Fermions: Particles for Which S = 1/2, 3/2,

Welcome back to PHY 3305

Today’s Lecture: Identical Particles Periodic Table

Wolfgang Ernst Pauli 1900 - 1958

Physics 3305 - Modern Physics Professor Jodi Cooley Anouncements

• Problem set 13 is due Tuesday, Nov 21st at 12:30 pm. • Regrade for problem set 12 is due Tuesday, Nov 21st at 12:30 pm. • Problem sets 14 will have no regrade attempt. • Final presentations are due MONDAY, NOVEMBER 20th at 3 pm. Email your pdf to [email protected] before due date and time. If your talk is late less than one hour, your maximum grade will drop one increment (i.e. A —> A- ). If your talk is late 1 - 2 hours, your maximum grade drops two increments (A —> B+) and so on. • PhysBowl teams for Thursday, November 16th: • Luke, Chris, Andrew • Rebecca, Hope, Ali • Robert, Connor, Gabriel Physics 3305 - Modern Physics Professor Jodi Cooley Review Question

The magnetic quantum number ms is most closely associated with what property of the electron in an atom?

a) Magnitude of the orbital angular momentum b) Energy c) z component of the spin angular momentum d) z component of the orbital angular momentum e) Radius of the orbit

Physics 3305 - Modern Physics Professor Jodi Cooley From Lecture Video: Identical Particles An identical particle is one for whom all quantum numbers and there mass are the same. The only thing that might distinguish them is their location in space. The SWE for this situation is

We need a TWO-PARTICLE WAVE FUNCTION to describe the system. ψ(x)=ψ(x1,x2)

With identical particles, the probability density and outcomes can not change if the particles are swapped. 2 2 |ψ(x1,x2)| = |ψ(x2,x1)|

Physics 3305 - Modern Physics Professor Jodi Cooley From Lecture Video:

Requirement: Probability density must be unchanged if the labels of indistinguishable particles are switched.

What will meet our requirement of the particles being indistinguishable?

Adding or subtracting the same function with the labels swapped.

Physics 3305 - Modern Physics Professor Jodi Cooley From Lecture Video:

Spin and the total wave function Wether or not the total wave function of a system of particles is symmetric or asymmetric depends ENTIRELY on the spin of the particles.

Bosons: particles for which s = 0, 1, 2, ... manifest a symmetric multiparticle state.

Fermions: particles for which s = 1/2, 3/2, ... manifest an antisymmetric multiparticle state.

Building blocks - electrons, protons, neutrons are fermions Particles that transmit forces are bosons.

Physics 3305 - Modern Physics Professor Jodi Cooley The Pauli exclusion Principle is obeyed by:

A) all particles B) all charged particles C) all Fermions D) all Bosons E) only electrons

Physics 3305 - Modern Physics Professor Jodi Cooley From Lecture Video:

Pauli Exclusion Principle: No two indistinguishable fermions may occupy the same individual particle state. Articulated by Wolfgang Pauli in 1924 and earned him the Nobel Prize in 1945.

Note: The Pauli Exclusion Principle does NOT apply to bosons. Bosons can all occupy the same state.

Physics 3305 - Modern Physics Professor Jodi Cooley What is the minimum possible energy for four (noninteracting) spin 1/2 particles of mass m in a one dimensional box of length L?

There may be 2 particles in the n=1 state, and 2 particles in the n=2 state.

(1)2~2⇡2 (2)2~2⇡2 E =2 +2 tot ⇥ 2mL2 ⇥ 2mL2

10~2⇡2 E = tot 2mL2

Physics 3305 - Modern Physics Professor Jodi Cooley What is the minimum possible energy for four (noninteracting) spin 1/2 particles of mass m in a one dimensional box of length L?

What if the particles where instead spin-1?

In this case the particles are bosons and they would not obey the Pauli Exclusion Principle. Hence, they all would be in the ground state.

(1)2~2⇡2 E =4 tot ⇥ 2mL2

4~2⇡2 E = tot 2mL2

Physics 3305 - Modern Physics Professor Jodi Cooley What is the minimum possible energy for four (noninteracting) spin 1/2 particles of mass m in a one dimensional box of length L?

What if the particles where instead spin 3/2? In this case the particles are fermions and they would obey the Pauli Exclusion Principle. However, in this case, the possible spins are -3/2, -1/2, 1/2 and 3/2 (4 in total). Thus, 4 particles could occupy the ground state.

(1)2~2⇡2 E =4 tot ⇥ 2mL2

4~2⇡2 E = tot 2mL2

Physics 3305 - Modern Physics Professor Jodi Cooley Whether a neutral whole atom behaves as a boson or a fermion is independent of Z, instead depends entirely on the number of neutrons in its nucleus. Why? What is it about this number that determines whether the atom is a boson or fermion?

Ans: There are as many electrons as protons in a neutral atom. Both are fermions, so together they will always add to an even number of fermions. Neutrons are also fermions. Thus, an odd number of neutrons means an odd total number of fermions, and thus fermonic behavior for the unit (known as composite fermion). An even neutron number gives an even total, and boson behavior (or composite boson).

Physics 3305 - Modern Physics Professor Jodi Cooley Lithium exists as two isotopes in nature: lithium-6 (3 neutrons) and lithium-7 (4 neutrons).

a) As independent atoms, would these behave as bosons or fermions?

Lithium-6 has an odd number of neutrons in its nucleus and hence an odd number of electrons. It behaves as a fermion. Lithium-7 has an even number of neutrons in its nucleus and hence an even number of electrons. It behaves as a boson.

Physics 3305 - Modern Physics Professor Jodi Cooley Lithium exists as two isotopes in nature: lithium-6 (3 neutrons) and lithium-7 (4 neutrons).

a) Might a gas of lithium-6 behave as a gas of bosons?

Yes, a gas of lithium-6 could behave as a gas of boson, provided that the atoms “pair up”. Each pair would then have an even number of electrons and hence, behave as a boson.

Physics 3305 - Modern Physics Professor Jodi Cooley When a lithium atom is made from a helium atom by adding a proton (and neutron) to the nucleus and an electron outside, the electron goes into an n = 2, l = 0 state rather than an n = 1, l = 0 state. This is an indication that electrons:

A) obey the Pauli exclusion principle B) obey the minimum energy principle C) obey the Bohr correspondence principle D) are bosons E) and protons are interchangeable

Physics 3305 - Modern Physics Professor Jodi Cooley From Lecture Video: The spacial states are distinguished by n, l, ml. The set of orbits with a certain value of n are known as an atomic shell.

The levels with a certain value of n and l are called subshells (i.e. 2s or 3d).

The Pauli Exclusion Principle tells us that the maximum number of electrons that can be placed in each subshell is 2(2l +1) = max electrons in subshell

number of different values two different of ml for each l values of ms

Physics 3305 - Modern Physics Professor Jodi Cooley From Lecture Video: From this you can ﬁnd the capacity of each of the subshells.

2(2l +1) = max electrons in subshell

number of different values of ml for each l two different values of ms The levels are ﬁlled using a rule of thumb: - the lowest value of n+l, - the state of n being lower in the case of equal n+l.

Physics 3305 - Modern Physics Professor Jodi Cooley From Lecture Video:

1s1 Periodic Table 1s22s1 1s2 1s22s2 1s22s22p1

Terminology: valence electrons are weakly bound, dangling at the periphery of the electron cloud. For Lithium (Z=3), there is one valence electron (2s).

Physics 3305 - Modern Physics Professor Jodi Cooley Chapter 8 Spin and Atomic Physics

8.45 Swapping only the spatial states in I gives ψψnn′′()xx12↑↓−↓↑ ()ψ nn () x 1ψ () x 2. This is neither the same nor the opposite of I, so its exchange symmetry is neither. For II, swapping n and n´ gives

ψψnn′′()xx12↓↑−↑↓ ()ψ nn () x 1ψ () x 2, which also has neither exchange symmetry.

(b) For I, swapped spins gives ψψnn()xx12↓↑−↑↓′′ ()ψ n () xx 12ψ n () and for II it gives

ψψnn()xx12↑↓−↓↑′′ ()ψ n () xx 12ψ n () . Both have neither symmetric nor antisymmetric exchange symmetry.

=ψψnn (xx12 )′′ ( )()↑↓ + ↓↑ −ψ nn ( xx 12 )ψ ( ) () ↑↓ + ↓↑ = ()ψψnn()xx12′′ ()−↑↓ψ nn () xx 12ψ ()()+ ↓↑ . Chapter 8 Spin and Atomic Physics (d) If the spatial states n and n´ are swapped, this changes sign—it is antisymmetric. If the arrows are swapped, it doesn’t change—it is symmetric. 8.45 Swapping only the spatial states in I gives ψψnn′′()xx12↑↓−↓↑ ()ψ nn () x 1ψ () x 2. This is neither the same nor (e)the opposite Yes, it of would I, so its changeexchange sign. symmetry is neither. For II, swapping n and n´ gives ψψ()xx↓↑−↑↓ ()ψ () xψ () x, which also has neither exchange symmetry. (f) nn′′ The12 sum would nn be 1 preferred, 2 for it is antisymmetric in the spatial state, so the particles’ locations will be

(b) farther For I, swapped apart. spins gives ψψnn()xx12↓↑−↑↓′′ ()ψ n () xx 12ψ n () and for II it gives ()xx () () xx () . Both have neither symmetric nor antisymmetric exchange Chapter 8 Spin andψψnn Atomic12↑↓−↓↑ Physics′′ ψ n 12ψ n symmetry. 8.46 ψψnn()xx12↑↓−↓↑′′ ()ψ n () xx 12ψ n () – ψψnn()xx12↓↑−↑↓′′ ()ψ n () xx 12ψ n () (c) ψψnn()xx12↑↓−↓↑′′ ()ψ n () xx 12ψ n () + ψψnn()xx12↓↑−↑↓′′ ()ψ n () xx 12ψ n () 8.45 Swapping only the spatial states in I gives ψψnn′′()xx12↑↓−↓↑ ()ψ nn () x 1ψ () x 2. This is neither the same nor =()()ψψnnxx12′′↑↓ − ↓↑ +ψ nn ()() xx 12ψ ↑↓ − ↓↑ = ()ψψnn()xx12′′ ()+ψ nn () xx 12ψ () ↑↓ − ↓↑ . If the the opposite =ψψ ofnn ( xxI,12 )so its′′ ( ()exchange )()↑↓ + symmetry↓↑ −ψ nn (is xx 12neither )ψ (. ) For () ↑↓ ()II,+ ↓↑swapping= ()ψψnn n() xxand12 n´′′ ()gives−↑↓ ψ nn () xx 12ψ ()()+ ↓↑ . () ψψspatial()xx↓↑−↑↓ states () n ψand () x n´ ψare () x swapped,, which alsothis has doesn’t neither exchange change symmetry. sign—it is symmetric. If the arrows are swapped, it does nn(d)′′ 12 If the spatial states nn 1 n and n 2´ are swapped, this changes sign—it is antisymmetric. If the arrows are swapped, change—it is antisymmetric. If both are swapped, it changes sign, so it is antisymmetric. (b) Forit I, doesn’t swapped change—it spins gives is ψψ symmetric.nn()xx12↓↑−↑↓′′ ()ψ n () xx 12ψ n () and for II it gives

(e) ψψnn Yes,()xx12↑↓−↓↑ it would′′ () changeψ n () xxsign. 12ψ n () . Both have neither symmetric nor antisymmetric exchange 8.47 Thesymmetry. individual particle states are , , and . Inserting these in each row and the three particle (f) The sum would− be preferred, for it isψ antisymmetric1,0,0 ↑ ψ 1,0,0 in↓ the spatialψ 2,0,0state,↑ so the particles’ locations will be (c) ψψnnfarther()xx12↑↓−↓↑ apart.′′ () ψ n () xx 12ψ n () + ψψnn()xx12↓↑−↑↓′′ ()ψ n () xx 12ψ n () ψψ1,0,0()rrr 1↑↓↑ 1,0,0 () 1ψ 2,0,0 () 1 =ψψnn (xx12 )′′ ( )()↑↓ + ↓↑ −ψ nn ( xx 12 )ψ ( ) () ↑↓ + ↓↑ = ()ψψnn()xx12′′ ()−↑↓ψ nn () xx 12ψ ()()+ ↓↑ . labels in each column gives ψψ()rrr↑↓↑ ()ψ () 1,0,0 2 1,0,0 2 2,0,0 2 8.46 (d)ψψ nn If()xx the12↑↓−↓↑ spatial′′ () states ψn and n () xx 12n´ areψ nswapped, () – thisψψ nnchanges()xx12↓↑−↑↓ sign—it′′ () is ψantisymmetric. n () xx 12ψ n () If the arrows are swapped, it doesn’t change—it is symmetric.ψψ 1,0,0()rrr 3↑↓↑ 1,0,0 () 3ψ 2,0,0 () 3 =()()ψψnnxx12′′()↑↓ − ↓↑ +ψ nn ()() xx 12ψ ()↑↓ − ↓↑ = ()ψψnn()xx12′′ ()+ψ nn () xx 12ψ ()()↑↓ − ↓↑ . If the (e)spatial Yes, statesit would n and change n´ are sign. swapped, this doesn’t change sign—it is symmetric. If the arrows are swapped, it does = ψψ1,0,0()rr 1↑↓↑ 1,0,0 () 2ψ 2,0,0 () r 3 – ψψ1,0,0()rrr 1↑↑↓ 2,0,0 () 2ψ 1,0,0 () 3 + ψψ1,0,0()rrr 1↓↑↑ 2,0,0 () 2ψ 1,0,0 () 3 (f)change—it The sum wouldis antisymmetric. be preferred, If for both it is are antisymmetric swapped, it inchanges the spatial sign, state, so it so is the antisymmetric. particles’ locations will be – ψψfarther1,0,0()rr 1apart.↓↑↑ 1,0,0 () 2ψ 2,0,0 () r 3 + ψψψ2,0,0()rrr 1↑↑↓ 1,0,0 () 2 1,0,0 () 3 – ψψψ2,0,0()rrr 1↑↓↑ 1,0,0 () 2 1,0,0 () 3

8.47 The individual−particle states are ψ 1,0,0 ↑, ψ 1,0,0 ↓, and ψ 2,0,0 ↑. Inserting these in each row and the three particle 8.46 ψψ()xx↑↓−↓↑ ()ψ () xxψ () – ψψ()xx↓↑−↑↓ ()ψ () xxψ () nn12′′ n 12 n()rrr nn12 ()′′ ()n 12 n ψψ1,0,01 1↑↓↑ 1,0,0a 1ψ 2,0,0 1 =()()xx ()() xx2 0 = ()xx () () xx () . If the 8.48 Fromψψnn12 equation′′()↑↓ (7-42), − ↓↑ +ψ nnra 12=ψ 1 ()↑↓ = − ↓↑ . ()ψψnn12′′+ψ nn 12ψ ()↑↓ − ↓↑ labels in each column gives ψψ101,0,0()rrr 2↑↓↑ 1,0,0 () 2ψ 2,0,0 () 2 spatial states n and n´ are swapped, this3 doesn’t change3 sign—it is symmetric. If the arrows are swapped, it does ψψ()rrr↑↓↑ ()ψ () change—it is antisymmetric. If both1,0,0 are 3swapped, 1,0,0 it changes 3 sign, 2,0,0 so 3 it is antisymmetric. 1 2 (b)= ψψ ra20()rr= ↑↓↑2 () = 4ψa0. () r – ψψ()rrr↑↑↓ ()ψ () + ψψ()rrr↓↑↑ ()ψ () 1,0,0 11 1,0,0 2 2,0,0 3 1,0,0 1 2,0,0 2 1,0,0 3 1,0,0 1 2,0,0 2 1,0,0 3 8.47 The individual−particle states are ψ ↑, ψ ↓, and ψ ↑. Inserting these in each row and the three particle – ψψ()rr↓↑↑ ()ψ () r1,0,0 + ψψψ1,0,0 ()rrr↑↑↓2,0,0 () () – ψψψ()rrr↑↓↑ () () (c) 1,0,0 We 1 see 1,0,0that 2 an n 2,0,0 = 2 3 orbit is2,0,0 very 1 much 1,0,0 farther 2 1,0,0 from 3 the 2,0,0origin 1 than 1,0,0 the 2 n 1,0,0= 1 3cloud. An electron in a more- ψψ1,0,0()rrr 1↑↓↑ 1,0,0 () 1ψ 2,0,0 () 1 circular 2p state will orbit essentially that whole cloud. The more-elliptical 2s, for which the orbit has a labels in each column gives ψψ1,0,0()rrr 2↑↓↑ 1,0,0 () 2ψ 2,0,0 () 2 1 a larger probability of()2rrr being0 () very close () to the origin, would pierce the inner cloud and partly feel the 8.48 From equation (7-42), ra10ψψ=1,0,01 3↑↓↑ = 1,0,0. 3ψ 2,0,0 3 Write the electronicattraction toconﬁguration more 3protons.3 of phosphorus. = ψψ1,0,0()rr 1↑↓↑ 1,0,0 () 2ψ 2,0,0 () r 3 – ψψ1,0,0()rrr 1↑↑↓ 2,0,0 () 2ψ 1,0,0 () 3 + ψψ1,0,0()rrr 1↓↑↑ 2,0,0 () 2ψ 1,0,0 () 3 1 2 (b) ra= 2 = 4a0. – ψψ()rr20↓↑↑ ()ψ () r + ψψψ()rrr↑↑↓ () () – ψψψ()rrr↑↓↑ () () 1,0,0 11 1,0,0 2 2,0,0 3 2 2 2,0,06 12 3 1,0,0 2 1,0,0 3 2,0,0 1 1,0,0 2 1,0,0 3 8.49 Phosphorus(c) We see that(Z =an 15): n = 2 1 orbits 2s is2 veryp 3 smuch3p farther from the origin than the n = 1 cloud. An electron in a more- circular 2p state will1 orbit2 a2essentially6 2 that6 10whole2 cloud.2 The more-elliptical 2s, for which the orbit has a 8.48 From equation (7-42), ra= 12 = 0 . Germaniumlarger probability (Z = 32):10 3 of 1 s being2s3 2 veryp 3 s close3p 3 tod the4s orig4pin, would pierce the inner cloud and partly feel the attraction to more 2protons.2 6 2 6 10 2 6 10 2 6 1 Write the electronic1 2 conﬁguration of germanium. (b)Cesium ra20= (2Z = =55): 4a0. 1s 2s 2p 3s 3p 3d 4s 4p 4d 5s 5p 6s 1 2 2 6 2 3 8.49 (c) Phosphorus We see that (Z an= 15):n = 2 1 orbits 2s 2isp very3s 3 muchp farther from the origin than the n = 1 cloud. An electron in a more- circular 2p state will orbit essentially that whole cloud. The more-elliptical 2s, for which the orbit has a 8.50 Starting at, say, element2 2 6 103,2 6 Lawrencium,10 2 2 and counting to the right in the periodic table, 117 would put us Germaniumlarger probability (Z = 32): of 1 s being2s 2 p very3s 3 closep 3d to4s the4p origin, would pierce the inner cloud and partly feel the directlyattraction under to more fluorine 2protons.2 6, in2 the6 10valence2 6 10negative2 6 1 one (or seven) column, with two 7s and five 7p electrons. Cesium (Z = 55): 1s 2s 2p 3s 3p 3d 4s 4p 4d 5s 5p 6s

2 2 6 2 3 Write8.49 the Phosphorus electronic (Z = 15): 1sconﬁguration2s 2p 3s 3p of cesium. 8.50 Starting at, say, element 103, Lawrencium, and counting to the right in the periodic table, 117 would put us 2 2 6 2 6 10 2 2 Germaniumdirectly under (Z = 32):fluorine 1s 2,s in2p the3s valence3p 3d 4 negatives 4p one (or seven) column, with two 7s and five 7p electrons. 122 2 2 6 2 6 10 2 6Copyright10 2 6 1 © 2008, Pearson Addison-Weslesy Cesium (Z = 55): 1s 2s 2p 3s 3p 3d 4s 4p 4d 5s 5p 6s

8.50122 Starting at, say, element 103, Lawrencium,Copyright and © counting2008, Pearson to the Addison-Weslesy right in the periodic table, 117 would put us directly under fluorine, in the valence negative one (or seven) column, with two 7s and five 7p electrons. Physics 3305 - Modern Physics Professor Jodi Cooley

122 Copyright © 2008, Pearson Addison-Weslesy A good electron thief needs a trap at low energy to entice its prey. A poor electron shepherd will have at least some of its ﬂock dangling out at high energy. Consider rows 2 and 5 in the periodic table. Why should ﬂuorine in row 2 be more reactive than iodine in row 5, while lithium in row 2 is less reactive than rubidium in row 5?

Outer electrons are less tightly bound. I is less tightly bound than F and Rb is less tightly bound than Li.

If the element’s role is to “give up” an electron (Li, Rb), it is more reactive if the electron is less tightly bound. If the role is to “seize” an electron (F, I), it is more reactive if it has a deeper hole (more tightly bound) to entice them.

Physics 3305 - Modern Physics Professor Jodi Cooley Which of the following statements concerning electromagnetic waves emitted from atoms is true?

A) A collection of atoms emits electromagnetic radiation only at speciﬁc wavelengths.

B) Atoms only emit radiation in the visible part of the electromagnetic spectrum.

C) Free atoms have 3n unique lines in their atomic spectra, where n is the number of electrons.

D) The wavelengths of electromagnetic radiation emitted by free atoms is speciﬁcally characteristic of the particular element.

Physics 3305 - Modern Physics Professor Jodi Cooley From Lecture Video: Electrons jump around within the atom when a hole is made in the inner-shell! Atomic energy levels are quantized --- So, only certain X-rays can be emitted by a given element.

Each element is different, so these x-rays can act as a ﬁngerprint by which we can distinguish different elements.

Notation: X-ray photons produced in the n=1 shell are referred to as K-shell, n=2 are referred to as L-shell, and so on.

Physics 3305 - Modern Physics Professor Jodi Cooley Notation:

n=1 —> K-shell This notation is also n=2 —> L-shell used to indicate the n=3 —> M=shell shell at which a transition terminates.

A subscript advancing from α (and going on through the Greek alphabet) designates how many shells higher the transition began.

Lβ is from the N-shell (n=4) down to the L-shell (n = 2)

Physics 3305 - Modern Physics Professor Jodi Cooley Rank the following lines according to increasing

wavelength: Kα, Kβ, and Lα.

Ans: Kβ, Kα and Lα.

The Kα comes from a transition from n=2 to n=1,the Kβ from n=3to n=1, and the Lα from n=3 to n=2. Energy levels tend to get closer together as n increases, so the 2 to 1 jump is bigger than the 3 to 2. Therefore, the highest

energy/shortest wavelength photon is the Kβ, next is the Kα, then the Lα is the lowest energy/longest wavelength.

Physics 3305 - Modern Physics Professor Jodi Cooley The end (for today)

Ever wonder what happens if you build a periodic table out of bricks, where each brick is made of the corresponding element?

http://www.wired.co.uk/magazine/archive/2014/10/play/xkd-does-science/viewgallery/338864

Physics 3305 - Modern Physics Professor Jodi Cooley