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2 õ>f�f�(�õ>f�fHamiltonianÚÅ¼êHartree-FockgU|nØ'é�AÚ|��p�^�{Ù¦p�ÇCq�{—þfzÆ�f�ÚÅ�

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Story about Helium

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The first evidence of helium was observed on August 18,1868 as a bright yellow line with a wavelength of 587.49 nmin the spectrum of the chromosphere of the Sun.

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Story about Helium

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Observation In 1882, Italian physicist Luigi Palmieridetected helium on Earth, for the first time, through its D3spectral line, when he analyzed the lava of Mount Vesuvius.

Isolation March 26, 1895; Scottish chemist Sir WilliamRamsay isolated helium, the mineral cleveite. Americangeochemist William Francis Hillebrand; prior to Ramsay’sdiscovery; mineral uraninite.

Liquefying In 1908; first liquefied; Dutch physicist HeikeKamerlingh Onnes; less than one kelvin.

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Story about Helium

In 1938, Russian physicist Pyotr Leonidovich Kapitsadiscovered that helium-4 has almost no viscosity attemperatures near absolute zero, a phenomenon now calledsuperfluidity. This phenomenon is related to Bose-Einsteincondensation.

In 1972, the same phenomenon was observed in helium-3,but at temperatures much closer to absolute zero, byAmerican physicists Douglas D. Osheroff, David M. Lee, andRobert C. Richardson. The phenomenon in helium-3 isthought to be related to pairing of helium-3 fermions to makebosons, in analogy to Cooper pairs of electrons producingsuperconductivity.

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Robert Richardson says:[The US should] get out ofthe business and let thefree market prevail. Theconsequence will be a risein prices. Unfortunately,party balloons will be 100$each rather than 3$ butwe’ll have to live with that.

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He�f�Hamiltonian{− ~2

2m∇2

1 −~2

2m∇2

2 +e2

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(− Z

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d?r12 = |r1 − r2|´ü�>f�m�ål"XJ�Ñ>f�m�ü½³§��Xe�/ª

(H1 + H2)ψ = E(0)ψ. (2)

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1 −e2

4πε0r1

3Z=2�a��f³²¥�ü�>f

E(0) = E1 + E2 = −13.6× 4× 2 = −108.8 eV

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ψ1s2 = RZ=21s (r1)RZ=2

1s (r2)× 14π

(3)

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0

∫ ∞

0ψ∗1s2

1r12ψ1s2 r2

1dr1r22dr2 = 34 eV (4)

XÚ�oUþµE(1) = −108.8 + 34 = −74.8 eV¶EEXP = −79.0 eV

He�f�1�>lK�ITH =74.8-54.4=20.4 eV¶IEXP = 24.6 eV"ØO(

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φ(r1, r2) = ψZeff1s (r1)ψ

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1s 〉. (6)

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〈φ|T1|φ〉 = −12〈ψZeff

1s |E1s|ψZeff1s 〉 =

12

Z2eff . (7)

Ón〈φ|T2|φ〉 = 〈φ|T1|φ〉 = 12 Z2

eff .

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〈φ| 1r1|φ〉 = 〈ψZeff

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〈φ| 1r2|φ〉 = 〈φ| 1

r1|φ〉 = Zeff , 〈φ|

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58

Zeff . (8)

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58

Zeff . (9)

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= 2Zeff − 2Z +58

= 0 (10)

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E(Zeff ) = −Z2 +58

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1s |1r1|ψZeff

1s 〉 = Zeff ,

〈φ| 1r2|φ〉 = 〈φ| 1

r1|φ〉 = Zeff , 〈φ|

1r12|φ〉 =

58

Zeff . (8)

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58

Zeff . (9)

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= 2Zeff − 2Z +58

= 0 −→ Zeff = Z − 516. (10)

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E(Zeff ) = −Z2 +58

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1s |1r1|ψZeff

1s 〉 = Zeff ,

〈φ| 1r2|φ〉 = 〈φ| 1

r1|φ〉 = Zeff , 〈φ|

1r12|φ〉 =

58

Zeff . (8)

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58

Zeff . (9)

∂E∂Zeff

= 2Zeff − 2Z +58

= 0 −→ Zeff = Z − 516. (10)

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E(Zeff ) = −Z2 +58

Z − 25256

= −(

Z − 516

)2

= −2.847 a.u. = −77.46 eV. (11)

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Hylleraas function for helium.-s = r1 + r2, t = r1 − r2, u = r12§�EÁ&Å¼êµ

φ(s, t, u) = e−ksN∑

l,m,n=0

Cl,2m,nslt2mun. (12)

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�N = 0�§=´©á>f�Á&Å¼ê¶Hylleraas�ù�L§§ÏLCþu = r12°(�Äü�>f�m$Ä�'é§�±��Ä��~O(�Uþ�"

�N = 6�§C©��E = −2.90324 a.u.=-78.97 eV§®²�©�C-79.0 eV�¢��"

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H− -1 -0.375 -0.475 -0.528He -4 -2.750 -2.848 -2.904Li+ -9 -7.125 -7.222 -7.280Be2+ -16 -13.50 -13.60 -13.66B3+ -25 -21.88 -21.97 -22.03C4+ -36 -32.25 -32.35 -32.41

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Ψ(1, 2) = ψ(r1, r2)χ(1, 2). (13)

du>f´¤�f§¤±oÅ¼ê´��é¡�§

g^Å¼ê´��é¡�ü�(S=0, MS=0)

Ψ(1, 2) = ψS(r1, r2)1√2

[α(1)β(2)− β(1)α(2)] . (14)

g^Å¼ê´��é¡�n�(S=1, MS=1,0,-1)

Ψ(1, 2) = ψA(r1, r2)×

α(1)α(2),1√2

[α(1)β(2) + β(1)α(2)] ,

β(1)β(2)

(15)

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ψS(r1, r2) =1√2

[ψa(1)ψb(2) + ψb(1)ψa(2)]

ψA(r1, r2) =1√2

[ψa(1)ψb(2)− ψb(1)ψa(2)] (16)

éuHe�f�Ä�§duψa = ψb = ψ1s§¤±d?��é¡Ü©Å¼êψA(r1, r2) = 0"=He�fÄ��3�mÅ¼ê��é¡Úg^Å¼ê��é¡�¹"

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L = l1 + l2, L = |l1 − l2|, · · · , l1 + l2ML = −L,−(L− 1), · · · ,L (17)

S = s1 + s2, S = 0, 1

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He�f�-u� II

•b�k��>f1s;�§,��>f3nl;�

u1s(1) = R1s(r1)× 14π

unl(2) = Rnl(r2)Ylm(θ2, φ2) (18)

NX�Schrödinger�§

(H0 + H′)ψ = Eψ

H0 = H1 + H2, H′ =e2

4πε0r12(19)

¿�H0ψ = E(0)ψ§�6E¤�Uþ?��∆E = E − E(0)§

H′ψ = ∆Eψ. (20)

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ψ = au1s(1)unl(2) + bu1s(2)unl(1). (21)

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(J KK J

)(ab

)= ∆E

(ab

). (22)

ùÙ¢´ª£20¤�Ý/ª§Ù¥J��È©(directintegral)§K��È©(exchange integral)"

J =1

4πε0

∫ ∫|u1s(1)|2 e2

r12|unl(2)|2d3r1d3r2

=1

4πε0

∫ ∫ρ1s(r1)ρnl(r2)

r12d3r1d3r2 (23)

K =1

4πε0

∫ ∫u∗1s(1)u∗nl(2)

e2

r12u1s(2)unl(1)d3r1d3r2 (24)

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He�f�-u� IV48 Helium

Fig. 3.1 The direct integral in a 1snsconfiguration of helium corresponds tothe Coulomb repulsion between twospherically-symmetric charge cloudsmade up of shells of charge like thoseshown.

Unlike the direct integral, this does not have a simple classical interpre-tation in terms of charge (or probability) distributions—the exchangeintegral depends on interference of the amplitudes. The spherical sym-metry of the 1s wavefunction makes the integrals straightforward toevaluate (Exercises 3.6 and 3.7).

Unperturbedconfiguration

Fig. 3.2 The effect of the direct andexchange integrals on a 1snl config-uration in helium. The singlet andtriplet terms have an energy separationof twice the exchange integral (2K).

The eigenvalues ∆E in eqn 3.14 are found from∣∣∣∣J − ∆E K

K J − ∆E

∣∣∣∣ = 0 . (3.17)

The roots of this determinantal equation are ∆E = J ± K. The directintegral shifts both levels together but the exchange integral leads to anenergy splitting of 2K (see Fig. 3.2). Substitution back into eqn 3.14gives the two eigenvectors in which b = a and b = −a. These correspondto symmetric (S) and antisymmetric (A) wavefunctions:

ψSspace =

1√2

{u1s(1)unl(2) + u1s(2)unl(1) } ,

ψAspace =

1√2

{u1s(1)unl(2) − u1s(2)unl(1) } .

The wavefunction ψAspace has an eigenenergy of E(0) + J − K, and this

is lower than the energy E(0) + J + K for ψSspace. (For the 1snl con-

figurations in helium K is positive.)9 This is often interpreted as the9It is easy to check which wavefunc-tion corresponds to which eigenvalue bysubstitution into the original equation.

electrons ‘avoiding’ each other, i.e. ψAspace = 0 for r1 = r2, and for this

wavefunction the probability of finding electron 1 close to electron 2 issmall (see Exercise 3.3). This anticorrelation of the two electrons makesthe expectation of the Coulomb repulsion between the electrons smallerthan for ψS

space.The occurrence of symmetric and antisymmetric wavefunctions has a

classical analogue illustrated in Fig. 3.3. A system of two oscillators,with the same resonance frequency, that interact (e.g. they are joinedtogether by a spring) has antisymmetric and symmetric normal modesas illustrated in Fig. 3.3(b) and (c). These modes and their frequenciesare found in Appendix A as an example of the application of degenerateperturbation theory in Newtonian mechanics.1010Another example is the classical

treatment of the normal Zeeman effect. The exchange integral decreases as n and l increase because of thereduced overlap between the wavefunctions of the excited electron and

48 Helium

Fig. 3.1 The direct integral in a 1snsconfiguration of helium corresponds tothe Coulomb repulsion between twospherically-symmetric charge cloudsmade up of shells of charge like thoseshown.

Unlike the direct integral, this does not have a simple classical interpre-tation in terms of charge (or probability) distributions—the exchangeintegral depends on interference of the amplitudes. The spherical sym-metry of the 1s wavefunction makes the integrals straightforward toevaluate (Exercises 3.6 and 3.7).

Unperturbedconfiguration

Fig. 3.2 The effect of the direct andexchange integrals on a 1snl config-uration in helium. The singlet andtriplet terms have an energy separationof twice the exchange integral (2K).

The eigenvalues ∆E in eqn 3.14 are found from∣∣∣∣J − ∆E K

K J − ∆E

∣∣∣∣ = 0 . (3.17)

The roots of this determinantal equation are ∆E = J ± K. The directintegral shifts both levels together but the exchange integral leads to anenergy splitting of 2K (see Fig. 3.2). Substitution back into eqn 3.14gives the two eigenvectors in which b = a and b = −a. These correspondto symmetric (S) and antisymmetric (A) wavefunctions:

ψSspace =

1√2

{u1s(1)unl(2) + u1s(2)unl(1) } ,

ψAspace =

1√2

{u1s(1)unl(2) − u1s(2)unl(1) } .

The wavefunction ψAspace has an eigenenergy of E(0) + J − K, and this

is lower than the energy E(0) + J + K for ψSspace. (For the 1snl con-

figurations in helium K is positive.)9 This is often interpreted as the9It is easy to check which wavefunc-tion corresponds to which eigenvalue bysubstitution into the original equation.

electrons ‘avoiding’ each other, i.e. ψAspace = 0 for r1 = r2, and for this

wavefunction the probability of finding electron 1 close to electron 2 issmall (see Exercise 3.3). This anticorrelation of the two electrons makesthe expectation of the Coulomb repulsion between the electrons smallerthan for ψS

space.The occurrence of symmetric and antisymmetric wavefunctions has a

classical analogue illustrated in Fig. 3.3. A system of two oscillators,with the same resonance frequency, that interact (e.g. they are joinedtogether by a spring) has antisymmetric and symmetric normal modesas illustrated in Fig. 3.3(b) and (c). These modes and their frequenciesare found in Appendix A as an example of the application of degenerateperturbation theory in Newtonian mechanics.1010Another example is the classical

treatment of the normal Zeeman effect. The exchange integral decreases as n and l increase because of thereduced overlap between the wavefunctions of the excited electron and

•�§22��±deª��§∣∣∣∣

J −∆E KK J −∆E

∣∣∣∣ = 0 (25)

¤±k

∆E = J ± K. (26)

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ψAspace =

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[u1s(1)unl(2)− u1s(2)unl(1)] (27)

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ψ = ψSspaceψ

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NOTE:

• Exchange degeneracy, exchange integrals, degenerateperturbation theory and symmetrised wavefunctions all occur inhelium and their interrelationship is not straightforward so thatmisconceptions abound.

• A common misinterpretation is to infer that because levels withdifferent total spin, S = 0 and 1, have different energies then thereis a spin-dependent interaction —this is not correct, butsometimes in condensed matter physics it is useful to pretend thatit is! (See Blundell 2001.)

• The interactions that determine the gross structure of helium areentirely electrostatic and depend only on the charge and positionof the particles.

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}

× 4Z3e−(Z/a02r2)r22dr2 =

58

Z = 34 eV (32)

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中国科学技术大学博士学位论文 25

图图图 3-1: He原子双电子激发的高分辨(1 meV)光电离谱线，来自文献 [22]。

个重要的研究方向 [27, 28, 29, 30, 31, 32, 33, 34, 35]。其中有几篇比较全面的综述文章，如对原子共振俄歇效应

的研究可参考文献 [30]，分子共振俄歇过程 [31]。另外，由于RA过程对于原子或分子的价环境很敏感，所以

近阈俄歇过程可以用来研究局域屏蔽环境 [36]，以及用来制备局域电子态 [37]。研究共振俄歇效应的理论方

法叫作辐射和非辐射共振拉曼散射理论(RRRRS—Radiative and radiationless resonant Raman Scattering)，

主要文献参见 [38, 39,40,41]。我们将在第§ 3.3节介绍共振俄歇效应的基本概念。

§ 3.2 Ar和Kr原子内壳层激发的光学振子强度和禁戒跃迁

使用电子能量损失谱仪，在2.5 keV的入射电子能量下，1× 10−2Pa的样品气体气压下( Kr原子和Ar原子

的纯度都是99.99%)，我们分别测量了0◦、4◦氪原子3d和氩原子2p 内壳层激发的能谱，如图3-2和3-3，其中能

量标定取自King的快电子碰撞数据 [13]。对于0◦的Kr原子内壳层3d电子激发谱，我们还用单点标定方法把它

转化成了绝对光学振子强度密度谱，其中标定值采用Chan等 [1]低分辨(1 eV)快电子碰撞实验在96 eV处的光

学振子强度密度1.138× 10−2eV−1。

在图3-2和图3-3中，可以看出，0◦谱线清楚的分成相应自旋–轨道分裂阈值(Kr: 3d−15/2和3d−1

3/2; Ar:

2p−13/2和2p−1

1/2)的两个系列，这些都对应光学允许的跃迁。从分辨率上看，我们的实验谱线可以和已有实

验数据中能量分辨最好的同步辐射光电离实验相比 [9, 8]。这说明，在内壳层激发态的测量中，高分辨快电子

能量损失谱仪仍然有一定的优势，它不用同步辐射一样庞大的设备，也不用昂贵的能量单色器，而仅使用传

统的电子枪和半球能量单色器就可以获得较大的稳定束流和很好的能量分辨。

对于Kr原子0◦的光学振子强度密度谱，我们通过减掉平滑的光电离本底，例如，首先减掉0.0051eV−1

(89 eV处的振子强度密度值)，再进行最小二乘拟合解谱，得到3d−15/2np (n=5,6,7)和3d−1

3/25p分立态的绝对光学

振子强度；同样的，减掉0.0095 eV−1的光电离本底之后，并通过最小二乘拟合，我们得到了分立态3d−13/2np

(n=6,7)的绝对光学振子强度(见表3-1)。

为了比较当前实验的振子强度和King等 [13]的相对振子强度密度，我们通过下式得到了光学振子强度密

度df/dE(原子单位)：

df

dE= N∗3 · fn. (3-1)

里德堡系列3d−15/2np和3d−1

3/2np, (n = 5, 6, 7) 绝对光学振子强度列于表3-1，并和King的相对振子强度

密度进行了比较。除了峰3d−13/26p，对于其他几个态的强度，两组实验在误差范围内是符合的。而且，每

He�f�V>f-uUÌ§V>f>lUI2 = 79 eV

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DI = 〈1s1s|e(r1 + r2)|1s2p〉= 〈1s|er1|1s〉 · 〈1s|2p〉+ 〈1s|1s〉 · 〈1s|er2|2p〉= 〈1s|er2|2p〉, (33)

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DI = 〈1s1s|e(r1 + r2)|2s2p〉= 〈1s|er1|2s〉 · 〈1s|2p〉+ 〈1s|2s〉 · 〈1s|er2|2p〉= 0. (34)

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• Configuration—|�§~Xψ: 1s1s, 1s2s, 1s3s, · · · ; 2s2s, 2s3s,2s4s, · · · ; 1s2p, 1s3p, 1s4p, · · · ; 1s3d, 1s4d, 1s5d, · · · ; 1sεs; 1sεp

• Term—Ì�§~X: 1s2s1S0, 1s2s3S1

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Ψ(E) = a(E)ψ(0)α +

∫b(E,E′)ψ(E′)dE′ (36)

•Ù¥ψ(0)α ´Ã�6�©á�U?§ ψ(E′)´Uþ�ENC�ëY

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• Ψ(E)�±ÏLguË�1fò-u§��±ÏL�Ñ��>fg>l�ëYÏ�"

hν + He(1s2 1S0) −→ He∗∗ −→{

He + hν′

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e− + He(1s2 1S0) −→ He∗∗ −→{

He + hν′

He+ + e−(38)

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þfêº�(Quantum defect)62 The alkalis

Fig. 4.1 The probability density of theelectrons in a sodium atom as a func-tion of r. The electrons in the n = 1and n = 2 shells make up the core,and the probability density of the un-paired outer electron is shown for then = 3 shell with l = 0, 1 and 2. Theprobability is proportional to |P (r)|2 =r2|R(r)|2; the r2 factor accounts for theincrease in volume of the spherical shellbetween r and r + dr (i.e. 4πr2 dr) asthe radial distance increases. The de-creasing penetration of the core as lincreases can be seen clearly—the 3d-electron lies mostly outside the corewith a wavefunction and binding en-ergy very similar to those for the 3dconfiguration in hydrogen. These wave-functions could be calculated by thesimple numerical method described inExercise 4.10, making the ‘frozen core’approximation, i.e. that the distribu-tion of the electrons in the core isnot affected by the outer electron—thisgives sufficient accuracy to illustratethe qualitative features. (The iterativemethod described in Section 4.4 couldbe used to obtain more accurate numer-ical wavefunctions.)

0

(b)

(a)

0

Core

Core

0

(c)

Core

Bohr’s formula works amazingly well for the energy levels of the alkalis:

E (n, l) = −hcR∞

(n − δl)2 . (4.1)

A quantity δl, called the quantum defect, is subtracted from the prin-cipal quantum number to give an effective principal quantum numbern∗ = n − δl.

7 The values of the quantum defects for each l can be esti-7This differs from the modificationused for X-ray transitions in Chapter1—hardly surprising since the physicalsituation is completely different for theinner and outer electrons.

mated by inspecting the energy levels shown in Fig. 4.2. The d-electronshave a very small quantum defect, δd � 0, since their energies are nearlyhydrogenic. We can see that the 3p configuration in sodium has com-parable energy to the n = 2 shell in hydrogen, and similarly for 4p andn = 3, etc.; thus δp ∼ 1. It is also clear that the quantum defect fors-electrons is greater than that for p-electrons. A more detailed analysisshows that all the energy levels of sodium can be parametrised by theabove formula and only three quantum defects:

62 The alkalis


0

(b)

(a)

0

Core

Core

0

(c)

Core


E (n, l) = −hcR∞

(n − δl)2 . (4.1)




62 The alkalis


0

(b)

(a)

0

Core

Core

0

(c)

Core


E (n, l) = −hcR∞

(n − δl)2 . (4.1)




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E(n, l) = −hcR∞

(n− δl)2 (39)

δs = 1.35, δp = 0.86, δd = 0.01, δl = 0 for l > 2.

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õ>f�f�Hamiltonian

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2m∇2

i −Ze2

4πε0ri

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N∑

i<j=1

e2

4πε0rij(40)

¤±§Schrödinger�§�

HΨ(q1, q2, · · · , qN) = EΨ(q1, q2, · · · , qN). (41)

Ù¥qi´1i�>f��m�IriÚg^�I�8Ü"

=

Ψ(q1, q2, · · · , qN) = ψ(r1, r2, · · · , rN)χ(1, 2, · · · ,N) (42)

Ù�mÅ¼êψ(r1, r2, · · · , rN)÷vSchrödinger�§"

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¥%|Cq(central-field approximation)

VCF(r) = − Ze2

4πε0r+ S(r) (43)

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i + VCF(ri)

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ψatom = ψ1ψ2 · · ·ψN . (45)

K��Xe�ü>fSchrödinger�§{− ~2

2m∇2

i + VCF(ri)

}ψi = Eiψi. (46)

�f�oUþ�Eatom = E1 + E2 + · · ·+ EN .

�z�Å¼ê�ψi = R(ri)Yl,m(θi, φi)§�Ä»�Ü©��

{− ~2

2md2

dr2 + VCF(r) +~2l(l + 1)

2mr2

}P(r) = EP(r). (47)

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éuk�³U¼êVCF(r)§kXe�5��r → 0�§¤ïÄ�>faÉ��·>|r

E(r) ∼ − Ze4πε0r2 r (48)

�r →∞§k

E(r) ∼ − e4πε0r2 r (49)

¤±k

ECF(r) ∼ − Zeff e4πε0r2 r (50)

? ��k�¥%³

VCF(r) = e∫ r

∞|ECF(r′)|dr′ (51)

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1

11

Fig. 4.3 The change-over from the short- to the long-range is not calculated but isdrawn to be a reasonable guess, using the following criteria. The typical radius ofthe 1s wavefunction around the nucleus of charge +Ze = +11e is about a0/11, andso Zeff will start to drop at this distance. We know that Zeff ∼ 1 at the distanceat which the 3d wavefunction has appreciable probability since that eigenstate hasnearly the same energy as in hydrogen. The form of the function Zeff (r) can befound quantitatively by the Thomas–Fermi method described in Woodgate (1980).

At large distances the other N − 1 electrons screen most of the nuclearcharge so that the field is equivalent to that of charge +1e:

E(r) → e

4πε0r2r . (4.9)

These two limits can be incorporated in a central field of the form

ECF(r) → Zeffe

4πε0r2r . (4.10)

The effective atomic number Zeff (r) has limiting values of Zeff(0) = Zand Zeff (r) → 1 as r → ∞, as sketched in Fig. 4.3.9 The potential9This is not necessarily the best way to

parametrise the problem for numericalcalculations but it is useful for under-standing the underlying physical prin-ciples.

energy of an electron in the central field is obtained by integrating frominfinity:

VCF(r) = e

∫ r

∞|ECF(r′)| dr′ . (4.11)

The form of this potential is shown in Fig. 4.4.So far, in our discussion of the sodium atom in terms of the wave-

function of the valence electron in a central field we have neglected

Fig. 4.4 The form of the potential en-ergy of an electron in the central-fieldapproximation (e2M = e2/4πε0). Thisapproximate sketch for a sodium atomshows that the potential energy crossesover from VCF(r) = −e2M/r at longrange to −11e2M/r + Voffset; the con-stant Voffset comes from the integrationin eqn 4.11 (if Zeff (r) = 11 for all r thenVoffset = 0 but this is not the case). Forelectrons with l > 0 the effective poten-tial should also include the term thatarises from the angular momentum, asshown in Fig. 4.5.

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66 The alkalis

1

11

Fig. 4.3 The change-over from the short- to the long-range is not calculated but isdrawn to be a reasonable guess, using the following criteria. The typical radius ofthe 1s wavefunction around the nucleus of charge +Ze = +11e is about a0/11, andso Zeff will start to drop at this distance. We know that Zeff ∼ 1 at the distanceat which the 3d wavefunction has appreciable probability since that eigenstate hasnearly the same energy as in hydrogen. The form of the function Zeff (r) can befound quantitatively by the Thomas–Fermi method described in Woodgate (1980).

At large distances the other N − 1 electrons screen most of the nuclearcharge so that the field is equivalent to that of charge +1e:

E(r) → e

4πε0r2r . (4.9)

These two limits can be incorporated in a central field of the form

ECF(r) → Zeffe

4πε0r2r . (4.10)

The effective atomic number Zeff (r) has limiting values of Zeff(0) = Zand Zeff (r) → 1 as r → ∞, as sketched in Fig. 4.3.9 The potential9This is not necessarily the best way to

parametrise the problem for numericalcalculations but it is useful for under-standing the underlying physical prin-ciples.

energy of an electron in the central field is obtained by integrating frominfinity:

VCF(r) = e

∫ r

∞|ECF(r′)| dr′ . (4.11)

The form of this potential is shown in Fig. 4.4.So far, in our discussion of the sodium atom in terms of the wave-

function of the valence electron in a central field we have neglected

Fig. 4.4 The form of the potential en-ergy of an electron in the central-fieldapproximation (e2M = e2/4πε0). Thisapproximate sketch for a sodium atomshows that the potential energy crossesover from VCF(r) = −e2M/r at longrange to −11e2M/r + Voffset; the con-stant Voffset comes from the integrationin eqn 4.11 (if Zeff (r) = 11 for all r thenVoffset = 0 but this is not the case). Forelectrons with l > 0 the effective poten-tial should also include the term thatarises from the angular momentum, asshown in Fig. 4.5.

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4.3 The central-field approximation 67

Fig. 4.5 The total potential in thecentral-field approximation includingthe term that is proportional to l(l +1)/r2 drawn here for l = 2 and thesame approximate electrostatic VCF(r)as shown in Fig. 4.4. The angularmomentum leads to a ‘centrifugal bar-rier’ that tends to keep the wavefunc-tions of electrons with l > 0 away fromr = 0 where the central-field potentialis deepest.

the fact that the central field itself depends on the configuration of theelectrons in the atom. For a more accurate description we must takeinto account the effect of the outer electron on the other electrons, andhence on the central field. The energy of the whole atom is the sum ofthe energies of the individual electrons (in eqn 4.6), e.g. a sodium atom inthe 3s configuration has energy E

(1s22s22p6 3s

)= 2E1s+2E2s+6E2p+

E3s = Ecore +E3s. This is the energy of the neutral atom relative to thebare nucleus (Na11+).10 It is more useful to measure the binding energy 10This is a crude approximation, espe-

cially for inner electrons.relative to the singly-charged ion (Na+) with energy E(1s22s22p6

)=

2E′1s+2E′

2s+6E′2p = E′

core. The dashes are significant—the ten electronsin the ion and the ten electrons in the core of the atom have slightlydifferent binding energies because the central field is not the same inthe two cases. The ionization energy is IE = Eatom − Eion = (Ecore −E′

core) + E3s. From the viewpoint of valence electrons, the differencein Ecore between the neutral atom and the ion can be attributed tocore polarization, i.e. a change in the distribution of charge in the coreproduced by the valence electron.11 To calculate the energy of multi-

11This effect is small in the alkalis andit is reasonable to use the frozen coreapproximation that assumes Ecore E′

core. This approximation becomesmore accurate for a valence electron inhigher levels where the influence on thecore becomes smaller.

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ΨA(q1, q2, · · · , qN) =1√N!

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ψ1(q1) ψ1(q2) · · · ψ1(qN)ψ2(q1) ψ2(q2) · · · ψ2(qN)· · ·· · ·

ψN(q1) ψN(q2) · · · ψN(qN)

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Φ =1√6!

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1sα(1) 1sα(2) · · · 1sα(6)1sβ(1) 1sβ(2) · · · 1sβ(6)· · ·· · ·

2pβ(1) 2pβ(2) · · · 2pβ(6)

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Jλµ = 〈uλ (qi) uµ (qj)|1rij|uλ (qi) uµ (qj)〉§ �� (56)

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Vg�£ã"Schrödinger introduced the word “entangled” todescribe the quantum state of two systems as “When twosystems, of which we know the states by their respectiverepresentatives, enter into temporary physical interactiondue to known forces between them, and when after a time ofmutual influence the systems separate again, then they canno longer be described in the same way as before, viz. byendowing each of them with a representative of its own. Iwould not call that one but rather the characteristic trait ofquantum mechanics, the one that enforces its entiredeparture from classical lines of thought. By the interactionthe two representatives (or ψ-functions) have becomeentangled”.

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ExtraordinaryHorizontal( )

OrdinaryVertical( )

A

B

NC

|ψ〉 = 1/√

2(|H〉1|V〉2 + eiφ|V〉1|H〉2) (57)

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ψSspin =

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| ↓↓〉ψA

spin =1√2

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Φ =1√6!

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1sα(1) 1sα(2) · · · 1sα(6)1sβ(1) 1sβ(2) · · · 1sβ(6)· · ·· · ·

2pβ(1) 2pβ(2) · · · 2pβ(6)

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1. Estimate of the binding energy of helium

Write down the Schrödinger equation for the helium atom andstate the physical significance of each of the terms.

Estimate the equilibrium energy of an electron bound to acharge +Ze by minimising

E(r) =~2

2mr2 −Ze2

4πε0r

Calculate the repulsive energy between the two electrons inhelium assuming that r12 ∼ r. Hence estimate the ionizationenergy of helium.

Estimate the energy required to remove a further electronfrom the helium-like ion Si12+, taking into account the scalingwith Z of the energy levels and the expectation value for theelectrostatic repulsion. The experimental value is 2400 eV.Compare the accuracy of your estimates for Si12+ andhelium. (IE(He) = 24.6 eV.)

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2. Direct and exchange integrals for an arbitrary system

Verify that for

ψA(r1, r2) =1√2

[uα(r1)uβ(r2)− uα(r2)uβ(r1)]

and H′ = e2/4πε0r12 the expectation value 〈ψA|H′|ψA〉 has theform J − K and give the expressions for J and K.

Write down the wavefunction ψS that is orthogonal to ψA

Verify that〈ψA|H′|ψS〉 = 0 so that H′ is diagonal in this basis.

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3. Quantum defects of sodiumThe binding energies of the 3s, 4s, 5s and 6s configurationsin sodium are 5.14 eV, 1.92 eV, 1.01 eV and 0.63 eV,respectively. Calculate the quantum defects for theseconfigurations and comment on what you find.

Estimate the binding energy of the 8s configuration and makea comparison with the n = 8 shell in hydrogen.

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4. Wavefunction of multi-electron atomsWrite down the wavefunction of Lithium atoms with the formof Slater determinants.

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M��§��§�÷�?ÍC��ÔnÆ£12�¤¥I�ÆEâ�ÆÑ��§2008§ISBN:978-7-312-01883-1.

B. H. Bransden and C. J. JoachainPhysics of atoms and molecules (2nd Edition)Pearson Education Limited, 2003, ISBN: 0-582-35692-X.

Christopher J. FootAtomic PhysicsOxford University Press, 2005, ISBN:978-0-19-850695-9.

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