for the common eigenstates of the operators 2
ˆ J and J ˆ :
z
Total
momentum:
eigenstates,
ˆ
J j, m m
j, m ,
2
J j,
2
m j( j )
1 j, m , 0 j, j m j,
(5.175)
z
j
j
j
j
j
j
and
eigenvalues where j and mj are new quantum numbers.49 Repeating the arguments just made for s and ms, we may conclude that j and mj may be either integers or half-integers.
Before we proceed, one remark on notation: it is very convenient to use the same letter m for numbering eigenstates of all momentum components participating in Eq. (171), with corresponding indices ( j, l, and s), in particular, to replace what we called m with ml. With this replacement, the main results of the last section may be summarized in a form similar to Eqs. (168), (169), (174), and (175): Orbital
3
ˆ ˆ
L , L
i
L
L L
,
(5.176)
j
j'
ˆ
,
j"
jj'j"
ˆ2 ˆ, j 0
momentum:
j" 1
basic
properties
ˆ
ˆ
(new notation)
L l, m m
l, m ,
2
L l,
2
m l( l )
1 l, m , 0 l, l m l.
(5.177)
z
l
l
l
l
l
l
In order to understand which eigenstates participating in Eqs. (169), (175), and (177) are compatible with each other, it is straightforward to use Eq. (172), together with Eqs. (168), (173), (174), and (176) to get the following relations:
ˆ 2 ˆ
J , 2
L 0, ˆ 2 ˆ
J , 2
S 0,
(5.178)
ˆ 2 ˆ
J , L
J S
(5.179)
z 0,
ˆ2 ˆ, z 0.
This result is represented schematically on the Venn diagram shown in Fig. 12, in which the crossed arrows indicate the only non-commuting pairs of operators. The color lines in this figure encircle two operator groups that commute with each other and hence may share their eigenstates. The first group (encircled red), consists of all operators but 2
ˆ J ; their shared eigenstates correspond to
definite values of the corresponding quantum numbers: l, ml, s, ms, and mj. Actually, only four of these numbers are independent, because due to Eq. (171) for these compatible operators, for each eigenstate of this group, their “magnetic” quantum numbers m have to satisfy the following relation: 49 Let me hope that the difference between the quantum number j, and the indices j, j’, j” numbering the Cartesian components in relations like Eqs. (168) or (174), is absolutely clear from the context.
Chapter 5
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m m m .
(5.180)
j
l
s
Hence the common eigenstates of the operators of this group are fully defined by just four quantum numbers, for example, l, ml, s, and ms. For some calculations, especially those for the systems whose Hamiltonians include only the operators of this group, it is convenient to use this set of eigenstates as the basis; frequently this approach is called the uncoupled representation. The most important example of such a situation is a non-relativistic particle moving in a spherically-symmetric potential (3.155), whose Hamiltonian does not depend on its spin. As we have seen in the previous section, its stationary states correspond to definite l and ml.
operators
2
ˆ L
2
ˆ S
2
ˆ J
diagonal in
the coupled
representation
operators
diagonal in
the uncoupled
L ˆ
S ˆ
J ˆ
z
z
z
Fig. 5.12. The Venn diagram of angular momentum
representation
operators, and their mutually-commuting groups.
However, in some situations, interactions between the orbital and spin degrees of freedom (in the common jargon, the spin-orbit coupling) cannot be ignored; this interaction leads in particular to splitting (called the fine structure) of the atomic energy levels even in the absence of external magnetic field. I will discuss these effects in detail in the next chapter and now will only note that they may be described by a term proportional to the product Lˆ Sîn the particle’s Hamiltonian. If this term is substantial, the uncoupled representation becomes inconvenient. Indeed, writing 2
2
2
2
2
2
2
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ ˆ
ˆ ˆ
ˆ
ˆ
ˆ
J (L S) L S 2L S, 2
that
so
L S J L S ,
(5.181)
and looking at Fig. 12 again, we see that the operator Lˆ Sˆ describing the spin-orbit coupling does not commute with operators L ând S ˆ . This means that stationary states of the system with such a term in z
z
the Hamiltonian do not belong to the uncoupled representation’s basis. On the other hand, Eq. (181) shows that the operator Lˆ Sˆ does commute with all four operators of another group, encircled blue in Fig. 12. According to Eqs. (178), (179), and (181), all operators of that group also commute with each other, so they have a group of common eigenstates, described by the quantum numbers l, s, j, and mj.
This group is the basis for the so-called coupled representation of particle states.
Excluding, for the notation briefness, the quantum numbers l and s that are common for both groups, it is convenient to denote the common ket-vectors of each group as, respectively, m , m ,
tion'
representa
uncolpled
for the
basis,
s
l
s
Coupled and
(5.182) uncoupled
j, m ,
tion'
representa
coupled
for the
basis.
s
bases
j
As we will see in the next chapter, for the solution of some important problems (e.g., the fine structure of atomic spectra and the Zeeman effect), we will need the relation between the kets j, mj and the kets
ml, ms. This relation may be represented as the usual linear superposition, Chapter 5
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Clebsch-
Jordan
coefficients:
j, m m , m m , m j, m .
(5.183)
j
l
s
l
s
j
definition
m , m
l
s
The short brackets in this relation, essentially the elements of the unitary matrix of the transformation between two eigenstate bases (182), are called the Clebsch-Gordan coefficients.
The best (though imperfect) classical interpretation of Eq. (183) I can offer is as follows. If the lengths of the vectors L and S (in quantum mechanics associated with the numbers l and s, respectively), and also their scalar product LS, are all fixed, then so is the length of the vector J = L + S – whose length in quantum mechanics is described by the number j. Hence, the classical image of a specific eigenket j, mj, in which l, s, j, and mj are all fixed, is a state in which L 2, S 2, J 2, and Jz are fixed.
However, this fixation still allows for an arbitrary rotation of the pair of vectors L and S (with a fixed angle between them, and hence fixed LS and J 2) about the direction of the vector J – see Fig. 13.
z
z
J
J
J
J
z
z
L
L
z
Lz
L
S
Fig. 5.13. A classical image of two
S
different quantum states with the
z
Sz
S
same quantum numbers l, s, j, and
0
mj, but different ml and ms.
0
Hence the components Lz and Sz in these conditions are not fixed, and in classical mechanics may take a continuum of values, two of which (with the largest and the smallest possible values of Sz) are shown in Fig. 13. In quantum mechanics, these components are quantized, with their states represented by eigenkets ml, ms, so a linear combination of such kets is necessary to represent every ket
j, mj. This is exactly what Eq. (183) does.
Some properties of the Clebsch-Gordan coefficients ml, ms j, mj may be readily established.
For example, the coefficients do not vanish only if the involved magnetic quantum numbers satisfy Eq.
(180). In our current case, this relation is not an elementary corollary of Eq. (171), because the Clebsch-Gordan coefficients, with the quantum numbers ml, ms in one state vector, and mj in the other state vector, characterize the relationship between different groups of the basis states, so we need to prove this fact; let us do that. All matrix elements of the null-operator
ˆ
ˆ
ˆ
J ( L S 0ˆ
)
(5.184)
z
z
z
should equal zero in any basis; in particular
ˆ
ˆ
ˆ
j, m J ( L S ) m , m .
0
(5.185)
j
z
z
z
l
s
Acting by the operator J ûpon the bra-vector, and by the sum ˆ
ˆ
( L S ) upon the ket-vector, we get
z
z
z
m ( m m ) j m m m
(5.186)
j
l
s
,
,
,
0
j
l
s
thus proving that
Chapter 5
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m , m j, m j, m m , m * ,
0
if m
m m .
(5.187)
l
s
s
s
l
s
j
l
s
As we will see in a minute, this property will enable us, in particular, to establish the range of possible values of the quantum number j, at fixed l and s.
For the most important case of spin-½ particles (with s = ½, and hence ms = ½), whose uncoupled representation basis includes 2(2 l + 1) states, the restriction (187) enables the representation of all non-zero Clebsch-Gordan coefficients on the simple “rectangular” diagram shown in Fig. 14.
Indeed, each coupled-representation eigenket j, mj, with mj = ml + ms = ml ½, may be related by nonzero Clebsch-Gordan coefficients to at most two uncoupled-representation eigenstates ml, ms. Since ml may only take integer values from – l to + l, mj may only take semi-integer values on the interval [– l – ½, l + ½]. Hence, by the definition of j as ( mj)max, its maximum value has to be l + ½, and for mj = l + ½, this is the only possible value with this j. This means that the uncoupled state with ml = l and ms = ½
should be identical to the coupled-representation state with j = l + ½ and mj = l + ½: j l ½, m l ½ m m ½, m ½
.
(5.188)
j
l
j
s
In Fig. 14, these two identical states are represented by the top-rightmost point (the uncoupled representation) and the sloped line passing through it (the coupled representation).
m
s
m l ½ m l 3 2
m l 3 2 m l ½
j
j
j
j
j l ½
j l ½
j l ½
j l ½
m l ½
j
½
j l ½
l
l 1
l 2
0
l 2
l 1
l
ml
m l ½
j
j l ½
½
Fig. 5.14. A graphical representation of possible basis states of a spin-½ particle with a fixed l. Each dot corresponds to an uncoupled-representation ket-vector ml, ms, while each sloped line corresponds to one coupled-representation ket-vector j, mj, related by Eq. (183) to the kets ml, ms whose dots it connects.
However, already the next value of this quantum number, mj = l – ½, is compatible with two values of j, so each ml, ms ket has to be related to two j, mj kets by two Clebsch-Gordan coefficients.
Since j changes in unit steps, these values of j have to be l ½. This choice, j l ½ ,
(5.189)
where the alternating sign is independent of the sign of ms, evidently satisfies all lower values of mj as well –
see Fig. 14.50 (Again, only one value, j = l + ½, is necessary to represent the state with the lowest mj = –
l – ½ – see the bottom-leftmost point of that diagram.)
50 Eq. (189) may be readily generalized to the case of arbitrary spin s: j may only take values that differ by 1, within the interval [ l - s , l + s]. This important result (whose proof is left for the reader’s exercise) allows a semi-quantitative classical interpretation in terms of the vector diagrams shown in Fig. 13: in them, the largest value of j corresponds to the parallel alignment of the vectors L and S, while its smallest value, to their antiparallel alignment.
Chapter 5
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Note that the total number of the coupled-representation states is 1 + 22 l + 1 2(2 l + 1), i.e. is the same as those in the uncoupled representation. So, for spin-½ systems, each sum (183), for fixed j and mj (plus the fixed common parameter l, plus the common s = ½), has at most two terms, i.e. involves at most two Clebsch-Gordan coefficients.
These coefficients may be calculated in a few steps, all but the last one rather simple even for arbitrary spin s. First, the similarity of the vector operators Jˆ
Sˆ
and
to the operator Lˆ , expressed by
Eqs. (169), (175), and (177), may be used to argue that the matrix elements of the operators S ˆ
J ˆ
and
,
defined similarly to L ˆ , have the matrix elements similar to those given by Eq. (164). Next, acting by
the operator J ˆ L ˆ
S ˆ
upon both parts of Eq. (183), and then inner-multiplying the result by the bra
vector ml, ms and using the above matrix elements, we may get recurrence relations for the Clebsch-Gordan coefficients with adjacent values of ml, ms, and mj. Finally, these relations may be sequentially applied to the adjacent states in both representations, starting from any of the two states common for them – for example, from the state with the ket-vector (188), corresponding to the top-rightmost point in Fig. 14.
Let me leave these straightforward but a bit tedious calculations for the reader’s exercise, and just quote the final result of this procedure for s = ½:51
l m ½ 1/2
j
m m ½, m ½ j l ½, m
,
l
j
s
j
Clebsch –
2 l 1
Gordan
(5.190)
coefficients
l m ½ 1/2
j
for s = ½
m m ½, m ½ j l ½, m
.
l
j
s
j
2 l 1
As a simple example, let an electron be in the p-state ( l = 1) with definite j = ½ and mj = ½, and we want to know the probability of its spin being directed down ( ms = –½). Since in this case, j = l – ½, the above formulas should be used with the upper signs, giving
1 ½ ½ 1/2
1 1/2
m ,
0 m ½ j ½, m ½
l
s
j
,
2 11
3
(5.191)
1 ½ ½ 1/ 2
2 1/2
m ,
1 m ½
j ½, m ½
l
s
j
,
2 l 1
3
so the general Eq. (183) takes the form
1 1/ 2
2 1/2
j ½, m ½
m
m
m
m
,
(5.192)
j
,
0
½
l
s
,
1
½
3
3
l
s
and the probability of the spin-down state with ms = –½ is W = 2/3.
In this course, Eqs. (190) will be used mostly in Sec. 6.4 for an analysis of the anomalous Zeeman effect. Also, the angular momentum addition rules described above are also valid for the addition of angular momenta of multiparticle system components, so we will revisit them in Chapter 8.
51 For arbitrary spin s, the calculations and even the final expressions for the Clebsch-Gordan coefficients are rather bulky. They may be found, typically in a table form, mostly in special monographs – see, e.g., A.
Edmonds, Angular Momentum in Quantum Mechanics, Princeton U. Press, 1957.
Chapter 5
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To conclude this section, I have to note that the Clebsch-Gordan coefficients (for arbitrary s) participate also in the so-called Wigner-Eckart theorem that expresses the matrix elements of spherical tensor operators, in the coupled-representation basis j, mj, via a reduced set of matrix elements. This theorem may be useful, for example, for the calculation of the rate of quantum transitions to/from high- n states in spherically symmetric potentials. Unfortunately, a discussion of this theorem and its applications would require a higher mathematical background than I can expect from my readers and more time/space than I can afford.52
5.8. Exercise problems
5.1. Use the discussion in Sec. 1 to find an alternative solution of Problem 4.18.
5.2. A spin-½ with a gyromagnetic ratio is placed into an external magnetic field, with a time-independent orientation, its magnitude B( t) being an arbitrary function of time. Find explicit expressions for the Heisenberg operators and the expectation values of all three Cartesian components of the spin as functions of time, in a coordinate system of your choice.
5.3. A two-level system is in the quantum state described by the ket-vector = + , with given (generally, complex) c-number coefficients . Prove that we can always select such a geometric c-number vector c = { cx, cy, cz} that would be an eigenstate of c σˆ , where σˆ is the Pauli vector operator. Find all possible values of c satisfying this condition, and the second eigenstate (orthogonal to ) of the operator c σˆ . Give a Bloch-sphere interpretation of your result.
5.4. Rewrite the key formulas of the solutions of Problems 4.27-4.29 in terms of the Bloch sphere angles, and verify at least one of them using the general relations of Sec. 5.1 of the lecture notes.
5.5. A spin-½ with a gyromagnetic ratio > 0 was placed into a time-independent magnetic field B0 = B0n z and let relax into the lowest-energy state. At t = 0, an additional field B1( t) is turned on; its vector has a constant magnitude but rotates within the [ x, y]-plane with an angular velocity . Calculate the expectation values of all Cartesian components of the spin at t 0, and discuss thе representation of its dynamics on the Bloch sphere.
5.6.* Analyze statistics of the spacing S E+ – E– between energy levels of a two-level system, assuming that all elements Hjj’ of its Hamiltonian matrix (2) are independent random numbers, with equal and constant probability densities within the energy interval of interest. Compare the result with that for a purely diagonal Hamiltonian matrix, with a similar probability distribution of its random diagonal elements.
5.7. For a periodic motion of a single particle in a confining potential U(r), the virial theorem of non-relativistic classical mechanics53 is reduced to the following equality:
52 For the interested reader, I can recommend either Sec. 17.7 in E. Merzbacher, Quantum Mechanics, 3rd ed., Wiley, 1998, or Sec. 3.10 in J. Sakurai, Modern Quantum Mechanics, Addison-Wesley, 1994.
53 See, e.g., CM Problem 1.12.
Chapter 5
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1
T r U ,
2
where T is the particle’s kinetic energy, and the top bar means averaging over the time period of motion.
Prove the following quantum-mechanical version of the theorem for an arbitrary stationary state, in the absence of spin effects:
1
T
r U ,
2
where the angular brackets denote (as usual in this course) the expectation values of the observables.
Hint: Mimicking the proof of the classical virial theorem, consider the time evolution of the following operator: ˆ
G rˆ pˆ .
5.8. A non-relativistic 1D particle moves in the spherically symmetric potential U( r) = C ln( r/ R).
Prove that for:
(i) v 2 is the same in each eigenstate, and
(ii) the spacing between the energy levels is independent of the particle’s mass.
5.9. Calculate, in the WKB approximation, the transparency T of the following saddle-shaped potential barrier:
xy
U ( x, y) U 1
,
0
2
a
where U 0 > 0 and a are real constants, for tunneling of a 2D particle with energy E < U 0.
5.10. In the WKB approximation, calculate the so-called Gamow factor 54 for the alpha decay of atomic nuclei, i.e. the exponential factor in the transparency of the potential barrier resulting from the following simple model for the alpha-particle’s potential energy as a function of its distance from the nuclear center:
U 0,
for
r R,
U r 0
2
ZZ'e
,
for R r,
4 r
0
where Ze = 2 e > 0 is the charge of the particle, Z’e > 0 is that of the nucleus after the decay, and R is the nucleus’ radius.
5.11. Use the WKB approximation to calculate the average time of ionization of a hydrogen atom, initially in its ground state, made metastable by the application of an additional weak, uniform, time-independent electric field E. Formulate the conditions of validity of your result.
5.12. For a 1D harmonic oscillator with mass m and frequency 0, calculate: (i) all matrix elements n x 3
ˆ n' , and
(ii) the diagonal matrix elements n x 4
ˆ n ,
where n and n’ are arbitrary Fock states.
54 Named after G. Gamow, who made this calculation as early as in 1928.
Chapter 5
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5.13. Calculate the sum (over all n > 0) of the so-called oscillator strengths, 2 m
f
,
n
E E
n
2
n ˆ x 0
2
0
(i) for a 1D harmonic oscillator, and
(ii) for a 1D particle confined in an arbitrary stationary potential.55
5.14. Prove the so-called Bethe sum rule,
2
2
2
E E
ˆ
n'
n
x
ik
k
n e
n'
n'
2 m
(where k is any c-number constant), valid for a 1D particle moving in an arbitrary time-independent potential U( x), and discuss its relation with the Thomas-Reiche-Kuhn sum rule whose derivation was the subject of the previous problem.
Hint: Calculate the expectation value, in a stationary state n, of the double commutator D ˆ ˆ
x
ik
H , e ˆ x
ik
, e
ˆ
in two ways: first, just by spelling out both commutators, and, second, by using the commutation relations between operators p ând
x
ik
e ˆ , and compare the results.
x
5.15. Spell out the commutator ˆ a,
†
exp ˆ a , where †
ˆ a and a âre the creation-annihilation
operators (5.65), and is a c-number.
5.16. Given Eq. (116), prove Eq. (117) by using the hint given in the accompanying note.
5.17. Use Eqs. (116)-(117) to simplify the following operators:
(i)
exp ia ˆ
x p êxp ˆ , and
x
ia
x
(ii)
exp iap ˆ x êxp iap ˆ ,
x
x
where a is a c-number.
5.18.* Derive the commutation relation between the number operator (5.73) and a reasonably defined quantum-mechanical operator describing the harmonic oscillator’s phase . Obtain the uncertainty relation for the corresponding observables, and explore its limit at N >> 1.
5.19. At t = 0, a 1D harmonic oscillator was in a state described by the ket-vector 1
31 32 ,
2
where n are the ket-vectors of the stationary (Fock) states of the oscillator. Calculate: (i) the expectation value of the oscillator’s energy, and
55 This Thomas-Reiche-Kuhn sum rule is important for applications because the coefficients fn describe, in particular, the intensity of dipole quantum transitions between the n th energy level and the ground state – see, e.g., Sec. 9.2 and also EM Sec. 7.2.
Chapter 5
Page 44 of 48
QM: Quantum Mechanics
(ii) the time evolution of the expectation values of its coordinate and momentum.
5.20.* Re-derive the London dispersion force’s potential of the interaction of two isotropic 3D
harmonic oscillators (already calculated in Problem 3.20), using the language of mutually-induced polarization.
5.21. An external force pulse F( t), of a finite time duration T, is exerted on a 1D harmonic oscillator, initially in its ground state. Use the Heisenberg-picture equations of motion to calculate: (i) the expectation values of the oscillator’s coordinate and momentum and their uncertainties, at an arbitrary moment,
(ii) its total energy after the end of the pulse.
5.22. Use Eqs. (144)-(145) to calculate the uncertainties x and p for a harmonic oscillator in its squeezed ground state, and in particular, to prove Eqs. (143) for the case = 0.
5.23. Calculate the energy of a harmonic oscillator in the squeezed ground state .
5.24.* Prove that the squeezed ground state described by Eqs. (142) and (144)-(145) may be sustained by a sinusoidal modulation of a harmonic oscillator’s parameter, and calculate the squeezing factor r as a function of the parameter modulation depth, assuming that the depth is small and the oscillator’s damping is negligible.
5.25. Use Eqs. (148) to prove that at negligible spin effects, the operators L ând 2
ˆ L commute
j
with the Hamiltonian of a particle placed in any central potential field.
5.26. Use Eqs. (149)-(150) and (153) to prove Eqs. (155).
5.27. Derive Eq. (164) by using any of the prior formulas.
5.28. Derive the expression L 2 = 2 l( l + 1) from basic statistics, by assuming that all (2 l + 1) values Lz = m of a system with a fixed integer number l have equal probability, and that the system is isotropic. Explain why this statistical picture cannot be used for proof of Eq. (5.163).
5.29. In the basis of common eigenstates of the operators L ând 2
ˆ L , described by kets l, m:
z
(i) calculate the matrix elements
ˆ
l, m L l, m and
2
ˆ
l, m L l, m ,
1
x
2
1
x
2
(ii) spell out your results for diagonal matrix elements (with m 1 = m 2) and their y-axis counterparts, and
(iii) calculate the diagonal matrix elements l m L ˆ
,
L ˆ l, m and l m L ˆ
,
L ˆ l, m .
x
y
y
x
5.30. For the state described by the common eigenket l, m of the operators L ând 2
ˆ L in a
z
reference frame { x, y, z}, calculate the expectation values L
2
z’ and Lz’ in the reference frame whose
z’-axis forms angle with the z-axis.
Chapter 5
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5.31. Write down the matrices of the following angular momentum operators: L ˆ L ˆ
,
L ˆ
,
L ˆ
and
,
, in the z-basis of the { l, m} states with l = 1.
x
y
z
5.32. Calculate the angular factor of the orbital wavefunction of a particle with a definite value of L 2, equal to 62, and the largest possible value of Lx. What is this value?
5.33. For the state with the wavefunction = Cxye– r, with a real positive , calculate: (i) the expectation values of the observables Lx, Ly, Lz, and L 2, and (ii) the normalization constant C.
5.34. An angular state of a spinless particle is described by the following ket-vector: 1
l ,3 m 0 l ,3 m 1 .
2
Calculate the expectation values of the x- and y-components of its angular momentum. Is the result sensitive to a possible phase shift between the component eigenkets?
5.35. A particle is in a quantum state with the orbital wavefunction proportional to the spherical harmonic 1
Y ( ,). Find the angular dependence of the wavefunctions corresponding to the 1
following ket-vectors:
(i) L ˆ ,
(ii) L ˆ ,
(iii) L ˆ , (iv) L ˆ L ˆ , and (v) 2
ˆ L .
x
y
z
5.36. A charged, spinless 2D particle of mass m is trapped in the potential well U( x, y) = m 2
0 ( x 2
+ y 2)/2. Calculate its energy spectrum in the presence of a uniform magnetic field B normal to the [ x, y]-
plane of the particle’s motion.
5.37. Solve the previous problem for a spinless 3D particle, placed (in addition to a uniform magnetic field B) into a spherically-symmetric potential well U(r) = m 2
0 r 2/2.
5.38. Calculate the spectrum of rotational energies of an axially symmetric rigid macroscopic body.
5.39. Simplify the double commutator r ˆ L ˆ
, 2, r ˆ
.
j
j'
5.40. Prove the following commutation relation:
L ˆ2 L ˆ
, 2 , r ˆ
2 2
r L ˆ
ˆ 2 L ˆ2 r ˆ .
j
j
j
5.41. Use the commutation relation proved in the previous problem and Eq. (148) to prove the orbital electric-dipole transition selection rules mentioned in Sec. 6.
5.42. Express the commutators listed in Eq. (179), J ˆ 2 L ˆ
,
and J ˆ 2 S ˆ
,
, via L ând S ˆ .
z
z
j
j
Chapter 5
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QM: Quantum Mechanics
5.43. Find the operator T ˆ describing a quantum state’s rotation by angle about a certain axis, by using the similarity of this operation with the shift of a Cartesian coordinate, discussed in Sec. 5.
Then use this operator to calculate the probabilities of measurements of spin-½ components of particles with z-polarized spin, by a Stern-Gerlach instrument turned by angle within the [ z, x] plane, where y is the axis of particle propagation – see Fig. 4.1.56
5.44. The rotation operator T ˆ analyzed in the previous problem and the linear translation operator T ˆ discussed in Sec. 5 have a similar structure:
X
ˆ
T
êxp iC / ,
where is a real c-number scaling the shift and C îs a Hermitian operator that does not explicitly depend on time.
(i) Prove that such operators are unitary.
(ii) Prove that if the shift by , induced by the operator T ˆ , leaves the Hamiltonian of some system unchanged for any , then C is a constant of motion for any initial state of the system.
(iii) Discuss what the last conclusion means for the particular operators T ând .
X
T ˆ
5.45. A particle with spin s is in a state with definite quantum numbers l and j. Prove that the observable LS also has a definite value and calculate it.
5.46. For a spin-½ particle in a state with definite quantum numbers l, ml, and ms, calculate the expectation value of the observable J 2 and the probabilities of all its possible values. Interpret your results in terms of the Clebsch-Gordan coefficients (190).
5.47. Derive general recurrence relations for the Clebsch-Gordan coefficients for a particle with spin s.
Hint: By using the similarity of the commutation relations discussed in Sec. 7, write the relations similar to Eqs. (164) for other components of the angular momentum, and then apply them to Eq. (170).
5.48. Use the recurrence relations derived in the previous problem to prove Eqs. (190) for the spin-½ Clebsch-Gordan coefficients.
5.49. A spin-½ particle is in a state with definite values of L 2, J 2, and Jz. Find all possible values of the observables S 2, Sz, and Lz, the probability of each listed value, and the expectation value for each of these observables.
5.50. Re-solve the Landau-level problem discussed in Sec. 3.2, now for a spin-½ particle.
Discuss the result for the particular case of an electron.
56 Note that the last task is just a particular case of Problem 4.18 (see also Problem 1).
Chapter 5
Page 47 of 48
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5.51. In the Heisenberg picture of quantum dynamics, find an explicit relation between the operators of velocity ˆv d ˆr / dt and acceleration ˆa d ˆv / dt of a nonrelativistic particle with an electric charge q, moving in an arbitrary external electromagnetic field. Compare the result with the corresponding classical expression.
Hint: For the orbital motion’s description, you may use Eq. (3.26).
5.52. One byproduct of the solution of Problem 47 was the following relation for the spin operators (valid for any spin s):
ˆ
m 1 S m
.
s m s m
s
s
s
1
s 1/ 2
Use this result to spell out the matrices S x, S y, S z, and S2 of a particle with s = 1, in the z-basis – defined as the basis in which the matrix S z is diagonal.
5.53.* For a particle with an arbitrary spin s, find the ranges of the quantum numbers mj and j that are necessary to describe, in the coupled-representation basis:
(i) all states with a definite quantum number l, and
(ii) a state with definite values of not only l but also ml and ms.
Give an interpretation of your results in terms of the classical vector diagram – see, e.g., Fig. 13.
5.54. For a particle with spin s, find the range of the quantum numbers j necessary to describe, in the coupled-representation basis, all states with definite quantum numbers l and ml.
5.55. A particle of mass m, with electric charge q and spin s, free to move along a planar circle of a radius R, is placed into a constant uniform magnetic field B directed normally to the circle’s plane.
Calculate the energy spectrum of the system. Explore and interpret the particular form the result takes when the particle is an electron with the g-factor g e 2.
Chapter 5
Page 48 of 48
QM: Quantum Mechanics
Chapter 6. Perturbative Approaches
This chapter discusses several perturbative approaches to problems of quantum mechanics, and their simplest but important applications starting with the fine structure of atomic energy levels, and the effects of external dc and ac electric and magnetic fields on these levels. It continues with a discussion of quantum transitions to continuous spectrum and the Golden Rule of quantum mechanics, which naturally brings us to the issue of open quantum systems – to be discussed in the next chapter.
6.1. Time-independent perturbations
Unfortunately, only a few problems of quantum mechanics may be solved exactly in an analytical form. Actually, in the previous chapters we have solved a substantial part of such problems for a single particle, while for multiparticle systems, the exactly solvable cases are even more rare.
However, most practical problems of physics feature a certain small parameter, and this smallness may be exploited by various approximate analytical methods giving asymptotically correct results – i.e. the results whose error tends to zero at the reduction of the small parameter(s). Earlier in the course, we explored one of them, the WKB approximation, which is adequate for a particle moving through a soft potential profile. In this chapter, we will discuss other techniques that are more suitable for other cases.
The historical name for these techniques is the perturbation theory, but it is fairer to speak about perturbative approaches because they are substantially different for different situations.
The simplest version of the perturbation theory addresses the problem of stationary states and energy levels of systems described by time-independent Hamiltonians of the type ˆ
ˆ (0)
ˆ )1
(
H H
H ,
(6.1)
where the operator
)
1
(
ˆ
H , describing the system’s “perturbation”, is relatively small – in the sense that its addition to the unperturbed operator
(0)
ˆ
H results in a relatively small change of the eigenenergies En
and the corresponding eigenstates of the system. A typical problem of this type is the 1D weakly anharmonic oscillator (Fig. 1), described by the Hamiltonian (1) with
Weakly
ˆ 2
2
p
m ˆ 2
x
anharmonic
ˆ (0)
0
ˆ
H
,
)
1
(
H
ˆ3
x ˆ 4
x ...
(6.2)
oscillator
2 m
2
with sufficiently small coefficients , , ….
U ( x)
2
2
m x
(0)
)
1
(
(0)
0
E
U
H
U
n
2
(0)
E
E
2
Fig. 6.1. The simplest application of
2
the perturbation theory: a weakly
(0)
E 1
anharmonic 1D oscillator. (Dashed
E 1
lines characterize the unperturbed
harmonic oscillator.)
x
0
© K. Likharev
QM: Quantum Mechanics
I will use this system as our first example, but let me start by describing the perturbative approach to the general time-independent Hamiltonian (1). In the bra-ket formalism, the eigenproblem (4.68) for the perturbed Hamiltonian, i.e. the stationary Schrödinger equation of the system, is
H (0)
ˆ
H )1
(
ˆ n E n .
(6.3)
n
Let the eigenstates and eigenvalues of the unperturbed Hamiltonian, which satisfy the equation (0) (0)
(0)
(0)
ˆ
H
n
E
n
,
(6.4)
n
be considered as known. In this case, the solution of problem (3) means finding, first, its perturbed eigenvalues En and, second, the coefficients n’ (0) n of the expansion of the perturbed state’s vectors n
in the following series over the unperturbed ones, n’ (0):
(0)
(0)
n n'
n'
n .
(6.5)
n'
Let us plug Eq. (5), with the summation index n’ replaced with n” (just to have a more compact notation in our forthcoming result), into both sides of Eq. (3):
n" (0) n H (0)
ˆ
n" ( 0) n" (0) n H )1(
ˆ
n" (0) n" (0) n E n" (0) .
(6.6)
n
n"
n"
n"
and then inner-multiply all terms by an arbitrary unperturbed bra-vector n’ (0) of the system. Assuming that the unperturbed eigenstates are orthonormal, n’ (0) n” (0) = n’n” , and using Eq. (4) in the first term on the left-hand side, we get the following system of linear equations
(0)
)
1
(
(0)
n" n H n' n E E ,
(6.7)
n'n"
(0)
n
n'
n"
where the matrix elements of the perturbation are calculated, by definition, in the unperturbed brackets: Perturbation’s
)
1
(
(0)
)
1
(
(0)
ˆ
H
n'
H
n"
.
(6.8)
matrix
n'n"
elements
The linear equation system (7) is still exact,1 and is frequently used for numerical calculations.
(Since the matrix coefficients (8) typically decrease when n’ and/or n” become sufficiently large, the sum on the left-hand side of Eq. (7) may usually be truncated, still giving an acceptable accuracy of the solution.) To get analytical results, we need to make approximations. In the simple perturbation theory we are discussing now, this is achieved by the expansion of both the eigenenergies and the expansion coefficients into the Taylor series in a certain small parameter of the problem: (0)
)
1
(
(2)
E E
E E ...,
(6.9)
n
n
n
n
(0)
)
1
(
(2)
(0)
(0)
(0)
(0)
n'
n n'
n
n'
n
n"
n
...,
(6.10)
where
( k )
( k)
(0)
k
E
n'
n
.
(6.11)
n
1 Please note the similarity of Eq. (7) to Eq. (2.215) of the 1D band theory. Indeed, the latter equation is just a particular form of Eq. (7) for the 1D wave mechanics, with a specific (periodic) potential U( x) considered as the perturbation Hamiltonian. Moreover, the whole approximate treatment of the weak-potential limit in Sec. 2.7 was essentially a particular case of the perturbation theory we are discussing now (in its 1st order).
Chapter 6
Page 2 of 36
QM: Quantum Mechanics
In order to explore the 1st-order approximation, which ignores all terms O(2) and higher, let us plug only the two first terms of the expansions (9) and (10) into the basic equation (7):
)
1
(
)
1
(
)
1
(
(0)
(0)
H n" n n' n
E
E E
.
(6.12)
n'n"
n"n
n'n
(0)
)
1
(
(0)
n
n
n'
n"
Now let us open the parentheses, and disregard all the remaining terms O(2). The result is
)
1
(
)
1
(
)
1
(
(0)
H
E n'
n
( (0)
(0)
E
E ),
(6.13)
n'n
n'n
n
n
n'
This relation is valid for any choice of the indices n and n’; let us start from the case n = n’, immediately getting a very simple (and practically, the most important!) result: Energy:
1st-order
)
1
(
)
1
(
(0)
)
1
(
(0)
ˆ
E
H
n
H
n
.
(6.14)
correction
n
nn
For example, let us see what this result gives for two first perturbation terms in the weakly anharmonic oscillator (2):
)
1
(
(0)
3
(0)
(0)
4
(0)
E
n
ˆ x n
n
ˆ x n
.
(6.15)
n
As the reader knows (or should know :-) from the solution of Problem 5.12, the first bracket equals zero, while the second one yields
)
1
(
3
4
E
x
n n .
(6.16)
n
0 2 2
2
1
4
Naturally, there should be some non-vanishing contribution to the energies from the (typically, larger) perturbation proportional to , so for its calculation, we need to explore the 2nd order of the theory.
However, before doing that, let us complete our discussion of its 1st order.
For n’ n, Eq. (13) may be used to calculate the eigenstates rather than the eigenvalues:
)
1
(
)
1
(
H
(0)
n'
n
n'n
,
for n' .
n
(6.17)
(0)
(0)
E
E
n
n'
This means that the eigenket’s expansion (5), in the 1st order, may be represented as
)
1
(
States:
H
1st-order
)
1
(
(0)
n'n
(0)
n
C n
result
n'
.
(6.18)
(0)
(0)
n'
n E
E
n
n'
The coefficient C n(0) n(1) cannot be found from Eq. (17); however, requiring the final state n to be normalized, we see that other terms may provide only corrections O(2), so in the 1st order we should take C = 1. The most important feature of Eq. (18) is its denominators: the closer the unperturbed eigenenergies of two states, the larger their mutual “interaction” due to the perturbation.
This feature also affects the 1st-order approximation’s validity condition, which may be quantified using Eq. (17): the magnitudes of the brackets it describes have to be much less than the unperturbed bracket n n(0) = 1, so all elements of the perturbation matrix have to be much less than the difference between the corresponding unperturbed energies. For the anharmonic oscillator’s energy corrections (16), this requirement is reduced to E (1)
n
<< 0.
Chapter 6
Page 3 of 36
QM: Quantum Mechanics
Now we are ready to go after the 2nd-order approximation to Eq. (7). Let us focus on the case n’
= n, because as we already know, only this term will give us a correction to the eigenenergies.
Moreover, since the left-hand side of Eq. (7) already has a small factor H(1) n’n” , the bracket coefficients in that part may be taken from the 1st-order result (17). As a result, we get
)
1
(
)
1
(
)
1
(
H H
(2)
(0)
)
1
(
E
n"
n
H
(6.19)
n
nn"
n"n nn" .
(0)
(0)
n"
n" n E
E
n
n"
Since
)
1
(
ˆ
H has to be Hermitian, we may rewrite this expression as
2
2
H )1
(
n' (0) H )1
(
ˆ
n(0)
Energy:
E (2)
.
(6.20)
n
n'n
2nd-order
(0)
(0)
(0)
(0)
n' n E
E
E
E
correction
n
n'
n' n
n
n'
This is the much-celebrated 2nd-order perturbation result, which frequently (in sufficiently symmetric problems) is the first non-vanishing correction to the state energy – for example, from the cubic term (proportional to ) in our weakly anharmonic oscillator problem (2). To calculate the corresponding correction, we may use another result of the solution of Problem 5.12: 3
x
n' ˆ3
0
x n
2
(6.21)
n( n )(
1 n )
2 1/2
3 3/2
n
(
3 n )
1 3/ 2
n n
n
n' , n3
n' , n 1
n' , n 1
(
)(
1
)(
2
)
3 1/2 n' , n3.
So, according to Eq. (20), we need to calculate
6
x
(2)
2
0
E
n
2
(6.22)
2
n( n )(
1 n )
2 1/2
3 3/2
n
(
3 n )
1 3/ 2
n n
n
n' , n3
n' , n 1
n' , n 1
(
)(
1
)(
2
)
3 1/2 n' , n3
.
n' n
( n n' )
0
The summation is not as cumbersome as may look because, at the curly bracket’s squaring, all mixed products are proportional to the products of different Kronecker deltas and hence vanish, so we need to sum up only the squares of each term, finally getting
x
(2)
15 2 60 2
11
E
.
(6.23)
n
n n
4 0
30
This formula shows that all 2nd-order energy level corrections are negative, regardless of the sign of .2
On the contrary, the 1st-order correction E (1)
n
given by Eq. (16), does depend on the sign of , so the net
correction, E (1)
(2)
n
+ En , may be of any sign.
The results (18) and (20) are clearly inapplicable to the degenerate case where, in the absence of perturbation, several states correspond to the same energy level, because of the divergence of their denominators.3 This divergence hints that in this case, the largest effect of the perturbation is the 2 Note that this is correct for the ground-state energy correction E (2) g
of any system, because for this state, the
denominators of all terms of the sum (20) are negative, while their numerators are always non-negative.
3 This is exactly the reason why this simple perturbation approach runs into serious problems for systems with a continuous spectrum, and other techniques (such as the WKB approximation) are often necessary.
Chapter 6
Page 4 of 36
QM: Quantum Mechanics
degeneracy lifting, e.g., some splitting of the initially degenerate energy level E(0) (Fig. 2), and that for the analysis of this case, we can, in the first approximation, ignore the effect of all other energy levels.
(A more detailed analysis shows that this is indeed the case until the level splitting becomes comparable with the distance to other energy levels.)
E
1
(0)
1
2(0)
...
(0)
N
(0)
E
E 2
...
Fig. 5.2. Lifting the energy
EN
level degeneracy by a
(0)
(0)
)
1
(
perturbation (schematically).
ˆ
ˆ
H H
ˆ
ˆ
ˆ
H H
H
Limiting the summation in Eq. (7) to a group of N degenerate states with equal E (0) n’
E(0), we
reduce it to
N
(0)
)
1
(
(0)
n" n H n' n
,
(6.24)
n'n"
(0)
E
E
n
n" 1
where now the indices n’ and n” number the N states of the group.4 For n = n’, Eq. (24) may be rewritten as
N
)
1
(
1
H E
n"
n'
E
E E
(6.25)
n'n"
n"
n'n"
(0)
,
0
where
)
1
(
(0) .
n
n
n" 1
For each n’ = 1, 2, … N, this is a system of N linear, homogenous equations (with N terms each) for N
unknown coefficients n” (0) n’ . In this problem, we may readily recognize the problem of diagonalization of the perturbation matrix H(1) – cf. Sec. 4.4 and in particular Eq. (4.101). As in the general case, the condition of self-consistency of the system is:
)
1
(
1
)
1
(
Initially
H
E
H
...
11
n
12
degenerate
system:
)
1
(
)
1
(
1
H
H
E
... 0 ,
(6.26)
21
22
n
energy levels
...
...
...
where now the index n numbers the N roots of this equation, in arbitrary order. According to the definition (25) of E (1)
(1)
n
, the resulting N energy levels En may be found as E(0) + En . If the perturbation matrix is diagonal in the chosen basis n(0), the result is extremely simple, (0)
)
1
(
)
1
(
E E
E H ,
(6.27)
n
n
nn
and formally coincides with Eq. (14) for the non-degenerate case, but now it may give a different result for each of N previously degenerate states n.
4 Note that here the choice of the basis is to some extent arbitrary because due to the linearity of equations of quantum mechanics, any linear combination of the states n” (0) is also an eigenstate of the unperturbed Hamiltonian. However, for using Eq. (25), these combinations have to be orthonormal, as was supposed in the derivation of Eq. (7).
Chapter 6
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Now let us see what this general theory gives for several important examples. First of all, let us consider a system with just two degenerate states with energy sufficiently far from all other levels. Then, in the basis of these two degenerate states, the most general perturbation matrix is
H
H
)
1
(
H
11
12
(6.28)
H
H
21
22
This matrix coincides with the general matrix (5.2) of a two-level system. Hence, we come to the very important conclusion: for a weak perturbation, all properties of any double-degenerate system are identical to those of the genuine two-level systems, which were the subject of numerous discussions in Chapter 4 and again in Sec. 5.1. In particular, its eigenenergies are given by Eq. (5.6), and may be described by the level-anticrossing diagram shown in Fig. 5.1.
6.2. The linear Stark effect
As a more involved example of the level degeneracy lifting by a perturbation, let us discuss the Stark effect 5 – the atomic level splitting by an external electric field. Let us study this effect, in the linear approximation, for a hydrogen-like atom/ion.6 Taking the direction of the external electric field E
(which is practically always uniform on the atomic scale) for the z-axis, the perturbation may be represented by the following Hamiltonian:
ˆ )1
(
H
ˆ z
F q Eˆ z q r
E cos .
(6.29)
(In the last form, the operator sign is dropped, because we will work in the coordinate representation.) As you (should :-) remember, energy levels of a hydrogen-like atom/ion depend only on the principal quantum number n – see Eq. (3.201); hence all the states, besides the ground 1 s state with n =
1 and l = m = 0, have some orbital degeneracy, which grows rapidly with n. Let us consider the lowest degenerate level with n = 2. Since, according to Eq. (3.203), 0 l n –1, at this level the orbital quantum number l may equal either 0 (one 2 s state, with m = 0) or 1 (three 2 p states, with m = 0, 1).
Due to this 4-fold degeneracy, H(1) is a 44 matrix with 16 elements:
l0
l
1
m
0 m
0 m
1 m 1
H
H
H
H
m ,
0
l ,
0
11
12
13
14
(6.30)
H
H
H
H
m ,
0
H )1
(
21
22
23
24
H
H
H
H
m ,
1 l .
1
31
32
33
34
H
H
H
H
m ,
1
41
42
43
44
5 This effect was discovered experimentally in 1913 by Johannes Stark and independently by Antonio Lo Surdo, so it is sometimes (and more fairly) called the “Stark – Lo Surdo effect”. Sometimes this name is used with the qualifier “dc” to distinguish it from the ac Stark effect – the energy level shift under the effect of an ac field – see Sec. 5 below.
6 An analysis of the quadratic Stark effect for the ground-state energy in the same system, changing with the field only as 2
E , is left for the reader’s exercise.
Chapter 6
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However, there is no need to be scared. First, due to the Hermitian nature of the operator, only 10 of these 16 matrix elements (4 diagonal and 6 off-diagonal ones) may be substantially different from each other. Moreover, due to the high symmetry of the problem, there are a lot of zeros even among these elements. Indeed, let us have a look at the angular components Y m l of the corresponding
wavefunctions, with l = 0 and l = 1, described by Eqs. (3.174)-(3.175). For the states with m = 1, the azimuthal parts of wavefunctions are proportional to exp{ i}; hence the off-diagonal elements H 34 and H 43 of the matrix (30), relating these functions, are proportional to
2
*
* ˆ
Ω
)
1
(
i
d Y H Y
d e
i
e
.
0
(6.31)
1
1
0
The azimuthal-angle symmetry also kills the off-diagonal elements H 13, H 14, H 23, H 24 (and hence their complex conjugates H 31, H 41, H 32, and H 42), because they relate states with m = 0 and m = 1, and hence are proportional to
2
0* ˆ
Ω
)
1
(
1
d Y H Y
i
d e
.
0
(6.32)
1
1
0
For the diagonal matrix elements H 33 and H 44, corresponding to l = 1 and m = 1, the azimuthal-angle integrals do not vanish, but since the corresponding spherical harmonics depend on the polar angle as sin, these elements are proportional to
1
1
* ˆ )1
(
1
d Y
H Y sin d sin cos sin cos
d
(6.33)
1
1
1 cos2 (cos ),
0
1
and hence are equal to zero – as any limit-symmetric integral of an odd function. Finally, for the states 2 s and 2 p with m = 0, the diagonal elements H 11 and H 22 are also killed by the polar-angle integration:
1
0* ˆ )
1
(
0
d Y H Y sin d cos cos d(cos ) 0
,
(6.34)
0
0
0
1
1
1* ˆ )
1
(
1
d Y H Y sin d cos3 cos3 d(cos ) 0.
(6.35)
0
0
0
1
Hence, the only non-zero elements of the matrix (30) are two off-diagonal elements H 12 and H 21, which relate two states with the same m = 0, but different l = {0, 1}, because they are proportional to
0*
0
3 2
2
1
d Y
cos Y
d sin d cos
0.
(6.36)
0
1
4 0
0
3
What remains is to use Eqs. (3.209) for the radial parts of these functions to complete the calculation of those two matrix elements:
H
q
H
E r drR ( r) rR ( r).
(6.37)
12
2
21
2,0
2 1
,
3 0
Due to the additive structure of the function R 2,0( r), the integral falls into a sum of two table integrals, both of the type MA Eq. (6.7d), finally giving
H H 3 q r
E ,
(6.38)
12
21
0
Chapter 6
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where r 0 is the spatial scale (3.192); for the hydrogen atom, it is just the Bohr radius r B – see Eq. (1.10).
Thus, the perturbation matrix (30) is reduced to
0
3 q r
E
0 0
0
3 q r
E
0
0 0
H )1
(
0
,
(6.39)
0
0
0 0
0
0
0 0
so the condition (26) of self-consistency of the system (25),
1
E
3 q r
E
0
0
2
0
1
3 q r
E
E
0
0
0
2
(6.40)
1
,
0
0
0
E
0
2
1
0
0
0
E 2
gives a very simple characteristic equation
E
E
q r
E
.
(6.41)
2 2
1
21
2
3
2
0
0
with four roots:
Linear
)1
(
E
E
q r
(6.42) Stark
2
,
0
,
1 2
)1(2
3 E .
3,4
0
effect
for n = 2
so the degeneracy is only partly lifted – see the levels in Fig. 3.
1
2 s 2 p
2
m 0
3 q E r
0
(0)
E
m 1
2
3 q E r 0
m 0
Fig. 6.3. The linear Stark effect for the
1
level n = 2 of a hydrogen-like atom.
2 s 2 p
2
Generally, in order to understand the nature of states corresponding to these levels, we should return to Eq. (25) with each calculated value of E (1)
2
, and find the corresponding expansion coefficients
n” (0) n’ that describe the perturbed states. However, in our simple case, the outcome of this procedure is clear in advance. Indeed, since the states with { l = 1, m = 1} are not affected by the perturbation at all (in the linear approximation in the electric field), their degeneracy is not lifted, and energy is not affected – see the middle line in Fig. 3. On the other hand, the partial perturbation matrix connecting the states 2 s and 2 p, i.e. the top left 22 part of the full matrix (39), is proportional to the Pauli matrix x, and we already know the result of its diagonalization – see Eqs. (4.113)-(4.114). This means that the upper and lower split levels correspond to very simple linear combinations of the previously degenerate states with m = 0,
Chapter 6
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1
2 s 2 p .
(6.43)
2
Finally, let us estimate the magnitude of the linear Stark effect for a hydrogen atom. For a very high dc electric field of E = 3106 V/m,7 q = e 1.610-19 C, and r 0 = r B 0.510-10 m, we get a level splitting of 3 q E r 0 0.810-22 J 0.5 meV. This number is much lower than the unperturbed energy of the level, E 2 = – E H/(222) –3.4 eV, so the perturbative result is quite applicable. On the other hand, the calculated splitting is much larger than the resolution limit imposed by the line’s natural width (~10-7 E 2, see Chapter 9), so the effect is quite observable even in substantially lower electric fields. Note, however, that our simple results are quantitatively correct only when the Stark splitting (42) is much larger than the fine-structure splitting of the same level in the absence of the field– see the next section.
6.3. Fine structure of atomic levels
Now let us use the same perturbation theory to analyze, also for the simplest case of a hydrogen-like atom/ion, the so-called fine structure of atomic levels – their degeneracy lifting even in the absence of external fields. Since the effective speed v of the electron motion in atoms is much smaller than the speed of light c, the fine structure may be analyzed as a sum of two independent relativistic effects. To analyze the first of them, let us expand the well-known classical relativistic expression8 for the kinetic energy T = E – m c 2 of a free particle with the rest mass m,9
1/ 2
2
1/ 2
p
T 2 4
m c 2 2
p c
2
m c
2
c
m 1
1 ,
(6.44)
2 2
m c
into the Taylor series in the small ratio ( p/ mc)2 ( v/ c)2:
1
2
p
p
2
1
4
2
4
p
p
T c
m
1
... 1
...,
(6.45)
2 m c
8
m c
2m 8 3 2
m c
and drop all the terms besides the two spelled-out ones. Of them, the first term is non-relativistic, while the second one represents the main relativistic correction to T.
Following the correspondence principle, the quantum-mechanical problem in this approximation may be described by Eq. (1) with the unperturbed Hamiltonian
(0
p ˆ2
)
C
H ˆ
U ˆ ( r
U ˆ
),
( r)
,
(6.46)
2m
r
(whose eigenstates and eigenenergies were discussed in Sec. 3.5) and the kinetic-relativistic perturbation 2
4
2
Kinetic-
ˆ p
1 ˆ p
)
1
(
ˆ
relativistic
H
.
(6.47)
3 2
2
perturbation
8m c
2
m c
2
m
Using Eq. (46), we may rewrite the last formula as
7 This value approximately corresponds to the threshold of electric breakdown in the air at ambient conditions, due to the impact ionization. As a result, experiments with higher dc fields are rather difficult.
8 See, e.g., EM Eq. (9.78).
9 This fancy font is used, as in Secs. 3.5-3.8, to distinguish the mass m from the magnetic quantum number m.
Chapter 6
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)
1
(
1
ˆ
ˆ
ˆ
H
H
U ( r) ,
(6.48)
2
(0
2
)
2 c
m
so its matrix elements participating in the characteristic equation (25) for a given degenerate energy level (3.201), i.e. a given principal quantum number n, are
)
1
(
1
nlm H ˆ
nl'm'
nlm
,
(6.49)
2
H ˆ(0) U ˆ( r H ˆ
)
(0)
U ˆ ( r) nl'm'
2 c
m
where the bra- and ket-vectors describe the unperturbed eigenstates, whose eigenfunctions (in the coordinate representation) are given by Eq. (3.200):
m
n,l,m = Rn,l( r) Yl (,).
It is straightforward (and hence left for the reader’s exercise) to prove that all off-diagonal elements of the set (49) are equal to 0. Thus we may use Eq. (27) for each set of the quantum numbers{ n, l, m}:
)
1
(
(0)
)
1
(
1
ˆ
E
E
E
nlm H
nlm
H
U r
n, l, m
n, l, m
n
2
ˆ (0 ˆ 2
)
( )
2 c
n, l, m
m
(6.50)
1
E
E
2
2
1
2
0
0
1
2
1
ˆ
ˆ
E 2 E U
U
C
C
n
n
,
2
2
c
n, l
n l
m
2
2
,
c
m
4 4
2
2
n
n
r
r
n, l
n, l
where the index m has been dropped, because the radial wavefunctions Rn,l( r), which affect these expectation values, do not depend on that quantum number. Now using Eqs. (3.191), (3.201) and the first two of Eqs. (3.211), we finally get
Kinetic-
2
m C
n
E
n
)
1
(
3
2 2 n
3
E
.
(6.51) relativistic
n, l
energy
2 2 2 4
c n l ½ 4
2
m c l ½
4
correction
Let us discuss this result. First of all, its last form confirms that the correction (51) is indeed much smaller than the unperturbed energy En (and hence the perturbation theory is valid) if the latter is much smaller than the relativistic rest energy m c 2 of the particle – as it is for the hydrogen atom. Next, since in the Bohr problem’s solution, n l + 1, the first fraction in the parentheses of Eq. (51) is always larger than 1, and hence than ¾ , so the kinetic relativistic correction to energy is negative for all n and l. (Actually, this fact could be predicted already from Eq. (47), which shows that the perturbation’s Hamiltonian is a negatively defined form.) Finally, for a fixed principal number n, the negative correction’s magnitude decreases with the growth of l. This fact may be interpreted using the second of Eqs. (3.211): the larger is l (at fixed n), the larger is the particle’s effective distance from the center, and hence the smaller is its effective velocity, i.e. the smaller is the magnitude of the quantum-mechanical average of the negative relativistic correction (47) to the kinetic energy.
The result (51) is valid for the Coulomb interaction U( r) = – C/ r of any physical nature. However, if we speak specifically about hydrogen-like atoms/ions, there is also another relativistic correction to energy, due to the so-called spin-orbit interaction (alternatively called the “spin-orbit coupling”). Its physics may be understood from the following semi-quantitative classical reasoning: from the “the point of view” of an electron rotating about the nucleus at distance r with velocity v, it is the nucleus, of the electric charge Ze, that rotates about the electron with the velocity (–v) and hence the time period T =
2 r/ v. From the point of view of magnetostatics, such circular motion of the electric charge Q = Ze , is Chapter 6
Page 10 of 36
QM: Quantum Mechanics
equivalent to a circular dc electric current I = Q/ T = ( Ze)( v/2 r). At the electron’s location, i.e. in the center of the current loop, it creates the magnetic field with the following magnitude:10
Zev Zev
0
0
0
B
I
.
(6.52)
a
2
2 r
2 r 2 r
4 r
The field’s direction n is perpendicular to the apparent plane of the nucleus’ rotation (i.e. that of the real rotation of the electron), and hence its vector may be readily expressed via the similarly directed vector L = m e vrn of the electron’s angular (orbital) momentum:
Zev
Ze
Ze
Ze
0
B
n
0
m
n
0
vr
L
L ,
(6.53)
a
2
3
e
3
3
2
4 r
4 r
m
4 r
m
4 r m c
e
e
0
e
where the last step used the basic relation between the SI-unit constants : 0 1/ c 20.
A more careful (but still classical) analysis of the problem11 brings both good and bad news. The bad news is that the result (53) is wrong by the so-called Thomas factor of two even for the circular motion, because the electron moves with acceleration, and the reference frame bound to it cannot be inertial (as was implied in the above reasoning), so the effective magnetic field felt by the electron is actually
Ze
B
L .
(6.54)
3
2
8 r m c
0
e
The good news is that this result is valid not only for circular but an arbitrary orbital motion in the Coulomb field U( r). Hence from the discussion in Sec. 4.1 and Sec. 4.4 we may expect that the quantum-mechanical description of the interaction between this effective magnetic field and the electron’s spin moment (4.115) is given by the following perturbation Hamiltonian12
Ze
Ze
)
1
(
1
2
1
ˆ
H
m
ˆ ˆ
B Sˆ
Lˆ
Sˆ Lˆ ,
(6.55)
e
8
3
2
r m c
2 2 2
m c 4
3
r
0
e
e
0
where at spelling out the electron’s gyromagnetic ratio e – g e e/2 m e, the small correction to the value g e
= 2 of the electron’s g-factor (see Sec. 4.4) is ignored, because Eq. (55) is already a small correction.
This expectation is confirmed by the fully-relativistic Dirac theory, to be discussed in Sec. 9.7 below: it yields, for an arbitrary central potential U( r), the following spin-orbit coupling Hamiltonian: Spin-dU r
)
1
(
1
1
( )
orbit
ˆ
H
Sˆ Lˆ .
(6.56)
coupling
2 2 2
m c r
dr
e
For the Coulomb potential U( r) = – Ze 2/40 r, this formula is reduced to Eq. (55).
10 See, e.g., EM Sec. 5.1, in particular, Eq. (5.24). Note that such an effective magnetic field is induced by any motion of electrons, in particular that in solids, leading to a variety of spin-orbit effects there – see, e.g., a concise review by R. Winkler et al., in B. Kramer (ed.), Advances in Solid State Physics 41, 211 (2001).
11 It was carried out first by Llewellyn Thomas in 1926; for a simple review see, e.g., R. Harr and L. Curtis, Am.
J. Phys. 55, 1044 (1987).
12 In the Gaussian units, Eq. (55) is valid without the factor 40 in the denominator; while Eq. (56), “as is”.
Chapter 6
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As we already know from the discussion in Sec. 5.7, the angular factor of this Hamiltonian commutes with all the operators of the coupled-representation group (inside the blue line in Fig. 5.12): 2
ˆ L , 2
ˆ S , 2
ˆ J , and J ˆ , and hence is diagonal in the coupled-representation basis with definite quantum z
numbers l, j, and mj (and of course s = ½). Hence, using Eq. (5.181) to rewrite Eq. (56) as 2
1
Ze
1 1
)
1
(
ˆ
ˆ
ˆ
ˆ
H
J L S ,
(6.57)
2 2
3
2 2 2
2 m c 4 r 2
e
0
we may again use Eq. (27) for each set { s, l, j, mj}, with common n:
)
1
(
1
Ze 2
1
1
E
J 2
ˆ L 2ˆ S 2
ˆ
,
(6.58)
n, j, l
2 m 2 c 2 4
r 3
2
j, s
e
0
n l,
where the indices irrelevant for each particular factor have been dropped. Now using the last of Eqs.
(3.211), and similar expressions (5.169), (5.175), and (5.177) for eigenvalues of the involved operators, we get an explicit expression for the spin-orbit corrections13
Ze
Spin-
j j l l
En
j j l l
)
1
(
1
2
2
(
)
1
(
)
1
¾
2
(
)
1
(
)
1
¾
E
n
,
(6.59) orbit
n, j, l
2 2 2
m c 4 2 3
3
r
n l( l ½)( l )
1
2
m c
l( l ½)( l )
1
energy
e
0
0
e
correction
with l and j related by Eq. (5.189): j = l ½.
The last form of its result shows clearly that this correction has the same magnitude scale as the kinetic correction (51).14 In the 1st order of the perturbation theory, they may be just added (with m =
m e), giving a surprisingly simple formula for the net fine structure of the n th energy level: 2
E
Fine
)
1
(
n
4 n
E
.
(6.60) structure
fine
3
2
2
m c
j
of atomic
e
½
levels
This simplicity, as well as the independence of the result of the orbital quantum number l, will become less surprising when (in Sec. 9.7) we see that this formula follows in one shot from the Dirac theory, in which the Bohr atom’s energy spectrum is numbered only with n and j, but not l. Let us recall that for an electron ( s = ½), according to Eq. (5.189) with 0 l n – 1, the quantum number j may take n positive half-integer values, from ½ to n – ½. Hence, Eq. (60) shows that the fine structure of the n th Bohr’s energy level has n sub-levels – see Fig. 4.
Please note that according to Eq. (5.175), each of these sub-levels is still (2 j + 1)-times degenerate in the quantum number mj. This degeneracy is very natural, because all m-numbers describe the state orientation in a certain direction, while in the absence of an external field, the system is still isotropic. Moreover, on each fine-structure level (besides the highest one with j = n – ½), each of the mj-
states is doubly degenerate in the orbital quantum number l = j ½ – see the labels of l in Fig. 4.
(According to Eq. (5.190), each of these states, with fixed j and mj, may be represented as a linear 13 The factor l in the denominator does not give a divergence at l = 0, because in this case j = s = ½, so j( j + 1) =
¾, and the numerator turns into 0 as well. A careful analysis of this case (see, e.g., G. Woolgate, Elementary Atomic Structure, 2nd ed., Oxford, 1983), including the so-called Darwin term not described by Eqs. (51) and (59), shows that the final Eq. (60), which does not include l, is valid even in this case.
14 This is natural because the magnetic interaction of charged particles is essentially a relativistic effect, of the same order (~ v 2/ c 2) as the kinetic correction (47) – see, e.g., EM Sec. 5.1, in particular Eq. (5.3).
Chapter 6
Page 12 of 36
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combination of two states with adjacent values of l, and hence different electron spin orientations, ms =
½, weighed with the Clebsch-Gordan coefficients.)
En
j n ½
l n 1
...
...
j
5 / 2
l ,
2 3
j 3/ 2
l ,
1 2
Fig. 6.4. The fine structure of a
j
hydrogen-like atom/ion’s level.
½
l ,
0 1
These details aside, one may crudely say that the relativistic corrections combined make the total eigenenergy grow with l, contributing to the effect already mentioned in the discussion of the periodic table of elements in Sec. 3.7. The relative scale of this increase may be quantified by the largest deviation from the unperturbed energy En, reached for the s-states (with l = 0):
)
1
(
E max
E
2
n
Ze
2 1
3
2
2 1
3
4 n 3
Z
.
(6.61)
2
E
2 m c
4 c
n
4 n
n
4 n
n
e
0
2
2
where is the fine-structure (“Sommerfeld’s”) constant,
2
e
1
,
(6.62)
4 c 137
0
(which was already mentioned in Sec. 4.4), which characterizes the relative strength (or rather weakness) of the electromagnetic effects in quantum mechanics – which in particular makes perturbative quantum electrodynamics possible.15 These expressions show that the fine-structure splitting is a very small effect (~2 ~ 10-6) for the hydrogen atom, but it rapidly grows (as Z 2) with the nuclear charge (i.e.
the atomic number) Z, and becomes rather substantial for the heaviest stable atoms with Z ~ 102.
6.4. The Zeeman effect
Now, we are ready to review the Zeeman effect – the atomic level splitting by an external magnetic field.16 Using Eq. (3.26), with q = – e, for the description of the electron’s orbital motion in the field, and the Pauli Hamiltonian (4.163) with = – e/ m e, for the electron spin’s interaction with the field, we see that even for a hydrogen-like (i.e. single-electron) atom/ion, neglecting the relativistic effects, the full Hamiltonian is rather involved:
1
2
2
Ze
e
ˆ
H
ˆ ˆ
p A
e
ˆ
B .
S
(6.63)
2 m
4 r
m
e
0
e
15 The expression 2 = E H/ m e c 2, where E H is the Hartree energy (1.13), i.e. the scale of the basic energies En, is also very revealing.
16 It was discovered experimentally in 1896 by Pieter Zeeman who, amazingly, was fired from the University of Leiden for unauthorized use of lab equipment for this work – just to receive a Nobel Prize for it in a few years!
Chapter 6
Page 13 of 36
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There are several simplifications we may make. First, let us assume that the external field is spatial-uniform on the atomic scale (which is a very good approximation for most cases), so we can take its vector potential in an axially symmetric gauge – cf. Eq. (3.132):
1
A B .
r
(6.64)
2
Second, let us neglect the terms proportional to B2, which are small in practical magnetic fields of the order of a few teslas.17 The remaining term in the effective kinetic energy, describing the interaction with the magnetic field, is linear in the momentum operator, so we may repeat the standard classical calculation18 to reduce it to the product of B by the orbital magnetic moment’s component mz = –
eLz/2 m e – besides that both mz and Lz should be understood as operators now. As a result, the Hamiltonian (63) reduces to Eq. (1), ˆ (0)
ˆ )1
(
H
H , where (0)
ˆ
H is that of the atom at B = 0, and
e
ˆ )1
(
H
B ˆ
ˆ
L 2 S
(6.65) Zeeman
z
z .
2 m
effect’s
e
perturbation
This expression immediately reveals the major complication with the Zeeman effect’s analysis.
Namely, in comparison with the equal orbital and spin contributions to the total angular momentum (5.170) of the electron, its spin produces a twice larger contribution to the magnetic moment, so the right-hand side of Eq. (65) is not proportional to J ˆ L ˆ S ˆ . As a result, the effect’s description is z
z
z
quite simple only in two limits.
If the magnetic field is so high that its effects are much stronger than the relativistic (fine-structure) effects discussed in the previous section, we may treat the two terms in Eq. (65) as independent perturbations of different (orbital and spin) degrees of freedom. Since each of the perturbation matrices is diagonal in its own z-basis, we can again use Eq. (27) to write e B
e
Paschen-
(0)
E E
ˆ
ˆ
,
n l, m L
,
n l, m 2 m S m
B
m
m
B m
(6.66) Back
l
z
l
s
z
s
2
l
s
(
).
1
2 m
2
B
l
m
effect
e
e
This result describes the splitting of each 2(2 l + 1)-degenerate energy level, with certain n and l, into (2 l +3) levels (Fig. 5), with the adjacent level distance of BB, of the order of 10-4 eV per tesla.
. . .
m ,
2 m ½
l
s
m ,
0 m ½
l
s
BB
m ,
1 m ½
(0)
E
l
s
n, l
m ,
1 m ½